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STD 12 - 3. Chemical kinetics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. Chemical kinetics4 Marks
Question
Consider the two different first order reactions given below

$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1$)$

$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction $2$)

The ratio of the half life of Reaction $1$ : Reaction $2$ is $5: 2$. If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {dd }}$ and $4 / 5^{\text {dd }}$ of Reaction $1$ and

Reaction $2$, respectively, then the value of the ratio $\mathrm{t}_1: \mathrm{t}_2$ is . . . .$\times 10^{-1}$ (nearest integer).

[Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]

✓

Answer

$\frac{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{I}}}{\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{II}}}=\frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{5}{2}$

$\therefore \mathrm{K}_1 \mathrm{t}_1=\ln \frac{1}{1-\frac{2}{3}}=\ln 3$

$\mathrm{~K}_2 \mathrm{t}_2=\ln \frac{1}{1-\frac{4}{5}}=\ln 5$

$\Rightarrow \frac{\mathrm{K}_1}{\mathrm{~K}_2} \times \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699}$

$\Rightarrow \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699} \times \frac{5}{2}=1.7=17 \times 10^{-1}$

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