JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 22)
The rate constant for a first order reaction is $20 \mathrm{~min}^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is _______ $\times 10^{-2} \mathrm{~min}$. (Nearest integer)
(Given : $\ln 10=2.303$ and $$ \log 2=0.3010 \text { )}$$
(Given : $\ln 10=2.303$ and $$ \log 2=0.3010 \text { )}$$
Answer
17
Explanation
$K=20 \mathrm{~min}^{-1}$
$$ \mathrm{t}_{1 / 2}=\frac{0.6932}{20}=\frac{\ln 2}{20} $$
Required time $=\mathrm{n} \times \mathrm{t}_{1 / 2}$
$$C = {{{C_0}} \over {{2^n}}} = {{{C_0}} \over {32}}$$
$$ \Rightarrow $$ $${2^n} = 32$$ = $${2^5}$$
$$ \Rightarrow $$ n = 5
$$ \begin{aligned} \text { Required time } & =\frac{5 \times 0.6932}{20} \\\\ & =0.173 \mathrm{~min} \\\\ & =17.3 \times 10^{-2} \mathrm{~min} \end{aligned} $$
$$ \mathrm{t}_{1 / 2}=\frac{0.6932}{20}=\frac{\ln 2}{20} $$
Required time $=\mathrm{n} \times \mathrm{t}_{1 / 2}$
$$C = {{{C_0}} \over {{2^n}}} = {{{C_0}} \over {32}}$$
$$ \Rightarrow $$ $${2^n} = 32$$ = $${2^5}$$
$$ \Rightarrow $$ n = 5
$$ \begin{aligned} \text { Required time } & =\frac{5 \times 0.6932}{20} \\\\ & =0.173 \mathrm{~min} \\\\ & =17.3 \times 10^{-2} \mathrm{~min} \end{aligned} $$
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