jee main chemistry important questions | Chemical Kinetics jee main questions FREE

Q 21 :

$A \to B$

The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ ${mol}^{- 1}$ and the frequency factor is $10^{20}$ , the time required for 50% molecules of A to become B is ______ picoseconds (nearest integer).

[R = 8.314 J $K^{- 1} {mol}^{- 1}$ ] [2025]

0%

(69)

Activation energy $E_{a} = 191.48 {kJ mol}^{- 1} = 191480 {J mol}^{- 1}$

Frequency factor $(A) = 10^{20}$

Temperature $(T) = 1000 K$

Gas constant R = 8.314 $J/mol K$

By Arrhenius equation,

$= A e^{\frac{- E_{a}}{R T}} = 10^{20} e^{\frac{- 191480}{8.314 \times 1000}}$

$= 10^{20} e^{- 23.03} = 10^{20} e^{- \ln 10 \times 10}$

$= 10^{20} e^{- \ln 10^{10}}$

$= \frac{10^{20}}{e^{\ln 10^{10}}} = \frac{10^{20}}{10^{10}} = 10^{10} s^{- 1}$

$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{10^{10}} s = 69.3 \times 10^{- 12} s$

$= 69.3 picosecond \approx 69 picosecond$

Q 22 :

A student has studied the decomposition of a gas ${AB}_{3}$ at 25°C. He obtained the following data.

$p$ (mm Hg) 50 100 200 400

relative $t_{1 / 2}$ (s) 4 2 1 0.5

The order of the reaction is: [2023]

1
0.5
0 (zero)
2

1

2

3

4

(4)

$T_{\frac{1}{2}} \propto {(C_{0})}^{1 - n}$

$\frac{{(T_{\frac{1}{2}})}_{1st}}{{(T_{\frac{1}{2}})}_{2nd}} = {(\frac{P_{1}}{P_{2}})}^{1 - n}$

$\Rightarrow \frac{4}{2} = {(\frac{50}{100})}^{1 - n} \Rightarrow 2 = {(\frac{1}{2})}^{1 - n}$

$2 = {(2)}^{n - 1}$

$n - 1 = 1 \Rightarrow n = 2$

$Order = 2$

Q 23 :

The graph which represents the following reaction is:

${(C_{6} H_{5})}_{3} C - Cl \to_{Pyridine}^{{OH}^{-}} {(C_{6} H_{5})}_{3} C - OH$ [2023]

1

2

3

4

(4)

Q 24 :

For the first order reaction $A \to B$ , the half-life is 30 min. The time taken for 75% completion of the reaction is ______ min. (Nearest integer)

Given: log 2 = 0.3010

log 3 = 0.4771

log 5 = 0.6989 [2023]

0%

(60)

$t_{75 %} = 2 t_{1 / 2} [For 1st order reaction]$

$t_{75 %} = 2 \times 30 = 60 min .$

Q 25 :

A first order reaction has the rate constant, $k = 4.6 \times 10^{- 3} s^{- 1}$ . The number of correct statement/s from the following is/are ______. Given: log 3 = 0.48 [2023]

A. Reaction completes in 1000 s.

B. The reaction has a half-life of 500 s.

C. The time required for 10% completion is 25 times the time required for 90% completion.

D. The degree of dissociation is equal to $(1 - e^{- kt})$ .

E. The rate and the rate constant have the same unit.

0%

(2)

(C & D)

Statement 4 is correct

$C_{t} = C_{0} e^{- k t}$

$C_{0} \to C_{0} - C_{t}$

$1 \to \frac{C_{0} - C_{t}}{C_{0}}$

Q 26 :

If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{- 3} s^{- 1}$ . The time taken by A (in seconds) to reduce from 7 g to 2 g will be ______. (Nearest integer) [log 5 = 0.698, log 7 = 0.845, log 2 = 0.301] [2023]

0%

(623)

$K = \frac{1}{t} \ln (\frac{a}{a - x})$

$K = \frac{1}{t} \ln (\frac{W_{0}}{W_{t}})$

$t = \frac{1}{K} \ln (\frac{7}{2})$

$t = \frac{2.303}{K} [\log 7 - \log 2]$

$= \frac{2.303}{K} [0.845 - 0.301]$

$= \frac{2.303}{2.011 \times 10^{- 3}} \times 0.544 = 622.989 sec$

Q 27 :

An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is ______ s. (Nearest integer).

(Given: ln 10 = 2.3; log 2 = 0.3) [2023]

0%

(1350)

First order reaction

$t = \frac{1}{k} \ln \frac{C_{0}}{C_{t}}$

$540 = \frac{1}{k} \ln \frac{100}{40}$ $. . . (1)$

$t = \frac{1}{k} \ln \frac{100}{10}$ $. . . (2)$

$From (1) and (2)$

$\frac{540}{t} = \frac{\ln \frac{100}{40}}{\ln \frac{100}{10}} = \frac{\log 100 - \log 40}{\log 100 - \log 10}$

$\frac{540}{t} = \frac{2 - 1.6}{2 - 1} = 0.4$

$t = \frac{540}{0.4} = 1350 s$

Q 28 :

The rate constant for a first order reaction is 20 $\min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ______ $\times 10^{- 2}$ min. (Nearest integer)

(Given: ln 10 = 2.303, log 2 = 0.3010) [2023]

0%

(17)

$t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{20} = 0.03465 min$

${(\frac{1}{2})}^{n} = (\frac{1}{32}) = {(\frac{1}{2})}^{5}$

$No. of half-life = 5$

$Time required = 0.03465 \times 5 = 0.17325 min$

$= 17.325 \times 10^{- 2} min$

Q 29 :

A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ______ min. [2023]

0%

(15)

$t_{1 / 2 (A)} = 15 min, t_{1 / 2 (B)} = 5 min$

${[B]}_{initial} = 4 {[A]}_{initial}$

${[B]}_{final} = {[B]}_{initial} {(\frac{1}{2})}^{t / 15}$

${[A]}_{final} = {[A]}_{_{initial}} {(\frac{1}{2})}^{t / 5}$

${[B]}_{final} = {[A]}_{final}$

$t = 15$

Q 30 :

$A \to B$

The above reaction is of zero order. Half-life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer) [2023]

0%

(75)

$Zero order$

$t_{1 / 2} = 50$

$t_{3 / 4} = 1.5$ $t_{1 / 2} = 75$

Topic Question Set

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