$[A]_{0} / {molL}^{- 1}$ $t_{1 / 2} / \min$
0.100 200
0.025 100

For a given reaction R P, t_1/2 is related to [A]₀ as given in table :
Given : log 2 = 0.30
Which of the following is true ?
A. The order of the reaction is $\frac{1}{2}$ .
B. If [A]₀ is 1M, 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 :

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A and B only

A and C only

C and D only

A, B and D only

answer is C.

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Detailed Solution

$\begin{array}{ll} t_{1 / 2} \propto \frac{1}{A_{0}^{n - 1}} \\ \frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = \frac{{(A_{0})}_{2}^{n - 1}}{{(A_{0})}_{1}^{n - 1}} \\ \frac{200}{100} = {(\frac{0.025}{0.100})}^{n - 1} \\ 2 = {(\frac{1}{4})}^{n - 1} & n - 1 = - \frac{1}{2} \\ \Rightarrow t_{1 / 2} \propto \sqrt{A_{0}} & n = \frac{1}{2} (order) \\ \frac{200}{t_{1 / 2}} = \frac{(0.1)^{1 / 2}}{(1)^{1 / 2}} & when A_{0} = 1 M \\ t_{1 / 2} = 200 \sqrt{10} min \end{array}$
Ist order kinetics have t_1/2 independent of their concentration. So upon changing the concentration t_1/2 should not change for first order reaction.
$\begin{array}{ll} \frac{200}{t_{1 / 2}} = \frac{(0.1)^{1 / 2}}{(1.6)^{1 / 2}} when A_{0} = 1.6 M \\ t_{1 / 2} = 800 \min \end{array}$