MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 3

If the rate with respect to

$O_2, NO, NO_2$ are

$\displaystyle \frac{- \Delta [O_2]}{\Delta t}$

$= \dfrac{-1}{2}\dfrac{\Delta [NO]}{\Delta t} = \dfrac{+1}{2} \dfrac{\Delta [NO_2]}{\Delta t}$ , then the corresponding chemical equation is

$2NO+O_2 \rightarrow NO_2$ .

0%
True
0%
False

An increase in the rate of a reaction for a rise in temperature is due to:

0%
increase in collision frequency
0%
shortening of mean free path
0%
increase in the number of activated molecules
0%
none of the above

For a first order reaction

$A\rightarrow B$ the rate constant is

$x\ min^{−1}$ . If the initial concentration of A is 0.01M , the concentration of A after one hour is given by the expression:

0%
$0.01e^{-x}$
0%
$1\times10^{-2}(1-e^{-60x})$
0%
$(1\times10^{-2}) e^{-60x}$
0%
None of these

Which statements are correct in terms of chemical kinetic stuides?

0%
The quenching of a reaction can be made by cooling the reaction mixture.
0%
The quenching of a reaction can be made by diluting the reaction mixture.
0%
The reaction is supposed to be completed if it is kept for long time or strongly heated.
0%
None of the above

All first order reactions are unimolecular.

0%
True
0%
False

To initiate a reaction, minimum energy which is required to break bonds is called

bond energy
activation energy
breaking energy
ionization energy

0%
1
0%
2
0%
3
0%
4

Radioactive disintegration follows _____ order kinetic.

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

Time required to complete half of the reactant in a zero order reaction is equal to a/(2k).

0%
True
0%
False

Explanation

The statement is true.
Time required to complete half of the reactant in a zero order reaction is equal to a/(2k).

The half life ( $t_{1/2}$ ) of a reaction is the time required for the concentration of the radioactive substance to decrease to one-half of its original value. The half-life of a zero-order reaction can be derived as follows:

For a reaction involving reactant $a$ and from the definition of a half-life, $t_{1/2}$ is the time it takes for half of the initial concentration of reactant $a$ to react. These new conditions can be substituted into the integrated rate law form to obtain the following:

$1/2[a] = -kt_{1/2}+ a$

Solving for $t_{1/2}$ gives the following:

$t_{1/2} = a / (2k)$

If 'a' is the initial concentration of a substance which reacts according to zero order kinetics and K is rate constant, the time for the reaction to go to completion is:

0%
a/K
0%
2/Ka
0%
K/a
0%
2K/a

Explanation

For zero order reaction,

$K=\frac {x}{t}$
if

$x=a$ (complete reactant to react);

$t=\frac {a}{K}$ .

Activation energy of a reaction is:

0%
the energy released during the reaction.
0%
the energy evolved when activated complex is formed.
0%
additional amount of energy needed by the reactants to overcome the potential barrier of reaction.
0%
the energy needed to form one mole of the product.

Explanation

Hint: Activation energy is the barrier energy required to initiate the reaction and reach transition state.

Correct Answer: Option

$C$ .

Explanation:

We know that activation energy is the minimum amount of energy required by the reactants to undergo a particular chemical reaction.

Final Answer : Hence, the option $C$ is the correct answer.

Calculate the partial pressure of reactants and products, when azomethane decomposed at an initial pressure of 200 mm for 30 minutes according to

$(CH_3)_2N_2\rightarrow C_2H_6+N_2$
The rate constant is

$2.5\times 10^{-4} sec^{-1}$ .

0%
$P'(CH_3)_2N_2=12.755 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$
0%
$P'(CH_3)_2N_2=127.55 mm, P'C_2H_6=72.45 mm, P'N_2=72.45 mm$
0%
$P'(CH_3)_2N_2=1275.5 mm, P'C_2H_6=72.45 mm, P'N_2=724.5 mm$
0%
None of these

Explanation

30 min corresponds to 1800 seconds.

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

$2.5 \times 10^{-4}=\displaystyle \frac {2.303}{1800}log \frac {200}{P-x}$

$\displaystyle 1.568 = log \frac {200}{P-x}$

$\displaystyle P-x =127.55 \: mm$
Hence, the partial pressure of azomethane is 127.55 mm Hg.
The partial pressures of ethane and nitrogen are

$\displaystyle200 \: mm-127.55 \: mm=72.45 \: mm$ respectively.

At room temperature, the reaction between

$NO$ and

$O_2$ to give

$NO_2$ is fast, while that between

$CO$ and

$O_2$ is slow. It is due to:

0%
CO is smaller in size than that of NO
0%
CO is poisonous
0%
the activation energy for the reaction, $2NO+O_2\rightarrow 2NO_2$ is less than $2CO+O_2\rightarrow 2CO_2$
0%
none of the above

Explanation

At room temperature, the reaction between $NO$ and $O_2$ to give $NO_2$ is fast, while that between $CO$ and $O_2$ is slow. It is due to the activation energy for the reaction, $2NO+O_2\rightarrow 2NO_2$ is less than $2CO+O_2\rightarrow 2CO_2$ .

Reactions having lower energy of activation occurs more fast under similar experimental conditions.

A first order reaction has a half-life period of 69.3 sec. At

$0.10\ M$ reactant concentration, rate will be:

0%
$10^{-3}\ M. sec^{-1}$
0%
$10^{-4}\ M. sec^{-1}$
0%
$10^{-2}\ M .sec^{-1}$
0%
None of these

Explanation

Given that a first order reaction has a half-life period of

$69.3$ seconds.

$K=\dfrac{0.693}{t_{1/2}}=\dfrac {0.693}{69.3}=10^{-2} \sec^{-1}$

Now,

$r=K[A]=10^{-2}\times 0.1=10^{-3} M sec^{-1}$ .

For a reaction for which the activation energies of forward and reverse reactions are equal:

0%
$\Delta H=0$
0%
$\Delta S=0$
0%
the order is zero
0%
there is no catalyst

Explanation

For a reaction for which the activation energies of forward and reverse reactions are equal, then both

$\Delta H$ and

$\Delta S$ are

$0$ .
The activation energy is the energy it takes to push the reactants up to the transition state.
For Exothermic reaction = negative

$\Delta H$
Endothermic reaction = positive

$\Delta H$

The rate constant for an isomerisation reaction,

$A\rightarrow B$ is

$4.5\times 10^{-3} min ^{-1}$ . If the initial concentration of A is 1 M, the rate of the reaction after 1 hr will be:

0%
$3.4354\times 10^{-3}$
0%
$4.4554\times 10^{-3}$
0%
$5.4364\times 10^{-3}$
0%
None of these

Explanation

$\underset {[A]_0=1 M}{A\rightarrow B} K=4.5\times 10^{-3} min^{-1}$
For Ist order reaction,

$K=\frac {2.303}{t}log \frac {a}{(a-x)}$ at

$t=60 min$

$4.5\times 10^{-3}=\frac {2.303}{60} log_{10}\frac {1}{(a-x)}$

$\therefore (a-x)=0.7634$
Thus, rate after 60 minute

$=K(a-x)$

$=4.5\times 10^{-3}\times 0.7634$

$=3.4354\times 10^{-3}$

The activation energy for a chemical reaction depends upon:

0%
reaction
0%
nature of reacting species
0%
frequency factor
0%
concentration of reacting species

$A\rightarrow B+2C$

$100^oC$ , in above gaseous reaction, is observed to be of the first order. On starting with pure A, at the end of

$14$ minutes, the total pressure found to be

$264$ mm of Hg. After a long time, the total pressure of the system was

$450$ mm of Hg.

Rate constant of the reaction is:

0%
$6.455\times 10^{-2} min^{-1}$
0%
$3.415\times 10^{-2} min^{-1}$
0%
$5.575\times 10^{-2} min^{-1}$
0%
none of these

Explanation

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)

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

$3a = 450 {\;};\ t=\infty$ .....(iii)

$\because$ By Eqs. (i), (ii) and (iii) we get

$X= 57; \ a=150.$

$K=\dfrac {2.303}{14}log \dfrac {150}{150-57}=3.415\times 10^{-2}\ min^{-1}$

Hence, the correct option is

$B$ .

For a first-order reaction,

$A\rightarrow Product$ , the initial concentration of A is 0.1M and after 40 minutes it becomes 0.025 M. Calculate the rate of reaction at reactant concentration of 0.01 M:

0%
$3.47\times 10^{-4} M .min^{-1}$
0%
$3.47\times 10^{-5} M. min^{-1}$
0%
$1.735\times 10^{-6} M. min^{-1}$
0%
$1.735\times 10^{-4} M. min^{-1}$

Explanation

As we know,

$K=\frac {2.303}{40}log\frac {[0.1]}{[0.025]}$

$\therefore K=0.03466\ min^{-1}$

$\therefore rate=K\times 0.01=0.03466\times 0.01=3.47\times 10^{-4} M min^{-1}$

The maximum value of activation energy is equal to:

0%
zero
0%
heat of reaction
0%
threshold energy
0%
none of these

Explanation

Option (D) is correct.
Activation energy may be greater than heat of reaction or lesser than threshold energy.At room temperature, most of the molecules have less than the threshold value. Hence, if energy is supplied to the reactant molecules in the form of light or heat, they absorb that energy and reach a higher energy level which is equal to or greater than the threshold energy level, is called as activation energy.

Activation energy (

$E_a$ ) = Threshold energy - Average kinetic energy of molecules

In a first order reaction the concentration of reactant decreases from 800

$mol/dm^3$ to

$50\ mol/dm^3$ in

$2\times 10^4 sec$ . The rate constant of reaction in

$sec^{-1}$ is:

0%
$2.000\times 10^{4}$
0%
$3.415\times 10^{-5}$
0%
$1.386\times 10^{-4}$
0%
$2.000\times 10^{-4}$

Explanation

As we know,

$t=\frac {2.303}{K} \times log \frac {N_0}{N}$

$\therefore K=\frac {2.303}{2\times 10^{4}}\times log \frac {800}{50}=1.386\times 10^{-4} sec^{-1}$

Which of the following is not correct reason for the substantially lower rate of reaction than the collision frequency?

0%
All the collisions do not attain threshold energy level
0%
The activated complex formed is short lived
0%
All the collisions do not have proper orientation
0%
Effective collision are lesser in number than all collisions

Effective collisions are those in which molecules must:

0%
have energy equal to or greater than the threshold energy
0%
have proper orientation
0%
acquire the energy of activation
0%
all of the above

The rate constant of a reaction is 0.0693

$min^{-1}$ . Starting with 10 mol, the rate of the reaction after 10 min is:

0%
0.0693 mol $min^{-1}$
0%
0.0693 $\times$ 2 mol $min^{-1}$
0%
0.0693 $\times$ 5 mol $min^{-1}$
0%
0.0693 $\times\, 5^2$ mol $min^{-1}$

Explanation

As we know,
for a first order reaction,

$\displaystyle t_{1/2}\, =\, \frac{0.693}{k}\, =\, \frac{0.693}{0.0693}\, =\, 10\, min$
Reactant after 10 min = 5 mol
Rate

$\displaystyle \left ( \frac{dx}{dt}\right )\, =\, k[A]\, =\, 0.0693\, \times\, 5\, mol\, min^{-1}$

The activation energy for a simple chemical reaction

$A\, \rightarrow\, B\, is\, E_a$ in the forward reaction. The activation energy of the reverse reaction:

0%
Is negative of $E_a$
0%
Is always less than $E_a$
0%
Can be less than or more than $E_a$
0%
Is always double of $E_a$

Explanation

As we know,
For exothermic reaction,

$\Delta H^{\ominus}$ = (PE of product) - (PE of reactant) < 0.

$(E_a)_f\, <\, (E_a)_b.$
For endothermic reaction:

$(E_a)_f\, >\, (E_a)_b.$

The decomposition of

$Cl_2O_7$ at

$400$ K in the gas phase to

$Cl_2$ and

$O_2$ is of

$1^{st}$ order. After

$55$ sec at 400 K, the pressure of

$Cl_2O_7$ falls from

$0.062$ to

$0.044$ atm.

The rate constant is :

0%
$6.24\times 10^{-3} sec^{-1}$
0%
$5.20\times 10^{-3} sec^{-1}$
0%
$7.34\times 10^{-3} sec^{-1}$
0%
none of the above

Explanation

The given reaction is:

$Cl_2O_7\rightarrow Cl_2+\dfrac{7}{2}O_2$

Mole at $t=0$ sec	$a$	0	0
Mole at $t=55$ sec	$a-x$	$x$	$7x/2$

Since, mole at any time

$\propto$ pressure at that time (at constant V and T)

$\therefore a\propto 0.062$

and

$(a-x)\propto 0.044$ at

$t=55 sec$

$\because K=\dfrac {2.303}{t}log \dfrac {a}{(a-x)}=\dfrac {2.303}{55}log \dfrac {0.062}{0.044}$

$\quad \quad =6.24\times 10^{-3} sec^{-1}$

Hence, the correct option is

$A$

For a first-order reaction, the ratio of times to complete 99.9% and half of the reaction is:

0%
$8$
0%
$9$
0%
$10$
0%
$12$

The activation energy for a simple chemical reaction,

$A\rightarrow{B}$ is

$E_a$ in forward direction. The activation energy for the reverse reaction:

0%
is negative of $E_a$
0%
is always less than $E_a$
0%
can be less than or more than $E_a$
0%
is always double of $E_a$

Explanation

As we know,

$\Delta H=E_{a_{(f)}}-E_{a_{(b)}}$ ; if

$\Delta H=-ve, E_{a_{(b)}} > E_{a_{(f)}}$
if

$\Delta H=+ve, E_{a_{(b)}} < E_{a_{(f)}}$

80% of a first order reaction was completed in 70 min. How much it will take for 90% completion of a reaction ?

0%
114 min
0%
140 min
0%
100 min
0%
200 min

Explanation

Hint: A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant.

Formula used:

For a first-order reaction,

$t=\dfrac{2.303}{k} \log \left[\frac{R_{0}}{R}\right]$

Here,

$R_{0}=$ initial concentration of reactant

$R=$ present concentration of reactant

STEP-1

Using

$k_{1}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$ ,

we can find out the value of

$\mathrm{K}$ . Using

$\mathrm{K}$ time required for

$90 \%$ reaction can be calculated.

By putting values in the above equation, we will get

$\mathrm{K}=\dfrac{2.303}{70} \log \dfrac{1}{0.2}=0.023$

STEP-2

Now, again using the same equation as above, but for

$90 \%$ completion, we can calculate the time.

$\mathrm{t}=\dfrac{2.303}{0.023} \log \dfrac{1}{0.1}$

So, we get time

$=100 \mathrm{~min}$ .

Final Answer:

Hence, the Correct answer is option $$C$$

For a given reaction of the first order, it takes 15 minutes for the concentration to drop from 0.8 M to 0.4 M. The time required for the concentration to drop from 0.1 M to 0.025 M will be:

0%
60 minutes
0%
15 minutes
0%
7.5 minutes
0%
30 minutes

Explanation

The reactant concentration drop from 0.8 to 0.4 M, i.e., 50% takes place in 15 minutes.

$K=\dfrac {2.303}{15} log \dfrac {0.8}{0.4}=\dfrac {0.693}{15}=0.0462\ min^{-1}$

Also,

$t=\dfrac {2.303}{K}log \dfrac {0.1}{0.025}$

$t=\dfrac {2.303}{0.0462} log \dfrac {0.1}{0.025}=30\ min$

The mathematical expression for

$t_{1/4}$ i.e., when

$(1/4)^{th}$ reaction is over following first-order kinetics can be given by:

0%
$t_{1/4}=\frac {2.303}{K}log 4$
0%
$t_{1/4}=\frac {2.303}{K}log 2$
0%
$t_{1/4}=\frac {2.303}{K}log \frac {4}{3}$
0%
$t_{1/4}=\frac {2.303}{K}log \frac {3}{4}$

Explanation

As we know,
for first order reaction,

$t=\frac {2.303}{K}log \frac {a}{(a-x)}$
if

$t=t_{1/4}; x=a/4$

$\therefore t_{1/4}=\frac {2.303}{K} log \frac {a}{a-a/4}$

$=\frac {2.303}{K} log\frac {4}{3}$ .

The rate of a chemical reaction generally increases rapidly even for small temperature increase because of a rapid increase in:

0%
Collision frequency
0%
Fraction of molecules with energies in excess of the activation energy
0%
Activation energy
0%
Average kinetic energy of molecules

In a reaction, 5 g ethyl acetate is hydrolyzed per litre in the presence of dil. HCl in 300 min.If the reaction is of the first order and the initial concentration of ethyl acetate is 22 g

$L^{-1}$ , the rate constant of the reaction is:

0%
$k \,=\, 8.6\,\times\, 10^{-4}\, min^{-1}$
0%
$k \,=\, 1.4\,\times\, 10^{-4}\, min^{-1}$
0%
$k \,=\, 6.9\,\times\, 10^{-4}\, min^{-1}$
0%
$None \:of \:these$

Explanation

a = 22 g

$L^{-1}$
x = 5 g

$L^{-1}$ , t = 300 min
k =

$\displaystyle \frac{2.3}{300\,min} \, log\, \frac{22\, g\, L^{-1}}{(22\,-\,5)g\,L^{-1}}\,=\, 8.6\,\times\, 10^{-4}\, min^{-1}$

What specific name can be given to the following sequence of steps: $Hg + hv \rightarrow\, Hg^*$

$Hg^*\, +\, H_2\, \rightarrow\, H_2^*\, +\, Hg$

0%
Fluorescence
0%
Phosphorescence
0%
Photosensitization
0%
Chemilumionescence

Explanation

$Hg + hv \rightarrow\, Hg^*$

$Hg^*\, +\, H_2\, \rightarrow\, H_2^*\, +\, Hg$

It is fluorescence as $Hg$ emitted the light which it absorbs in first step. (Fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a form of luminescence. In most cases, the emitted light has a longer wavelength and therefore, lower energy than the absorbed radiation)

In a multistep reaction such as

$A + B$

$\rightarrow\, Q\, \rightarrow\, C.$ The potential energy diagrame is shown below. What is

$E_a$ for the reaction

$Q\, \rightarrow\, C\, ?$

0%
$3 kcal$ $mol^{-1}$
0%
$5 kcal$ $mol^{-1}$
0%
$8 kcal$ $mol^{-1}$
0%
$11 kcal$ $mol^{-1}$

Explanation

For

$Q\, \rightarrow\, C$

$E_a$

$= 23 - 20$

$= 3 kcal$

$mol^{-1}$

If the activation energies of the forward and backward reactions of a reversible reaction are

$E_a(f)$ and

$E_a(b)$ , respectively. The

$\Delta E$ of the reaction is ...........

0%
$E_a(F)-E_a(b)$
0%
$E_a(F)+E_a(b)$
0%
$E_a(F)=E_a(b)$
0%
$-E_a(F)+E_a(b)$

Explanation

As we know,

$\Delta E = E_a(F)-E_a(b)$

The rate of reaction increase by the increase of temperature because:

0%
collision frequency is increased.
0%
energy of products decreases.
0%
the fraction of molecules possessing energy $\geq\, E_T$ (threshold energy) increases.
0%
mechanism of a reaction is changed.

The reaction

$A\, \rightarrow\, B$ follows first-order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hr. What is the time taken for the conversion of 0.9 mol of A to produce 0.675 mol of B?

0%
1 hr
0%
0.5 hr
0%
0.25 hr
0%
2 hr

Explanation

Time for the conversion of 0.8 mol of A to 0.6 mol of B :

$\displaystyle B\, =\, \frac{0.6}{0.8}\, =\, 0.75\, =\, t_{75\%}$

Similarly, time for the conversion of 0.9 mol of A to 0.675 mol of B =

$\displaystyle \frac{0.675}{0.9}\, =\, 0.75\, =\, t_{75\%}$

Hence,

$t_{75\%}$ in both case

$= 1 hr$ .

The activation energy of reactant molecules in a reaction depends upon:

0%
Temperature
0%
Nature of the reactants
0%
Collision per unit time
0%
Concentration of reactants

What can you say about the existence of A if the potential energy diagram for the reaction

$A\, \rightarrow\, B$ looks like :

0%
A will exist
0%
A will not exist
0%
B will not exist
0%
A and B are in equilibeium

In a certain reaction.

$10$ % of the reactant decomposes in one hour,

$20$ % in two hours,

$30$ % in three hours, and so on. The dimension of the velocity constant (rate constant) are:

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

Explanation

here,

$t_{1/2}\, \propto$ a. Hence, zero order reaction. so, dimension of k = mol

$L^{-1}\, hr^{-1}$
For zero order: t =

$\displaystyle \frac{x}{k}\, or\, k\, =\, \frac{x}{t}$
If t =

$t_{10\%}\, =\, 10\, min,\, x\, =\, 10,$
Then, k =

$\displaystyle \frac{10}{10}\, =\, 1\, mol\, L^{-1}\, hr^{-1}$
If t =

$t_{20\%}\, =\, 20\, min,\, x\, =\, 20,$
Then, k =

$\displaystyle \frac{20}{20}\, =\, 1\, mol\, L^{-1}\, hr^{-1}.$
Similarly, for

$30$ % and so on.
Thus, reaction is of zero order.

For a hypothetical reaction: A + B

$\rightarrow$ Products, the rate law is r = k[A]

$[B]^0$ . The order of reaction is:

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

Explanation

r =

$[A][B]^0$
OR = 1 + 0 = 1

A chemical reaction occurs as a result of collisions between reacting molecules. Therefore, the reaction rate is given by:

0%
total number of collision occuring in a unit volume per second
0%
fraction of molecules which possess energy less than the threshold energy
0%
total number of effective collisions
0%
none of the above

Explanation

According to collision theory, the reaction occurs when molecules collide with each other. The rate is given by:

Rate= ${ Z }_{ AB }\times f$

Where $Z_{AB}$ =collision frequency of reactants $A$ & $B$ .

i.e. total number of collisions occurring in a unit volume per second. &

$f$ =fraction of effective collisions.

So, the rate depends on both (A) & (C)

$k_1=X\:hr^{-1};k_1:k_2=1:10$ . Calculate

$\displaystyle \frac{[C]}{[A]}$ after one hour from the start of the reaction. Assuming only A was present in the beginning.

0%
$\displaystyle \frac{[C]}{[A]}=\frac{10}{11}(e^{11x}-1)$
0%
$\displaystyle \frac{[C]}{[A]}=\frac{10}{11}(e^{11x}+1)$
0%
$\displaystyle \frac{[C]}{[A]}=\frac{10}{11}(e^{22x}-1)$
0%
$\displaystyle \frac{[C]}{[A]}=\frac{10}{11}(e^{22x}+1)$

Explanation

$\displaystyle \dfrac {[C]}{[A]} = \dfrac {k_2}{k_1+k_2} [\dfrac {1-e^{-(k_1+k_2)t}}{exp(-(k_1+k_2)t)}]$

But

$\displaystyle k_1=X$ and

$\displaystyle \dfrac {k_1}{k_2}=\dfrac {1}{10}$

$\displaystyle k_2=10X$

Substitute values in the above equation.

$\displaystyle \dfrac {[C]}{[A]} = \dfrac {10X}{X+10X} [\dfrac {1 - e^{-(X+10X)t}}{e^{-(k_1+k_2)t}}]$

$\displaystyle \dfrac {[C]}{[A]} = \dfrac {10}{11}[e^{11X}-1]$

The incorrect statement is:

0%
all the collisions between reactant molecules do not lead to a chemical change
0%
a zero order reaction proceeds at a constant rate independent of concentration or time
0%
fast reactions have low activation energies
0%
in a first order reaction, the reaction ideally takes finite time to be completed

Explanation

For a reaction to take place, the reacting molecules must colloid together but only those collisions, in which colliding molecules possess certain minimum energy is called threshold energy.
Zero-order reaction,

$rate = k[A]^0 = k$
So, a zero order reaction proceeds at a constant rate independent of concentration or time.
Every first reaction takes infinite time for completion.

A first order reaction has a rate constant

$1.5\times10^{-3}\:sec^{-1}$ . How long will 5.0 g of this reactant take to reduce to 1.25 g.

0%
924 sec
0%
462 sec
0%
1848 sec
0%
none of these

Explanation

As we know, for a first order reaction,

$t_{1/2} = 0.693/k$ and time taken for

$5$ gm to reduce to

$1.25$ gm is

$t = 2t_{1/2} = 2\times 0.693/1.5\times 10^{-3} = 924$ sec

The rate constant for a zero order reaction is

$2\times10^{-2}\:mol\:L^{-1}\:sec^{-1}$ , if the concentration of the ractant after

$25 sec$ is

$0.25 M$ , calculate the initial concentration.

0%
$0.75 M$
0%
$1.5 M$
0%
$0.375$
0%
None of these

Explanation

using the above formula we can get

${A}_{0}$ . given the values of K and

${\epsilon}$ in the question.

For a reaction

$\:A\overset{K_1}{\longrightarrow}B\overset{K_2}{\longrightarrow}C.$ If the reaction are of

$1^{st}$ order then

$\:\displaystyle\frac{d\left[B \right]}{d\,t}\:$ is equal to:

0%
$\;-k\left[B \right]$
0%
$\;+k_1\left[A \right]$
0%
$\;k_1\left[A \right]-k_2\left[B \right]$
0%
$\;k_1\left[A \right]+k_2\left[B \right]$

Explanation

For the first reaction

$\displaystyle A \rightarrow B$

$\displaystyle \frac {d[B]}{dt} = k_1A$ .....(1)
For the second reaction

$\displaystyle B \rightarrow C$

$\displaystyle -\frac {d[B]}{dt} = k_2[B]$
or

$\displaystyle \frac {d[B]}{dt} = -k_2[B]$ ......(2)
Total

$\displaystyle \frac {d[B]}{dt}$ can be obtained by adding equations (1) and (2)

$\displaystyle \frac {d[B]}{dt} = k_1[A] -k_2[B]$

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as ;

$k_1=1.26\times10^{-4}\:sec^{-1}$ and

$k_2=3.6\times10^{-5}\:sec^{-1}$ . Calculate the % distribution of B & C.

0%
77.7, 22.3
0%
22.3, 77.7
0%
67.7, 32.3
0%
None of these

Explanation

Rate law equation for two parallel reactions is,

$Rate= k_1[A]+k_2[A]$

$Percentage\ of\ B= \frac{k_1}{k_1+k_2} \times 100= \frac{12.6 \times 10^{-5}}{ (12.6 + 3.6) \times 10^{-5}} = 77.7$

$Percentage\ of\ C= \frac{k_2}{k_1+k_2} \times 100 = \frac {3.6 \times 10^{-5}} { (12.6 + 3.6) \times 10^{-5}}= 22.3$

Thus the answer should be an option (A).

In a first order reaction the concentration of reactant decreases from 800

$mol/dm^3$ to

$50 mol/dm^3$ in

$2\times 10^4 sec$ . The rate constant of reaction in

$sec^{-1}$ is:

0%
$2\times 10^{4}$
0%
$3.415\times 10^{-5}$
0%
$1.386\times 10^{-4}$
0%
None of these

Explanation

As we know,

$t=\dfrac {2.303}{k} log \left(\frac {N_0}{N}\right)$

$k=\dfrac {2.303}{2\times 10^{4}}log \left(\dfrac {800}{50}\right)=1.386\times 10^{-4} sec^{-1}$

In a two step reaction, the rate determining step has the lowest energy of activation.

0%
True
0%
False

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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