MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 5

Here is another look at the reaction of crystal violet with sodium hydroxide, a first-order reaction (ln

$A$ v time).
What is the significance of the slope?

0%
The slope represents the change in concentration over time.
0%
The slope represents the inverse of change in concentration over time.
0%
The slope represents the negative value of the rate constant.
0%
The slope represents the negative rate of product concentration over time.

A graph of concentration versus time data for a first-order reaction gives a straight line in which of the following plots of the data?

0%
$[A]_{t} = -kt + [A]_{0}$
0%
$ln [A]_{t} = -kt + ln [A]_{0}$
0%
$\dfrac {1}{[A]_{t}} = kt + \dfrac {1}{[A]_{0}}$
0%
All of the above
0%
None of the above

Explanation

For the first order reaction,

$\ln { \cfrac { [{ A }_{ o }] }{ [{ A }_{ t }] } } =kt\\ \therefore \ln { [{ A }_{ o }] } =kt+\ln { { [A] }_{ t } } \\ \therefore \ln { [{ A }]_{ t } } =-kt+\ln { { [A }]_{ o } }$

In a reaction mechanism consisting of elementary reaction steps where the relative rate of each is given, which of the following is most likely to be the rate-determining step?

0%
A step labeled fast
0%
A step labeled moderate
0%
A step labeled slow
0%
It is not possible to tell which step is rate determining from this information.

Reacting

$Cl$ with

$NO_{2}Cl$ molecules produces

$NO_{2}$ and

$Cl_{2}$ gases.
Which of the following will require the lowest activation energy based on proper orientation for this reaction?

Explanation

The atom collides to

$Cl$ atom present in

$N{ O }_{ 2 }Cl$ . molecule with proper orientation & head or collision.

Therefore, very low activation energy is required.

An experiment was performed to study the kinetics of the decomposition of substance

$X$ . The concentration of

$X$ ,

$[X]$ , was monitored over time as the reaction proceeded. The data are shown in the table above. The time interval between concentration measurements was

$1.0\ minutes (60 s)$ .

Time (s)	$[X]$	$ln[X]$	$\dfrac {1}{[X]}$
$0$	$20.0$	$3.00$	$0.050$
$60$	$10.0$	$2.30$	$0.100$
$120$	$6.67$	$1.90$	$0.150$
$180$	$5.00$	$1.61$	$0.200$
$240$	$4.00$	$1.39$	$0.250$

Which of the following best describes the reaction?

0%
The decomposition of substance $X$ is a zero-order reaction.
0%
The decomposition of substance $X$ is a first-order reaction.
0%
The decomposition of substance $X$ is a second-order reaction.
0%
The decomposition of substance $X$ is a third-order reaction.

Explanation

Form the given data,

$\cfrac { 1 }{ [X] } \propto Time(t)\\ \therefore \cfrac { 1 }{ [X] } =Kt\longrightarrow$ K=constant

$\therefore K=\cfrac { 1 }{ [X]t }$

$\therefore$ Unit of K are

${ M }^{ -1 }{ time }^{ -1 }$ which is possible only for second order reaction

In a given reaction under standard conditions in a closed container, which type of reaction would show no real increase in the rate of the reaction when the concentration of each reactant is doubled?

0%
A zero order
0%
A first order
0%
A second order
0%
A third order

A first order reaction is 60% complete in 20 minutes. How long will the reaction take to be 84% complete?

0%
54 mins
0%
68 mins
0%
40 mins
0%
76 mins

Explanation

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

The reaction is 60% complete in 20 minutes.

$k = \dfrac {2.303}{20}log \dfrac {100}{100-60}$ .....(1)

The reaction is 84% complete

$k = \dfrac {2.303}{t}log \dfrac {100}{100-84}$ .....(2)

But (1) = (2),

$\dfrac {2.303}{20}log \dfrac {100}{100-60}=\dfrac {2.303}{t}log \dfrac {100}{100-84}$

$\dfrac {1}{20}log \dfrac {100}{100-60}=\dfrac {1}{t}log \dfrac {100}{100-84}$

$t=\dfrac {20}{log \dfrac {100}{100-60}}log \dfrac {100}{100-84}$

$t=\dfrac {20}{ 0.3979 } \times 0.79588$

$t=40$ minutes.
The reaction will take 40 minutes to be 84% complete.

A plot of

$\dfrac { 1 }{ T }$ vs.

$logk$ for a reaction gives the slope

$(-1\times { 10 }^{ 4 }\ K)$ . The energy of activation for the reaction is :
(Given

$R=8.314\ J{ K }^{ -1 }{ mol }^{ -1 }$ )

0%
$1.202\ kJ{ mol }^{ -1 }$
0%
$83.14\ kJ{ mol }^{ -1 }$
0%
$8314\ J { mol }^{ -1 }$
0%
$191.47 \ kJ mol^{-1}$

Explanation

$log\ k=log\ A-\dfrac{E_a}{2.303RT}$

Slope

$=\dfrac{-E_a}{2.303R}=-1\times 10^{4}$

$\Rightarrow E_a=2.303\times 8.314\times 10^4=191.47\ kJmol^{-1}$

The correct statement regarding the following energy diagrams is

0%
Reaction M is faster and less exothermic than Reaction N
0%
Reaction M is slower and less exothermic than Reaction N
0%
Reaction M is faster and more exothermic than Reaction N
0%
Reaction M is slower and more exothermic than Reaction N

The activation energy of a chemical reaction can be determined by :

0%
evaluating rate constants at two different temperatures
0%
changing the concentration of reactants
0%
evaluating the concentration of reactants at two different temperatures
0%
evaluating rate constant at standard temperature

Explanation

Explanation:

Activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction.

The formula to be used to find the activation energy is :

log

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

$\dfrac{{E}_{a}}{2.303R}$ [

$\dfrac{1}{{T}_{1}}$ -

$\dfrac{1}{{T}_{2}}$ ]

where,

${k}_{1}$ and

${k}_{2}$ are the rate constants for the reaction at two different temperatures

${T}_{1}$ and

${T}_{2}$ .

Thus, the activation energy can be calculated by evaluating the value of rate constants at two different temperatures.

Hence, The correct answer is option $(A)$ .

A given sample of milk turns sour at room temperature

$(27 ^oC)$ in 5 hours. In a refrigerator at

$-3 ^oC$ , it can be stored 10 times longer. The energy of activation for the souring of milk is :

0%
$2.303 \times 10 R \ \ kJ \cdot mol^{-1}$
0%
$2.303 \times 5 R \ \ kJ \cdot mol^{-1}$
0%
$2.303 \times 3 R \ \ kJ \cdot mol^{-1}$
0%
$2.303 \times 2.7 R \ \ kJ \cdot mol^{-1}$

Explanation

The energy of activation is given by the expression

$\displaystyle E_a = \dfrac {2.303RTT'}{T'-T}log\dfrac {k'}{k}$

$\displaystyle E_a = \dfrac {2.303 \times R \times 300 \times 270}{300-270}log\dfrac {10k}{k}$

$\displaystyle E_a = 2.303 \times 2699 R \ \ J \cdot mol^{-1}$

$\displaystyle E_a =2.303 \times 2.7 R \ \ kJ \cdot mol^{-1}$

When the concentration of a reactant of first order reaction is doubled, the rate becomes ________ times, but for ___________ order reaction, the rate remains same.

0%
Four; zero
0%
Two; zero
0%
Zero; four
0%
Zero; three

Explanation

When the concentration of a reactant of the first-order reaction is doubled, the rate becomes two times, but for zero-order reaction, the rate remains the same.

For first order reaction,
rate

$\displaystyle = k[A]$
When

$\displaystyle [A]$ is doubled, the rate is doubled.

For zero order reaction,

rate

$\displaystyle = k[A]^0$
When

$\displaystyle [A]$ is doubled, the rate is unchanged.

So, the correct option is

$B$

The unit of zero order rate constant is:

0%
$litre$ $mol^{-1} sec^{-1}$
0%
$mol$ $litre^{-1} sec^{-1}$
0%
$sec^{-1}$
0%
$litre^{2} sec^{-1}$

Explanation

The unit of zero order rate constant is

$\displaystyle mol \ litre^{-1} sec^{-1}$ .

For zero order reaction rate

$\displaystyle = k[A]^0=k$ .

The unit of the rate constant is equal to the unit of rate which is concentration per unit time.

The excess energy which a molecule must posses to become active is known as:

0%
kinetic energy
0%
threshold energy
0%
potential energy
0%
activation energy

The unit of rate constant for zero order reaction is:

0%
$s^{-1}$
0%
$mol\, L^{-1}s^{-1}$
0%
$L\, mol^{-1}s^{-1}$
0%
$L^2mol^{-2}s^{-1}$

Explanation

$Rate=\dfrac{dC}{dt} = k[R_0]^o = k$ or

$k=\dfrac{dC}{dt} = \dfrac{conc.}{time}$

$=\dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}$ .

Which graph shows zero activation energy for reaction?

For a zero-order reaction with a rate constant k, the slope of the plot of reactant concentration against time is:

0%
$\dfrac{ k} {2.303}$
0%
$k$
0%
$-\dfrac {k}{2.303}$
0%
$-k$

Explanation

For a Zero order reaction,

$A_t=A_o-kt$

Slope

$=-k$

Hence option

$D$ is correct.

The relationship between rate constant and half-life period of zero-order reaction is given by:

0%
${ t }_{ \frac { 1 }{ 2 } }={ \left[ A \right] }_{ 0 }2k$
0%
${ t }_{ \frac { 1 }{ 2 } }=\dfrac { 0.693 }{ k }$
0%
${ t }_{ \frac { 1 }{ 2 } }=\dfrac { { \left[ A \right] }_{ 0 } }{ 2k }$
0%
${ t }_{ \frac { 1 }{ 2 } }=\dfrac { 2{ \left[ A \right] }_{ 0 } }{ k }$

Explanation

${ t }_{ \frac { 1 }{ 2 } }=\frac { { [A] }_{ 0 } }{ 2k }$

Where

${ t }_{ \frac { 1 }{ 2 } }$ is half life

$[A]_0$ is initial concentration and k is rate constant.

The half-life of

$^{14}C$ is 5570 yr. How many years will it take 90% of a sample to decompose?

0%
5,570 yr
0%
17,700 yr
0%
18,510 yr
0%
50,100 yr

Explanation

The half-life of

$\displaystyle ^{14}C$ is

$5570 yr.\$

$\displaystyle t_{1/2} = 5570$ years

The decay constant

$\displaystyle \lambda = \dfrac {0.693}{t_{1/2}}$

$\displaystyle \lambda = \dfrac {0.693}{ 5570 \: yr}$

Also

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

$\displaystyle t = \dfrac {2.303}{\dfrac {0.693}{ 5570 \: yr}} log\left( \dfrac {100}{100-90}\right)$

$\displaystyle t=18,510 \: yr$

So, the correct option is

$C$

Collision theory is applicable to

0%
First order reactions
0%
Zero order reactions
0%
Bimolecular reactions
0%
Intramolecular reactions

$3A\longrightarrow B+C$

It would be a zero-order reaction when:

0%
The rate of reaction is proportional to square of concentration of $A$
0%
The rate of reaction remains the same at any concentration of $A$
0%
The rate remains unchanged at any concentration of $B$ and $C$
0%
The rate of reaction doubles if concentration of $B$ is increased to double

Explanation

For reaction,

$3A\longrightarrow B+C$

If it is zero order reaction, therefore the rate remains same at any concentration of '

$A$ ' or

$\dfrac { dx }{ dt } =k {[ A }]^{ 0 }$ .

It means that the rate is independent of the concentration of reactants.

Hence, option B is correct.

A certain radioactive isotope decay has

$\alpha$ -emission,

$_{ { Z }_{ 1 } }^{ { A }_{ 1 } }{ X }\longrightarrow$

$_{ { Z }_{ 1 }-2 }^{ { A }_{ 1 }-4 }{ Y }$
half life of

$X$ is

$10$ days. If

$1$

$mol$ of

$X$ is taken initially in a sealed container, then what volume of helium will be collected at STP after

$20$ days?

0%
$22.4 L$
0%
$11.2 L$
0%
$16.8 L$
0%
$33.6 L$

Explanation

$_{Z_1}^{A_1}X\rightarrow _{Z_1-2}^{A_1-4}Y+_2^4He$

$1\quad\quad 0\quad\quad 0$ .......initially

$0.25\quad\quad 0.75\quad\quad 0.75$ .... after two half lives(20 days)

After

$20$ days

$0.75$

$mol$ helium will be formed.

$\therefore$ Volume of helium at STP

$=0.75\times 22.4=16.8\ L$

One gram of

$^{ 226 }{ Ra }$ has an activity of nearly

$1 Ci$ . The half life of

$^{ 226 }{ Ra }$ is

0%
$1582$ yrs
0%
$12.5$ hrs
0%
$140$ days
0%
$4.5\times { 10 }^{ 9 }$ yrs

Explanation

Use the following relation for calculation of activity:
Activity

$=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } \times \dfrac { w }{ At.wt } \times 6.023\times { 10 }^{ 23 }$

$3.7\times { 10 }^{ 10 }=\dfrac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } \times \dfrac { 1 }{ 226 } \times 6.023\times { 10 }^{ 23 }$ =

$0.5\times 10^{8}\ sec$ .

In years, it will be

$\dfrac{0.5\times 10^{8}}{86400\times 365}=1582\ years$

The order of reaction for which half-life period is independent of initial concentration is:

0%
zero
0%
first
0%
second
0%
third

Explanation

For first order reaction, the half-life period is independent of initial concentration.

For first order reaction, the half-life period expression

$\displaystyle (t_{1/2})$ is given by the expression

$\displaystyle t_{1/2} = \dfrac {0.693}{k}$ , where k is the rate constant.

For nth order reaction,

$\displaystyle t_{1/2} \propto \dfrac {1}{a^{n-1}}$ .

A freshly prepared radio medicine has half life

$2$ hours. Its activity is

$64$ times the permissible safe value. The minimum time after which it would be possible to treat the patients with the medicine is:

0%
$3$ hrs
0%
$9$ hrs
0%
$24$ hrs
0%
$12$ hrs

Explanation

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

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

$\dfrac { 1 }{ 64 } ={ \left( \dfrac { 1 }{ 2 } \right) }^{ n }$

$n=6$ half lives

$\therefore$ time

$=2\times 6=12$ hrs

Hence, the correct option is

$\text{D}$

For an endothermic reaction, where

$\Delta H$ represents the enthalpy of reaction, the minimum value for the energy of activation will be:

0%
less than $\Delta H$
0%
zero
0%
equal to $\Delta H$
0%
more than $\Delta H$

Explanation

Plot of energy v/s reaction progress for a typical endothermic reaction :-

The minimum value of the activation energy is the difference between the energy content of reactant and product.

$\Delta H$ also gives the difference between the reactant and product.

Hence, the minimum value for the activation energy will be equal to

$\triangle H$ .

A first-order reaction has a half-life of

$14.5\ hrs$ . What percentage of the reactant will remain after

$24\ hrs$ ?

0%
$18.3$ %
0%
$31.8$ %
0%
$45.5$ %
0%
$68.2$ %

Explanation

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

$\dfrac { 0.693 }{ 14.5 } =\dfrac { 2.303 }{ 24 } \log { \dfrac { 100 }{ \left( a-x \right) } }$
On solving,

$\left( a-x \right) =31.8$ %

The rates of a certain reaction at different time intervals are as follows:

$\begin{matrix} Time(second) & 0 & 10 & 20 \\ Rate(mol\ L^{-1}s^{-1}) & 1.8\times { 10 }^{ -2 } & 1.82\times { 10 }^{ -2 } & 1.79\times { 10 }^{ -2 } \end{matrix}$
The reaction is of:

0%
zero order
0%
first order
0%
second order
0%
third order

Explanation

There is no effect of reactant concentration of the rate. this is only possible in a

$0$ order reaction.

Which one of the following is not a first order reaction?

0%
$C{ H }_{ 3 }COOC{ H }_{ 3 }+{ H }_{ 2 }O\xrightarrow [ ]{ \quad { H }^{ + }\quad } C{ H }_{ 3 }COOH+C{ H }_{ 3 }OH$
0%
$C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+NaOH\longrightarrow C{ H }_{ 3 }COONa+{ C }_{ 2 }{ H }_{ 5 }OH$
0%
$2{ H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }$
0%
$2{ N }_{ 2 }{ O }_{ 5 }\longrightarrow 4N{ O }_{ 2 }+{ O }_{ 2 }$

Explanation

The hydrolysis of ethyl acetate (

$CH_3COOC_2H_5$ ) by sodium hydroxide yields ethanol (

$C_2H_5OH$ ) and sodium acetate (

$NaO_2CCH_3$ ) by the reaction showed follows second order kinetics.

Reaction,

$A+B\longrightarrow C+D+38kcal$ has activation energy

$20kcal$ . Activation energy for the reaction,

$C+D\longrightarrow A+B\quad$ is:

0%
$20kcal$
0%
$-20kcal$
0%
$18kcal$
0%
$58kcal$

Explanation

Reaction,

$A+B\longrightarrow C+D+\text {38 kcal }$ has activation energy

$\text {20 kcal }$ . Activation energy for the reaction,

$C+D\longrightarrow A+B\quad$ is

$20\text { kcal }+ 38 \text { kcal }=58 \text { kcal }$ .
Note:
The reaction is exothermic as for exothermic reaction (heat liberated during reaction),

$\displaystyle E_{a, r} = E_{a, f} + \Delta H_{rxn}$

$\displaystyle E_{a, f} =$ activation energy for forward reaction

$\displaystyle E_{a, r} =$ activation energy for reverse reaction

$\displaystyle \Delta H_{rxn} =$ enthalpy change for the reaction.

$500$

$K$ , the half-life period of a gaseous reaction at an initial pressure of

$80$

$kPa$ is

$350$

$sec$ . When the pressure is

$40$

$kPa$ , the half life period is

$175$

$sec$ ; the order of the reaction is:

0%
Zero
0%
One
0%
Two
0%
Three

Explanation

$\dfrac { { \left( { t }_{ { 1 }/{ 2 } } \right) }_{ 1 } }{ { \left( { t }_{ { 1 }/{ 2 } } \right) }_{ 2 } } ={ \left( \dfrac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ n-1 }$

$\dfrac { 350 }{ 175 } ={ \left( \dfrac { 40 }{ 80 } \right) }^{ n-1 }$

$2={ \left( \dfrac { 1 }{ 2 } \right) }^{ n-1 }$

$n-1=-1$

$n=0$ (zero order reaction)

Which of the following is a first order reaction?

0%
$2HI\longrightarrow { H }_{ 2 }+{ I }_{ 2 }$
0%
$2N{ O }_{ 2 }\longrightarrow 2NO+{ O }_{ 2 }$
0%
$2NO+{ O }_{ 2 }\longrightarrow 2N{ O }_{ 2 }$
0%
$N{ H }_{ 4 }N{ O }_{ 2 }\longrightarrow { N }_{ 2 }+2{ H }_{ 2 }O$

Explanation

Correct Answer: Option D

Explanation:

$aA + bB + cC \rightarrow dD + eE$
Or, $\dfrac{\mathrm{d} r}{\mathrm{d} t} =$ Rate of reaction
$K=$ constant of proportionality or Rate constant
$\left [ A \right ]^{\alpha},\left [ B \right ]^{\beta},\left [C \right ]^{\gamma}=$ concentrations of reactants raised to power of order of reaction in respect to that particular reactant.
For elementary or one-step reactions, the order of reaction with respect to each reactant is equal to their coefficients in a balanced equation.
Since all given reactions are elementary, the order can be determined for each of these.

1. $2HI \rightarrow H_{2} + I_{2}$
$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{1}\left [ HI \right ]^{2}$
Therefore, it’s a second order reaction.

2. $2NO_{2} \rightarrow 2NO + O_{2}$
$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{2}\left [ NO_{2} \right ]^{2}$
Therefore, it’s a second order reaction.

3. $2NO + O_{2} \rightarrow 2NO_{2}$
$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{3}\left [ NO \right ]^{2} \left [ O_{2} \right ]^{1}$
Therefore, it’s a third order reaction.

4. $NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O$
$\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{4} \left [ NH_{4}NO_{2} \right ]^{1}$
Therefore, it’s a first order reaction.

Hence, $NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O$ is a first order reaction.

A substance reacts with initial concentration of

$a$

$mol$

${ dm }^{ -3 }$ according to zero order kinetics. The time it takes for the completion of the reaction is: (

$k=$ rate constant)

0%
$\dfrac { k }{ a }$
0%
$\dfrac { a }{ 2k }$
0%
$\dfrac { a }{ k }$
0%
$\dfrac { 2k }{ a }$

Explanation

Since rate constant of zero order reaction is $[A]=[A]_o−kt$
$\implies t = \frac{[A]_o−[A]}{k}$
$\implies t = \frac{a - 0}{k}$
$\implies t =\frac{a}{k}$

The decomposition of

$HI$ on the surface of gold is:

0%
pseudo first order
0%
zero order
0%
first order
0%
second order

For a zero order reaction which of the following relation is correct, where

$t_{1/2}\ and\ a$ are half-life and initial concentration respectively.

0%
${ t }_{ { 1 }/{ 2 } }\propto a$
0%
${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ a }$
0%
${ t }_{ { 1 }/{ 2 } }\propto { a }^{ 2 }$
0%
${ t }_{ { 1 }/{ 2 } }\propto \dfrac { 1 }{ { a }^{ 2 } }$

Explanation

When zero order kinetic rate law is followed, let $[A]$ is the current concentration, $[A]_0$ is the initial concentration, and $k$ is the reaction constant and $t$ is time
Then relation is given by $[A]=[A]_0−k$

In order to find the half life we need to isolate t on its own, and divide it by 2. We would end up with a formula as such depict how long it takes for the initial concentration to dwindle by half given by -:
$\implies t_{1/2}=\frac{[A]_0}{2k}$

A drop of solution (volume

$0.05$

$mL$ ) contains

$3\times { 10 }^{ -6 }$ mole of

${ H }^{ + }$ . If the rate constant of disappearance of

${ H }^{ + }$ is

${ 10 }^{ 7 }$

$mol$

${ litre }^{ -1 }$

${ sec }^{ -1 }$ , how long would it take for

${ H }^{ + }$ in the drop to disappear?

0%
$6\times { 10 }^{ -8 }\ sec$
0%
$6\times { 10 }^{ -9 }\ sec$
0%
$6\times { 10 }^{ -10 }\ sec$
0%
$6\times { 10 }^{ -12 }\ sec$

Explanation

$k={ 10 }^{ 7 }{ molL }^{ -1 }{ sec }^{ -1 }\\ [{ H }^{ + }]=\cfrac { 3\times { 10 }^{ -6 } }{ 0.05\times { 10 }^{ -3 } } { molL }^{ -1 }=\cfrac { 3 }{ 5 } \times { 10 }^{ -1 }{ mol }L^{ -1 }\\ \therefore \ln { \left( \cfrac { { A }_{ o } }{ A } \right) } =kt\\ { t }_{ 99\% }=\cfrac { \ln { \left( \cfrac { { A }_{ o } }{ A } \right) } }{ k } \\ =\cfrac { \ln { \left( \cfrac { 3/5\times { 10 }^{ -1 } }{ 3/5\times { 10 }^{ -3 } } \right) } }{ { 10 }^{ 7 } } \\ =9.212\times { 10 }^{ -3 }sec\\ So,\\ v=k[{ H }^{ + }]=0.6\times { 10 }^{ -1 }\times { 10 }^{ 7 }\\ =0.6\times { 10 }^{ 6 }\\ =6\times { 10 }^{ 5 }{ mol }^{ 2 }{ L }^{ -1 }{ s }^{ -1 }\\ \therefore t=6\times { 10 }^{ -9 }sec$ for

$\cfrac { 3 }{ 5 } \times { 10 }^{ -1 }mol{ L }^{ -1 }$

The specific rate constant of a first order reaction depends on:

0%
concentration of the reactants
0%
concentration of the products
0%
time
0%
temperature

Explanation

$k={ Ae }^{ \cfrac { -{ E }_{ a } }{ RT } }$

$\therefore$

$k$ depends on temperature.

The rate constant of a reaction is found to be

$3\times { 10 }^{ -3 }$

$mol$

${ L }^{ -1 }$

${ min }^{ -1 }$ . The order of the reaction is:

0%
$0$
0%
$1$
0%
$2$
0%
$1.5$

Explanation

Since the rate constant of a zero-order reaction is Rate = K.

The unit of rate constant will be $mol . L^{-1} min^{-1}.$

The hydrolysis of ethyl acetate is a reaction of:

$C{ H }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O\xrightarrow [ ]{ \quad { H }^{ + }\quad } C{ H }_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH$

0%
zero order
0%
first order
0%
second order
0%
third order

Explanation

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { H }^{ + }\\ { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O\longrightarrow { CH }_{ 3 }COOH+{ C }_{ 2 }{ H }_{ 5 }OH$

In the above reaction medium of reaction is water which will be in excess, hence it can be avoided in rate expression, thus rate will be given as

$Rate=K\left[ { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 } \right]$

Hence it is a

${ 1 }^{ st }$ order reaction.

Graph between the concentration of the product '

$x$ ' and time '

$t$ ' for

$A\rightarrow B$ is given above.
The graph between

$-\dfrac { d\left[ A \right] }{ dt }$ and time

$'t'$ will be of the type:

Explanation

For a zero order reaction

Rate of reaction=-

$\frac{d[A]}{dt}$ =

$K(a-x)^0$ =K

So the correct answer is (C) because there is no change in rate w.r.t time.

For a zero order reaction,

$A\longrightarrow P$ ,

${ t }_{ { 1 }/{ 2 } }$ is:

(

$k$ is the rate constant,

$[A]_0$ is the initial concentration of

$A$ )

0%
$\dfrac { { \left[ A \right] }_{ 0 } }{ 2k }$
0%
$\dfrac { \ln { 2 } }{ k }$
0%
$\dfrac { 1 }{ k{ \left[ A \right] }_{ 0 } }$
0%
$\dfrac { \ln { 2 } }{ { \left[ A \right] }_{ 0 }k }$

Explanation

For zero order

$kt=a\\ { t }_{ 1/2 }=\cfrac { { a }_{ o } }{ 2k } =\cfrac { { \left[ A \right] }_{ o } }{ 2k }$

Following is the graph between

$\log{{ t }_{ { 1 }/{ 2 } }}$ and

$\log { a }$ (

$a=$ initial concentration) for a given reaction at

${ 27 }^{ o }C$ . Hence, order of the reaction is:

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$\cfrac { 1 }{ { a }^{ n-1 } } =\cfrac { 1 }{ { a }_{ 0 }^{ n-1 } } +(n-1)kt$

for

$n=0$

$\cfrac { 1 }{ { a }^{ -1 } } =\cfrac { 1 }{ { a }_{ 0 }^{ -1 } } +(-1)kt\\ a={ a }_{ o }-kt\\ { kt }_{ 1/2 }=\cfrac { { a }_{ o } }{ 2 } \\ \log { { (t }_{ 1/2 })= } \log { ({ 0 }_{ o }) } -\log { \left( \cfrac { 1 }{ 2k } \right) }$

So,

$\log { { (t }_{ 1/2 }) } =\log { ({ 0 }_{ o }) } +\log { (2k) }$

Which is a straight line as given.

When the initial concentration of the reaction is doubled, the half-life is also doubled. The order of the reaction will be:

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

Half life

$\alpha$ initial concentration

This corresponds to zero order reaction where

$t_{1/2}=\frac{a}{2k}$

A substance having rate constant

$k$ and initial concentration

$a$ reacts according to zero order kinetics. What will be the time for the reaction to go to completion?

0%
$\dfrac { a }{ k }$
0%
$\dfrac { k }{ a }$
0%
$\dfrac { a }{ 2k }$
0%
$\dfrac { 2k }{ a }$

Explanation

For zero order reaction,

$t=\frac{a-at}{k}$

where at=concentration of reactant at time t

For completion of reaction, at=0

$t=\frac{a}{k}$

The time for half-life period of a certain reaction,

$A\rightarrow$ Product, is

$1$ hour, when the initial concentration of the reactant '

$A$ ' is

$2$

$mol$

${ L }^{ -1 }$ . How much time does it take for its concentration to come from

$0.50$ to

$0.25$

$mol$

${ L }^{ -1 }$ if it is a zero order reaction?

0%
$0.25\ hr$
0%
$1\ hr$
0%
$4\ hrs$
0%
$0.5\ hr$

Explanation

$\dfrac { { t }_{ 1 } }{ { t }_{ 2 } } ={ \left( \dfrac { { a }_{ 2 } }{ { a }_{ 1 } } \right) }^{ n-1 }$

$\dfrac { 1 }{ { t }_{ 2 } } ={ \left( \dfrac { 0.5 }{ 2 } \right) }^{ 0-1 }$

${ t }_{ 2 }=0.25 hr$

Raw milk sours in about 4 h at

$27^ o C$ , but in about 48 h in a refrigerator at

$17^ o C$ . What is the activation energy for souring of milk ?

0%
78.3 kJ $mol^{-1}$
0%
46.21 kJ $mol^{-1}$
0%
23.5 kJ $mol^{-1}$
0%
80.8 kJ $mol^{-1}$

Explanation

$K\propto \cfrac {1}{t}$

$log \cfrac{K_2}{K_1}=log \cfrac {t_1}{t_2}=\cfrac {E_a}{R}\left ( \cfrac {T_2-T_1}{T_1 \times T_2} \right )$

$log \cfrac {48}{4}=\cfrac {E_a}{8.314}\left ( \cfrac {10}{300\times 290} \right )$

$E_a=78.3 \, kJ\, mol^{-1}$

Which of the following is not correct for zero order reaction?

In the reaction,

$A+B\longrightarrow C+D$ , the rate

$\left( \dfrac { dx }{ dt } \right)$ when plotted against time '

$t$ ' gives a straight line parallel to time axis and at some time

$'t'$ ,

$\dfrac{dx}{dt}=k$ . The order and rate of reaction will be:

0%
$1, k+1$
0%
$0, k$
0%
$\left( 1+k \right) , 1$
0%
$k, k+1$

Explanation

A straight line parallel to time axis indicate that the rate is constant. This is the case for zero order reaction.

$\frac{dx}{dt} =k$

rate of the reaction=k

$N_2+3H_2\rightleftharpoons 2NH_3 +22$ kcal
The activation energy for the forward reaction is

$50$ kcal. What is the activation energy for the backward reaction?

0%
$-72$ kcal
0%
$-28$ kcal
0%
$+28$ kcal
0%
$+72$ kcal

Explanation

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) +22$ kcal

As there is 22Kcal heat is released so its exothermic.

$\because$ The activation energy for the forward reaction

$=50$ kcal

$\therefore$ The activation energy for the backward reaction

$=50+22=+72$ kcal.

For the first order reaction,

$t_{99}$ %

$= x \times t_{90}$ %, the value of

$x$ will be:

0%
10
0%
6
0%
3
0%
2

Explanation

For

$t_{90\% }$ ,

$a_t=\frac{10a_0}{100} = \frac{a_0}{10}$

$t_{90\% }=\frac{1}{k}In\frac{a-0}{a_0/10}=\frac{In10}{k}$ ....(1)

For

$t_{99\% }$ ,

$a_t=\frac{1}{100}a-0$

$t_{99\% }=\frac{1}{K} In\frac{a_0}{a_0/100}=\frac{2In10}{k}$ ....(2)

From (1) and (2),

$t_{99\% }=2\times t_{90\% }=> x=2$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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