For a given reaction , is related to as given in the table. Given: log 2 = 0.30 Which of the following is true? (A) The order of the reaction is 1/2. (B) If is 1 M, then is min. (C) The order of the reaction changes to 1 if the concentration

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Solution

Q.

${[A]}_{0} / {mol L}^{- 1}$ $t_{1 / 2} / min$

0.100 200

0.025 100

For a given reaction $R \to P$ , $t_{1 / 2}$ is related to ${[A]}_{0}$ as given in the table.
Given: log 2 = 0.30

Which of the following is true

(A) The order of the reaction is 1/2.

(B) If ${[A]}_{0}$ is 1 M, then $t_{1 / 2}$ is $200 \sqrt{10}$ min.

(C) The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.

(D) $t_{1 / 2}$ is 800 min for ${[A]}_{0}$ = 1.6 M.

Choose the correct answer from the options given below: [2025]

1 (C) and (D) Only

2 (A) and (B) Only

3 (A), (B) and (D) Only

4 (A) and (C) Only

Ans.
(3)

(A) For $n^{th}$ order reaction, $t_{1 / 2}$ is related to initial concentration $(a_{0})$ as:

$t_{1 / 2} \propto a_{0}^{1 - n}$

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{{(a_{0})}_{1}}{{(a_{0})}_{2}})}^{1 - n}$

$\frac{200}{100} = {(\frac{0.10}{0.025})}^{1 - n}$

$2 = 4^{1 - n}$

$n = \frac{1}{2}$

(B) $\frac{{(t_{1 / 2})}_{for a_{0} = 1 M}}{{(t_{1 / 2})}_{1}} = {(\frac{1}{{(a_{0})}_{1}})}^{1 - n}$

$\frac{{(t_{1 / 2})}_{for a_{0} = 1 M}}{200} = {(\frac{1}{0.1})}^{1 - \frac{1}{2}}$

${(t_{1 / 2})}_{for a_{0} = 1 M} = 200 \sqrt{10} min$

(D) $\frac{{(t_{1 / 2})}_{for a_{0} = 1.6 M}}{{(t_{1 / 2})}_{1}} = {(\frac{1.6}{{(a_{0})}_{1}})}^{1 - n}$

$\frac{{(t_{1 / 2})}_{for a_{0} = 1.6 M}}{200} = {(\frac{1.6}{0.1})}^{1 - \frac{1}{2}}$

${(t_{1 / 2})}_{for a_{0} = 1.6 M} = 800 min$

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