MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 8

For a first - order reaction, the concentration of reactant:

0%
is independent of time
0%
varies linearly with time
0%
varies exponentially with time
0%
None

For the first order decomposition of ${ SO }_{ 2 }{ Cl }_{ 2 } (g), { SO }_{ 2 }{ Cl }_{ 2 } \longrightarrow { SO }_{ 2 }(g) + { Cl }_{ 2 }(g)$
A graph of $log ({ a } -x)\ vs\ t$ is shown in figure. What is the rate constant $({ sec }^{ -1 })$ ?

0%
0.2
0%
$4.3 \times { 10 }^{ -5}$
0%
$7.7 \times { 10 }^{ -3 }$
0%
$1.15 \times { 10 }^{ -2 }$

The conversion of vinyl allyl ether to pent-4-enol follows a certain kinetics. The following plot is obtained for such a reaction.

0%
zero
0%
-1
0%
1
0%
2

Explanation

Since the plot of

$\log(a-x)$ vs

$t$ is a straight line,

$\therefore \log(a-x)\propto t$

This is followed by

$1^{st}$ order reaction where

$\log (a-x)=\log a_{\circ}-\cfrac{Kt}{2.303}$

Acid hydrolysis of ester in first order reaction and rate constant is given by

$k=\dfrac { 2.303 }{ t } log\dfrac { { V }_{ \infty }-{ V }_{ 0 } }{ { V }_{ \infty }-{ V }_{ t } }$ where

${ V }_{ 0 }$ ,

${ V }_{ t }$ and

$V_{ \infty }$ are the volume of standard

$NaOH$ required to neutralise acid present at a given time, if ester is

$50%$ neutralised then:

0%
${ V }_{ \infty }={ V }_{ t }$
0%
${ V }_{ \infty }=\left( { V }_{ t }-{ V }_{ 0 } \right)$
0%
${ V }_{ \infty }=\left( { 2V }_{ t }-{ V }_{ 0 } \right)$
0%
${ V }_{ \infty }=\left( { 2V }_{ t }+{ V }_{ 0 } \right)$

Explanation

$\;\;\;\;\;RCOOR'+H_2O\; \;\;\xrightarrow { H^{ + } } \;\;\;RCOOH+R'OH$

$t=0$

$a$

$0$

$t$

$a-x$

$x$

$t=\infty$

$a-a$

$a$

Let

$V_0=$ Volume of

$NaOH$

$\left( \;at\;\;t=o\right),$

$V_t=x+V_0$

$V_{\infty}=a+V_0$

If ester is

$50\%$ hydrolyzed,

$x=\dfrac{a}{2}$

$\therefore V_t=\dfrac{a}{2}+V_0$

$a=2V_t-2V_0$

$\therefore V_{\infty} =2V_t-2V_0+V_0$

$=2V_t-V_0$

Hence, the answer is

$2V_t-V_0.$

The reaction

${ N }_{ 2 }{ O }_{ 5 }\quad \left( In\quad { CCl }_{ 4 } \right) \rightarrow 2{ NO }_{ 2 }+{ 1 }/{ 2 }\quad { O }_{ 2 }\left( g \right)$ is first order in

${ N }_{ 2 }{ O }_{ 5 }$ with rate constant

$6.2\times { 10 }^{ -4 }{ S }^{ -1 }$ . What is the value of rate of reaction when

$\left[ { N }_{ 2 }{ O }_{ 5 } \right] =1.25\quad mole\quad { L }^{ -1 }$

0%
$7.75\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$
0%
$6.35\times { 10 }^{ -3 }\quad \quad { L }^{ -1 } { S }^{ -1 }$
0%
$5.15\times { 10 }^{ -5 }\quad \quad { L }^{ -1 } { S }^{ -1 }$
0%
$3.85\times { 10 }^{ -4 }\quad \quad { L }^{ -1 } { S }^{ -1 }$

Explanation

Its a 1st Order Reaction.

$r = k[A]$

$r = 6.2 \times {10}^{-4} \times [1.25]$

$r = 7.75 \times {10}^{-4}\space {L}^{-1}{S}^{-1}$

An endothermic reaction

$A \rightarrow B$ have an activation energy 15 kcal/mol and the heat of reaction is 5 kcal/mol. The activation energy of the reaction

$B \rightarrow A$ is:

0%
20 kcal/mol
0%
15 kcal/mol
0%
10 kcal/mol
0%
zero

Explanation

Its a Endothermic Reaction, So

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

Given,

$[{E}_{a(f)}] = 15kcal/mol$ and

$\Delta H = 5\space {kcal}/mol$

$[{E}_{b(f)}] = 15 - 5 = 10\space kcal/mol$

The reaction

$A(s) \rightarrow 2 B(g)+C(g)$ is the first order. The pressure after 20 min. and after very long time are 150 mm Hg and 225 mm Hg. The value of rate constant and pressure after 40 min. are:

0%
0.05 in 5 $min^{-1}$ ,200mm
0%
0.5 in 2 $min^{-1}$ ,300mm
0%
0.05 in 3 $min^{-1}$ ,300mm
0%
0.05 in 3 $min^{-1}$ ,200mm

In a first order reaction, the concentration of the reactant decreases from 0.8M to 0.4M in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is:

0%
60 min
0%
15 min
0%
7.5 min
0%
30 min

Explanation

$k = \dfrac{2.303}{t} log \dfrac{[A]}{[A - x]}$

$k = \dfrac{2.303}{15} log \dfrac{0.8}{0.4}$

$k = 0.046\space {min}^{-1}$

The time required

$0.046 = \dfrac{2.303}{t} log \dfrac{0.1}{0.025}$

$t = 30\space min$

Let there be as first-order reaction of the type,

$A\rightarrow B+C$ . Let us assume that only A is gaseous. We are required to calculate the value of rate constant based on the following data.

Time	0	T	$\infty$
Partial pressure of A	$P_{0}$	$P_{t}$	-

0%
$k=\frac{1}{t}In(\frac{P_{0}}{P_{t}})$
0%
$k=\frac{1}{t}In(\frac{P_{t}}{P_{0}})$
0%
$k=\frac{1}{t}In(\frac{2P_{0}}{P_{t}})$
0%
$k=\frac{1}{t}In(\frac{P_{t}}{2P_{0}})$

Explanation

Its a 1st Order Reaction.

At t = 0 , Pressure is

${P}_{0}$

At t = T , Pressure is

${P}_{t}$

So,

$k = \dfrac{2.303}{t} log \dfrac{{P}_{0}}{{P}_{t}}$

$k = \dfrac{1}{t} log \dfrac{{P}_{0}}{{P}_{t}}$

K for a zero order reaction is

$2\times { 10 }^{ -2 }\quad { L }^{ -1 }{ Sec }^{ -1 }$ . If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration must have been.

0%
0.5 M
0%
1.25 M
0%
12.5 M
0%
1.0 M

Explanation

It is a Zero Order Reaction

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

$2 \times {10}^{-2} = \dfrac{[{A}_{0}] - [0.5]}{25}$

$[{A}_{0}] = 1\space M$

Hence, the correct option is

$D$

Two identical balls A & B having velocity of 0.5 m/s & -0.3 m/s respectively. colloid elastically in 1-The velocity of A & B after collision will be?

0%
-0.5 m/s, 0.3 m/s
0%
-0.3 m/s,0.5 m/s
0%
0.5 m/s, 0.3 m/s
0%
0.3 m/s,0.5 m/s

$99\%$ of a first - order reaction was completed in

$32$ min When will

$99.9\%$ of the reaction complete?

0%
$50$ Min
0%
$46$ Min
0%
$49$ Min
0%
$48$ Min

Explanation

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

Substituting the values,

$k = \dfrac{2.303}{32} log \dfrac{100}{1}$

$k = 0.144\space {min}^{-1}$

$t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}$

$t = 48\space min$

Hence, the correct option is

$D$

$A + B \longrightarrow C; \Delta H = + 60 KJ/mol$

$E_{at}$ is

$150$ kj. What is the activation energy for the backward reation?

0%
$210$ kJ
0%
$105$ kJ
0%
$90$ kJ
0%
$145$ kJ

Explanation

The Reaction is a Endothermic Reaction. So,

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

$[{E}_{b(f)}] = 150 - 60\space kJ$ =

$90\space kJ$

The half year period for a zero order reaction is equal to:

0%
$2K/[A]o$
0%
$\cfrac{[A]_o}{2k}$
0%
$\cfrac{0.693}{k}$
0%
$\cfrac{0.693}{k[A]_o}$

Explanation

Half Year Period for a Zero order reaction is

$\dfrac{[{A}_{0}]}{2k}$

Where

$[A_0]=$ initial concentration

$k=$ rate constant

What fraction of a reactant showing first order remains after 40 minute if

$t_{1/2}$ is 20 minute?

0%
1/4
0%
1/2
0%
1/8
0%
1/6

Explanation

Its a 1st Order Reaction,

Taking the Initial Concentration as 100 (For Convenience)

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

$k = 0.03465\space {min}^{-1}$

Now,

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

$0.03465=\dfrac{2.303}{40}log{\dfrac{a}{a-x}}$

$\dfrac{a}{a-x}=\dfrac{1}{4}$

A reaction is of first order. After

$100\ minutes\ 75\ g$ of the reactant A are decomposed when

$100\ g$ are taken initially. Calculate the time required when

$150\ g$ of the reactant A are decomposed, the initial weight taken is

$200\ g$ :

0%
$100\ minutes$
0%
$200\ minutes$
0%
$150\ minutes$
0%
$175\ minutes$

Explanation

From the first order reaction system we know that,

$K = \dfrac{2.303 log\dfrac{A_0}{A}}{t}$

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

For 200 g,

$\dfrac{2.303 log\dfrac{100}{25}}{100} =\dfrac{2.303log\dfrac{200}{50}}{t}$

$0.602/100 = 0.602/t$

$t = 100 minutes$

Option A is correct.

Select the correct statement regarding activation energy?

0%
Activation energy may be greater than heat of reaction
0%
Activation energy is less than threshold energy
0%
Rate of reaction is inversely proportional to the activation energy
0%
All of these

$2A \rightarrow B + C$ It would be a zero order reaction when:

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

What is the activation energy for a reaction if its rate doubles when the temperature is raised from

$20^0C$ to

$35^0C$ ? (R = 8.314 J

$mol^{-1}K^{-1}$ )

0%
269 kJ $mol^{-1}$
0%
34.7 kJ $mol^{-1}$
0%
15.1 kJ $mol^{-1}$
0%
342 kJ $mol^{-1}$

Explanation

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

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

Substituting the values

$log\space {2}$ =

$\dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{293} - \dfrac{1}{308})$

${{E}_{a}} = 34.7\space kJ\space {mol}^{-1}$

For a first order reaction involving decomposition of

$N_2O_5$ the following information is available

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

$=k[N_2O_5]$

$N_2O_5(g)\rightarrow 2NO_2(g)+1/2 O_2(g)$ rate

$=k^\prime[N_2O_5]$

0%
$k=k^\prime$
0%
$k^\prime=2k$
0%
$k^\prime=1/2k$
0%
$k>k^\prime$

Explanation

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

$\Rightarrow k=\cfrac {[NO_2]^4[O_2]}{[N_2O_5]^2}\longrightarrow (1)$

${N_2O_5}_{(g)} \longrightarrow {2NO_2}_{(g)}+{1/2O_2}_{(g)}$

$\Rightarrow k'= \cfrac {[NO_2]^2[O_2]^{1/2}}{[N_2O_5]}$

$\Rightarrow [N_2O_5]=\cfrac {[NO_2]^2[O_2]^{1/2}}{k'}\longrightarrow (2)$

$(2)$ in

$(1) \Rightarrow k=\cfrac {[NO_2]^4[O_2]}{\cfrac {[NO_2]^4[O_2]}{k'^2}}=k'^2$

$\Rightarrow k >k'$

In a 1st order reaction the fraction of molecules at

$450^oC$ having sufficient energy (or fraction of effective collisions) is

$1.92\times10^{-16}$ .What is activation energy value of this reaction?

0%
$27.757\times10^2 J mole^{-1}$
0%
$21.757\times10^3 J mole^{-1}$
0%
$21.75\times10^4 J mole^{-1}$
0%
None

For the first order consecutive reaction

$P\rightarrow Q\rightarrow R$ under steady state approximation to

$\left[ P \right] ,\left[ Q \right]$ and

$\left[ R \right]$ with time are best represented by:

The rate of first order reaction,

$A \rightarrow$ Products, is

$7.5$ M/s. If the concentration of

$A$ is 0.5 M. The rate constant is:

0%
3.75
0%
2.51
0%
15.0
0%
8.01

Explanation

As we know that,

$r\ (rate) = K \left[ A \right]$ ------ 1

where,

$r =$ Rate of reaction

$= 7.5 \; {mol}/{(L.s)}$

$K =$ Rate constant

$= \ ?$

$\left[ A \right] =$ Initial concentration of reactant

$= 0.5 \; {mol}/{L}$

Therefore,

$K = \cfrac{7.5}{0.5} = 15.0 \; {s}^{-1}$

Hence the rate constant for the given reaction is

$15.0\; {s}^{-1}$ .

Milk turns sour at

$40^0C$ three times as faster as at

$0^0C$ . The energy of activation for souring of milk is:

0%
4.693 kcal
0%
2.6 kcal
0%
6.6 kcal
0%
none of these

Explanation

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

Milk turns sour at 313 K 3 times as faster as at 273 K.Substituting the values

$log{1}{3} = \dfrac{{E}_{a}}{2.303 \times 2}(\dfrac{1}{313} - \dfrac{1}{273})$

${E}_{a} = 4.693\space kcal$

A catalyst lower the energy of activation by

$25\%$ . The temperature at which rate of uncatalysed reaction would be equal to that of the catalyzed one at

$27^oC$ is:

0%
$400^oC$
0%
$127^oC$
0%
$300^oC$
0%
$227^oC$

Explanation

$\dfrac{{E}_{a(1)}}{{E}_{a(2)}}$ =

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

$\dfrac{4}{3}$ =

$\dfrac{{T}_{1}}{300}$

${T}_{1} = 400 K$

${T}_{1} = {127}^{o}C$

$1^{st}$ order reaction is

$50\%$ complete in

$30$ minutes at

$27^oC$ and in

$10$ minutes at

$47^oC$ . Calculate energy of activation for the reaction.
(Assume log

$3$ =

$0.48$ )

0%
$46.87 \, KJ \, mol^{-1}$
0%
$42.21 \, KJ \, mol^{-1}$
0%
$44.11 \, KJ \, mol^{-1}$
0%
$49.20 \, KJ \, mol^{-1}$

Explanation

${k}_{1} = \dfrac{0.693}{10}$

${k}_{2} = \dfrac{0.693}{30}$

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

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

$log3 = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{300} - \dfrac{1}{320})$

${E}_{a} = 44.11\space KJ\space {mol}^{-1}$

The decomposition

$NH_3$ gas on a heated tungsten surface gave the following results:

Initial pressure (mm)	65	105	y	185
Half-life (sec)	290	x	670	820

Calculate approximately the values of x and y.

0%
x = 410 sec, y = 115 mm
0%
x = 467 sec, y = 150 mm
0%
x = 490 sec, y = 120 mm
0%
x = 430 sec, y = 105 mm

Explanation

${t}_{1/2}\space \alpha\space {P}^{1-n}$

$\dfrac{290}{820} = \dfrac{65}{185}$ , Which Implies

$1 -n = 1$

$n = 0$

$k =\dfrac{[{P}_{0}]}{2{t}_{1/2}}$

$k = 0.112\space mol{L}^{-1}\space {t}^{-1}$

$x = \dfrac{105}{2 \times {0.112}}$ =

$467\space {sec}$

$y = 0.112 \times 2\times {670}$ =

$150\space mm$

For the zero order reaction

$A\rightarrow 2B$ , the rate constant is

$2\times 10^{-6}M min^{-1}$ the reaction is started with 10 M of A.

0%
what will be concentration of A after 2 days
0%
what is intial half-line one of reaction?
0%
in what time, the reaction will complete?
0%
none

The time elapsed between

$33\%$ and

$67\%$ completion of a first order reaction is

$30$ minutes. What is the time needed for

$25\%$ completion?

0%
$15.5\ min$
0%
$12.5\ min$
0%
$18.5\ min$
0%
$16.5\ min$

Explanation

As per the question,

$30 = -\dfrac{2.303}{k} log \dfrac{[100]}{[100 - 33]}$ +

$\dfrac{2.303}{k} log \dfrac{[100]}{[100 - 67]}$

$\Rightarrow k = 2.36 \times {10}^{-2}\space {min}^{-1}$

$t = \dfrac{2.303}{2.36 \times {10}^{-2}} log \dfrac{[100]}{[75]}$

$t = 12.5\space min$

In a particular reaction the time required to complete half of the reaction was found to increase 16 times when the initial concentration of the reactant was reduced to one-fourth. What is the order of the reaction?

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

Which among the following plots are liner (a-x) is the concentration of the reactant remaining after time, t?
(1):-(a-x) vs t, for a first order reaction
(2):- (a-x) vs t, for a zero order reaction
(3):-(a-x) vs t, for a second order reaction
(4):-1/ (a-x) vs t, for a second order reaction

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

A first order reaction follows the expression:

0%
${ C }_{ t }{ e }^{ { k }_{ 1 }t }=C_0$
0%
$C_t={ C }_{ t }{ e }^{ { k }_{ 1 }t }$
0%
In $\frac { { C }_{ O } }{ { C }_{ t } } =-k_1t$
0%
In $\frac { { C }_{ t } }{ { C }_{ o } } =k_1t$

Explanation

1st Order Reaction,

${k}_{1} = \dfrac{1}{t} ln\dfrac{[{C}_{0}]}{[{C}_{t}]}$

${C}_{t}{e}^{{k}_{1}t} = {C}_{0}$

Which of the following represents the expression for

$\cfrac{3}{4}$ th life of first order reaction?

0%
$\cfrac{2.303}{k}\log{4/3}$
0%
$\cfrac{2.303}{k}\log{3/4}$
0%
$\cfrac{2.303}{k}\log{4}$
0%
$\cfrac{2.303}{k}\log3$

Explanation

Answer

$C.$

The expression for rate of first order reaction,

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

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

$x=\cfrac{3}{4}a$

$t=\cfrac{2.303}{k}\log \cfrac{a}{a-\cfrac{3}{4}a}$

$=\cfrac{2.303}{k}\log \cfrac{a}{\cfrac{1}{4}a}$

$=\cfrac{2.303}{k}\log 4$

The reaction

$v_1A + v_2B \rightarrow$ products is first order with respect to

$A$ and zero- order with respect to

$B$ .If the reaction is started with

$[A]_0$ and

$[B]_0$ , the integrated rate expression of this reaction would be:

0%
$ln \dfrac { [A]_ 0 }{ [A]_ 0-x } =k_ 1t$
0%
$ln\dfrac { [A]_ 0 }{ [A]_ 0-v_ 1x } =k_ 1t$
0%
$ln\dfrac { [A]_ 0 }{ [A]_ 0-v_ 1x } =v_ 1k_ 1t$
0%
$ln\dfrac { [A]_0 }{ [A]_ 0-v_ 1x } = -v_ 1k_ 1t$

Explanation

Its 1st Order wrt A and Zero Order wrt B.

So, the integrated rate equation of this reaction is,

$ln \dfrac{[{A}_{0}]}{[{A}_{0}] - {v}_{1}x}$ =

${v}_{1}{k}_{1}t$

If a I-order reaction is completed to the extent of

$60\%$ and

$20\%$ in time intervals,

$t_1$ and

$t_2$ what is the ratio,

$t_1 : t_2$ ?

0%
6.32
0%
5.587
0%
4.11
0%
8.33

Explanation

In 1st Order Reaction,

$\dfrac {{t}_{1}}{{t}_{2}} = \dfrac {log({\dfrac{100}{100-60}})}{log({\dfrac{100}{100-20}})}$

$\dfrac{{t}_{1}}{{t}_{2}} = 4.11$

What is the half life of a radioactive substance if

$75%$ of its given amount disintegrate in

$60\ min$ ?

0%
$30\ min$
0%
$45\ min$
0%
$75\ min$
0%
$90\ min$

Explanation

$t=\cfrac {2.303}{k}\log \left(\cfrac {N_o}{N}\right)$

Case- (i)

$60= \cfrac {2.303}{k} \log \left(\cfrac {4}{3}\right)$

Case- (ii)

$t= \cfrac {2.303}{k} \log (2)$

Dividing we get

$\cfrac {60}{t}= \cfrac {\log 4-\log 3}{\log 2}$

$\Rightarrow \cfrac {60}{t}= \cfrac {1}{0.415}$

$\Rightarrow t= 60 \times 0.415= 24.9 min= 30 min$

Fora reaction,

$A \rightarrow B +C,$ it was found that at the end of 10 minutes from the start, the total optical rotation of the system was

$50^o$ and when the reaction is complete, it was

$100^o C$ Assuming that only B and C are optically active and dextro rotatory, the rate constant of this first order reaction would be ?

0%
$0.069 min^{-1}$
0%
$0.69 min^{-1}$
0%
$6.9 min^{-1}$
0%
$6.9 \times 10^{-2} min^{-1}$

Explanation

Let the rotation of Sucrose be $r_1^o$ per mole and the initial moles of Sucrose be a.
$\implies r_0 = ar_1^0$

Let the moles converted be x.
$\implies r_t = (a-x)r_1^0$
$\implies \frac{a}{a-x} = \frac{r_o}{r_t}$
$\implies k = \frac{1}{t} \ln \frac{a}{a-x} = \frac{1}{t} \ln\frac{r_o}{r_t}$
$\implies k = \frac{2.303}{10}\log\frac{100}{50}$
$\implies k = 0.069 \space min^{-1}$

A reaction proceeds 5 times more at

${ 60 }^{ o }$ C as it does at

${ 30 }^{ o }$ C the energy of activation is

$(\log { 5=0.69990 } )$

0%
12 K cal / mole
0%
11 K cal / mole
0%
13 K cal / mole
0%
16 K cal / mole

Explanation

Given:

$T_2=60+273=333K$ ,

$T_1=30+273=303K$ ,

$R=1.987\times 10^{-3}kcal$

We know that at a temperature

$T$

$\dfrac{r_2}{r_1}=\dfrac{K_2}{K_1}$

$\dfrac{r_2}{r_1}=5$ (at Temperature

$T_2\ and\ T_1$ )

therefore,

$\dfrac{K_2}{K_1}=5$

Hence:

$2.303log_{10}\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\dfrac{T_2-T_1}{T_1\times T_2}$

$2.303log_{10}\ 5=\dfrac{E_a}{10^{-3}\times 1.987}\dfrac{333-300}{333\times 300}$

$E_a=10.757\ kcalmol^{-1}$

Option B is correct.

The rate of reaction increases with rise in temperature because of :

0%
increase in the number of activated molecules
0%
increase in the activation energy
0%
decrease in the activation energy
0%
increase in the number of molecular collisions

A graph of

$\ln {k}$ vs

$\left(\dfrac {1}{T}\right)$ has slope equal to

0%
$-\dfrac {E}{2.303R}$
0%
$+\dfrac {E_{a}}{R}$
0%
$-\dfrac {E_{a}}{2.303R}$
0%
$-\dfrac {E_{a}}{R}$

Explanation

We know that

$lnk=\dfrac{-E_a}{RT}+lnA$

Option D is correct.

$3/4\ th$ of first order reaction was completed in

$32\ min, 15/16$ the part will be completed in ?

0%
$24\ min$
0%
$64\ min$
0%
$16\ min$
0%
$32\ min$

When ln k is plotted against

${ 1 }/{ T }$ , the slope was found to be -10.7

$\times$

${ 10 }^{ 3 }$ K, activation energy for the reaction would be :

0%
-78.9 kJ ${ mol }^{ -1 }$
0%
2.26 kJ ${ mol }^{ -1 }$
0%
88.9 kJ ${ mol }^{ -1 }$
0%
10.7 kJ ${ mol }^{ -1 }$

Explanation

From the Arrhenius equation, we get

$lnk=lnA-\dfrac{E_a}{RT}$

Given,

$-\dfrac{E_a}{R}=-10.7\times10^3$

$\Rightarrow E_a=8.314\times10.7\times10^3\ Jmol^{-1}$

$\Rightarrow E_a=88.9\ kJmol^{-1}$

The decomposition of

$NH_3$ on platinum surface is zero order reaction. What are the rates of production of

$N_2$ and

$H_2$ if

$K= 2.5\times 10^{-4} mol^{-1} L s^{-1}$ ?

0%
$2.5 \times 10^{-4}$ and $7.5 \times 10^{-4}$ respectively
0%
$1.25 \times 10^{-4}$ and $3.75 \times 10^{-4}$ respectively
0%
$3.75 \times 10^{-3}$ and $1.25 \times 10^{-3}$ respectively
0%
$1.25 \times 10^{-3}$ and $3.25 \times 10^{-3}$ respectively

Explanation

The reaction is zero order. Hence, Rate of the reaction is equal to rate constant.

The reaction is,

$2NH_3 \rightarrow N_2 + 3H_2$

Therefore,

$\dfrac {d[{N_2}]}{dt}=2.5\times 10^{-4}$

$\dfrac{d[H_2]}{dt}=2.5\times 10^{-4}\times 3=7.5\times 10^{-4}$

In a first order reaction the amount of reactant decayed in three half lives (let a be is initial amount) would be:

0%
7a/8
0%
a/8
0%
a/6
0%
5a/6

Explanation

The amount of reactant left after

$n$ half lives for a first order reaction is

$(1/2)n$

times the Initial concentration.

Here $n=3$ . So Amount of Reactant left is $\dfrac{1}{8}$

So Amount of reactant decayed is $a-\dfrac{1}{8}=7a/8$

Option A is correct.

The rate of a reaction at 200 K 100 times less than the rate at 400 K.The activation energy of the reaction is

0%
184 R
0%
1840 R
0%
460 R
0%
4600 R

Rate constant of a first order reaction is

$1.15\times 10^{-5}$ . The percentage of initial concentration that remained after 1 hour is?

0%
$89.15\%$
0%
$95.96\%$
0%
$55.86\%$
0%
$83.45\%$

Explanation

Given:

$I^{st}$ order reaction,

$k=1.15\times 10^{-5}$

$t=1\ hour=60\times 60=3600\ seconds$

We have :

$k=\dfrac{2.303}{t}\dfrac{Initial\ conc}{final\ Conc}$

On putting all given values:

$1.15\times 10^{-5}=\dfrac{2.303}{3600}\dfrac{I_c}{F_c}$

$I_c$ = initial concenmtration,

$f_c$ = final concenmtration

On solving we will get

$F_c=95.96%$ %

Option B is correct.

The activation energy for the bimolecular reaction

$A+BC \rightarrow AB+C$ is

$E_{0}$ in the gas phase. If the reason carried out in a confined volume of

$\lambda^{3}$ , the activation energy is expected to:

0%
remain unchanged
0%
increase with decreasing $\lambda$ .
0%
decrease with decreasing $\lambda$ .
0%
oscillate with decreasing $\lambda$ .

Radioactive decay follows order kinetics

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

In a first order reaction ,if

$25\%$ of the reaction is completed in

$50$ minutes

$50\%$ of the total reaction is completed in _________ minutes (approx):

0%
$120$
0%
$100$
0%
$80$
0%
$150$

Explanation

Given:

First order reaction-

$25$ % complete in

$50\ minutes$

Then

$50$ % of reaction will complete in ?

For the first order reaction,

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

Here

$a=100$ %,

$x=25$ %

$t=50\ minutes$

$k=\dfrac{2.303}{50}\dfrac{log\ 100}{100-25}$

$k=\dfrac{2.303}{50}\dfrac{log\ 100}{75}$

$k=\dfrac{2.303}{50}\dfrac{log\ 4}{3}$ ------------(1)

In Second case

$a=100$ %,

$x=50$ %

$t=? minutes$

$k=\dfrac{2.303}{t}\dfrac{log\ 100}{100-50}$

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

so so

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

Compare (1) and (2) we will get:

$\dfrac{2.303}{50}\dfrac{log\ 4}{3}$ =

$\dfrac{2.303}{t}\ log\ 2$

on solving we will get

$t=120\ minutes$ approximately

Option A is correct.

For a first order reaction, the plot of

$'t'$ against

$log C$ gives a straight line with slope equal to:

0%
$(k/2.2303)$
0%
$(-k/2.303)$
0%
$(In k/2.303)$
0%
$-k$

Explanation

For a first-order reaction,

$log(a-x)=-\dfrac{kt}{2.303}+loga$

It is similar to straight-line equation

$y=mx+c$

where intercept is equal to

$loga$ and slope equal to

$-\dfrac{k}{2.303}$

Option B is the correct answer.

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 8 - 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 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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