Class 12 Engineering Chemistry - Chemical Kinetics

For the two parallel reactions $\displaystyle A\overset{k_{1}}{\rightarrow}B$ and $\displaystyle A\overset{k_{2}}{\rightarrow}C,$ show that the activation E' for the disappearance of A is given in terms of activation energies $\displaystyle E_{1}$ and $\displaystyle E_{2}$ for the two paths by $\displaystyle E'=\frac{k_{1}E_{1}+k_{2}E_{2}}{k_{1}+k_{2}}$

Let' s consider that one mole of A is getting converted into B and C.

Now let x moles of A get converted into B and (1-x) moles of A get converted into C.

We know that

$\dfrac{x}{1-x}=\dfrac{k_1}{k_2}$ - - - - - - - 1

Also E'(1) = E

$_1$ (x)+E

$_2$ (1-x)---------2

Using equation 1 and equation 2, the required relation can be easily deduced.

What is threshold energy?

Threshold energy:
The minimum energy that all colliding molecules must possess in order to make the collisions effective and successfull is called threshold energy.

A radioactive isotope is being produced at a constant rate $x$ . Half life of the radioactive substance is ' $y$ '. After sometime, the number of radioactive nuclei becomes constant the value of this constant is ________

At the stage of radioactive equilibrium, rate of formation of nuclei

$=$ rate of decay of nuclei

$x=\lambda N$

$N=\dfrac { x }{ \lambda } =\dfrac { x }{ { \left( \ln { 2 } \right) }/{ y } } =\dfrac { xy }{ \ln { 2 } }$

The gaseous reaction ${A_2} \to {\rm{ }}2A$ is first order in $A_2$ . After $12.3$ minutes $65\%$ of $A_2$ remains undecompensed. How long will it take to decompose $90\%$ of $A_2$ ? What is the half life of the reaction?

The given reaction is first order.

$\begin{array}{l}\ln A = - kt + \ln {A_0}\\\ln \left( {0.65} \right) = - k\left( {738s} \right) + \ln 1.00\\k = 0.000584010\,{s^{ - 1}}\end{array}$
For decomposing

$90\%$ means

$10\%$ remains.
Therefore,

$\begin{array}{l}\ln A = - kt + \ln {A_0}\\\ln 0.100 = - \left( {0.000584010\,{s^{ - 1}}} \right)t + \ln 1.00\\t = 3943\,s\end{array}$
Half life of reaction
=

$\dfrac{{0.693}}{k}$
=

$\dfrac{0.693}{0.000584010}$ =

$1187$ seconds.

For a zero order reaction at $200K$ reaction complete in 5 minutes while at $300 K$ , same reaction, completes in $2.5$ minutes. What will be the activation energy in calorie? $(R = 2 Cal/(mol.K); \ ln 2 = 0.7)$

For zero order reaction,

$A=A_{o}-Kt$

When reaction completes, then A becomes zero.

So,

$A_{o}=Kt$ or

$K=\cfrac{A_{o}}{t}$

$K_{1}=\cfrac{A_{o}}{t_{1}}=\cfrac{A_{o}}{5}$ ,

$T_{1}$ =200 K

$K_{2}=\cfrac{A_{o}}{t_{2}}=\cfrac{A_{o}}{2.5}$ ,

$T_{1}$ =300 K

By using equation,

$\log { \left( \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\cfrac { { E }_{ q } }{ 2.303R } \left( \cfrac { 1 }{ { T }_{ 1 } } -\cfrac { 1 }{ { T }_{ 2 } } \right) \\ \Longrightarrow \ln { \left( \cfrac { \cfrac { A_{ o } }{ 2.5 } }{ \cfrac { A_{ o } }{ 5 } } \right) } =\cfrac { { E }_{ q } }{ R } \left( \cfrac { 1 }{ 200 } -\cfrac { 1 }{ 300 } \right) \\ \Longrightarrow \ln { 2 } =\cfrac { { E }_{ q } }{ 2cal } \left( \cfrac { 1 }{ 200 } -\cfrac { 1 }{ 300 } \right) \\ \Longrightarrow { E }_{ q }=\cfrac { 0.7\times 2cal{ mol }^{ -1 }{ K }^{ -1 }[300\times 200]{ K }^{ 2 } }{ 300-200 } =0.7\times 2\times 600=840cal{ mol }^{ -1 }$

At $300 \,K, 50%$ of molecule collide with energy grater than or equal to $E_a$ . At what temperature, $25%$ molecule will have energy greater than or equal to $E_a$ .

From

$K=Ae^{-Ea/RT}$ , We know that

$\cfrac {K_1}{K_2}=\cfrac {T_1}{T_2}$ (

$K$ = number of molecules,

$T$ = Temperature)

$\Rightarrow \cfrac {50}{25}=\cfrac {300}{T_2}$

$\Rightarrow T_2=150K$

Inversion of cane sugar is pseudo first order reaction. Give reason.

Here, Breaking down of sugar into glucose and fructose is not a first order reaction. But, it is made one by increasing the concentration of one of the reactant. Hence, Inversion of cane sugar is a pseudo first order reaction

Define (i) Rate Constant
(ii) Energy of Activation of a reaction

1) Rate constant :

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.

2) Energy of activation of a reaction :

The minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation is called energy of activation.

What is activation energy?

The minimum quantity of energy which the reacting species must possess in order to undergo a specified reaction is known as activation energy.

After five half-life periods for a first order reaction, what fraction of reactant remains?

After first half periods,

$\dfrac{1}{32}$ of the reaction remains.

Explain why?
Rate of reaction is change in concentration in a given time interval. The change in concentration is (final concentration - initial concentration). As the reaction progresses the change in concentration of reactants decreases and that of products increases. D C gets a negative sign in case of reactants and
positive sign in case of products.

A negative sign placed before a reaction rate symbol signifies a decrease in concentration of the reactant with increase of time and a positive sign before the rate symbol signifies that the concentration of the product increases with increase in time.

$Rate = \pm \frac{\Delta [C]}{\Delta t}$
$+ \frac{\Delta [C]}{\Delta t}$ = Rate at which concentration of product increases
$- \frac{\Delta [C]}{\Delta t}$ = Rate at which concentration of reactants decreases

Calculate the rate of reaction for the change $2A \rightarrow Products$ , when rate constant of the reaction is $2.1 \times 10^{-5} time^{-1}$ and $[A]^{0} = 0.2\ M$ .

The unit of rate constant suggest it to be first order.
Thus rate

$= K[A]$

$= 2.1\times 10^{-5} \times 0.2 = 4.2\times 10^{-6} mol\ litre^{-1} \sec^{-1}$ .

The time required for a first-order reaction for $99$ % completion is $x$ times for the time required for $90$ % completion. $x$ is:

Time required of a first-order reaction for 99% completion is x times for the time required for 90% completion. x is 2.

$t=\dfrac {2.303}{k}log \dfrac {a}{a-x}$

For 90% completion,

$t_{90}=\dfrac {2.303}{k}log \dfrac {100}{100-90}=\dfrac {2.303}{k}$

For 99% completion,

$t_{99}=\dfrac {2.303}{k}log \dfrac {100}{100-99}=\dfrac {4.605}{k}$

Hence,

$\dfrac {t_{99}}{t_{90}} = \dfrac {\dfrac {4.605}{k}}{\dfrac {2.303}{k}}=2=x$

$\therefore \ \ \ \ x=2$

The rate of reaction is fast when the activation energy is _______ (small/large).

The activation energy of a chemical reaction is related to its rate of reaction.Specifically, the higher the activation energy, the slower the chemical reaction.

For reactions having small activation energy, a large number of molecules will posses energy higher than the activation energy.

Therefore, the rate of reaction is fast when the activation energy is small.

A reaction is said to be of zero if the rate is entirely independent of the concentration of the reactants.
If true enter 1, if false enter 0.

A reaction is said to be of zero if the rate is entirely independent of the concentration of the reactants.
Thus the reaction having the rate law expression

$rate = k [A]^0$ is a zero order reaction as the rate of the reaction is independent of the concentration of A.The photochemical reaction for the formation of HCl is a zero order reaction.

$H_2(g)+Cl_2(g) \xrightarrow {sunlight} 2HCl(g)$

Collision frequency depends upon concentration and temperature.
If true enter 1, if false enter 0.

Collision frequency depends upon concentration and temperature. Higher is the concentration of the reactants and higher is the temperature, higher will be the collision frequency.
Collision rate

$=Z \times$ reactant concentration.
Here, Z is the collision frequency.

Molecules must collide before they react.
If true enter 1, if false enter 0.

For a chemical reaction to occur, the molecules must collide. However all the collisions do not result in a chemical reaction. Only those collisions, in which the collision molecules have energy equal to or more than the activation energy and proper orientation are fruitful.

State whether the statement is True or False:

Two molecules can react only if they have a correct relative orientation.

Two molecules can react only if they have a correct relative orientation. This is especially true for reactions involving complex molecules. If the orientation is not proper, then the reaction will not occur even if the molecules have energy greater than or equal to kinetic energy.

The molecules must be so oriented relative to each other that the groups reacting or bonds to be shifted are relatively close. Hence, the correct statement is true.

______ (Kinetics/Ionic equilibrium) studies of reactions involve molecular reactions.

Kinetics studies of reactions involve the study of the rate of the reaction.

The various factors such as concentration, temperature, pressure, catalyst effects on the rate of reaction.

So, the correct answer is kinetic.

Initial pressure of $A$ in mm is___________.

To provide a long time or heating to a reaction mixture means that reaction has gone to completion.

$A\rightarrow B+2C$

Mole before dissociation	a	0	0
Mole after dissociation	a-x	x	2x
Mole after complete dissociation	0	a	2a

$\because$ Total mole at a time

$\propto$ pressure at that time

$a\propto P_0 {\;} t=0$ ......(i)

$a+2X\propto 264 {\;} t=14 \ minute$ ....(ii)

$3a\propto 264 {\;} t=\infty$ .....(iii)

$\therefore$ By Eqs. (i) and (ii)

$P^0=150 \ mm$

Ethylene is produced by, $C_4H_8\xrightarrow {\Delta} 2C_2H_4$ . The constant is $2.48\times 10^{-4} sec^{-1}$ . Time in hours (nearest integer) in which the molar ratio of the ethylene to cyclobutane in reaction mixture will attain the value of 100 is___________.

The reaction is

$C_4H_8\rightarrow 2C_2H_4$
Mole at

$t=0 {\;} a {\;} 0$
Mole at

$t=t {\;} (a-X) {\;} 2X$
If

$\frac {2X}{(a-X)}=100$ i.e.,

$X=\frac {100a}{102}$

$\therefore t=\frac {2.303}{2.48\times 10^{-4}}log \frac {a}{\left (a-\frac {100a}{102}\right )}$

$=15856.9\ second =264.3 \ min\approx 4 \ hr$

A first order reaction is $20\%$ complete in 10 min. The specific rate constant of the reaction is_________ ( $in\:\: min^{-1}$ ).

$\displaystyle k\, =\, \frac{2.303}{t} log \frac{a}{a\, -\, x}$

For

$20\%$ completion,

$\displaystyle k\, =\, \frac{2.303}{10} log \frac{100}{80}\, =\, 0.0223\, min^{-1}$

Derive the relation between half-life period and rate constant for first order reaction

For a first order reaction differential rate law can be written as:

$-\dfrac{dA}{dt}\ =\ k[A]$ where

$k$ is the rate constant of first order reaction

Integrating the above equation we can write the first order rate law expression as:

$[A]_t=[A]_oe^{-kt}$ or

$kt=2.303\ log\dfrac{[A]_o}{[A]_t}$

For, half life period,

$[A]_t=[A]_o/2$

Therefore,

$t_{1/2}=\dfrac{2.303\ log\ 2}{k}$ or

$t_{1/2}=\dfrac{ln\ 2}{k}$

Half life period of a _ order reaction is of the concentration of the reactant.

Half life period of a first order reaction is given by the formula:

$t_{1/2}=\dfrac{ln\ 2}{k}$

As, there is no concentration term in the first order

$t_{1/2}$ formula, half life is independent of the concentration of the reactant.

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with $t_{1/2}=3$ hours. What fraction of the sample of sucrose remains after $8$ hours?

Given:

$t_{1/2}=3$ hours,

$t=8$ hours
As source decomposes according to first order rate law,

$k=\displaystyle\frac{2.303}{t}log\frac{[A]_0}{[A]_t}$
Using the formula

$k=\displaystyle\frac{0.693}{t_{1/2}}$

$\therefore k=\displaystyle\frac{0.693}{t_{1/2}}=\frac{0.693}{3}=0.231hr^{-1}$
From formula,

$k=\displaystyle\frac{2.303}{t}log\frac{[A]_0}{[A]_t}$

$\therefore 0.231=\displaystyle\frac{2.303}{8}log\frac{[A]_0}{[A]_t}$

$\therefore log\displaystyle\frac{[A]_0}{[A]_t}=0.8024$
Taking antilog on both sides,

$\therefore \displaystyle\frac{[A]_0}{[A]_t}=$ Antilog

$(0.8024)=6.345$

$\therefore \displaystyle\frac{[A]_0}{[A]_t}=\frac{1}{6.345}=0.158$ .
Hence, the fraction of the sample of sucrose that remains after

$8$ hours is

$0.158$ .

What is half-life of a reaction? Derive formula for finding out half-life from first order rate reaction.

The half-life period of reaction is the time needed for the reactant concentration to fall to one half of its initial value.
The rate equation for first order reaction is

$\displaystyle k = \dfrac {2.303}{t} log \dfrac {[A]_0}{[A]_t}$
When time

$\displaystyle t=t_{1/2}=$ half life period, the initial concentration decreases from

$\displaystyle [A]_0$ to

$\displaystyle \dfrac {[A]_0}{2}$

$\displaystyle k = \dfrac {2.303}{t_{1/2}} log \dfrac {[A]_0}{ \dfrac {[A]_0}{2}}$

$\displaystyle k = \dfrac {2.303}{t_{1/2}} log\ 2$

$\displaystyle k = \dfrac {0.693 }{k}$

The half-life period for such reaction is independent of the initial concentrations of the reactants.

Write the characteristics of first order reaction.

(i) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is

$sec^{-1}$ or

$time^{-1}$

$k_1=\displaystyle\frac{rate}{(a-x)}=\frac{mol.lit^{-1}sec^{-1}}{mol.lit^{-1}}=sec^{-1}$
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if

$t_{1/u}$ is the time of one

$'u'$ the fraction of reaction to take place then from equation.

$k_1=\displaystyle\frac{2.303}{t}log\frac{a}{a-x}$

$x=\displaystyle\frac{a}{u}$ and

$t_{1/u}=\displaystyle\frac{2.303}{k_1}log\frac{a}{\displaystyle a-\frac{a}{u}}$

$t_{1/u}=\displaystyle\frac{2.303}{k_1}loh\frac{u}{(u-1)}$
since

$k_1$ -rate constant

$t_{1/u}$ is independent of initial concentration 'a'.

Derive the integrated rate equation for the rate constant of a zero order reaction.

Consider a zero order reaction

$\displaystyle A \rightarrow$ products.
The rate law is rate

$\displaystyle = -\dfrac {d[A]}{dt} = k[A]^0=k$
Rearrange above equation

$\displaystyle d[A]=-kdt$
Integrate between limits

$\displaystyle [A]=[A]_0$ at

$\displaystyle t=0$ and

$\displaystyle [A]=[A]_t$ at

$\displaystyle t=t$

$\displaystyle \int _{[A]_0}^{[A]} d[A]=-k \int_{0}^{t}dt$
Hence,

$\displaystyle [A]^{[A]_t}_{[A]_0}=-k(t)^t_0$

$\displaystyle [A]_t-[A]_0=-kt$

$\displaystyle [A]_t=-kt+[A]_0$

$\displaystyle k=\dfrac {[A]_0-[A]_t}{t}$

Derive an expression for the rate constant for a first order reaction.

Rate constant of a first order reaction: The reaction in which, the overall rate of the reaction is proportional to the first power of concentration of one of the reactants only are called first order reaction.

$A\overset{k_{1}}{\rightarrow}$ products
Rate of reaction

$=\dfrac{-d\left [ A \right ]}{dt}=k_{1}\left [ A \right ]^{10}$
where

$k_{1}$ is the rate constant of the first order reaction.
At the beginning of the reaction, time

$'t'=0$ , let the concentration of

$A$ be '

$a$ ' mole lit

$^{-1}$ . After the reaction has proceeded for some time '

$t$ ', let the concentration of

$A$ that has reacted be

$x$ mole lit

$^{-1}$ . The concentration of unreacted

$A$ remaining at time '

$t$ ' will be

$(a-x)$ mole lit

$^{-1}$ . The rate of the reaction will be

$dx/dt$ . For a first order reaction,
rate

$=dx/dt = k_{1}\left ( a-x \right )$ ...... (1)
Integrating

$(1)$ , both sides

$\int \dfrac{dx}{\left ( a-x \right )}=k_{1}\int dt$
Which is

$-ln\left ( a-x \right )=k_{1}t+c$ ...... (2)

$C=$ integration constant.
at time,

$t=0, x=0$
In equation

$(2)$

$-ln\left ( a-0 \right )=k_{1}\times 0+c$
or

$c=ln$

$a$
Substituting

$c$ value in equation

$(2)$

$-ln\left ( a-x \right )=k_{1}t-ln$

$a$
rearranging,

$k_{1}=\dfrac{1}{t}ln\dfrac{a}{a-x}$

$k_{1}=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$
Unit of

$k_{1}$ is

$\sec ^{-1}$
This equation is known as the first order rate constant equation.

Show that for a first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion.

$t=\frac{2.303}{k}log \frac{a}{a-x}$

$t_{99.9}=\frac{2.303}{k}log \frac{100}{100-99.9}=\frac{2.303}{k}log_{10}\frac{100}{1}$

$=\frac{2.303}{k}log_{10}1000 .....(i)$

$t_{50}=\frac{2.303}{k}log \frac{100}{100-50}$

$=\frac{2.303}{k}log_{10}\frac{100}{50}$

$=\frac{2.303}{k}log_{10}^2 .....(ii)$

$\frac{i}{ii}=\frac{99.9}{50}=\frac{\frac{2.303}{K}log_{10}1000}{\frac{2.303}{K}log_{10}^2}$

$=\frac{3.000}{0.3010}=10$

What happens to the half life period for a first order reaction, if the intial concentration of the reactants is increased?

If the initial concentration of the reactants is increased, the half life period for a first order reaction remains unchanged. This is because the half life period for a first order reaction is independent of the initial concentration.

Give a detailed account of the collision theory of reaction mils of Bimolecular gaseous reactions.

b) The detailed account of the collision theory of reaction mils of Bimolecular gaseous reactions:
(i) The reaction molecules are assumed to be hard spheres
(ii) When molecules collide with each other, the reaction will occur.
(iii) Collision frequency (Z) is the number of collisions per second per unit volume of the reaction mixture.
(iv) For a bimolecular elementary reaction

$\displaystyle A+B \rightarrow$ products
Rate

$\displaystyle = Z e^{-E_a/RT}$
(v) All collisions do not result in reaction.
(vi) The probability factor (p) also called steric factor accounts for effective collisions.
Rate

$\displaystyle = pZ \cdot e^{-E_a/RT}$
(vii) The proper orientation of reactant molecules result in bond formation. No product is formed with improper orientation.

Give two examples of gaseous first-order reactions.

The following are the examples of gaseous first-order reactions.
(1) The decomposition of dinitrogen pentoxide

$\displaystyle N_2O_5$

$\displaystyle 2N_2O_5(g) \rightarrow 4NO_2(g)+O_2(g)$

(2) The decomposition of azomethane

$\displaystyle CH_3N_2CH_3$

$\displaystyle CH_3N_2CH_3(g) \rightarrow CH_3CH_3(g)+N_2(g)$

Define collision frequency. Give an example for Pseudo-first order reaction.

Collisional Frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system.

Under certain conditions, the 2nd order kinetics can be well approximated as first order kinetics. These are pseudo first order. These are the reactions that appear to be second order in nature but are approximated as a first order reaction on close analysis. For example a second order reaction given by the equation : hydrolysis of ethyl acetate.

Write short notes on:
(a) Threshold Energy, (b) Energy of Activation.

(a) Threshold energy: The minimum amount of energy which the reactant molecules should possess for effective collision.
(b) Energy of Activation: According to Arrhenius any chemical reaction is only possible when reacting molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.

For a zero order reaction $t_{1/2}$ is proportional to what?

For a zero order reaction

$\displaystyle t_{1/2}$ is proportional to initial concentration

$\displaystyle [A]_0$

$\displaystyle t_{1/2} \propto [A]_0$

$\displaystyle t_{1/2} = \dfrac {[A]_0}{2k}$

The half life of $^{212}Pb$ is $10.6\ hour$ . It undergoes decay to its daughter (unstable) element $^{212}Bi$ of half life $60.5$ minute. Calculate the time at which the daughter element will have maximum activity.

${ \lambda }_{ Pb }=\cfrac { 0.693 }{ 10.6\times 60 } =1.0896\times { 10 }^{ -3 }{ min }^{ -1 }$

${ \lambda }_{ Bi }=\cfrac { 0.693 }{ 60.5 } =11.45\times { 10 }^{ -3 }{ min }^{ -1 }$

${ t }_{ max }=\cfrac { 2.303 }{ { \lambda }_{ Bi }-{ \lambda }_{ Pb } } \log { \dfrac { { \lambda }_{ Bi } }{ { \lambda }_{ Pb } } }$

$=\cfrac { 2.303 }{ \left( 11.45\times { 10 }^{ -3 }-1.0896\times { 10 }^{ -3 } \right) } \times \log { \dfrac { 11.45\times { 10 }^{ -3 } }{ 1.0896\times { 10 }^{ -3 } } } =227.1 min$

$^{ 131 }{ I }$ has half life period $133$ hour. After $79.8$ hour, what fraction of $^{ 131 }{ I }$ will remain?

$k=\dfrac{0.693}{133}$

$N=N_oe^{(-\dfrac{0.693}{133}\times 79.8)}$

$\dfrac{N}{N_o}=0.66$

$^{ 90 }{ Sr }$ shows $\beta$ -activity and its half life period is $28$ years. What is the activity of a sample containing $1 g$ of $^{ 90 }{ Sr }$ ?

Activity

$=$ Number of atoms disintegrating per second

$=\lambda \times$ total number of atoms

$\lambda =\dfrac { 0.693 }{ 28\times 365\times 24\times 60\times 60 }$

Total number of atoms in

$1 g$ of

$^{ 90 }{ Sr }=\dfrac { 6.023\times { 10 }^{ 23 } }{ 90 }$

Activity

$=\dfrac { 0.693 }{ 28\times 365\times 24\times 60\times 60 } \times \dfrac { 6.023\times { 10 }^{ 23 } }{ 90 }$

$=5.25\times { 10 }^{ 12 }$ disintegrations per second

$=\dfrac { 5.25\times { 10 }^{ 12 } }{ 3.7\times { 10 }^{ 10 } } =141.89$ curie

A chemist prepares $1.00 g$ of pure $_{ 6 }^{ 11 }{ C }$ . This isotope has half life of $21$ minutes, decaying by the equation:
$_{ 6 }^{ 11 }{ C } \rightarrow _{ 5 }^{ 11 }{ B } + _{ 1 }^{ 0 }{ e }$
(a) What is the rate of disintegration per second (dps) at start?
(b) What are the activity and specific activity of $_{ 6 }^{ 11 }{ C }$ at start?
(c) How much of this isotope $_{ 6 }^{ 11 }{ C }$ is left after $24$ hours of its preparation?

(a) Applying,

$-\dfrac { dN }{ dt } =\lambda { N }_{ 0 }$

$=\dfrac { 0.693 }{ 21\times 60 } \times \dfrac { 1\times 6.02\times { 10 }^{ 23 } }{ 11 }$

$=3\times { 10 }^{ 19 } dps$
(b) Activity

$=\dfrac { 3\times { 10 }^{ 19 } }{ 3.7\times { 10 }^{ 10 } }$

$\left( 1curie=3.7\times { 10 }^{ 10 }dps \right)$

$=8.108\times { 10 }^{ 8 }curie$
Specific activity

$=3\times { 10 }^{ 19 }\times { 10 }^{ 3 }=3\times { 10 }^{ 22 }{ dis }/{ kg }s$

$=8.108\times { 10 }^{ 11 } curie$
(c) Applying,

$N={ N }_{ 0 }{ \left( \dfrac { 1 }{ 2 } \right) }^{ n }$

$\left[ n=\dfrac { t }{ { t }_{ { 1 }/{ 2 } } } =\dfrac { 24\times 60 }{ 21 } =68.57 \right]$

$N=1\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 68.57 }=2.29\times { 10 }^{ -21 } g$

Binding energy per nucleon of $_{ 1 }^{ 2 }{ H }$ and $_{ 2 }^{ 4 }{ He }$ are $1.1 MeV$ and $7 MeV$ respectively. Calculate the amount of energy released in the following process:
$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\longrightarrow$ $_{ 2 }^{ 4 }{ He }$

Amount of energy released

$=\displaystyle\sum { Binding\ energy\ of\ products } -\displaystyle\sum { Binding\ energy\ of\ reactants }$

$=\left[ 4\times 7 \right] -\left[ 4\times 1.1 \right] =23.6\ MeV$

For the dissociation of gaseous $HI$ , the energy of activation is $44.3kcal$ . Calculate the energy of activation for the reverse reaction. Given, $\Delta H$ for the formation of $1$ mole of $HI$ from ${H}_{2}$ and ${I}_{2}$ is $-1.35kcal$ .

For the dissociation of gaseous HI, the energy of activation is 44.3kcal.

$HI(g)\rightleftharpoons \cfrac { 1 }{ 2 } { H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { I }_{ 2 }(g);{ E }_{ a }=44.3kcal$

For the reverse reaction, the enthalpy change is

$-$ 1.35kcal.

$\cfrac { 1 }{ 2 } { H }_{ 2 }(g)+\cfrac { 1 }{ 2 } { I }_{ 2 } (g)\rightleftharpoons HI(g);\Delta H=-1.35kcal\quad$

$\Delta H={ E }_{ a }(f)-{ E }_{ a }(b)$

$-1.35=44.3-{E}_{a}(b)$

${E}_{a}(b)=(44.3+1.35)$

${E}_{a}(b)=45.65 \: kcal$

The energy of activation for the reverse reaction is

$=45.65 \: kcal$

Note :

${E}_{a}(f)$ and

${E}_{a}(b)$ represents the energy of activation for forward and reverse reaction respectively.

Thermal decomposition of a compound is of first order. If $50\%$ sample of the compound is decomposed in $120$ minute, how long will it take for $90\%$ of the compound to decompose?

Half life of reaction

$=120$ min
We know that,

$k=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } =\dfrac { 0.693 }{ 120 } =5.77\times { 10 }^{ -3 }{ min }^{ -1 }$
Applying first order reaction equation,

$t=\dfrac { 2.303 }{ k } \log _{ 10 }{ \dfrac { a }{ a-x } }$
If

$a=100$ ,

$x=90$ or

$\left( a-x \right) =10$ ,
So,

$t=\dfrac { 2.303 }{ 5.77\times { 10 }^{ -3 } } \log _{ 10 }{ 10 } =\dfrac { 2.303 }{ 5.77\times { 10 }^{ -3 } } =399 min$ .

A drop of solution (volume $0.05\ mL$ ) contains $3\times { 10 }^{ -6 }\ mole$ ${ H }^{ + }$ ions. If the rate of disappearance of the ${ H }^{ + }$ ions is $1\times { 10 }^{ 7 }$ $mol$ ${ litre }^{ -1 }$ ${ sec }^{ -1 }$ , how long would it take for ${ H }^{ + }$ ions in the drop to disappear?

Concentration of drop

$=\dfrac { mole }{ volume\quad in\quad mL } \times 1000$

$=\dfrac { 3\times { 10 }^{ -6 } }{ 0.05 } \times 1000=0.06$

$mol$

${ litre }^{ -1 }$
Rate of disappearance

$=\dfrac { conc.change }{ time }$

$1\times { 10 }^{ 7 }=\dfrac { 0.06 }{ time }$
Time

$=6\times { 10 }^{ -9 }sec$

For a reaction $X\rightarrow Y)$ , heat of reaction is $+83.68kJ$ , energy of reactant $X$ is $167.36kJ$ and energy of activation is $209.20kJ$ . Calculate (i) threshold energy (ii) energy of product $Y$ and (iii) energy activation of the reverse reaction $(Y\rightarrow X)$ .

(i) Given, the energy of activation for forward reaction

${ E }_{ a }(f) =209.20 \: kJ$
The heat of reaction

$\Delta E=+83.68 \: kJ$
Energy of reactant,

$X=167.36 \: kJ$
Threshold energy

$=$ Energy of reactant

$X+$ Activation energy for
forward reaction.
Threshold energy

$=167.36+209.20=376.56 \: kJ$

(ii) Energy of reaction,

$\Delta E=$ Energy of product

$-$ Energy of
reactant
Energy of product

$=$ Energy of reaction,

$\Delta E +$ Energy of
reactant
Energy of product

$=83.68+167.36=251.04 \: kJ$

(iii) Activation energy for backward reaction

$=$ Threshold energy

$-$ Energy of product
Activation energy for backward reaction

$=376.56-251.04=125.52kJ$

Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with $t_{ {1}/{2}} = 3\ Hrs$ . What fraction of the sample of sucrose remains after $8$ hours.

Half-life of a reaction is the time required for the reaction to be half completed.

Amount of substance left unreacted after one half life=

$\cfrac {[A]_0}{2}$ ;

$[A]_0$ = initial amount of substance at

$t=0$

Amount of substance left after

$2$ half lives=

$\cfrac {[A]_0}{2^2}$

In general,

amount of substance left after

$n$ half lives=

$\cfrac {[A]_0}{2^n}$

Now, given half life=

$t_{1/2}=3$ hrs

Time after which amount left is asked=

$8$ hours

No. of half lives=

$\cfrac {8}{3}$

Amount of sucrose left=

$\cfrac {[A]_0}{2^{8/3}}$

Fraction of sucrose left=

$\cfrac {1}{2^{2/3}}=\cfrac {1}{6.35}=0.16$

Give the definition of $(X)$ denoted in graph:

$(X)$ represents the activation energy. It is the difference between the energy of transition state and the energy of reactants. Activation energy is defined as the minimum energy required for a molecular collision to lead to reaction.

For the decomposition of ethylene oxide into $CH_4$ and CO, the variation of total pressure (P) of the reaction mixture with time is as given below:
t/s 0 300 600 900
P/mm 120 127.2 134.0 140.3
Show that the reaction is a first order reaction.

t/s

$0$

$300$

$600$

$900$

p/mm

$120$

$127.2$

$134.0$

$140.3$

$(0, 120) , \,\, (300 , 127.2)$

slope =

$\dfrac{127.2 - 120}{300 - 0} = \dfrac{7.2}{300}$

$(600, 134.0) \,\,\, (300, 127.2)$

slope =

$\dfrac{134 - 127.2}{300} = \dfrac{6.8}{300}$

$(900, 140.3), \,\,(600 , 134.0)$

slope =

$\dfrac{140.3 - 134.0}{900 - 600} = \dfrac{6.3}{300}$

slope is almost equal so reaction is first order if we draw the graph between P and f it will be linear

The half-life of a first order reaction is $900$ min at $820\ K$ . Estimate its half-life at $720$ K if the energy of activation of the reaction is $250\ kJ\ mol^{-1}$ .

From Arrhenius equation

$\displaystyle log_{10} \dfrac { t_{1/2}}{t_{1/2}' }= log_{10} \dfrac { k'}{ k} = \dfrac { E_a}{2.303 R} [\dfrac { T'-T}{TT' }]$

$\displaystyle log_{10} \dfrac { 900 \ min}{t_{1/2}' }= log_{10} \dfrac { k'}{ k} = \dfrac { 250 \ kJ/mol \times 1000 \ J/kJ}{2.303 \times 8.314 \ J/mol/K } \ \times \ [\dfrac { 720 \ K-820 \ K}{820 \ K \times 720 \ K }]$

$\displaystyle log_{10} \dfrac { 900 \ min}{t_{1/2}' }= 13057 \ \times \ [-1.694 \times 10^{-4}]$

$\displaystyle log_{10} \dfrac { 900 \ min}{t_{1/2}' } =-2.212$

$\displaystyle \dfrac { 900 \ min}{t_{1/2}' } =0.006145$

$\displaystyle t_{1/2}'=146473 \ min$

The half life at 720 K is 146473 min.

Calculate the activation energy for a reaction of which rate constants becomes 4 times when temperature changes from $30^0C$ to $50^0C$ [R= 8.314 $J mol^{-1}K^{-1}$ ]

$\displaystyle ln \dfrac {k'}{k}=\dfrac {E_a}{R} \times [\dfrac {T'-T}{TT'}]$

$\displaystyle T=30 \ ^oC=30+273 \ K =303 \ K$

$\displaystyle T'=50 \ ^oC=50+273 \ K =323 \ K$

$\displaystyle ln \dfrac {k'}{k}=\dfrac {E_a}{R} \times [\dfrac {T'-T}{TT'}]$

$\displaystyle ln \dfrac {4k}{k}=\dfrac {E_a}{8.314 \ J/mol/K} \times [\dfrac { 323-303}{303 \times 323}]$

$\displaystyle 1.386=2.46 \times 10^{-5} \times E_a$

$\displaystyle E_a=56400 \ J/mol$

$\displaystyle E_a=56.4 \ kJ/mol$
because

$\displaystyle 1 \ kJ=1000 \ J$
Note: Since rate constant becomes 4 times,

$\displaystyle k'=4k$

Write the two steps involved in the mechanism of the enzyme-catalyzed reaction.

Step

$1$ : Binding of enzyme to substrate to form an activated complex.

$E+S\rightarrow ES^{\ast}$
Step

$2$ : Decomposition of the activated complex to form product

$ES^{\ast}\rightarrow E+P$

Derive an integrated rate equation for rate constant of a zero order reaction.

Consider a zero order reaction

$R\rightarrow P$
Rate

$=\dfrac{-d[R]}{dt}=k[R]^o$

$\dfrac{-d[R]}{dt}=k[R]^o$

$\dfrac{-d[R]}{dt}=k$

$d[R]=-k\cdot dt$
Integrating on both sides

$[R]=-kt+I$

$(1)$

$k=\dfrac{[R]_o-[R]}{t}$

For a first order reaction $A\to P$ , the temperature $(T)$ and dependent rate constant $(k)$ , was found to follow the equation $\log k = -(2000)\dfrac{1}{T} + 6.0$ . What is activation energy?

Equate with this formula,

$logK = logA -\dfrac{E_a}{RT}$ , Now

$-\dfrac{E_a}{RT} = -\dfrac{2000}{T}$ or

$E_a = 2000\times 8.314 = 16.6KJ$

A first-order reaction is $40\%$ complete in $50$ min. How long(minutes) will it take to $80\%$ completion?

Since we know $k = \frac{2.303}{50} \log(\frac{a}{a-2a/5})$
$\implies k = \frac{2.303}{50} \log\frac{5}{3}$

Now , $t = \frac{2.303 \times 50}{2.303} \frac{log(5)}{log(\frac{5}{3})}$
$\implies t = 50 \times \frac{0.699}{0.222}$
$\implies t = 50 \times 3.15$
$\implies t = 157.57 \space min$

Sucrose decomposes in solution into glucose and fructose according to the first order rate law. With $t^{1/2} 3$ Hr. What fraction of sample of sucrose remains after $8$ hour?

${ t }_{ \frac { 1 }{ 2 } }=3hrs$

$or,\frac { 0.93 }{ K } =3$

$or,\quad K=0.231$

let,

${ A }_{ 0 }$ be the initial concentration.

$=\frac { 2.303 }{ t } \log { \left[ \frac { { A }_{ 0 } }{ { A }_{ 8 } } \right] }$

$or,0.231=\frac { 2.303 }{ t } \log { \left[ \frac { { A }_{ 0 } }{ { A }_{ 8 } } \right] }$

$or,\log _{ 10 }{ \left[ \frac { { A }_{ 0 } }{ { A }_{ 8 } } \right] } =0.802$

$or,\frac { { A }_{ 0 } }{ { A }_{ 8hours } } =6.33$

$\therefore \\$ concentration remaining after 8 hours

${ A }_{ 8hours }\\= \frac { { A }_{ 0 } }{ 6.33 }$

For a first order reaction, time required for $99.0$ % completion is $x$ times for the first time required for the completion of $90$ % of the reaction $x$ is:

For the first order reaction,

$K=\cfrac { 2.303 }{ k } log\cfrac { a }{ a-x }$

x=0.999(given)

$K=\cfrac { 2.303 }{ k } log\cfrac { a }{ a-0.99 }$

$K=\cfrac { 2.303 }{ k } log\cfrac { a }{ 0.001a }$

$K=\cfrac { 2.303 }{ k } log1000$

$K=\cfrac { 2.303 }{ k } 3$ ....(i)

the half life is given by,

${ t }_{ 1/2 }=\cfrac { 2.303 }{ k } log\cfrac { a }{ (a-0.5a) } \\ \quad \quad \quad \quad \quad =\cfrac { 2.303 }{ k } log2\\ \quad \quad \quad \quad \quad =\cfrac { 2.303 }{ k } \times 0.3010.....(ii)$

divide(i) by (ii)

$\cfrac { { t }_{ 0.999 } }{ { t }_{ 0.5 } } =\cfrac { 2.303 }{ k } 3\times \cfrac { k }{ 2.303\times 0.3010 } =10$

${ t }_{ 0.999 }=10\times { t }_{ 0.5 }$ .

Derive an expression for the rate constant of a First order reaction.

Any reaction is called a first order reaction if a change in concentration of just one reactant determines the rate of reaction. For a reaction as follows

$\rightarrow$ Product or products

${ [A] }_{ o }=$ Initial concentration

${ [A] }_{ t }=$ concentration at time t.

since reaction in first order-

Rate

$=-\cfrac { d[A] }{ dt } =k[A]\\ -\cfrac { d[A] }{ dt } =-k.dt$

On integrating, we get

$\int _{ { [A] }_{ o } }^{ [A] }{ \cfrac { d[A] }{ [A] } = } -k\int _{ 0 }^{ t }{ dt } \\ \ln { [A] } -\ln { { [A] }_{ o } } =-k[t-0]\\ \ln { \cfrac { [A] }{ { [A] }_{ o } } = } -kt\\ \ln { \cfrac { { [A] }_{ o } }{ [A] } } =kt\\ \log { \cfrac { 2.303 }{ t } } \cfrac { { [A] }_{ o } }{ [A] } =k$

For the zero order reaction $A\rightarrow 2B$ , the rate constant is $2\times {10}^{-6}M$ ${min}^{-1}$ . The reaction is started with $10M$ $A$ .
(i) What will be the concentration of A after 2 days (ii) What is the initial half-life of the reaction
(iii) In what time, the reaction will complete?

Here

$K_t=a_o-a_t$

$a_o=10M, K=2\times 10^{-6} M min^{-1}$

(i)

$t= 2$ days=

$42\times 24 \times 60= 2880$ mins

$\therefore 2\times 10^{-6}\times 2880=10-a_t$

$\therefore a_t= 10-0.00576= 9.99424M$

(ii) At half life time,

$a_t= \cfrac {a_o}{2}$

$\therefore t_{1/2}=\cfrac {a_o}{2K}=\cfrac {10}{2\times 1\times 10^{-6}}=2.5\times 10^{6}min$

(iii) At completion,

$a_t=0$

$\therefore t=\cfrac {a_o}{K}=\cfrac {10}{2\times 10^{-6}}= 5\times 10^{6}$ min

How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?

The rate of chemical reaction can be increased by using catalysts. For hydrogen peroxide decomposition,

${ MnO }_{ 2 }$ is used as a catalyst.

For a homogeneous gaseous reaction, $A \longrightarrow B+C+D$ , the initial pressure was $P_{0}$ while pressure at time $t$ was $P$ . Derive an expression for rate constant $K$ in terms of $P_{0},P$ and $t$ .
[Assume first-order reaction]

$A\rightarrow$

$B +C +D$

$t=0$

${P}^{0}$ 0 0 0

$t=t$

${P}^{0}-{P}_{1}$

${P}_{1}$

P is the pressure

${P}_{1}$ is the supposed quantity

$P={ P }^{ 0 }-{ P }_{ 1 }+{ P }_{ 1 }+{ P }_{ 1 }+{ P }_{ 1 }=2{ P }_{ 1 }+{ P }_{ 0 }$

${P}_{1}=\dfrac { P-{ P }_{ 0 } }{ 2 }$

${P}^{0}={ P }^{ 0 }-\dfrac { P }{ 2 } +\dfrac { { P }^{ 0 } }{ 2 }$

$K=\dfrac { 2.303 }{ t } log\dfrac { { P }^{ 0 } }{ { { P }^{ 0 }-P }_{ 1 } }$

$K=\dfrac { 2.303 }{ t } log\dfrac { 2{ P }^{ 0 } }{ { { 3P }^{ 0 }-P } }$

A reaction occurs in 3 steps having rate constant ${ K }_{ 1 },{ K }_{ 2 },{ K }_{ 3 }$ the overall rate constant of the reaction $1\triangle$
$K=\frac { { K }_{ 1 }{ K }_{ 2 } }{ { K }_{ 3 } }$
The ${ E }_{ a }$ energies of first, second and third steps are 30, 60, 40 $kJ{ mol }^{ -1 }$ respectively. Determine activation energy of the reaction.

$R_1=A_1e^{-E_{a_1}/{RT}}=A_1e^{-30KJ/RT}$

$R_2=A_2e^{-EoRJ/RT}$

$R_3=A_3e^{-40KJ/RJ}$

$K_3=A_3e^{-40KJ/RJ}$

$K=\dfrac{K_1K_2}{K}=\dfrac{A_1e^{-30KJ/RT}\times A_2e^{-60KJ/RT}}{A_3e^{-40KJ/RT}}$

$=\left(\dfrac{A_1A_2}{A_3}\right)e^{-50KJ/RT}$

$\therefore E_{ net} =50\ RJ$

A and B are two different chemical species undergoing 1st order decomposition with half lives equal to $5$ sec. and $7.5$ sec. respectively. If the initial concentration of A and B are in the ratio $3:2$ , calculate $\cfrac{{C}_{A}}{{C}_{B}}$ , after three half lives of A. Report your answer after multiplying it with $100$ .

${ t }_{ 1/2 }(A)=5sec=\cfrac { 0.693 }{ { k }_{ A } } \\ { k }_{ A }=0.1386\quad { sec }^{ -1 }\\ { t }_{ 1/2 }(B)=1.5sec=\cfrac { 0.693 }{ { k }_{ B } } \\ { k }_{ B }=0.0924\quad { sec }^{ -1 }$

Let initial concentration of A be 150.

Then initial concentration of B is 100.

Three half lives of

$A=5\times 3=15sec.\\ { C }_{ A }=\cfrac { 150 }{ { 2 }^{ 3 } } =18.75\quad mol/L$

$15sec=2$ half lives of B

${ C }_{ B }=\cfrac { 100 }{ { 2 }^{ 2 } } =25\quad mol/L\\ \cfrac { { C }_{ A } }{ { C }_{ B } } =\cfrac { 18.75 }{ 25 } =0.75$

Multiplying it by 100 we get

$0.75\times 100=75$

The rate of most of the reaction double when their temperature is raised from $298 \ K$ to $308 \ K$ . Calculate the activation energy of such a reaction.

Using Arrhenius Equation,

$k=A_oe^{\cfrac {-E_a}{RT}}$

$log \cfrac {K_2}{K_1}=\cfrac {E_a}{2.303} \left ( \cfrac 1{T_1}- \cfrac 1{T_2}\right )$

Given, rate of reaction doubles when temperature is raised from

$208 - 308 \ K$

$\therefore \cfrac {K_2}{K_1}=2$

$E_a=\cfrac {2.303 \times 8.314 \times 298 \times 308 \times 0.3014}{1000}$

$=52.89 \ KJ/mol$

In gaseous reaction important for understanding the upper atmosphere, $H_2O$ and $O$ react bimolecularly to form two $OH$ radicals. $\Delta H$ for this reaction is $72 \,J$ at $500 \,K$ and $E_a = 77 \,kj \,mol^{-1}$ , then calculate $E_a$ for the bimolecular recombination of $2OH$ radicals to form $H_2O$ and $O$ at $500 \,K$ .

$H_2O+O \longrightarrow 2OH \Rightarrow \Delta H= 72 kJ$

$E_{\alpha}=77 kJ /mole$ (given)

$\Rightarrow 2OH \longrightarrow H_2O+O$

$\Rightarrow 77= 72+E_{\alpha}$

$\Rightarrow E_{\alpha}=5kJ/mol$

Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$ . Calculate activation energy for the saponification of ethyl acetate.

Given that

$K_2/K_1=1.75$

$T_1=25^oC=298$

$T_2=35^oC=308$

$\Rightarrow 2.303 \log \cfrac {K_2}{K_1}=\cfrac {E_{\alpha}}{R}\left[\cfrac {T_2-T_1}{T_1T_2}\right]$

$\Rightarrow 2.303 \log 1.75=\cfrac {E_{\alpha}}{1.987}\left[\cfrac {308-298}{308 \times 298}\right]$

$\Rightarrow E_{\alpha}=10.207 K cal/mole$

A first order reaction takes $69.3$ minutes for $50\%$ completion. How much time will be needed for $80\%$ completion?

$t_{1/2}=69.3 \ min = \cfrac {ln \ 2}{K}$

$K= \cfrac {ln \ 2}{69.3} min^{-1}$

For $80$ % conversion, if we assume initial concentration
to be $a_o$ , concentration left would be $\cfrac {a_o}5$

$t \times \cfrac {ln \ 2}{69.3}= ln \left (\cfrac {a_o}{a_o/5} \right)$

$t= \cfrac {69.3 \ ln \ 5}{ln \ 2}= 161 \ min^{-1}$

The rate constant for decomposition of $COCl_2(g)$ according to following reaction $COCl_2(g) \rightarrow CO(g) + Cl_2 (g); \Delta H = 19 \,kcal/mol$ , expressed as $ln \,k = 15 - \dfrac{5000}{T}$
Calculate activation energy for given reaction. $(in \,\,K cal/mol)$ .

$ln\space k = 15 - \dfrac{5000}{T}$

$ln\space k - ln(3.27 \times {10}^{6}) = -\dfrac{5000}{T}$

$ln \dfrac{k}{3.27 \times {10}^{6}} = -\dfrac{5000}{T}$

$k = 3.27\times {10}^{6}\space \times {e}^{-10/RT}$

Here the activation energy is

${10}\space K\space cal/mol$

Calculate the age of a vegetarian beverage whose tritium content is only $15\%$ of the level in living plants. Given $t_{\frac{1}{2}}$ for $_{1}H^{3} = 12.3$ years. ( $log 2 = 0.3, log3 = 0.48$ )

Its a 1st Order Reaction

${t}_{1/2} = \dfrac{0.693}{k}$

$k = 0.056\space {years}^{-1}$

$k = \dfrac{2.303}{t} log\dfrac{100}{15}$

$t = \dfrac{2.303}{0.056} log \dfrac{100}{15}$

$t = 33.88\space years$

In gaseous reaction important for the understanding of the upper atmosphere ${ H }_{ 2 }O$ and $O$ react bimolecularly to form two $OH$ radicals. $\Delta H$ for this reaction is 72 kJ at 500 K and ${ E }_{ a }$ is 77 kJ ${ mol }^{ -1 }$ ${ E }_{ a }$ for the bimolecular recombination of two OH radicals to form ${H}_{2}O$ and $O$ is:
(in kJ ${ mol }^{ -1 }$ )

The reaction is:

$H_2O+O\longrightarrow 2OH$ ,

$\Delta H= -72 kJ$ ;

$E_a=77kJ/mol$

$2OH\longrightarrow H_2O+O$ ,

$E_a=?$ (Refer to Image)

$E_a$ for the reverse reaction

$=(E_a)_{forward}+\Delta H$

$=77+(-72)=5kJ$

1206516_1100025_ans_4b2f4df9b9e741f3b011ae03a0057c98.JPG

For a zero order chemical reaction,
$2N{ H }_{ 3 }(g)\rightarrow { N }_{ 2 }(g)+3{ H }_{ 2 }(g)$
rate of reaction $=0.1$ atm/sec. Initially only $N{ H }_{ 3 }$ is present and its pressure $=3$ atm. Calculate total pressure at $t=10$ sec.

For zero order reaction,

rate = rate constant

$k$

$[A]=-kt+[A_o]$

$(Rate=k[A]^\hat{0})$

Given

${2NH_3}_{(g)}\longrightarrow {N_2}_{(g)}+{3H_2}_{(g)}$

$k=rate=0.1 atm$

$\Rightarrow [NH_3]= 3 atm, t=0,$

$\Rightarrow [N_2]=x$

$\Rightarrow [H_2]=3x$

Total moles=

$3-2x+x+3x=3+2x$

Concentration of

$NH_3$ at

$t=10$ ,

$\Rightarrow [NH_3]=(-0.1\times 10)+3$

$=-1+3=2 atm$

$\therefore 3-2x=2$

$\Rightarrow 3-2=2x$

$\Rightarrow 2x=1$

$\Rightarrow x=0.5$

$\therefore$ Total pressure=

$3+2x=3+2(0.5)$

$=3+1$

$=4 atm$ .

Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

In the given case:

$E_a=209.5kJ$

$mol^{-1}$ =

$209500J$

$mol^{-1}$

$T=581 K$

$R=8.314J$

$K^{-1}$

$mol^{-1}$
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

$x=e^{-Ea/RT}$

$ln^{x}=-E_a/RT$

$log x=-\frac{E_a}{2.303RT}$

$log x=-\frac{209500Jmol^{-1}}{2.303\times 8.314JK^{-1}mol^{-1}\times 581}$

$=18.8323$
Now, x= Anti log

$(18.8323)$

$=Antilog \bar { 19} .1677$

The rate of the reaction quadruples when the temperature changes from $293K$ to $313K$ .Calculate the energy of activation?

$log\dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

$log\space {4} = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{293} - \dfrac{1}{313})$

${E}_{a} = 52.86\space kJ\space {mol}^{-1}$

A first order reaction is 75% completed in 72 min. How long time will it take for
(i) 50% completion (ii) 87.5% completion

$k = \dfrac{2.303}{t} log \dfrac{100}{100 - x}$

$k = \dfrac{2.303}{72} log \dfrac{100}{25}$

$k = 0.019\space {min}^{-1}$

(i)

$t = \dfrac{2.303}{0.019} log \dfrac{100}{50}$

$t = 36 min$

(ii)

$t = \dfrac{2.303}{0.019} log \dfrac{100}{12.5}$

$t = 109.4 min$

For a zero order chemical reaction,
$2NH_{3}(g)\rightarrow N_{2}(g) + 3H_{2}(g)$
rate of reaction = 0.1 atm /sec. Initially only $NH_{3}$ is present and its pressure = 3atm. Calculate total pressure at t = 10 sec.

$2NH_3(g) \longrightarrow N_2(g)+3H_2(g)$

$t=0$ ,

$3atm$

$0$

$0$ rate=

$0.1 atm/sec$

$t=10$ ,

$3-2x$

$x$

$3x$

Given order= zero

$\therefore$ Rate=

$k$ = rate constant=

$0.1$

$\Rightarrow k= \cfrac {[R_o]-[R]}{t}$

$\Rightarrow 0.1= \cfrac {3-[R]}{10} \Rightarrow [R]= 2 atm \longrightarrow (1)$

Total pressure=

$3-2x+x+3x$

$=3+2x \longrightarrow (2)$

$t=10, [R]= 3-2x= 2$ (from (1))

$\Rightarrow x= 0.5$

Substitute

$x$ in equation

$(2) \Rightarrow$ Total pressure=

$3+2(0.5)= 4 atm$ at

$t=10$

The rate of r $\times$ n becomes $4$ times when temp. changes from $293 K$ to $313 K$ Calculate energy of activation (Ea) of the rxn assuming that it does not change with temp. $(R= 8.3/4 JK^-mol^{-1} )$

$log \dfrac{{k}_{2}}{{k}_{1}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

$log(4) = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{293} - \dfrac{1}{313})$

${{E}_{a}} = 52.85\space KJ\space {mol}^{-1}$

The $t_{1/2}$ of first order reaction is $60$ minutes. What percentage will be left after $240$ minutes?

$k = \dfrac{2.303}{t} log \dfrac{A_o}{A_t}$

$t = \dfrac{2.303}{0.693/60} log \dfrac{A_o}{A_t}$

$240 = \dfrac{2.303}{0.693/60} log \dfrac{A_o}{A_t}$

$\dfrac{240 \times 0.639}{2.303 \times 60} = log \dfrac{A_o}{A_t}$

$log \dfrac{A_o}{A_t} = 1.2036$
Suppose

$A_o = 100$

$A_t = 6.25%$ .

Which reaction will have the greater temperature dependence tor the rate constant - one with a small value of energy of activation (E) or one with a large value of activation energy(E)?

The rate of a reaction depends on the temperature at which it is run. As the temperature increases, the molecules move faster and therefore collide more frequently.Thus, the proportion of collisions that can overcome the activation energy for the reaction increases with temperature.

Thus the greater temperature dependence For the rate constant - one with a small value of energy of activation (E).

What is zero order reaction?

In some reactions, the rate is apparently independent of the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant

Write two applications of integrated rate equation.

$(1)$ The integrated rate law helps us to know the amount of reactant or product present after a certain period of time.

$(2)$ It helps to estimates the time required for a reaction to proceed to a certain extent.

$3/4^{th}$ of a first-order reaction is completed in 32 min. Calculate the time required to complete $15/16^{th}$ part of the reaction.

$k =\dfrac{2.303}{32} log \dfrac{1}{1 -3/4}$

$k = 0.043\space {min}^{-1}$

$t = \dfrac{2.303}{0.043} log \dfrac{1}{1 - 15/16}$

$t = 64\space min$

If A $\rightarrow$ products is a first order reaction then, write the integrated law equation.

Image result for 1st order integrated rate law equation

Derive the integrated rate equation for expressing the rate constant of a first order :
$R\rightarrow P$

In this type of reaction, the sum of the powers of concentrations of reactants in rate law is equal to 1, that is the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider the reaction

$R \rightarrow P$ again. Therefore, the rate law for this reaction is,

Rate is directly proportional to [R]

We know that

$[R] = -kt + [R]_0$ ( from equation II). Taking log of both sides, we get

$ln[R] = -kt + ln[R]_0$ ………………………….(IV)

$ln[R]/[R]_0 = -kt$ …………………………….(V)

$k = (1/t) ln [R]_0 /[R]$ ………………………(VI)

Now, consider equation II again. At time t1 and time t2, the equation II will be [R]1 = -kt1 + [R]0 and [R]2 = -kt2 + [R]0 respectively, where [R]1 and [R]2 are concentrations of the reactants at time t1 and t2 respectively. Subtracting second equation from first one, we get

$ln [R]_1– ln[R]_2 = -kt_1 – (- kt_2 )$

$ln[R]_1 /[R]_2 = k (t_2 – t_1)$

$k = [1/(t_2 – t_1)] ln[R]_1 /[R]_2$

Now, taking antilog of both sides of equation V, we get

$[R] = [R]_0e^{-kt}$

Comparing this equation with equation of a straight line y = mx + c, if we plot ln [R] against t, we get a straight line with slope = -k and intercept

$= ln[R]_0$

On removing natural logarithm from equation VI, the first-order reaction can also be written as,

$k = 2.303/t log[R]_0 /[R]$ …………..(VII)

If we plot a graph of

$log[R]_0 /[R]$ against t, we get slope = k/2.303

1236637_1122495_ans_9d42a6cdda3b4c08bf56b56a18b6f167.JPG

A certain reaction is $50\%$ complete in $20\ minutes$ at $300k$ ; the reaction is $50\%$ complete in $5$ minutes at $350k$ calculate the activeity energy, of it is a first order reaction.

${k}_{1} = \dfrac{0.693}{20}$

${k}_{2} = \dfrac{0.693}{5}$

$log \dfrac{{k}_{2}}{{k}_{1}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

$log(4) = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{300} - \dfrac{1}{350})$

${E}_{a} = 24.2 kJ\space {mol}^{-1}$

Activation energy of a reaction is 100 kJ/mol. By using a catalyst $E_a$ decreases by 75%. What will be the effect on rate of reaction at $20^0C$ ?
Consider all other things are equal.

According to Arrhenius equation,

$K=Ae^{-Ea/RT}$

In the absence of catalyst

$K_1=Ae^{-100/RT}$

in the presence of catalyst,

$K_2=Ae^{-25/RT}$

$\therefore \cfrac {K_1}{K_2}=\cfrac {e^{-100/RT}}{e^{-25/RT}}=e^{-75/RT}$

$2303 \log \cfrac {K_1}{K_2}=\cfrac {75}{RT}=\cfrac {75}{8.314\times 10^{-3}\times 298}=30.27$

$\cfrac {K_1}{K_2}=1.38 \times 10^{13}$

A rain in $50\%$ completed om shown and $75\%$ in d $4$ hours. Find the order of reaction :

50% reaction completes in 2 hours. That is, half life of reaction is 2 hours

$t_{(1/2)}$ = 2 hours. Also given that,

$t_{(3/4)}$ = 4 hours

We can observe that,

$t_{(3/4)}=2\times t_{(1/2)}$

Thus relation holds true only in first order

In the equation $K$ = $A.e^ {\dfrac{-Ea}{RT}}$ , what does the term 'A' represent?

'A' represents the Frequency Factor.

In chemical kinetics, the pre-exponential factor or A factor is the pre-exponential constant in the Arrhenius equation, an empirical relationship between temperature and rate coefficient

The solution of ${ H }_{ 2 }{ O }_{ 2 }$ of normality 0.73 is catalytically decomposed. What will be the concentration at the end of 45mins, assuming the decompostion to follow first order rate law if half life is 15 minute?

$k = \dfrac{0.693}{{t}_{1/2}}$

$k = \dfrac{0.693}{15}$ =

$0.0462\space {min}^{-1}$

$0.0462 = \dfrac{2.303}{45} log \dfrac{0.73}{[A]}$

$[A] = 0.091\space {N}$

Derive an expression for integrated rate law for a zero order reaction.

That's the expression for integrated rate law for zero order reaction.
1079261_1125293_ans_d0cf812bd6874b36a7d3fde2efdba5ce.png

Draw energy profile diagram of potential energy with reaction progress. Show that position for the following in the diagram.
(a) Activated complex
(b) Energy of activition

1382325_1131495_ans_f65c0453c37e4e4a82855e019c05827e.png

For a substitution reaction, rate constant increases by factor $10.6$ when temperature changes from ${25}^{o}C$ to ${35}^{o}C$ . Calculate energy of activation of the reaction.

$log\space \dfrac{{{k}_{2}}}{{{k}_{1}}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

Rate increases by a factor 10.6. Substituting the values,

$log\space {10.6}$ =

$\dfrac{{E}_{a}}{2.303 \times 8.314}(\dfrac{1}{298} - \dfrac{1}{308})$

${E}_{a} = 180.18\space kJ\space {mol}^{-1}$

The activation energy for the reaction, 2 $HI(g) \rightarrow H_2 + I_2 (g)$ is $209.5 k J mol^{-1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

Activation energy,

$Ea=209.5KJ/mol=209500J/mol$

Temperature

$=581K, R=8.314JK^{-1}mol^{-1}$

We know that according to Arrhenius equation,

$K=Ae^{-\frac{Ea}{RT}}$

Here,

$e^{-\frac{Ea}{RT}}$ represents the number of molecules which have energy equal or more than activation energy,

So,

$K=Ae^{-\frac{Ea}{RT}}$

$e^{-\frac{Ea}{RT}}=1.47\times 10^{-19}$

For a first order reaction, $A\longrightarrow$ Product, the temperature dependent rate constant (k) was found to follow the equation $logk=-\frac { 2000 }{ T } +6$ calculate the pre-exponential factor (A) and activation energy for the reaction.

Multiply the Equation by 2.303

$2.303 logk = -\dfrac{4606}{T} + 13.818$

$lnk - ln(1.002 \times {10}^{6}) = -\dfrac{4606}{T}$

$k = {1.002} \times {10}^{6} \times {e}^{-4606/T}$

$k = 1.002 \times {10}^{6}$

${e}^{-9212/RT}$

A =

$1.002 \times {10}^{6}$ and

${E}_{a} = 9212$ in Cal

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate ${ E }_{ a }$ .

It is given that T

$_1$ = 298 K

∴T

$_2$ = (298 + 10) K

= 308 K

We also know that the rate of reaction doubles when the temperature is increased by 10 K.

Therefore, let us take the value of k1 = k and that of k2 = 2k

Also, R = 8.314 J K - 1 mol - 1

Now, substituting these values in the equation:

$log\dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

$log \dfrac{2}{1} = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{298} - \dfrac{1}{308})$

${E}_{a} = 52.9 kJ\space {mol}^{-1}$

Derive the relation between half life and rate constant for a first order reaction.

Derivation:-

According to integrated law of rate:-

$k=\dfrac{2.303}{t}log_{10}\dfrac{[A]_o}{[A]_t}$

where

$[A]_o$ is at

$t=0$

the concentration of reactant falls to

$[A_o]/2$ at

$t_{1/2}$

$\therefore t=t_{1/2}$

$[A]_t=A_o/2$

$\therefore$ Equation for first order can be written as:-

$k=\dfrac{2.303}{t_{1/2}}log_{10}\dfrac{[A]_o}{\dfrac{[A]_o}{2}}$

$k=\dfrac{2.303}{t_{1/2}}log_{10}2=\dfrac{0.693}{t_{1/2}}$

$t_{1/2}=\dfrac{0.693}{k}$ .

In a reaction $A\longrightarrow B+C$ , The following data were obtained:
t in seconds 0 900 1800
conc. of A 50.8 19.7 7.62
Prove that it is a first order reaction.

For a 1st order reaction

$k =\dfrac{2.303}{t} log \dfrac{[{A}_{0}]}{[A]}$

Taking the 1st 2

$k = \dfrac{2.303}{900} log \dfrac{[50.8]}{[19.7]}$

$k = 1.05 \times {10}^{-3} {s}^{-1}$

Taking 1 and 3

$k = \dfrac{2.303}{1800} log \dfrac{[50.8]}{[7.62]}$

$k = 1.05 \times {10}^{-3} {s}^{-1}$

So it is a 1st Order Reaction

Sucrose is converted to glucose and fructose of acidic solution which is first order reaction.

Its a

$1$ st Order Reaction.
1074055_1158556_ans_7fe225701a684bb88d936409872df26d.png

The activation energy for the reaction ${ 2HI }_{ (g) }\rightarrow { H }_{ 2 }{ I }_{ 2(g) }$ is 209.5 $209.5\quad { mol }^{ -1 }$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

The Arrenhius Equation is given by,

$k=Ae^{\dfrac{-{E_a}}{RT}}$

here in this equation

$E_a$ is the activation energy and the term

$e^{\dfrac{-{E_a}}{RT}}$ is the fraction of molecules of reactants having energy equal to or greater than activation energy.

Therefore,

$e^{\dfrac{_{E_a}}{RT}}=e^{\dfrac{-209500}{8.314\times 581}}$

$e^{-43.4}$

$\dfrac{1}{e^{43.4}}$

finding the value of

$\dfrac{1}{antilog(43.4)}$ we get

$e^{\dfrac{_{E_a}}{RT}}=1.47\times 10^{-19}$

The rate constant of a reaction is ${ 1.5\times }10^{ 7 }{ S }^{ -1 }{ at\quad 50 }^{ o }$ and ${ 4.5\times }10^{ 7 }{ S }^{ -1 }{ at\quad 100 }^{ o }$ , calculate value of activation energy (Ea) for the reaction. ${ 8.314\quad JK }^{ -1 }{ mol }^{ -1 }$

$log\space \dfrac{{{k}_{2}}}{{{k}_{1}}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

Substituting the values, we get

$log\space {3}$ =

$\dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{323} - \dfrac{1}{373})$

${E}_{a} = 22.01 kJ\space {mol}^{-1}$

The rate constants of a reaction at 500 K and 700 K are ${ 0.02s }^{ -1 } and { 0.07 }s^{ -1 }$ respectively. Calculate the values of Ea and A.

$log\space \dfrac{{{k}_{2}}}{{{k}_{1}}}$ =

$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

Substituting the values,

$log\space \dfrac{0.07}{0.02}$ =

$\dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{500} - \dfrac{1}{700})$

${E}_{a} = 18.23\space kJ\space {mol}^{-1}$

$k = A \times {e}^{-{E}_{a}/RT}$

$0.02 = A \times {e}^{-{18230}/8.314 \times 500}$

$A = 1.6\space {s}^{-1}$

What is the hybridisation of ${{\text{C}}_2}{{\text{H}}_4}$ . Draw its structure also.

hybridisation of

$C$ is

$s{p^2}$

1380916_1246713_ans_c44a9a0e39d64cac9c70a65b2b7b8c30.png

In the $1^{st}$ order reaction $A \longrightarrow$ products $80\%$ of given sample of compound decomposes in $40\ mins$ . What is half life period of reaction ?

1351806_1267667_ans_d2bd26557d5649d9afddc6a1a5e0793d.jpg

Derive the relationship between half life period and rate constant of a zero order reaction.

1352612_1288064_ans_f04f5e2a318e41cb917c5c772aad9bf8.jpeg

For a chemical reaction, what is the effect of a catalyst on the following?
(i) Activation energy of the reaction
(ii) Rate of the reaction.

1350744_1228771_ans_9faef8451a184fabbb657aa8bb6b97e7.jpg

One mole of an organic compound(A) with molecular formula $C_{3}H_{8}O$ reacts completely with two moles of hydroiodic acid at high temperature to form 'x' & 'y'. Identify 'A', 'X' & 'Y' in the reaction.

1321521_1346067_ans_982121566eee444289884a37b6f76dc3.jpg

According to collision theory, write name of two factors which increases the rate of reaction as temperature increase.

As temperature increases

(i) Collision frequency increases and therefore rate of reaction increases

$\implies r\propto Z$

(ii) No. of activated molecules increases.

As, r

$\propto$ No. of activated molecules

$\implies \dfrac{N_a}{N_0}=e^{\dfrac{-E_a}{RT}}$

Write any two characteristics of first order reaction.

Characteristics of first order reaction

$\rightarrow$

The rate constant does not depends upon the concentration of reactant.
Half life period for a first order reaction is constant.

Show that, in a first order reaction, time required for completion of $99.9 \%$ is $10$ times of half life. $(log 10 =1)$

$t_{1/2}=\dfrac{2.303}{K}log2$ ......

$(1)$

$t_{99.9\%}=\dfrac{2.303}{K}log\dfrac{100}{0.1}$

$=\dfrac{2.303}{K}log10^3$

$=\dfrac{2.303}{K}3\,log\,10$ ......

$(2)$

From equation

$(1)$ &

$(2)$

$t_{99.9\%}=10\times t_{1/2}$

A drug is known to be ineffective after it has decomposed $30\%$ . The original concentration of one sample was $500\ units/mL$ . When analysed $20$ months later, the concentration was found to be $420\ units/mL$ . Assuming that decomposition is of $1$ order, what will be the expiration time of the drug. What is the half life of drug.

1759230_1684613_ans_3995a87c530e4e7f9a30d965249e8595.jpg

Define activation energy of a reaction .

The energy required to form the intermediate called activated complex is known as activation energy .

Activation energy = Threshold energy - Average energy of the reactants.

The equation $A + B \rightarrow C$ has zero order . What is the rate equation?

Rate =

$k[A]^0 [B]^0$ or Rate

$= k$ .

Can a reaction have zero Activation Energy ?

$E_a = 0$ , then according to Arrhenius equation

$k = Ae^{-E_a / RT} = Ae^0 = A$ . This implies that every collision results into a chemical reaction which cannot be true. So, a reaction cannot have zero activation energy.

For the reaction $A \rightarrow$ products, the following data is given for a particular run.
time(min): 0 5 15 35
$\dfrac {1}{[A]}(M^{-1}) :$ 1 2 4 8
Determine the order of the reaction.

For the reaction

$A \rightarrow$ products, the following data is given for a particular run.
time(min): 0 5 15 35

$\dfrac {1}{[A]}(M^{-1}) :$ 1 2 4 8
Determine the order of the reaction.
t : 0 5 15 35
[A]: 1 0.5 0.25 0.125
Second order reaction

For the assumed reaction $X_2 + 3Y_2 \rightarrow 2XY_3$ , write the rate of equation in terms of rate of disappearance of $Y_2$ .

Rate

$= - \dfrac{d[X_2]}{dt} = - \dfrac{1}{3} \dfrac{d[Y_2]}{dt} = + \dfrac{1}{2} \dfrac{d[XY_3]}{dt}$

rate of disappearance of

$Y_2 = - \dfrac{d[Y_2]}{dt} = - 3 \dfrac{d[X_2]}{dt} = + \dfrac{3}{2} \dfrac{d[XY_3]}{dt}$

For an elementary reaction $2A + B \rightarrow 3C$ the rate of appearance of $C$ at time 't' is $1.3 \times 10^{-4} \, mol \, L^{-1} \,s^{-1}$ .
Calculate at this time rate of the reaction .

Given that ,

$the \ rate \ of \ appearance \ of$ C

$at \ time \ 't' \ is \ 1.3 \times 10^{-4} \, mol \, L^{-1} \,s^{-1}$

Rate

$= \dfrac{1}{3} \dfrac{d[C]}{dt}$

$= \dfrac{1}{3} \times 1.3 \times 10^{-4} \, mol \, L^{-1}\,s^{-1} = 0.43 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$

From the graph, What is the activation energy for forward reaction ?

Activation energy for the forward reaction

$= 300 - 150$

$= 150 \, kJ \, mol^{-1}$

The decomposition of $NH_{3}$ on platinum surface
$2NH_{3}(g) \overset{Dt}{\rightarrow} N_{2}(g) + 3H_{2}(9)$
is zero order with $k = 2.5 \times 10^{-4} mol L^{-1} S^{-1}$ What are the rates of production of $N_{2}$ and $H_{2}$

$2NH_{3} \rightarrow N_{2} + 3H_{2}$

Rate

$= -1/2 \cfrac {d\left [ NH_{3} \right ]}{dt} = \dfrac{+d\left [ N_{2} \right ]}{dt} = +1/3 \dfrac{d\left [ H_{2} \right ]}{dt}$

$-1/2 \dfrac{d\left [ NH_{3} \right ]}{dt} = \dfrac{d\left [ N_{2} \right ]}{dt} = 1/3\dfrac{d\left [ H_{2} \right ]}{dt}$

$= 2.5 \times 10^{-4} mol L^{-1}S^{-1}$

Write the rate law for a first order reaction . Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant .

Consider the first order reaction ,

$R \rightarrow P$

For this reaction , rate law which relates the rate of reaction to the concentration of reactants can be given as

Rate

$= \dfrac{-d[R]}{dt} = k[R]$

For a first order reaction ,

$t = \dfrac{2.303}{k} log \dfrac{[R]_0}{[R]}$ , where

$[R]_0$ = initial concentration ,

$[R] =$ concentration at time

$t$ .

At

$t_{1/2} , [R] = [R]_0 / 2$

So , the above equation becomes

$t_{1/2} = \dfrac{2.303}{k} log \dfrac{[R]_0}{[R]_0 / 2}$

$t_{1/2} = \dfrac{2.303}{k} log\,2$ or

$t_{1/2} = \dfrac{2.303}{k} \times 0.3010$

$t_{1/2} = \dfrac{0.693}{k}$

The rate law for a reaction is expressed as : rate $= k[Cl_2][NO]^2$
Find out the orders of the reaction with respect to $Cl_2$ and with respect to $NO$ and also the overall order of the reaction .

If Rate Law for a reaction is expressed as

$r=k[A]^p[B]^q$

then

order of reaction with respect to A is 'p'

order of reaction with respect to B is 'q'

overall order of the reaction is 'p+q'.

Hence, for the Rate Law :

$r=k[Cl_2][NO]^2$

order of reaction with respect to

$Cl_2$ is

$'1'$

order of reaction with respect to

$NO$ is

$'2'$

overall order of the reaction is

$'3'$ .

Define Rate equation with example.

An expression that relates the rate (velocity) of the reaction with change in concentration of a reactant in a given interval of time.

$+ \dfrac{d[A]}{dt} = \dfrac{+1}{2} \dfrac{d[B]}{dt} = \dfrac{-1}{3} \dfrac{d[C]}{dt} = \dfrac{-1}{4} \dfrac{d[D]}{dt}$

Define Threshold Energy of a Reaction.

Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecules.

Write the rate equation for the reaction $A_2 + 3B_2 \rightarrow 2C$ , if the overall order of the reaction is zero .

Since overall order of the reaction is Zero. Rate of reaction is given by,

Rate

$= k[A_2]^0 [B_2]^0$ or Rate

$= k$

Explain why $H_2$ and $O_2$ do not react at room temperature .

$H_2 \ \& \ O_2$ do not react at room temperature because Activation Energy of the Reaction is very hgh.

Observation No. Time (in minute) $P_x$ (in mm of Hg)
1 0 800
2 100 400
3 200 200
At constant temperature and volume $X$ decomposes as $2\mathrm{X}(\mathrm{g})\rightarrow 3\mathrm{Y}(\mathrm{g})+2\mathrm{Z}(\mathrm{g})$ , where $\mathrm{P}_{x}$ is the partial pressure of $X$ . What is the order of reaction with respect to $X$ ?
(i) Find the rate constant.
(ii) Find the time for 75% completion of the reaction.
(iii) Find total pressure when pressure of $X$ is 700 mm of Hg.

Applying integrated rate law of first order kinetics

$\mathrm{k}_{1}=\dfrac{2.303}{\mathrm{t}}\log(\dfrac{\mathrm{P}_{\mathrm{o}}}{\mathrm{P}_{\mathrm{t}}})$

$=\dfrac{2.303}{100}\log(\dfrac{800}{400})$

$=\displaystyle \dfrac{2.303}{100} log\ 2 =6.932\displaystyle \times 10^{-3}min^{-1}$

In the two observations the values of rate constant remain same. Therefore it must be following the first order kinetics. Hence order of reaction with respect to

$X$ is one.

(i) Rate constant,

$\displaystyle \mathrm{k}=6.932\times 10^{-3}min^{-1}$

(ii)

$6.932\displaystyle \times 10^{-3}=\dfrac{2.303}{\mathrm{t}}\log(\dfrac{100}{25})$

$\therefore\ \displaystyle \mathrm{t}_{75\%}=200\min$ .

(iii)

$2\mathrm{X}(\mathrm{g})\rightarrow 3\mathrm{Y}(\mathrm{g})+2\mathrm{Z}(\mathrm{g})$

In:

$800\ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ 0$

Fi:

$(800-2P) \ \ \ 3 \mathrm{P} \ \ \ \ \ \ \ 2\mathrm{P}$

When the pressure of

$X$ is 700 mm of Hg then

$800-2\mathrm{P}=700$

$2\mathrm{P}=100$

$\mathrm{P}=50$ mm of Hg

Total pressure

$=800-2\mathrm{P}+3\mathrm{P}+2\mathrm{P}$

$=800+3\mathrm{P}$

$=800+3\times 50=950$ mm of Hg

In case of unimolecular reaction, the time required for 99.9% of the reaction to take place is x times that required for half of the reaction. The value of x is_________.

As we know,
for first order reaction

$t = (2.303/k)log(a/a-x)$

$t_1 = (2.303/k)log(a/a-a/2)$

$t_2 = (2.303/k)log(1/1-0.1)$

$t_2 = 10t_1$

The specific rate constant of the decomposition of $N_2O_5$ is $0.008 min^{-1}$ . The volume of $O_2$ collected after 20 minute is 16 mL. The volume in ml that would be collected at the end of reaction. $NO_2$ formed is dissolved in $CCl_4$ is

$k=0.008 \: /min, t=20 \: min, x=16 \: mL$
The unit of the rate constant indicates that it is a first order reaction.

$k=\frac {2.303}{t}log\frac {a}{a-x}$

$0.008=\frac {2.303}{20}log \frac {a}{a-16}$

$log \frac {a}{a-16}=1.17$

$a=108.23$
Henc, 108.23 mL would b collectd at the end of reaction.

The acid catalysed hydrolysis of an organic A at $30^oC$ has a half-life of 100 minutes when carried out in a buffer solution of $pH=5$ and 10 minutes when carried out at $pH=4$ . Both the times the half-life are independent of the initial concentration of A. If the rate of reaction is given by $rate=K[A]^m[H^+]^n$ , $m+n$ is ___________.

The half life period is independent of the initial concentration. This is characteristic of first order reaction. Thus, the reaction is first order w.r.t [A] and first order w.r.t

$[H^+]$ .
Hence,

$rate=k[A]^1[H^+]^1$
Thus,

$m=1,n=1$

$m+n=1+1=2$ .

Dimethyl ether gaseous phase decomposition is: $CH_3OCH_3\rightarrow CH_4+H_2+CO$ at 750 K having rate constant $6.72\times 10^{-3} min^{-1}$ . The time in which initial pressure of 400 mm in closed container becomes 750 mm in hours (to the nearest integer) is________.

$CH_3OCH_3\rightarrow CH_4+H_2+CO$

$a=400 \: mm, x=\frac {750 \: mm-400 \: mm}{2}=175 \: mm, a-x=400 \: mm-175 \: mm=225 \: mm$

$t=\frac {2.303}{t}log \frac {a}{a-x}=\frac {2.303}{6.72 \times 10^{-3}} log\frac {400 \: mm}{225 \: mm}=85.64$ .
Hence, the time in which the initial pressure of 400 mm in closed container becomes 750 mm is 85.64 min.

A reaction $2A\rightarrow Products$ , has a rate constant $3.5\times 10^{-4} mol^{-1} litre^{-1} sec^{-1}$ . If the time required for the concentration of A to change from 0.260 M to 0.011 M $x\times 10^5 sec$ then $x$ is___________.(to the nearest integer)

$k=3.5 \times 10^4 \: Lit/mol/s$ ,

$a=0.260 \: M, \: a-x=0.011 \: M, \: x=0.249 \: M, \: t=x \times 10^5 \: s$
The units of the rate constant indicates that the reaction is second order reaction. For second order reaction,

$t=\dfrac {x}{ka(a-x)}=\dfrac {0.249}{3.5 \times 10^{-4} \times 0.260 \times 0.011}=2.49 \times 10^5 \: s =x \times 10^5 \: s$ .
Hence,

$x=2.49 \approx 2$ .

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. The value of $\frac {[t_{1/8}]}{[t_{1/10}]}\times 10$ is ______ (take $log_{10}2 = 0.3$ )

$t=\dfrac {2.303}{k}log\left( \frac {a}{a-x}\right)$

For decomposition to 1/8 of its initial concentration

$t_{1/8}=\dfrac {2.303}{k}log \left(\frac {1}{1/8}\right)=\dfrac {2.08}{k}$

For decomposition to 1/10 of its initial concentration

$t_{1/10}=\dfrac {2.303}{k}log\left( \frac {1}{1/10}\right)=\dfrac {2.303}{k}$

Hence,

$\dfrac {[t_{1/8}]}{[t_{1/10}]}\times 10 = \dfrac {\dfrac {2.08}{k}}{\frac {2.303}{k}} \times 10 = 9$

The gas phase decomposition of dimethyl ether follows first order kinetics,

$CH_3OCH_3(g)\rightarrow CH_4(g)+H_2(g)+CO(g)$

The reaction is carried out in a constant volume container at $500^oC$ and has a half-life of 14.5 minute. Initially only dimethyl ether is present at a pressure of 0.40 atmosphere. The total pressure of the system after 12 minutes assuming ideal gas behaviour in atm is__________. (write answer as the first integer after decimal, e.g., if answer is $2.56$ then answer should be $5$ )

$CH_3OCH_3(g)\rightarrow CH_4(g)+H_2(g)+CO(g)$
At

$t=0$

$0.4$

$0$

$t=12 min$

$(0.4 -P)$

$P$

For ideal gas behaviour mole

$\propto pressure$

$\therefore a\propto 0.40, (a-X)\propto (0.04-P)$

$\because K=\dfrac {2.303}{t}log \dfrac {a}{(a-X)}$

$\dfrac {0.693}{14.5}=\dfrac {2.303}{12} log \dfrac {0.40}{(0.40-P)}$

$\therefore P=0.175 atm$

Thus, total pressure after 12 minutes

$=0.04-P+P+P+P$

$=0.40+2P$

$=0.40+2\times 0.175$

$=0.75 atm$

Answer is 7.

If the percentage decompostion in $100$ minutes at the same temperature but at an initial pressure of $600$ mm is $(70+x)$ find the value of $x$ __________.

$\because K=\dfrac {2.303}{t}log \frac {a}{(a-X)}$

Case I:

$a\propto 200mm$ ;

$X\propto 200\times (50/100)mm$ and

$t_{1/2}=53 minute$

$\therefore K_1=\dfrac {2.303}{53} log \dfrac {200}{200-100}$

$=1.307\times 10^{-2} minute^{-1}$

Case II:

$a\propto 200 mm; X\propto 200\times (73\times 100)\propto 146 mm$

$\therefore K_2=\dfrac {2.303}{100}log \dfrac {200}{200-146}$

$=1.309\times 10^{-2} \ minute^{-1}$

For a Ist order reaction,

$t_{(1/n)}\propto (a)^0$ and therefore if initial pressure is 600 mm, the decomposition in 100 minutes will be 73%.

Half-life period of reation in minutes is:
(Give answer to the nearest integer)

To provide a long time or heating to a reaction mixture means that reaction has gone to completion:

$A\rightarrow B+2C$

Mole before dissociation	a	0	0
Mole after dissociation	a-x	x	2x
Mole after complete dissociation	0	a	2a

$\because$ Total mole at a time

$\propto$ pressure at that time

$a\propto P_0 {\;} t=0$ ......(i)

$a+2X\propto 264 {\;} t=14$ minute ....(ii)

$3a\propto 264 {\;} t=\infty$ .....(iii)

$t_{1/2}=\dfrac {0.693}{K}=\dfrac {0.693}{3.415\times 10^{-2}}=20.29$ min

So, the answer is

$20$ mins.

If pressure of $Cl_2O_7$ after $100$ sec in atm is $x$ , then $1000x$ is__________.

For first order reaction,

$k=\dfrac {2.303}{t}log \dfrac {a}{a-x}$

At 55 seconds,

$k=\dfrac {2.303}{55}log \dfrac {0.062}{0.044}=6.2365 \times 10^{-3}$

At 100 seconds,

$6.2365 \times 10^{-3} = \dfrac {2.303}{100}log \dfrac {0.062}{x}$

$log \dfrac {0.062}{x}=0.2708$

$\dfrac {0.062}{x}=1.8655$

$x=0.033$

$1000x=33$

For a gaseous reaction $A\, \rightarrow\, B\, +\, 2C$ at $250^\circ$ C, following data were observed. The partial pressure of the gases at 100th min will be (in 100x mm)
Time (min) 0 20 40 60
Total pressure (mm Hg) 400 500 587.6 664

$A\, \rightarrow\, B\, +\, 2C$
t = 0

$a\, \quad\, 0\, \quad\, 0$
t = t

$a\, -\, x\, \quad\, x\, \quad\, 2x$

$P_{total}\, \alpha\, a\, -\, x\, +\, 2x\, \alpha\, a\, +\, 2x$
a. i.

$a\, \alpha\, 400$

$400\, +\, 2x\, \alpha\, 500$

$x\, \alpha\, 50$

$k\, =\, \displaystyle \frac{2.303}{20}log\displaystyle \frac{400}{400\, -\, 50}\, =\, 0.00666\, min^{-1}$
ii.

$a\, \alpha\, 400$

$400\, +\, 2x\, \alpha\, 664$

$2x\, \alpha\, 264$

$x\, \alpha\, 132$

$k\, =\, \displaystyle \frac{2.303}{60}log\displaystyle \frac{400}{400\, -\, 132}\, =\, 0.00667\, min^{-1}$
Since the value of k is constant, hence first order.
b. 0.00667 =

$\displaystyle \frac{2.303}{100}log\displaystyle \frac{400}{400\, -\, x}$
log x = 0.2896
x = 1.949 mm
Partial pressure of gases at 100 min = a + 2x
= 400 + 2

$\times$ 1.949
= 403.89 mm

The decomposition of a compound is found to follow a first-order rate law. If it takes 15 min for 20% of the original material to react, the specific rate constant is ________. ( $in \:\:min^{-1}$ )
[If the answer is X, write the value of 1000X in nearest integer value possible]

We know for the first order,

$k\, =\, \displaystyle \frac{2.303}{t}log\displaystyle \frac{a}{a\, -\, x}$

Given that,

$t=15;\ x=0.2a$ where

$a=$ initial concentration.

$k=\, \displaystyle \frac{2.3}{15\, min}log \left (\displaystyle \frac{a}{a\, -\, 0.20a} \right )$

$= 0.01488$

$min^{-1}$

$0.01488\times 1000 \approx 15$

Ans-

$15$

Catalytic decomposition of nitrous oxide by gold at $900^{0}C$ at an initial pressure of 200 mm was 50 % in 53 min and 73 % in 100 min. The amount that will compose in 100 min at the same temperature but at an initial pressure of 600 mm is (multiply the answer by 100) :

We have, $K = (\frac{2.303}{t})log10 [\frac{a}{(a – x)}]$ --- (1)
Case I : $t_{1/2} = 53 \space minutes$
$\therefore \space \space \space \space \space K_1 = (\frac{2.303}{53}) log10 [\frac{200}{(200 – 100)}] = 1.307×10^2 \space minute^{-1}$

Case II : $t_{73} = 100 \space minutes$
$\therefore \space \space \space \space \space K_2 = (\frac{2.303}{100})log10 [\frac{200}{(200 – 146)}] = 1.309 \times 10^2 minute^{-1}$

Since the value of K is constant. Hence it is first order reaction, the time required to complete any fraction is independent of its initial concentration of reactant.

$\therefore \space$ 73% of $N_2O$ will decomposes when the initial concentration is 600 mm which corresponds to a pressure of
$\implies \frac{600 \space mm}{100} = 4.38 mm$

If the initial concentration of reactants in a certain reaction is doubled, the half-life period of the reaction doubles. The order of the reaction is:

1958358_355082_ans_db950f8698d244beac1e1f97483643ac.jpg

Reaction $A+B\longrightarrow C+D$ follow's following rate law rate $=k={ \left[ A \right] }^{ \frac { 1 }{ 2 } }{ \left[ B \right] }^{ \frac { 1 }{ 2 } }$ . Starting with initial conc. of one mole of A and B each, what is the time taken for amount of A of become 0.25 mole. Given $k = 2.31\times 10^{-3} sec^{-1}$ . (in sec). (Answer in the form of 100X)

$A+B\longrightarrow C+D\\ rate=k\left[ A \right] ^{ 1/2 }\left[ B \right] ^{ 1/2 }\\ \displaystyle \frac { dx }{ dt } =k\sqrt { (a-x)(a-x) } \\ \displaystyle \frac { dx }{ dt } =k(a-x)\\ \Rightarrow t= \displaystyle \frac { 2.303 }{ k } \log_ { 10 } \left( \displaystyle \frac { a }{ a-x } \right) \\ t= \displaystyle \frac { 2.303 }{ 2.31\times 10^{ -3 } } \log_ { 10 } \left( \displaystyle \frac { 1 }{ 0.25 } \right) \\ t=600\quad sec.$

The rate for the decomposition of $NH_{3}$ on platinum surface is zero order. What are the rate of production of $N_{2}$ and $H_{2}$ if $K = 2.5 \times 10^{-4} mol\ litre^{-1} s^{-1}$ .

$\displaystyle 2NH_3 \rightarrow N_2 + 3H_2$
Rate of reaction

$\displaystyle = -\frac {1}{2}\frac {d}{dt}[NH_3] =\frac {d}{dt}[N_2]=\frac {1}{3}\frac {d}{dt}[H_2]=k$

For a zero order reaction,
Rate

$\displaystyle = k = 2.5 \times 10^{-4} \: M/s$

Rate of production of

$\displaystyle N_2 = \frac {d}{dt}[N_2] = 2.5 \times 10^{-4} \: M/s$
Rate of production of

$\displaystyle H_2 = \frac {d}{dt}[H_2] = 3 \times 2.5 \times 10^{-4} \: M/s = 7.5 \times 10^{-4} \: M/s$

The recombination of methyl radicals has no activation barrier. The activation energy values for the first step in the thermal thermal decomposition of acetaldehyde given and $E_{photochemical}$ are $309.32\;kJ\;mol^{-1}$ and $41.8\;kJ\;mol^{-1}$ respectively.
Calculate the overall activation energy for the thermal decomposition of acetaldehyde.

${ E }_{ thermal }={ E }_{ 2 }+\frac { 1 }{ 2 } { E }_{ 1 }-\frac { 1 }{ 2 } { E }_{ 4 }=1/2{ E }_{ 1 }+{ E }_{ photo }\\ (as,{ E }_{ photo }={ E }_{ 2 }-1/2{ E }_{ 4 })\\ { E }_{ thermal }=1/2\times 309.32+41.8=196.46kJ/mol$

The rate constant for a first order reaction is $60\ s^{-1}$ . How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

Given that the rate constant for a first order reaction is

$60\ s^{-1}$ .

Let

$a$ M be the initial concentration.

Final concentration will be

$\displaystyle \frac {a}{16}$ M.

Rate constant,

$\displaystyle k = 60 \: /s$

$\displaystyle t = \frac {2.303}{k} log\left( \frac {[A]_0}{[A]}\right)$

$\displaystyle t = \frac {2.303}{60} log \left(\frac {a}{\dfrac {a}{16}}\right)$

$\displaystyle t = 4.6 \times 10^{-2}$ seconds

For the decomposition of azo-isopropane to hexane and nitrogen at $543 K$ , the following data are obtained.
$t (sec)$ $P$ (mm of $Hg$ )
$0$ $35.0$
$360$ $54.0$
$720$ $63.0$
Calculate the rate constant.

Let

$\displaystyle p_i$ be the initial pressure of azoisopropane and at time

$t,\ p$ atm of azoisopropane decomposes.

$(CH_3)_2CHNH=NHCH(CH_3)_2 \longrightarrow N_2 + C_6H_{14}$

At time,

$t=0$

$\displaystyle P_{azoisopropane} = P_i \ \ \ \ \: P_{N_2} = 0 \ \ \ \ \: P_{C_6H_{14}}=0$

At time,

$t$

$\displaystyle P_{azoisopropane} = P_i-p \ \ \ \: P_{N_2} = p \ \ \ \: P_{C_6H_{14}}= p$

(The pressure is in atm.)

Total pressure,

$P_t = (P_i - p) + p + p$

$\displaystyle p = P_t - P_i$

$\displaystyle P_{azoisopropane} = 2P_i - P_t$

$\displaystyle k = \dfrac {2.303}{t} log \dfrac {[A]_0}{[A]}$

$\displaystyle k - \dfrac {2.303}{t}log \dfrac {P_i}{2P_i-P_t}$

When time is

$360$ s,

$\displaystyle k =\dfrac {2.303}{360} log \dfrac {35.0}{2 \times 35.0-54}$

$\displaystyle k = 2.17 \times 10^{-3} /s$

When time is

$720$ s,

$\displaystyle k = \dfrac {2.303}{720} log \dfrac {35.0} {2 \times 35.0 - 63.0}$

$\displaystyle k = 2.24 \times 10^{-3} /s$

Average value of k

$\displaystyle = 2.20 \times 10^{-3} /s$

The following data were obtained during the first order thermal decomposition of $SO_{2}Cl_{2}$ at a constant volume.
$SO_{2}Cl_{2}(g)\rightarrow SO_{2}(g) + Cl_{2} (g)$
Experiment $Time/s^{-1}$ Total pressure/ atm
$1$ $0$ $0.5$
$2$ $100$ $0.6$
Calculate the rate of the reaction when total pressure is $0.65$ atm.

For a first order reaction, we have:

$\displaystyle k = \dfrac {2.303}{t} log \dfrac {P_o}{2P_o-P_t}$

$\displaystyle k = \dfrac {2.303}{100} log \dfrac {0.5}{2 \times 0.5 - 0.6}$

$\displaystyle k = 2.23 \times 10^{-23} /s$

Rate

$\displaystyle = kP_{SO_2Cl_2}$

The total pressure is

$0.65$ atm.

$\displaystyle P_{SO_2Cl_2} = 2p_o-p_t = 2 \times 0.5 - 0.65 = 0.35 \: atm$

Hence, rate

$\displaystyle = 2.23 \times 10^{-3} \times 0.35$

Rate

$\displaystyle = 7.805 \times 10^{-4} \: atm/s$

Consider a certain reaction $A\rightarrow$ Products with, $k = 2.0 \times 10^{-2}s^{-1}$ . Calculate the concentration of A remaining after $100$ s, if the initial concentration of A is $1.0 mol L^{-1}$ .

The integrated rate expression for the first order reaction is as follows:

$\displaystyle k = \dfrac {2.303}{t} log \dfrac {[A]_0}{[A]}$

$\implies \displaystyle 2.0 \times 10^{-2} = \dfrac {2.303}{t} log\dfrac {1.0}{[A]}$

$\implies \displaystyle log \dfrac {1.0}{[A]} = 0.868$

$\implies \displaystyle [A] = 0.135 M$
Hence, the concentration of

$A$ remaining after

$100$ seconds is

$0.135$ M.

$10\ g$ atoms of an $\alpha$ -active radioisotope are disintegrating in a sealed container. In one hour, the helium gas collected at STP is $11.2{ cm }^{ 3 }$ . Calculate the half-life of the radioisotope.

Amount of radioactive isotope

$=10 g$ atoms

${ N }_{ 0 }=10\times 6.023\times { 10 }^{ 23 }$ atoms

$=6.023\times { 10 }^{ 24 }$ atoms

$22400{ cm }^{ 3 }$ of helium contains

$=6.023\times { 10 }^{ 23 }$ atoms

$11.2{ cm }^{ 3 }$ of helium will contain

$=\dfrac { 6.023\times { 10 }^{ 23 } }{ 22400 } \times 11.2$ atoms

$=3.01\times { 10 }^{ 20 }$ atoms

As one helium atom is obtained by the disintegration of one atom of radioisotope, the total number of atoms of the radioactive isotope which have disintegrated in one hour.

$=3.01\times { 10 }^{ 20 }$ or

$\times 0.0003\times { 10 }^{ 24 }$

The number of atoms of the radioactive isotope left after one hour,

$N=\left( 6.023\times { 10 }^{ 24 }-0.0003\times { 10 }^{ 24 } \right)$

$=6.0227\times { 10 }^{ 24 }$

Using,

$\lambda =\dfrac { 2.303 }{ t } \log { \dfrac { { N }_{ 0 } }{ N } }$

$\lambda =\dfrac { 2.303 }{ t } \log { \dfrac { 6.023\times { 10 }^{ 24 } }{ 6.0227\times { 10 }^{ 24 } } }$

$=2.303\times 2.1632\times { 10 }^{ -5 }=4.982\times { 10 }^{ -5 }{ hr }^{ -1 }$

${ t }_{ { 1 }/{ 2 } }={ 0.693 }/{ \left( 4.982\times { 10 }^{ -5 }\times 24\times 365 \right) }=1.58$ years.

In hydrogenation reaction at ${ 25 }^{ o }C$ , it is observed that hydrogen gas pressure falls from $2$ $atm$ to $1.2$ $atm$ in $50$ $min$ . Calculate the rate of reaction in molarity per $sec$ . $\left( R=0.0821\ litre\cdot atm\ { K}^{ -1 }\ { mol }^{ -1 } \right)$

Rate

$=\dfrac { dP }{ dt } =\dfrac { 2-1.2 }{ 50\times 60 }$

$=2.666\times { 10 }^{ -4 }$

$atm$

${ s }^{ -1 }$

$PV=nRT$

$\dfrac { P }{ sec } =\left( \dfrac { n }{ V } \right) \dfrac { RT }{ sec }$

$\left[ \dfrac { n }{ V } \right] \dfrac { 1 }{ sec } =\left[ \dfrac { P }{ sec } \right] \dfrac { 1 }{ RT }$
Rate (molarity/sec)

$=\dfrac { 2.666\times { 10 }^{ -4 } }{ 0.0821\times 298 }$

$=1.09\times { 10 }^{ -5 }mol{ litre }^{ -1 }{ s }^{ -1 }$

The first order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete 80%.

For first order reaction we have:-

$K=\cfrac {1}{t} ln \left(\cfrac {a}{a-x}\right)$

$- (i)$

where,

$K$ = rate constant

$t$ = time

$a$ = initial concentration of the reactant

$x$ = amount of reactant reacted in time

$t$

Case I:

$t=20$ min

$x=\cfrac {20}{100}\times a$

$\Rightarrow x=0.2a$

$(\because 20$ % reaction is complete

$)$

$\Rightarrow$

$a-x=0.8a$

Putting the values in

$eq^n1$ :-

$K=\cfrac {1}{20} ln \left(\cfrac{a}{0.8x}\right)$

$- (ii)$

Case II:

$t=?$ ,

$x=0.8a$

$\Rightarrow a-x=0.2a$

Putting the values in (i):-

$\Rightarrow K= \cfrac {1}{t} ln \left(\cfrac {x}{0.2x}\right)$

$- (iii)$

From

$(ii)$ and

$(iii)$ :-

$\cfrac {1}{20} ln \left(\cfrac {10}{8}\right)=\cfrac {1}{t} ln \left(\cfrac {10}{2}\right)$

$\Rightarrow t=20\times \cfrac {ln S}{ln(S/4)}=144.25$ min

The reaction
$2{ N }_{ 2 }{ O }_{ 5 }\left( g \right) \rightarrow 4N{ O }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right)$
Follows first order kinetics. The pressure of a vessel containing only ${ N }_{ 2 }{ O }_{ 5 }$ was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be :(Assume temperature remains constant)

$2N_2O_5(g)\longrightarrow 4NO_2(g)+O_2(g)$

$t=0$

$P_0$

$0$

$t=t$

$P_0-P$

$2P$

$P/2$

Total pressure after time

$t=P_0-P+2P+P/2$

Total=

$P_0+3P/2$

Given,

$P_0=50mm$

$Hg$ &

$P_{total}=87.5mm$

$Hg$ at

$t=30$ min

$\Rightarrow 87.5=50+3P/2$

$\Rightarrow 3P/2=37.5 \Rightarrow P=\cfrac {37.5\times 2}{3}=25mm$

$Hg$

Since the reaction follows first order kinetics,

so,

$Kt=ln\cfrac {P_0}{P_0-P}$

$\Rightarrow Kt=ln\left(\cfrac {50}{25}\right)$ (at

$t=30$ min)

$\Rightarrow K=\cfrac {1}{30}ln2$

$- (i)$

After

$60$ min,

$\Rightarrow K=\cfrac {1}{60}ln\left(\cfrac {50}{50-P'}\right)$

$- (ii)$

$\Rightarrow \cfrac {1}{30}ln2=\cfrac {1}{60}ln\left(\cfrac {50}{50-P'}\right)$ (From

$(i)$ )

$\Rightarrow 2ln^2=ln(50/50-P')$

$\Rightarrow ln2^2=ln (50/50-P')$

$\Rightarrow 4=\cfrac {50}{50-P'} \Rightarrow 200-4P'=50$

$\Rightarrow 4P'=150$

$\Rightarrow P'=37.5mm$

$Hg$

So, total pressure at

$t=60$ min=

$P_0+3P'/2$

$=50+\cfrac {3\times 37.5}{2}=106.25mm$

$Hg$

For a first order reversible reaction $A\underset {K_{b}}{\xleftarrow {K_{f}}}B$ , the initial concentration of $A$ and $B$ are $[A]_{0}$ and zero respectively. If concentrations at equilibrium are $[A]_{eq.}$ and $[B]_{eq.}$ , derive an expression for the time taken by $B$ to attain concentration equal to $[B]_{eq/2}$ .

The given reaction is

$K_f$

$A\quad \rightleftharpoons\quad B$

$K_b$

Initial concentration of

$A=[A]_0$ &

$B=0$

Concentration at equilibrium are

$[A]_{eq}$ &

$[B]_{eq}$ respectively.

Now,

Rate=

$\cfrac {d[A]}{dt}=K_f[A]-K_b[B]$

$- (i)$

At,

$t=0, [B]=0$

$\Rightarrow [B]=[A]_0-[A]$

$-(ii)$

Using

$(ii)$ in

$(i)$ :-

$\therefore$

$\cfrac {d[A]}{dt}=K_f([A])-K_b([A]_0-[A])$

$\cfrac {d[A]}{dt}=(K_f+K_b)[A]-K_b[A]_0$

$-(iii)$

At equilibrium,

$\cfrac {d[A]}{dt}=0$ &

$[A]=[A]_{eq}$

$\Rightarrow (K_f+K_b)[A]_{eq}-K_b[A]_0=0$

$\Rightarrow [A]_{eq}=\cfrac {K_b}{K_f+K_b}[A]_0$

$- (iv)$

Now, from

$eq^n (i)$ , at equilibrium, we have:-

$K_f[A]_{eq}-K_b[B]_{eq}=0$

$\Rightarrow [B]_{eq}=\cfrac {K_f}{K_b}[A]_{eq}$

$\Rightarrow [B]_{eq}=K_{eq}[A]_{eq}$

$- (v)$

$\left(\because K_{eq}=\cfrac {K_f}{K_b}\right)$

Now, Integrating

$eq^n (iii)$ after substituting

$[A]_{eq}$ from

$(iv)$ :-

$\Rightarrow [A]=[A]_{eq}+C_1e^{-(K_f+K_b)t}$

$- (vi)$

;

$C$ is integration constant

At,

$t=0,[A]=[A]_0$

$\Rightarrow [A]=[A]_{eq}+([A]_0-[A]_{eq})e^{-(K_f+K_b)t}$

$- (vii)$

Now, when

$[B]=[B]_{eq}/2$ (From

$(ii)$ ) -

$(viii)$

Putting

$(viii)$ in

$(vii)$ :-

$\therefore$

$[A]_0-\cfrac {[B]_{eq}}{2}=[A]_{eq}+([A]_0-[A]_{eq})e^{-(K_f+K_b)t}$

$\Rightarrow e^-(K_b+K_f)t=\cfrac {2[A]_0-[B]_{eq}-2[A]_{eq}}{2([A]_0-[A]_{eq})}$

Taking natural log on both sides:-

$\Rightarrow -(K_f+K_b)t=ln \left(\cfrac {2[A]_0-[B]_{eq}-2[A]_{eq}}{2([A]_0-[A]_{eq})}\right)$

$\Rightarrow t= \cfrac {-1}{(K_f+K_b)} ln \left(\cfrac {2[A]_0-[B]_{eq}-2[A]_{eq}}{2([A]_0-[A]_{eq})}\right)$

$A$ and $B$ are two different chemical species undergoing 1st order decomposition with half lives equal to $5$ sec and $7.5$ sec respectively. If the initial concentration of $A$ and $B$ are in the ratio $3:2$ . Calculate $\cfrac { { C }_{ { A }_{ 1 } } }{ { C }_{ { B }_{ 1 } } }$ after three half lives of $A$ . Report your answer after multiplying it with $100$ .

Half life of a reaction is the time required for reaction to be half completed.

Amount of substance left unreacted after one half life=

$\cfrac {[A]_0}{2}$

Amount of substance left unreacted after two half lives=

$\cfrac{[A]_0}{2n}$

Half life of

$A=5$ sec.

So, after

$3$ half lives of

$A, t=15$ sec

Half life of

$B=7.5$ sec

No. of half lives

$B$ undergoes=

$\cfrac {15}{7.5}=2$

Now, Amount of

$A$ left after

$3$ half lives,

$A_1=\cfrac {[A]_0}{23}=\cfrac {[A]_0}{8}$

Amount of

$B$ left after

$2$ half lives,

$B_1=\cfrac {[B]_0}{2^2}=\cfrac{[B]_0}{4}$

So,

$\cfrac {C_{A1}}{C_{B1}}=\cfrac {[A]_0}{8}\times \cfrac {4}{[B]_0}$

$=\cfrac {1}{2} \cfrac {[A]_0}{[B]_0}$

Now,

$\cfrac {[A]_0}{[B]_0}=\cfrac {3}{2}$ (given)

$\Rightarrow \cfrac {C_{A1}}{C_{B1}}=\cfrac {3}{2}\times \cfrac {1}{2}=\cfrac {3}{4}$

As stated in the question,

$100\times \cfrac {C_{A1}}{C_{B1}}= \cfrac {3}{4}\times 100= 75$ (after multiplying by 100)

In a zero order reaction, the time taken to reduce the concentration of reactant from 50% to 25% is 30 minutes.What is the time required to reduce the concentration from 25% to 12.5% ?

For a zero order reaction we have:-

$K=\cfrac {1}{t}\left([A]_o-[A]_t\right)$

$- (i)$

where,

$K$ = rate constant

$t$ = time

$[A_o]$ = concentration of reactant at

$t=0$

$[A]_t$ = concentration of reactant at

$t=t$

Case I:- Let,

$[A]$ be initial concentration

$t=30$ min

$[A]_o=0.5[A]$

$[A]_t=0.25[A]$

Putting the value in

$eq^n$ (i):-

$K=\cfrac {1}{30}\left(0.5[A]-0.25[A]\right)$

$K=\cfrac {1}{30}\times 0.25[A]$

$- (ii)$

Case II:-

$t=?$ ,

$[A]_o=0.25[A]$ ,

$[A]_t=0.125[A]$

Putting the values in

$eq^n (i)$

$\Rightarrow K=\cfrac {1}{t} \left(0.25[A]-0.125[A]\right)$

$\Rightarrow K=\cfrac {1}{t}\times 0.125[A]$

$-(iii)$

From

$(ii)$ and

$(iii)$ :-

$\Rightarrow \cfrac {1}{30}\times 0.25[A]=\cfrac {1}{t}\times 0.125 [A]$

$\Rightarrow t=15$ min

$\Rightarrow -\log(1-x)=0.117$

$\Rightarrow (1-x)= 10^{-0.117}=0.7638$

Now,

$Rate=K[A]$ [at 60 min]

$4.5\times 10^{-3}\times 0.7638$

$3.4354\times 10^{-3} mol L^{-1} min^{-1}$

The activation energy of the reaction is $75kJ/mol$ in absence of a catalyst and it lowers to $50.14kJ/mol$ with a catalyst. How many times will the rate of reaction grow in presence of a catalyst if the reaction proceeds at $25^\circ C$ ?

We know that Arrhenius for calculation of activation energy is given by:-

$K=A$

$e^{-Ea/RT}$

$- (i)$

$K_1$ and

$K_2$ are two rate constants at activation energy

$Ea_1$ and

$Ea_2$ . Then,

$K_1=A$

$e^{-Ea/RT}$

$-(ii)$

$K_2=A$

$e^{-Ea/RT}$

$-(iii)$

Now,

$(iii)/(ii)$

$\Rightarrow \cfrac {K_2}{K_1}=\cfrac {e^{-Ea_2/RT}}{e^{-Ea_1/RT}}$

$\Rightarrow K_2=K_1$

$e^{+\cfrac {1}{RT}(Ea_1-Ea_2)}$

Given,

$T=25^\circ C=25+273=298K$

$R=8.314$

$JK^{-1}mol^{-1}$

$Ea_1=75KJ/mol$

$Ea_2=50.14KJ/mol$

$\Rightarrow \cfrac {K_2}{K_1}= e^{+(75-50.14)\times 10^3/8.314\times298}$

$=e^{+10.034}$

$\Rightarrow \cfrac {K_2}{K_1}=222789$

For a chemical reaction the specific rate constant at 600K is $1.60\times { 10 }^{ 6 }{ s }^{ -1 }$ , calculate the rate constant for the reaction at 700K, if energy of activation for the reaction is $2.209\ KJ\ { mol }^{ -1 }$ .

Arrhenius equation for calculation of rate constant at different temperatures is given by

$K=A$

$e^{-Ea/RT}$

where,

$K$ = rate constant at temperature

$TK$

$A$ = Arrhenius constant

$Ea$ = activation energy of the reaction

$R$ = universal gas constant

$T$ = temperature in

$K$

Let, us suppose, at temperature

$T,K$ and

$T_2K$ the rate constants are

$K_1$ and

$K_2$ respectively,so,

$K_1=A$

$e^{-Ea/RT_1}$

$- (i)$

$K_2=A$

$e^{-Ea/RT_22}$

$-(ii)$

$(ii)/(i)$

$\Rightarrow \cfrac {K_2}{K_1}=\cfrac {e^{-Ea/RT_2}}{e^{-Ea/Rt_1}}$

Taking log on both sides:-

$\Rightarrow \log \left(\cfrac {K_2}{K_1}\right)= \log \left(e^{\cfrac {-Ea}{R}\left(\cfrac {1}{T_2} - \cfrac {1}{T_1}\right)}\right)$

$\Rightarrow \log \cfrac {K_2}{K_1}=\cfrac {Ea}{R} \left(\cfrac {1}{T_1} - \cfrac {1}{T_2}\right)$

Now, according to question,

$T_1=600K$ ,

$K_1=1.60\times 10^{6} s^{-1}$

$K_2=?$ ,

$T_2=700K$ ,

$Ea=2.209\times 10^{3}$

$J/mol$

$R=8.314$

$JK^{-1} mol^{-1}$

$\Rightarrow \log \left(\cfrac {K_2}{K_1}\right)=\cfrac {+2209}{8.314} \left[\cfrac {1}{600}-\cfrac {1}{700}\right]$

$\Rightarrow \log \left(\cfrac {K_2}{K_1}\right)= \cfrac {2209\times 100}{83140\times 6\times7}$

$\Rightarrow \log \left(\cfrac {K_2}{1.6\times 10^{6}}\right)=0.63$

Taking antilog on both sides:-

$\Rightarrow \cfrac {K_2}{1.6\times 10^6}=$

$Antilog (0.63)$

$\Rightarrow K_2= 1.87\times 1.6\times 10^{6}$

$\Rightarrow K_2= 3\times 10^{6} s^{-1}$

A first-order reaction is $50\%$ completed in $40$ minutes at $300\ K$ and in $20$ minutes at $320\ K$ . Calculate the activation energy of the reaction.

[Given: log $2=0.3010$ , log $4=0.6021$ , R= $8.314$ $JK^{-1}$ $mol^{-1}$ ]

When a first order reaction is

$50\%$ completed, the time is equal to half life period.

A first order reaction is

$50\%$ completed in

$40$ minutes at

$300$ K and in

$20$ minutes at

$320$ K.

$t_{1/2}=40$ minutes

$t'_{1/2}=20$ minutes

The Arrhenius equation and temperature variation is given by the expression.

$log\left( \dfrac {k'}{k} \right)= \dfrac {E_a}{2.303R}[\dfrac {T'-T}{TT'}]$

Also,

$t_{1/2}=\dfrac {0.693}{k}$ or

$t_{1/2} \propto \dfrac {1}{k}$

Hence,

$log \left(\dfrac {t_{1/2}}{t_{1/2}'}\right) = \dfrac {E_a}{2.303R}[\dfrac {T'-T}{TT'}]$

$log\left( \dfrac {40}{20} \right)= \dfrac {E_a}{2.303 \times 8.314 J/mol/K}[\dfrac {320K-300K}{300K \times 320K}]$

$0.3010 = \dfrac {E_a}{19.147 J/mol/K}[ 0.0002083 /K]$

$E_a=27664 J/mol$

For a first order reaction
Percentage of $B$ in the product is $30$ %. Calculate the value of $k_{1}$ and $k_{2}. (t_{1/2})_{over\ all} = 100\ hr$ .

$[B] = 2[A_0] \, e^{-k_1 t}$

$[C] = 3[A_0] \, e^{-k_2 t}$

$[A] = [A_0] \, e^{-(k_1 + k_2) t}$

% of

$[B] = \dfrac{2[A_0] e^{-k_1 t} \times 100}{2[A_0] e^{-k_1 t} + 3[A_0] e^{-k_2t}}$

$\dfrac{30}{100} = \dfrac{2e^{-k_1 t}}{2e^{-k_1 t} + 3e^{-k_2 t}}$

$\dfrac{3}{10} = \dfrac{1}{1 + \dfrac{3}{2} e^{(k_2 - k_1) t}}$

$3 + \dfrac{9}{2} e^{-(k_2 - k_1)t} = 10$

$e^{-(k_2 - k_1)t} . \dfrac{14}{9}$

$k_1 - k_2 = ln \dfrac{14}{9}$

$t = 100 hr$

$k_1 - k_2 = 0.4418 \times 10^{-2}$

$\dfrac{ln^2}{k_1 + k_2} = 100 hr$

$k_1 + k_2 = 0.693 \times 10^{-2}$

by solving

$k_1 = 0.5674 \,\, k_2 = 0.1256$

1065477_879179_ans_0d5fcbe725584926ba35fe6d42d39f7a.png

The reaction $A(aq) \rightarrow B(aq) + C (aq)$ is monitored by measuring optical rotation of reaction mixture at different time interval. The species $A,\ B$ and $C$ are optically active with specific rotations $20^0, 30^0$ and $-40^0$ respectively. Starting with pure $A$ if the value of optical rotation was found to be $2.5^0$ after $6.93$ minutes and optical rotation was $-5^0$ after infinite time. Find the rate constant for first order conversion of $A$ into $B$ and $C$ .

moles

$A(aq) \longrightarrow B(aq)+C(aq)$ Rotation

$t=0$

$90$ - -

$r_o= r_A=20^o$

$t=6.93$ min

$90-x$

$x$

$r_t= 2.5^o$

$t= \infty$ -

$90$

$r_{\infty}= -5^o$

as,

$x\propto r_o-r_t$

$(a_o-(a_o-x)\propto r_o-r_t)$

$a_o \propto r_o-r_{\infty} r_o- r_{\infty}$

$(a_O-O \propto r_o-r_{\infty})$

$K= \cfrac {2.303}{t} \log \cfrac {a_o}{a_o-x}$

$\Rightarrow K=\cfrac {2.303}{6.93} \log \cfrac {r_o-r_{\infty}}{(r_o-r_{\infty})-(r_o-r_t)}$

$\Rightarrow K=\cfrac {2.303}{6.93} \log \cfrac {r_o-r_{\infty}}{r_t-r_{\infty}}$

$\Rightarrow K= \cfrac {2.303}{6.93} \log \cfrac {20-(-5)}{2.5-(-5)}$

$\Rightarrow K= \cfrac {2.303}{6.93}\log \cfrac {25}{7.5}$

$\Rightarrow K= 0.173 min^{-1} \Rightarrow$ Rate constant

a) Derive an integrated rate equation for rate constant of a first order reaction.
b) Draw a graph of potential energy V/S reaction co-ordinates showing the effect of catalyst on activation energy $(E_a)$ of a reaction.

a) Consider a general first order reaction
R

$\rightarrow$ P
The differential rate equation for given reaction can be written as
Rate

$=-\dfrac{d[R]}{dt}=K[R]^1$
Rearrange above equation.

$\dfrac{d[R]}{[R]}=-K\times dt$
Integrating on both sides of the given equation

$\displaystyle\int\dfrac{d[R]}{[R]}=-k\displaystyle\int dt$
ln[R]

$=-Kt+I$ ..

$(1)$
Where I is Integration constant
At

$t=0$ the concentration of reactant [R]

$=[R]_0$ where

$[R]_0$ is initial concentration of reactant
Substituting in equation

$(1)$ we get
ln

$[R]_0=(-K\times 0)+I$
ln

$[R]_0=I$

$(2)$
Substitute I value in equation

$(1)$
ln

$[R]=-Kt+ln [R]_0$

$Kt=ln[R]_0-ln R$

$Kt =ln\dfrac{[R]_o}{[R]}$
or

$K=\dfrac{1}{t}ln\dfrac{[R]_0}{[R]}$ .
or

$K=\dfrac{2.303}{t}log\dfrac{[R]_0}{[R]}$ .

In a first order reaction, $x$ is converted into $y$ . $40\%$ of the sample decomposes in $112$ min. What is the half life of this reaction?

After 112 mins, reactant left

$=a_{t}=0.6a_{0}$

where

$a_{0}=$ initial reactant concentration

For 1st order reaction ,

$K=\cfrac{1}{t}\ln\cfrac{a_{0}}{a_{t}}$ &

$K=\cfrac{\ln 2}{t_{\frac{1}{2}}}$

$\therefore \cfrac{\ln 2}{t_{\frac{1}{2}}}=\cfrac{1}{112}\times \ln\cfrac{a_{0}}{0.6a_{0}}$

$\therefore t_{\frac{1}{2}}=151.9\,mins$

The rate constants of a reaction at $500K$ and $700K$ are $0.02{s}^{-1}$ and $0.07{s}^{-1}$ respectively.
Calculate the values of $Ea$ and $A$ :

$500K$ ,

$K=0.02{ s }^{ -1 }$

$700K,\quad k=0.07{ s }^{ -1 }$

Calculation of

${ E }_{ a }$ and A can be done using Arrhenius equation

$k=A{ e }^{ { -E }_{ a }/RT }\\ \ln { k } =\ln { A } +\left( \cfrac { -{ E }_{ a } }{ RT } \right)$

$At\ 500\ K,\\ \ln { 0.02 } =\ln { A } +\cfrac { -{ E }_{ a } }{ 500R } \\ \ln { A } =\ln { (0.02) } +\cfrac { { E }_{ a } }{ 500R } \longrightarrow (1)\\ At\quad 700K\\ \ln { A } =\ln { (0.07) } +\cfrac { { E }_{ a } }{ 700R } \longrightarrow (2)$

Equating (1) and (2)

$\ln { (0.02) } +\cfrac { { E }_{ a } }{ 500R } =\ln { (0.07)+\cfrac { { E }_{ a } }{ 700R } } \\ \cfrac { 1 }{ 100R } \left[ \cfrac { { E }_{ a } }{ 5 } -\cfrac { { E }_{ a } }{ 7 } \right] =\ln { \left[ \cfrac { 0.07 }{ 0.02 } \right] } \\ { E }_{ a }=\cfrac { 35\times 100\times 8.314\times 1.2528 }{ 2 } \\ { E }_{ a }=18227.6\quad J\\ { E }_{ a }=18.2\quad KJ$

Substituting value of

${ E }_{ a }$ in (1),

$\ln { A } =\ln { (0.02) } +\cfrac { 18227.6 }{ 500\times 8.314 } \\ =(-3.9120)+4.3848\\ \ln { A } =0.4728\\ \log { A } =1.0889\\ A=antilog\quad (1.0889)\\ A=12.27$

Thus

$E_a$ is 18.2 KJ and A is 12.27

What is zero order reaction? Explain with suitable example.

Reaction in which concentration of the reactants do not change with time and the concentration rates remain constant throughout are called zero order reactions.

$A \rightarrow Product$

Example:

$H_2 + Cl_2 \xrightarrow {hv} 2HCl$

$Rate= k[H_2]^0[Cl_2]^0$

$\boxed {Rate=k}$

The half life for the reaction, ${ N }_{ 2 }{ O }_{ 5 }\rightleftharpoons 2{ NO }_{ 2 }+\frac { 1 }{ 2 } { O }_{ 2 }$ in 24 hour at ${ 30 }^{ 0 }C$ starting with 10g of $N_{ 2 }{ O }_{ 5 }$ how many grams of $N_{ 2 }{ O }_{ 5 }$ will remain after a period of 96 hours.

For the first order reaction,

$t_1$

$_2$ = 0.693/ k

K = 0.693 /

$t_1$

$_2$ =

$\frac{0.693}{24}$ = 0.0289

$hr^{-1}$

kt = 2.303 * log

$\frac{a}{a-x}$

0.0289 * 98 = 2.3038 * log

$\frac{10}{10-x}$

log

$\frac{10}{10-x}$ = 1.2

$\frac{10}{10-x}$ = antilog 1.2

$\frac{10}{10-x}$ = 15.85

10 = 158.5 - 15.85 x

15.85 x = 148. 5

x = 9.36

a-x =10 - 9.36 = 0.63 g

Therefore the answer is 0.63 g

A hydrogenation reaction is carried out at 5000 k. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is $400 K$ $. What would be the value of the activation energy of the reaction if the catalyst lowers the activation barrier by$ 20 kJ { mol\quad L }^{ -1}$$?

We have if

${E}_{a}$ and

${{E}_{a}}'$ be the energy of activation in presence and absence of catalyst for hydrogenation reaction then,

$K = A {e}^{-{{E}_{a}}/{RT}}$

In presence of catalyst-

Required temperature

$\left( T \right) = 400 \; K$

${K}_{1} = A {e}^{-{{E}_{a}}/{\left( R \times 400 \right)}}$

In absence of catalyst-

Required temperature

$\left( T \right) = 5000 \; K$

${K}_{2} = A {e}^{-{{{E}_{a}}'}/{\left( R \times 5000 \right)}}$

The two rates are given to be same, i.e.,

${r}_{1} = {r}_{2}$

$\Rightarrow {K}_{1} = {K}_{2}$

$\therefore A {e}^{-{{E}_{a}}/{\left( R \times 400 \right)}} = A {e}^{-{{{E}_{a}}'}/{\left( R \times 5000 \right)}}$

$\Rightarrow \cfrac{{E}_{a}}{R \times 400} = \cfrac{{{E}_{a}}'}{R \times 5000}$

Given that catalyst lowers the activation energy by

$20 \; {KJ}/{mol}$ .

$\therefore \cfrac{{{E}_{a}}' - 20}{400} = \cfrac{{{E}_{a}}'}{5000}$

$\Rightarrow {{E}_{a}}' = 21.73 \; {KJ}/{mol}$

Hence the required answer is

$21.73 \; {KJ}/{mol}$ .

A first order reaction is 90% completed in 25 seconds. Find the rate constant and half-life period.

(A)

Rate constant

Formula for rate constant

$K=\dfrac{2.303}{t}\log_{10}\left(\dfrac{a_{o}}{a_{o}-x}\right)$

Given

$t=25\ sec$

let,

$a_{o}=100$ (since initial value not given)

$x=90$ (

$90\%$ reaction completed)

$a_{o}-x=100-90=10$

Substituting the values

$K=\dfrac{2.303}{25}\log_{10}\left(\dfrac{100}{10}\right)$

$K=\dfrac{2.303}{25}\times 1 \left\{\log_{10}10=1\right\}$

$k=0.09212\ sec^{-1}$

(B)

Half life period

Formula for

$t 1/2=\dfrac{0.693}{k}$

Substituting value of

$k$ from above

$t 1/2=\dfrac{0.693}{0.09212}$

$t 1/2=7.5228\ sec$

The rate of reaction becomes four times when the temperature changes from $300 \ K$ to $320 \ K$ Calculate the energy of activation of the reaction, assuming that it does not change with temperature.

1350742_1228758_ans_1339201fb84b4cf3887f92fbe1a06a4b.jpg

A first Order reaction has a rate constant value of $0.00510\, min^{-1}$ . If we begin with $0.10M$ Concentration of the reactant , how much of the reactant will remain after $300h$ ?

1323217_1326586_ans_7e91c009bd1544f9a62cf25634aa2f5f.jpg

The temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$ . Calculate the activation energy.

Given :

$\dfrac {K_{2}}{K_{1}} = 1.75$

$T_{1} = 25^{\circ}C = 25 + 273 = 298\ K, T_{2} = 35^{\circ}C = 35 + 273 = 308\ K$

(Since temperature coefficient is ratio of rate constants at

$35^{\circ}C$ and

$25^{\circ}C$ respectively.)

$\because 2.303\log \dfrac {K_{2}}{K_{1}} = \dfrac {E_{a}}{R} \left [\dfrac {T_{2} - T_{1}}{T_{1}T_{2}}\right ]$

$\therefore 2.303\log 1.75 = \dfrac {E_{a}}{1.987}\left [\dfrac {308 - 298}{308\times 298}\right ]$

$E_{a} = 10.207\ kcal\ mol^{-1}$ .

A certain reaction is 50% complete in 20 minutes at 300K and the same reaction is again 50% complete minutes at 350K. Calculate the activation energy if it is a first order reaction. $[R = 8.314\,J{K^{ - 1}}\,\,mo{l^{ - 1}},\log \,\,4 = 0.602 ]$

1657235_1384470_ans_57206f5f7c5246a1a67cb891aac4e83b.jpg

The rates of most reactions double when their temperature is raised from $298\ K$ to $308\ K$ . Calculate their activation energy.

$2.303\log \dfrac {K_{2}}{K_{1}} = \dfrac {E_{a}}{R} \left [\dfrac {T_{2} - T_{1}}{T_{1}T_{2}}\right ]$

$\dfrac {K_{2}}{K_{1}} = 2; T_{2} = 308K, T_{1} = 298\ K$

$\therefore 2.303\log 2 = \dfrac {E_{a}}{8.314} \times \dfrac {10}{308 \times 298}$

$E_{a} = 52.903 \times 10^{3} J$ or

$E_{a} = 52.903\ kJ$ .

Calculate the energy of activation of a reaction for which rate constant becomes doubles by Increase of 10 $\mathrm { K }$ temperature from 298 $\mathrm { K }$

It is given that T1 = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k1 = k and that of k2 = 2k

Also, R = 8.314 J K - 1 mol - 1

Now, substituting these values in the equation:

= 52897.78 J mol - 1

= 52.9 kJ mol - 1

Let the most probable velocity of hydrogen molecules at a temp. 1 degree c is Vsuppose all the molecules dissociate into atoms when temperature is raised to (2t+273) degree c then the new rms velocity is

2/\sqrt3 v

The reaction $A\rightarrow B$ follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole B is 1 hour. What is the time taken for conversion, 0.9 mole of A to produce 0.675 mole of B?

1655496_1528883_ans_c2f4d124d03e4c12a60198c70b5d016c.jpeg

For the first order reaction the time taken to reduce the initial concentration by a factor of 1/4 is 20 minutes The time required to reduce initial concentration by a factor of 1/16 is
(a) 20 min
(b) 10 min
(c) 80 min
(d) 40 min
(e) 5 min

1356696_1589845_ans_c2aebe17aa3442a9b6ff4625e32e8164.jpg

The half life period of a first order reaction is $6.0h$ . Calculate the rate constant.

For

$1^{st}$ order

Rate constant

$K = \dfrac{0.693}{t_{1/2}}$

$=\dfrac{0.693}{6}$

$= 0.01925 \, hr^{-1}$

Show that the time required for the completion of three-fourth of a twice first-order reaction time required for the completion of half-reaction.

1413369_1560323_ans_7e8145f51fd24dddae28404e92979946.jpg

A certain physiologically important first-order reaction has an activation energy equal to $45.0\ kJ/ mol$ at normal body temperature $(37^oC)$ . Without a catalyst the rate constant for the reaction is $5.0\times 10^{-4}s^{-1}$ . To be effective in the human body, where the reaction is catalysed by an enzyme, the rate constant must be at least $2.0\times 10^{-2}s^{-1}$ . If the activation energy is the only factor affected by the presence of the enzyme, by how much must the enzyme lower the activation energy of the reaction to achieve the desired rate?

$Ea=45.0\ KJ/mole$

$K=$ Rate constant without catalyst

$=5.0\times 10^{-4}\ s^{-1}$

$K'=$ Rate constant with catalyst

$=2.0\times 10^{-2}\ s^{-1}$

$K=Ae^{-Ea/RT}$ &

$K'=Ae^{-Ea}/RT$

$\dfrac{K'}{K}=\dfrac{e^{-Ea/RT}}{e^{-Ea/RT}}$

$2.303\times \log \left( \dfrac{K'}{K}\right)=\dfrac{1}{RT}(Ea-Ea')$

$(Ea-Ea')=2.303RT\log \left( \dfrac{K'}{K}\right)$

$(Ea-Ea')=2.303\times 8.314\times 10^{-3}\times 310\log \dfrac{2\times 10^{-2}}{5\times 10^{-4}}$

$(Ea-Ea')=5.9356\times \log \dfrac{200}{5}=9.51\ kJ$

For the reaction $A\to B+C$ the following data were obtained:
$t(s)$ $0$ $900$ $1800$
$[A]$ $50.8$ $19.7$ $8.62$
Prove that the reaction is of first order.

$A \rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,B\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C$

$t(s)$

$0$

$900$

$1800$

$[A]$

$50.8$

$19.7$

$7.62$

$\log [A]$

$1.7058$

$1.29446$

$0.882$

$\therefore$ slope of

$\log [A]$ vs

$t$

$=\dfrac{1.7058-1.26446}{900-0}=4.57\times 10^{-4}$

$\dfrac{1.29446-0.882}{1800-900}=4.58\times 10^{-4}$

Slope of

$\log P$ vs

$t$ is same, & hence graph of

$\log P$ vs

$t$ is same, & hence graph of

$\log P$ vs

$t$ is linear, which is the case of first order reaction.

A drop $(0.05\ mL)$ of a solution contains $3.0\times 10^{-6}$ mole of $H^+$ ions. If the rate constant of disappearance of $H^+$ is $1.0\times 10^7\ mol/L /s$ , how long would it take for $H^+$ ions in drop to disappear?

Rate constant of disappearance of

$H^+=1.0\times 10^7\ mol/L/s$

Unit of the rate constant is same as the unit rate of reaction for

$n=0$

For zero order reaction,

$A_t=A_0-Kt$

$0=\dfrac{3\times 10^{-6}}{0.05\times 10^{-3}}-1\times 10^7\ t$

$t=\dfrac{3\times 10^{-6}}{0.5\times 10^{-4}}\times \dfrac{1}{1\times 10^7}=\dfrac{6\times 10^{-2}}{10^7}$

$t=6\times 10^{-9}\ sec$

The rate constant of the first order reaction, that is, decomposition of ethylene oxide into $CH_4$ and $CO$ , may be described by the following equation
$\log k(s^{-1})=14.34-\dfrac {1.25\times 10^4}{T}K$ .
Find energy of activation and rate constant at $397^oC$ .

$\log k(s^{-1})=14.34-\dfrac{1.25\times 10^4}{T}K$ (1)

Comparing with:

$\log k=\log A-\dfrac{Ea}{2.303RT}$

$\dfrac{Ea}{2.303R}=1.25\times 10^4$

(a)

$Ea=239.34\ KJ$

(1)

$\log k=14.34-\dfrac{1.25\times 10^4}{670}=14.34-18.65=-4.3167$

$k=4.8\times 10^{-5}\ s^{-1}$

What will be the time taken to decrease the reactant concentration from $2\times 10^{-2}$ mol $lit^{-1}$ to $1\times 10^{-2}$ mole $lit^{-1}$ ?

For Zero-Order reaction:

$\mathrm{t=\cfrac{A_o- A_t}{k}}$

$\mathrm{t=\cfrac{2 \times 10^{-4}- 1 \times 10^{-4}}{5 \times 10^{-8}}}$

$\mathrm{t=2000 \ sec}$

From the following data for the reaction between A and B:

$[A]$ , mol $L^{-1}$ $[B]$ , mol $L^{-1}$ Initial rate, mol $L^{-1}s^{-1}$ at Initial rate, mole $L^{-1}s^{-1}$ , at
$300$ K $320$ K
$2.5\times 10^{-4}$ $3.0\times 10^{-5}$ $5.0\times 10^{-4}$ $2.0\times 10^{-3}$
$5.0\times 10^{-4}$ $6.0\times 10^{-5}$ $4.0\times 10^{-3}$ -
$1.0\times 10^{-3}$ $6.0\times 10^{-5}$ $1.6\times 10^{-2}$ -

Calculate the energy of activation.

1614949_1739958_ans_eca9f7f2838a459a9283d632fd320f9c.jpg

What is the energy of activation for this reaction?

1614967_1740587_ans_3a514ba9310e4668a4ec4ba246a99690.jpg

How long does it take for reactant concentration to decrease from $4\times 10^{-4}$ mole $lit^{-1}$ to $2\times 10^{-4}$ mole $lit^{-1}$ ?

For Zero-Order reaction:

$\mathrm{t=\cfrac{A_o- A_t}{k}}$

$\mathrm{t=\cfrac{4 \times 10^{-4}- 2 \times 10^{-4}}{5 \times 10^{-8}}}$

$\mathrm{t=4000 \ sec}$

Shw that in case of a first-factor reaction, the time required for $93.75\%$ of the reaction to take place is four times that required for half of the reaction.

1614953_1739964_ans_7c070d709b144d3982d32b434ecd71e4.jpg

The decomposition of $N_2O$ into $N_2$ and O in the presence of gaseous argon follows second-order kinetics, with
$k=(5.0\times 10^{11}L mol^{-1}s^{1-})e^{-29000k/T}$
What is the energy of activation of this reaction?

1614971_1740604_ans_add40ec1de4f47bfba04a55a2f2a463c.jpg

A given sample of milk turns sour at room temperature $20^{\circ}C)$ in $64\ hour$ . In a refrigerator at $3^{\circ}C$ , milk can be stored three times as long before it sours. Estimate
(a) The activation energy for souring of milk,
(b) How long it take milk to sour at $40^{\circ}C$ ?

1751177_1742465_ans_3649a87cfea14a00b4e36cd9753f4d12.jpg

Calculate the rate constant and half life period for first order reaction having activation energy $39.3\ kcal\ mol^{-1}$ at $300^{\circ}C$ and the frequency constant $1.11 \times 10^{11} \sec^{-1}$ .

Given :

$A = 1.11\times 10^{11}\sec^{-1}; E_{a} = 39.3\times 10^{3} cal\ mol^{-1}$ ;

$R = 1.987\ cal; T = 573\ K$ .

$\because K = Ae^{-E_{a}/ RT}$

$\therefore ln\ K = ln\ A - \dfrac {E_{a}}{RT}$

$\log_{10} K = \log_{10} A - \dfrac {E_{a}}{2.303\ RT}$

$\log_{10} K = \log_{10} 1.11\times 10^{11} - \left \{\dfrac {39.3\times 10^{3}}{2.303\times 1.987\times 573}\right \}$

$\therefore K = 1.14\times 10^{-4} \sec^{-1}$

Also for first order,

$t_{1/2} = \dfrac {0.693}{K} = \dfrac {0.693}{1.14\times 10^{-4}} = 6078\sec$ .

For a reversible first order reaction, $A\underset {K_{2}}{\overset {K_{1}}{\rightleftharpoons}} B; K_{f} = 10^{-2} s^{-1}$ and $\dfrac {B_{eq.}}{A_{eq.}} = 4$ ; If $A_{0} = 0.01\ ML^{-1}$ and $B_{0} = 0$ , what will be concentration of $B$ after $30\sec$ .

Given ,

$k_{f} = 10^{-2}s^{-1}$ and

$A_{0}=0.01 ML^{-1}$ and

$B_{0} = 0$

As we know,

$k_{1} + k_{2} = \frac{1}{t}log\frac{x_{e}-x_{0}}{x_{e}-x}$

and

$k_{e} = \frac{k_{1}}{k_{2}}$

And from given data, conc. of B after 30 sec.

$k_{1} + k_{2} = \frac{1}{30}log\frac{4-0}{4-x}$

$\therefore$ x = 0.0025m

Hence , concentration of B after 30 sec is 0.0025 m.

The oxidation of certain metal is found to obey the equation $A = \alpha t + \beta$ , where $A$ is the thickness of the oxide film at time $t, \alpha$ and $\beta$ are constants. What is the order of this reaction.

1762455_1742521_ans_0c851d9d0a884b538541edcb712cd487.jpg

Match order of the reaction (in List -I) with the corresponding rate constant (in List -II) :

Zero order

$= x = kt$

First order

$K = \dfrac {1}{t}ln \dfrac {1}{(a - x)}$

Second order

$k = \dfrac {1}{t}\left [ \dfrac {1}{(a - x)} - \dfrac {1}{a} \right ]$

Third order

$k = \dfrac {1}{2t}\left [ \dfrac {1}{(a - x)^2} - \dfrac {1}{a^2} \right ]$

A graph between ln K and $\dfrac{1}{T}$ for a reaction is given . Here k is rate constant and $T$ is temperature in Kelvin .
If $OA = a$ and $OB = b$ , answer the following :
What is the activation energy $(E_a)$ of the reaction ?

According to Arrhenius equation ,

$ln \, k = - \dfrac{E_a}{RT} + ln \, A$

Slope

$= - \dfrac{OB}{OA} = - \dfrac{b}{a} = - \dfrac{E_a}{R}$ or

$E_a = \dfrac{b}{a} R$

From the graph, What is the activation energy for backward reaction?

Activation energy for the backward reaction

$= 300 - 100$

$= 200 \, kJ \, mol^{-1}$

The decomposition of $NH_{3}$ on platinum surface is zero order reaction.

What are the rates of production $N_{2}$ and $H_{2}$ If $K = 2.5 \times10 ^{-4} mol L^{-1}s^{-1}$

1910696_1828108_ans_190df87864154b1caafb47102a318a3e.JPG

The rate constant of a first order reaction increases from $2 \times 10^{-2}$ to $4 \times 10^{-2}$ when the temperature changes from $300 \,K$ to $310\,K$ . Calculate the activation energy $(E_a)$ .
$(log \, 2 = 0.301 \ , \ log \, 3 = 0.4771 \ , \ log \, 4 = 0.6021 )$

Substituting

$k_1 = 2 \times 10^{-2} , k_2 = 4 \times 10^{-2} , T_1 = 300 \,K , T_2 = 310\,K , R = 8.314\, J \, K^{-1}\, mol^{-1}$ in the expression

$log \dfrac{k_1}{k_2} = \dfrac{E_a}{2.303 \times R} \left ( \dfrac{T_2 - T_1}{T_1T_2} \right )$ , we get

$log \dfrac{4 \times 10^{-2}}{2 \times 10^{-2}} = \dfrac{E_a}{2.303 \times 8.314} \left ( \dfrac{310 - 300}{300 \times 310} \right )$

$log \, 2 = \dfrac{E_a}{19.147} \times \dfrac{10}{300 \times 310}$

$E_a = 0.3010 \times 19.147 \times 300 \times 31 = 53598 \, J \, mol^{-1}$

$E_a = 53.598 \, kJ \, mol^{-1}$

Define half-life of a reaction.
Write the expression of half-life for :
i. zero-order reaction
ii. first-order reaction.

The half life

$(t_{1/2})$ of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration.

i.

$t_{1/2}$ for a zero order reaction

$= \dfrac{[R]_0}{2k}$ where

$[R]_0 =$ initial concentration ,

$k =$ rate constant.

ii.

$t_{1/2}$ for a first reaction

$= \dfrac{0.693}{k}$

The decomposition of $NH_{3}$ on a platinum surface takes place as follows:

$2NH_{3}(g) \overset{Pt}{\rightarrow} N_{2}(g) + 3H_{2}(g)$

The reaction is of zero order with $k = 2.5 \times 10^{-4} mol\ L^{-1}\ s^{-1}$ . What are the rates of production of $N_{2}$ and $H_{2}$ ?

1962562_1828385_ans_cf812bada44f4b61a20c040ea1bb98ef.jpg

Give an example of a pseudo first order reaction.

Acid catalysed hydrolysis of ethyl acetate

$CH_{3} COOC_{2} H_{5} + H_{2} O \overset{H^+}{\rightarrow} CH_{3} COOH + C_{2}H_{5}OH$

Rate

$= K \left [ CH_3COOC_2H_5 \right ] \left [ H_2O \right ]^{0}$

For an elementary reaction $2A + B \rightarrow 3C$ the rate of appearance of $C$ at time 't' is $1.3 \times 10^{-4} \, mol \, L^{-1} \,s^{-1}$ .
Calculate at this time rate of disappearance of A .

Rate

$= \dfrac{-d[A]}{dt} = \dfrac{2}{3} \times \dfrac{d[C]}{dt}$

$= \dfrac{2}{3} \times 1.3 \times 10^{-4} \, mol \, L^{-1} \, s^{-1} = 0.86 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that is does not change with temperature.

$T_{1} = 293 K$

$T_{2} = 313 K$

$k_{2} = 4k_1$

$\log \dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{2.303R} \left [ \dfrac{T_{2} - T_{1}}{T_{1}T_{2}} \right ]$

$\log 4 = \dfrac{E_{a}}{2.303 \times 8.314}\left [ \dfrac{20}{313 \times 293} \right ]$

$E_{a} = \dfrac{0.6021 \times 2.303 \times 8.314 \times 313 \times 293}{20}$

$E_{a} = 52.86\ kJ / mol^{-1}$

All energetically effective collisions do not result in a chemical change. Explain with the help of an example.

All energetically effective collisions do not result in a chemical change. For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown below:

$\mathrm{CH}_{3} \mathrm{Br}+\overline{0} \mathrm{H} \rightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{Br}^{-}$

The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply bounce back and no products are formed.

1675450_1796434_ans_0cf78405453a45cc91238e92be38d782.png

The activation energy for the acid-catalyzed hydrolysis of sucrose is $6.22\ kJ\ mol^{-1}$ , while the activation energy is only $2.15\ kJ\ mol^{-1}$ when hydrolysis catalyzed by the enzyme sucrase. Explain.

Enzymes are needed only in small quantities for the progress of a reaction. Similar to the action of chemical catalysts, enzymes are said to reduce the magnitude of activation energy. For example, the activation energy for acid hydrolysis of sucrose is

$6.22\ kJ\ mol^{-1}$ , while the activation energy is only

$2.15\ kJ\ mol^{-1}$ when hydrolyzed by the enzymes, sucrase.

The activation energy for the reaction $2HI_{(g)}\rightarrow H_2{(g)}+I_2{(g)}$ is $209.5 kJmol^{-1}$ at 581 K . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Fraction of molecules having energy equal to greater than activation energy

$x=\dfrac{n}{N}=e^{-E_{a}/RT}$

$ln x = -\dfrac{E_{a}}{RT}$

$log x = - \dfrac{E_{a}}{2.303 RT}$

$log x = - \dfrac{209.5\times 10^{3}Jmol^{-1}}{2.303\times 8.314JK^{-1}mol^{-1}\times 581K}$

$log x=-18.8323$

$x$ = Antilog (- 18.83)

$x=1.471\times 10^{-19}$

The decomposition of A into products has value of k as $4.5 \times 10^{3} s^{-1}$ at $10^{\circ} C$ and energy of activation is $60$ kJ $mol^{-1}$ . At what temperature (in kelvin) would k be $15 \times 10^{3} s^{-1}$ ?

$k_{1} = 4.5 \times 10^{3} s^{-1}$

$T_{1} = 10 + 273 = 283 K$

$k_{2} = 1.5 \times 10^{4} s^{-1}$

$T_{2} = ?$

$E_{a} = 60 kJ/ mol = 60 \times 10^{3} J / mol$

According to Arrhenius equation,

$\log (\dfrac{k_{2}}{k_{1}}) = \dfrac{Ea}{2.303R} \left [ \dfrac{T_{2} - T_{1}}{T_{1}T_{2}} \right ]$

$\log \dfrac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = \dfrac{60000}{2.303 \times 8. 314} \left [ \dfrac{T_{2} - 283}{283T_{2}} \right ]$

$T_2 = 297 K = 24^0C$

The rate of a particular reaction quadruples on increasing temperature from $293$ K to $313$ K. For this reaction, Calculate activation energy.

According to question,

$\dfrac{k_{2}}{k_{1}} = 4$

$\log \dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{2.303R} \left [ \dfrac{T_{2} - T_{1}}{T_{1}T_{2}} \right ]$

$\log 4 = \dfrac{E_{a}}{2.303 \times 8.314} \left [ \dfrac{313 - 293}{313\times 293} \right ]$

$E_{a} = \dfrac{0.6020 \times 2.303 \times 8.314 \times 91709}{20}$

$E_{a} = 52854.55 Joule = 52.85 kJ$

Gaseous cyclobutene isomerizes to butadiene in a first order process which has 'k' value of $3.3 \times 10^{-4} s^{-1}$ at $153^{\circ} C$ . The time in minutes it takes for the isomerization to proceed $40\%$ to completion at this temperature is __________.
(Rounded off to the nearest integer)

Integrated rate law for

$1^{st}$ order reaction,

$Kt = \ln \dfrac{[A]_{0}}{[A]_{t}}$

$3.3 \times 10^{-4} \times t = \ln \left (\dfrac{100}{60} \right )$

$t = 1547.956$ sec

$t = 25.799$ min

$t= 26$ min

2079357_1978261_ans_53e3f06020194a59a5d4e5ddf89b0d2f.png

The decomposition of $NH_3$ on plantinum surface is a zero order reaction. if rate constant (k) is $4\times 10^{-3}Ms^{-1}$ . how long will it take to reduce the initial concentration of $NH_3$ from $0.1M$ to $0.064M$ .

$t=\dfrac{[R]_o-[R]_t}{k}$

$=\dfrac{[0.1-0.064]}{4\times 10^{-3}}$

$=9s$

Observation No.	Time (in minute)	$P_x$ (in mm of Hg)
1	0	800
2	100	400
3	200	200

Experiment	$Time/s^{-1}$	Total pressure/ atm
$1$	$0$	$0.5$
$2$	$100$	$0.6$

$[A]$ , mol $L^{-1}$	$[B]$ , mol $L^{-1}$	Initial rate, mol $L^{-1}s^{-1}$ at	Initial rate, mole $L^{-1}s^{-1}$ , at
		$300$ K	$320$ K
$2.5\times 10^{-4}$	$3.0\times 10^{-5}$	$5.0\times 10^{-4}$	$2.0\times 10^{-3}$
$5.0\times 10^{-4}$	$6.0\times 10^{-5}$	$4.0\times 10^{-3}$	-
$1.0\times 10^{-3}$	$6.0\times 10^{-5}$	$1.6\times 10^{-2}$	-

Chemical Kinetics - Class 12 Engineering Chemistry - Extra Questions

What is threshold energy?

A radioactive isotope is being produced at a constant rate xx. Half life of the radioactive substance is 'yy'. After sometime, the number of radioactive nuclei becomes constant the value of this constant is ________

The gaseous reaction A2→2A{A_2} \to {\rm{ }}2A is first order in A2A_2. After 12.312.3 minutes 65%65\% of A2A_2 remains undecompensed. How long will it take to decompose 90%90\% of A2A_2? What is the half life of the reaction?

For a zero order reaction at 200K200K reaction complete in 5 minutes while at 300K300 K, same reaction, completes in 2.52.5 minutes. What will be the activation energy in calorie? (R=2Cal/(mol.K); ln2=0.7)(R = 2 Cal/(mol.K); \ ln 2 = 0.7)

At 300K,50300 \,K, 50% of molecule collide with energy grater than or equal to EaE_a. At what temperature, 2525% molecule will have energy greater than or equal to EaE_a.

Inversion of cane sugar is pseudo first order reaction. Give reason.

Define (i) Rate Constant (ii) Energy of Activation of a reaction

What is activation energy?

After five half-life periods for a first order reaction, what fraction of reactant remains?

Calculate the rate of reaction for the change 2A→Products2A \rightarrow Products, when rate constant of the reaction is 2.1×10−5time−12.1 \times 10^{-5} time^{-1} and [A]0=0.2 M[A]^{0} = 0.2\ M.

The time required for a first-order reaction for 9999% completion is xx times for the time required for 9090% completion. xx is:

The rate of reaction is fast when the activation energy is _______ (small/large).

A reaction is said to be of zero if the rate is entirely independent of the concentration of the reactants.If true enter 1, if false enter 0.

Collision frequency depends upon concentration and temperature.If true enter 1, if false enter 0.

Molecules must collide before they react.If true enter 1, if false enter 0.

State whether the statement is True or False:Two molecules can react only if they have a correct relative orientation.

______ (Kinetics/Ionic equilibrium) studies of reactions involve molecular reactions.

Initial pressure of AA in mm is___________.

A first order reaction is 20\%20\%complete in 10 min. The specific rate constant of the reaction is_________ (in\:\: min^{-1}in\:\: min^{-1}).

Derive the relation between half-life period and rate constant for first order reaction

Half life period of a _________ order reaction is ________ of the concentration of the reactant.

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_{1/2}=3t_{1/2}=3hours. What fraction of the sample of sucrose remains after 88hours?

What is half-life of a reaction? Derive formula for finding out half-life from first order rate reaction.

Write the characteristics of first order reaction.

Derive the integrated rate equation for the rate constant of a zero order reaction.

Derive an expression for the rate constant for a first order reaction.

Show that for a first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion.

What happens to the half life period for a first order reaction, if the intial concentration of the reactants is increased?

Give a detailed account of the collision theory of reaction mils of Bimolecular gaseous reactions.

Give two examples of gaseous first-order reactions.

Define collision frequency. Give an example for Pseudo-first order reaction.

Write short notes on:(a) Threshold Energy, (b) Energy of Activation.

For a zero order reaction t_{1/2}t_{1/2} is proportional to what?

The half life of ^{212}Pb^{212}Pb is 10.6\ hour10.6\ hour. It undergoes decay to its daughter (unstable) element ^{212}Bi^{212}Bi of half life 60.560.5 minute. Calculate the time at which the daughter element will have maximum activity.

^{ 131 }{ I }^{ 131 }{ I } has half life period 133133 hour. After 79.879.8 hour, what fraction of ^{ 131 }{ I }^{ 131 }{ I } will remain?

^{ 90 }{ Sr }^{ 90 }{ Sr } shows \beta\beta-activity and its half life period is 2828 years. What is the activity of a sample containing 1 g1 g of ^{ 90 }{ Sr }^{ 90 }{ Sr }?

For the dissociation of gaseous HIHI, the energy of activation is 44.3kcal44.3kcal. Calculate the energy of activation for the reverse reaction. Given, \Delta H\Delta H for the formation of 11 mole of HIHI from {H}_{2}{H}_{2} and {I}_{2}{I}_{2} is -1.35kcal-1.35kcal.

Thermal decomposition of a compound is of first order. If 50\%50\% sample of the compound is decomposed in 120120 minute, how long will it take for 90\%90\% of the compound to decompose?

Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with t_{ {1}/{2}} = 3\ Hrst_{ {1}/{2}} = 3\ Hrs. What fraction of the sample of sucrose remains after 88 hours.

Give the definition of (X)(X) denoted in graph:

For the decomposition of ethylene oxide into CH_4CH_4 and CO, the variation of total pressure (P) of the reaction mixture with time is as given below:t/s 0 300 600 900P/mm 120 127.2 134.0 140.3Show that the reaction is a first order reaction.

The half-life of a first order reaction is 900900 min at 820\ K820\ K. Estimate its half-life at 720720K if the energy of activation of the reaction is 250\ kJ\ mol^{-1}250\ kJ\ mol^{-1}.

Calculate the activation energy for a reaction of which rate constants becomes 4 times when temperature changes from 30^0C30^0C to 50^0C50^0C [R= 8.314 J mol^{-1}K^{-1}J mol^{-1}K^{-1}]

Write the two steps involved in the mechanism of the enzyme-catalyzed reaction.

Derive an integrated rate equation for rate constant of a zero order reaction.

For a first order reaction A\to PA\to P, the temperature (T)(T) and dependent rate constant (k)(k), was found to follow the equation \log k = -(2000)\dfrac{1}{T} + 6.0\log k = -(2000)\dfrac{1}{T} + 6.0. What is activation energy?

A first-order reaction is 40\%40\% complete in 5050 min. How long(minutes) will it take to 80\%80\% completion?

Sucrose decomposes in solution into glucose and fructose according to the first order rate law. With t^{1/2} 3t^{1/2} 3 Hr. What fraction of sample of sucrose remains after 88 hour?

For a first order reaction, time required for 99.099.0% completion is xx times for the first time required for the completion of 9090% of the reaction xx is:

Derive an expression for the rate constant of a First order reaction.

How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?

For a homogeneous gaseous reaction, A \longrightarrow B+C+DA \longrightarrow B+C+D, the initial pressure was P_{0}P_{0} while pressure at time tt was PP. Derive an expression for rate constant KK in terms of P_{0},PP_{0},P and tt. [Assume first-order reaction]

The rate of most of the reaction double when their temperature is raised from 298 \ K298 \ K to 308 \ K308 \ K. Calculate the activation energy of such a reaction.

Given that the temperature coefficient for the saponification of ethyl acetate by NaOHNaOH is 1.751.75. Calculate activation energy for the saponification of ethyl acetate.

A first order reaction takes 69.369.3 minutes for 50\%50\% completion. How much time will be needed for 80\%80\% completion?

Calculate the age of a vegetarian beverage whose tritium content is only 15\%15\% of the level in living plants. Given t_{\frac{1}{2}}t_{\frac{1}{2}} for _{1}H^{3} = 12.3_{1}H^{3} = 12.3 years. (log 2 = 0.3, log3 = 0.48log 2 = 0.3, log3 = 0.48)

Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

The rate of the reaction quadruples when the temperature changes from 293K293K to 313K313K .Calculate the energy of activation?

A first order reaction is 75% completed in 72 min. How long time will it take for(i) 50% completion (ii) 87.5% completion

For a zero order chemical reaction, 2NH_{3}(g)\rightarrow N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)\rightarrow N_{2}(g) + 3H_{2}(g) rate of reaction = 0.1 atm /sec. Initially only NH_{3}NH_{3} is present and its pressure = 3atm. Calculate total pressure at t = 10 sec.

The rate of r\times\timesn becomes 4 4 times when temp. changes from 293 K 293 K to 313 K 313 K Calculate energy of activation (Ea) of the rxn assuming that it does not change with temp. (R= 8.3/4 JK^-mol^{-1} ) (R= 8.3/4 JK^-mol^{-1} )

The t_{1/2}t_{1/2} of first order reaction is 6060 minutes. What percentage will be left after 240240 minutes?

Which reaction will have the greater temperature dependence tor the rate constant - one with a small value of energy of activation (E) or one with a large value of activation energy(E)?

What is zero order reaction?

Write two applications of integrated rate equation.

3/4^{th}3/4^{th} of a first-order reaction is completed in 32 min. Calculate the time required to complete 15/16^{th}15/16^{th} part of the reaction.

If A \rightarrow\rightarrow products is a first order reaction then, write the integrated law equation.

Derive the integrated rate equation for expressing the rate constant of a first order :R\rightarrow PR\rightarrow P

A certain reaction is 50\%50\% complete in 20\ minutes20\ minutes at 300k300k; the reaction is 50\%50\% complete in 55 minutes at 350k350k calculate the activeity energy, of it is a first order reaction.

Activation energy of a reaction is 100 kJ/mol. By using a catalyst E_aE_a decreases by 75%. What will be the effect on rate of reaction at 20^0C20^0C? Consider all other things are equal.

A rain in 50\%50\% completed om shown and 75\%75\% in d44 hours. Find the order of reaction :

In the equation KK=A.e^ {\dfrac{-Ea}{RT}}A.e^ {\dfrac{-Ea}{RT}}, what does the term 'A' represent?

The solution of { H }_{ 2 }{ O }_{ 2 }{ H }_{ 2 }{ O }_{ 2 } of normality 0.73 is catalytically decomposed. What will be the concentration at the end of 45mins, assuming the decompostion to follow first order rate law if half life is 15 minute?

Derive an expression for integrated rate law for a zero order reaction.

Draw energy profile diagram of potential energy with reaction progress. Show that position for the following in the diagram.(a) Activated complex(b) Energy of activition

For a substitution reaction, rate constant increases by factor 10.610.6 when temperature changes from {25}^{o}C{25}^{o}C to {35}^{o}C{35}^{o}C. Calculate energy of activation of the reaction.

The activation energy for the reaction, 2 HI(g) \rightarrow H_2 + I_2 (g)HI(g) \rightarrow H_2 + I_2 (g) is 209.5 k J mol^{-1}209.5 k J mol^{-1} at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

For a first order reaction, A\longrightarrowA\longrightarrow Product, the temperature dependent rate constant (k) was found to follow the equation logk=-\frac { 2000 }{ T } +6logk=-\frac { 2000 }{ T } +6 calculate the pre-exponential factor (A) and activation energy for the reaction.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate { E }_{ a }{ E }_{ a }.

A radioactive isotope is being produced at a constant rate $x$ . Half life of the radioactive substance is ' $y$ '. After sometime, the number of radioactive nuclei becomes constant the value of this constant is ________

The gaseous reaction ${A_2} \to {\rm{ }}2A$ is first order in $A_2$ . After $12.3$ minutes $65\%$ of $A_2$ remains undecompensed. How long will it take to decompose $90\%$ of $A_2$ ? What is the half life of the reaction?

For a zero order reaction at $200K$ reaction complete in 5 minutes while at $300 K$ , same reaction, completes in $2.5$ minutes. What will be the activation energy in calorie? $(R = 2 Cal/(mol.K); \ ln 2 = 0.7)$

At $300 \,K, 50%$ of molecule collide with energy grater than or equal to $E_a$ . At what temperature, $25%$ molecule will have energy greater than or equal to $E_a$ .

Define (i) Rate Constant
(ii) Energy of Activation of a reaction

Calculate the rate of reaction for the change $2A \rightarrow Products$ , when rate constant of the reaction is $2.1 \times 10^{-5} time^{-1}$ and $[A]^{0} = 0.2\ M$ .

The time required for a first-order reaction for $99$ % completion is $x$ times for the time required for $90$ % completion. $x$ is:

A reaction is said to be of zero if the rate is entirely independent of the concentration of the reactants.
If true enter 1, if false enter 0.

Collision frequency depends upon concentration and temperature.
If true enter 1, if false enter 0.

Molecules must collide before they react.
If true enter 1, if false enter 0.

State whether the statement is True or False:

Two molecules can react only if they have a correct relative orientation.

Initial pressure of $A$ in mm is___________.

A first order reaction is $20\%$ complete in 10 min. The specific rate constant of the reaction is_________ ( $in\:\: min^{-1}$ ).

Half life period of a _ order reaction is of the concentration of the reactant.

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with $t_{1/2}=3$ hours. What fraction of the sample of sucrose remains after $8$ hours?

Write short notes on:
(a) Threshold Energy, (b) Energy of Activation.

For a zero order reaction $t_{1/2}$ is proportional to what?

The half life of $^{212}Pb$ is $10.6\ hour$ . It undergoes decay to its daughter (unstable) element $^{212}Bi$ of half life $60.5$ minute. Calculate the time at which the daughter element will have maximum activity.

$^{ 131 }{ I }$ has half life period $133$ hour. After $79.8$ hour, what fraction of $^{ 131 }{ I }$ will remain?

$^{ 90 }{ Sr }$ shows $\beta$ -activity and its half life period is $28$ years. What is the activity of a sample containing $1 g$ of $^{ 90 }{ Sr }$ ?

For the dissociation of gaseous $HI$ , the energy of activation is $44.3kcal$ . Calculate the energy of activation for the reverse reaction. Given, $\Delta H$ for the formation of $1$ mole of $HI$ from ${H}_{2}$ and ${I}_{2}$ is $-1.35kcal$ .

Thermal decomposition of a compound is of first order. If $50\%$ sample of the compound is decomposed in $120$ minute, how long will it take for $90\%$ of the compound to decompose?

Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with $t_{ {1}/{2}} = 3\ Hrs$ . What fraction of the sample of sucrose remains after $8$ hours.

Give the definition of $(X)$ denoted in graph:

For the decomposition of ethylene oxide into $CH_4$ and CO, the variation of total pressure (P) of the reaction mixture with time is as given below:
t/s 0 300 600 900
P/mm 120 127.2 134.0 140.3
Show that the reaction is a first order reaction.

The half-life of a first order reaction is $900$ min at $820\ K$ . Estimate its half-life at $720$ K if the energy of activation of the reaction is $250\ kJ\ mol^{-1}$ .

Calculate the activation energy for a reaction of which rate constants becomes 4 times when temperature changes from $30^0C$ to $50^0C$ [R= 8.314 $J mol^{-1}K^{-1}$ ]

For a first order reaction $A\to P$ , the temperature $(T)$ and dependent rate constant $(k)$ , was found to follow the equation $\log k = -(2000)\dfrac{1}{T} + 6.0$ . What is activation energy?

A first-order reaction is $40\%$ complete in $50$ min. How long(minutes) will it take to $80\%$ completion?

Sucrose decomposes in solution into glucose and fructose according to the first order rate law. With $t^{1/2} 3$ Hr. What fraction of sample of sucrose remains after $8$ hour?

For a first order reaction, time required for $99.0$ % completion is $x$ times for the first time required for the completion of $90$ % of the reaction $x$ is:

For a homogeneous gaseous reaction, $A \longrightarrow B+C+D$ , the initial pressure was $P_{0}$ while pressure at time $t$ was $P$ . Derive an expression for rate constant $K$ in terms of $P_{0},P$ and $t$ .
[Assume first-order reaction]

The rate of most of the reaction double when their temperature is raised from $298 \ K$ to $308 \ K$ . Calculate the activation energy of such a reaction.

Given that the temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$ . Calculate activation energy for the saponification of ethyl acetate.

A first order reaction takes $69.3$ minutes for $50\%$ completion. How much time will be needed for $80\%$ completion?

Calculate the age of a vegetarian beverage whose tritium content is only $15\%$ of the level in living plants. Given $t_{\frac{1}{2}}$ for $_{1}H^{3} = 12.3$ years. ( $log 2 = 0.3, log3 = 0.48$ )

The rate of the reaction quadruples when the temperature changes from $293K$ to $313K$ .Calculate the energy of activation?

A first order reaction is 75% completed in 72 min. How long time will it take for
(i) 50% completion (ii) 87.5% completion

For a zero order chemical reaction,
$2NH_{3}(g)\rightarrow N_{2}(g) + 3H_{2}(g)$
rate of reaction = 0.1 atm /sec. Initially only $NH_{3}$ is present and its pressure = 3atm. Calculate total pressure at t = 10 sec.

The rate of r $\times$ n becomes $4$ times when temp. changes from $293 K$ to $313 K$ Calculate energy of activation (Ea) of the rxn assuming that it does not change with temp. $(R= 8.3/4 JK^-mol^{-1} )$

The $t_{1/2}$ of first order reaction is $60$ minutes. What percentage will be left after $240$ minutes?

$3/4^{th}$ of a first-order reaction is completed in 32 min. Calculate the time required to complete $15/16^{th}$ part of the reaction.

If A $\rightarrow$ products is a first order reaction then, write the integrated law equation.

Derive the integrated rate equation for expressing the rate constant of a first order :
$R\rightarrow P$

A certain reaction is $50\%$ complete in $20\ minutes$ at $300k$ ; the reaction is $50\%$ complete in $5$ minutes at $350k$ calculate the activeity energy, of it is a first order reaction.

Activation energy of a reaction is 100 kJ/mol. By using a catalyst $E_a$ decreases by 75%. What will be the effect on rate of reaction at $20^0C$ ?
Consider all other things are equal.

A rain in $50\%$ completed om shown and $75\%$ in d $4$ hours. Find the order of reaction :

In the equation $K$ = $A.e^ {\dfrac{-Ea}{RT}}$ , what does the term 'A' represent?

The solution of ${ H }_{ 2 }{ O }_{ 2 }$ of normality 0.73 is catalytically decomposed. What will be the concentration at the end of 45mins, assuming the decompostion to follow first order rate law if half life is 15 minute?

Draw energy profile diagram of potential energy with reaction progress. Show that position for the following in the diagram.
(a) Activated complex
(b) Energy of activition

For a substitution reaction, rate constant increases by factor $10.6$ when temperature changes from ${25}^{o}C$ to ${35}^{o}C$ . Calculate energy of activation of the reaction.

The activation energy for the reaction, 2 $HI(g) \rightarrow H_2 + I_2 (g)$ is $209.5 k J mol^{-1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?

For a first order reaction, $A\longrightarrow$ Product, the temperature dependent rate constant (k) was found to follow the equation $logk=-\frac { 2000 }{ T } +6$ calculate the pre-exponential factor (A) and activation energy for the reaction.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate ${ E }_{ a }$ .

In a reaction $A\longrightarrow B+C$ , The following data were obtained:
t in seconds 0 900 1800
conc. of A 50.8 19.7 7.62
Prove that it is a first order reaction.

The rate constants of a reaction at 500 K and 700 K are ${ 0.02s }^{ -1 } and { 0.07 }s^{ -1 }$ respectively. Calculate the values of Ea and A.

What is the hybridisation of ${{\text{C}}_2}{{\text{H}}_4}$ . Draw its structure also.

In the $1^{st}$ order reaction $A \longrightarrow$ products $80\%$ of given sample of compound decomposes in $40\ mins$ . What is half life period of reaction ?

For a chemical reaction, what is the effect of a catalyst on the following?
(i) Activation energy of the reaction
(ii) Rate of the reaction.

One mole of an organic compound(A) with molecular formula $C_{3}H_{8}O$ reacts completely with two moles of hydroiodic acid at high temperature to form 'x' & 'y'. Identify 'A', 'X' & 'Y' in the reaction.

Show that, in a first order reaction, time required for completion of $99.9 \%$ is $10$ times of half life. $(log 10 =1)$

The equation $A + B \rightarrow C$ has zero order . What is the rate equation?

For the reaction $A \rightarrow$ products, the following data is given for a particular run.
time(min): 0 5 15 35
$\dfrac {1}{[A]}(M^{-1}) :$ 1 2 4 8
Determine the order of the reaction.

For the assumed reaction $X_2 + 3Y_2 \rightarrow 2XY_3$ , write the rate of equation in terms of rate of disappearance of $Y_2$ .

For an elementary reaction $2A + B \rightarrow 3C$ the rate of appearance of $C$ at time 't' is $1.3 \times 10^{-4} \, mol \, L^{-1} \,s^{-1}$ .
Calculate at this time rate of the reaction .

The rate law for a reaction is expressed as : rate $= k[Cl_2][NO]^2$
Find out the orders of the reaction with respect to $Cl_2$ and with respect to $NO$ and also the overall order of the reaction .

Write the rate equation for the reaction $A_2 + 3B_2 \rightarrow 2C$ , if the overall order of the reaction is zero .

Explain why $H_2$ and $O_2$ do not react at room temperature .

The specific rate constant of the decomposition of $N_2O_5$ is $0.008 min^{-1}$ . The volume of $O_2$ collected after 20 minute is 16 mL. The volume in ml that would be collected at the end of reaction. $NO_2$ formed is dissolved in $CCl_4$ is

If the percentage decompostion in $100$ minutes at the same temperature but at an initial pressure of $600$ mm is $(70+x)$ find the value of $x$ __________.

Half-life period of reation in minutes is:
(Give answer to the nearest integer)

If pressure of $Cl_2O_7$ after $100$ sec in atm is $x$ , then $1000x$ is__________.

For a gaseous reaction $A\, \rightarrow\, B\, +\, 2C$ at $250^\circ$ C, following data were observed. The partial pressure of the gases at 100th min will be (in 100x mm)
Time (min) 0 20 40 60
Total pressure (mm Hg) 400 500 587.6 664

The decomposition of a compound is found to follow a first-order rate law. If it takes 15 min for 20% of the original material to react, the specific rate constant is ________. ( $in \:\:min^{-1}$ )
[If the answer is X, write the value of 1000X in nearest integer value possible]

Catalytic decomposition of nitrous oxide by gold at $900^{0}C$ at an initial pressure of 200 mm was 50 % in 53 min and 73 % in 100 min. The amount that will compose in 100 min at the same temperature but at an initial pressure of 600 mm is (multiply the answer by 100) :

The rate for the decomposition of $NH_{3}$ on platinum surface is zero order. What are the rate of production of $N_{2}$ and $H_{2}$ if $K = 2.5 \times 10^{-4} mol\ litre^{-1} s^{-1}$ .

The rate constant for a first order reaction is $60\ s^{-1}$ . How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

For the decomposition of azo-isopropane to hexane and nitrogen at $543 K$ , the following data are obtained.
$t (sec)$ $P$ (mm of $Hg$ )
$0$ $35.0$
$360$ $54.0$
$720$ $63.0$
Calculate the rate constant.

Consider a certain reaction $A\rightarrow$ Products with, $k = 2.0 \times 10^{-2}s^{-1}$ . Calculate the concentration of A remaining after $100$ s, if the initial concentration of A is $1.0 mol L^{-1}$ .

$10\ g$ atoms of an $\alpha$ -active radioisotope are disintegrating in a sealed container. In one hour, the helium gas collected at STP is $11.2{ cm }^{ 3 }$ . Calculate the half-life of the radioisotope.