Chemical Kinetics - Result Question 71
70. A first order reaction, $A \rightarrow B$, requires activation energy of $70$ $ kJ $ $mol^{-1}$. When a $20 \%$ solution of $A$ was kept at $25^{\circ}$ $ C$ for $20$ $ min, 25 \%$ decomposition took place. What will be the percentage decomposition in the same time in a $30 \%$ solution maintained at $40^{\circ} $ $C$ ? Assume that activation energy remains constant in this range of temperature.
$(1993,4 M)$
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Answer:
Correct Answer: 70. $(67 \%)$
Solution:
- $\ln \{\dfrac{k\left(40^{\circ} C\right)}{k\left(25^{\circ} C\right)} \}=\dfrac{E _a}{R}\left(\dfrac{15}{298 \times 313}\right)$
$ \begin{aligned} & = \quad \dfrac{70 \times 1000}{8.314} \times \dfrac{15}{298 \times 313}=1.35 \\ & \Rightarrow \quad \dfrac{k\left(40^{\circ} C\right)}{k\left(25^{\circ} C\right)}=3.87 \\ & \text { Also } k\left(25^{\circ} C\right)=\dfrac{1}{20} \ln \dfrac{100}{75}=\dfrac{1}{20} \ln \dfrac{4}{3} \\ & \Rightarrow \quad k\left(40^{\circ} C\right)=3.87 \times k\left(25^{\circ} C\right) \\ & =\quad 3.87 \times \dfrac{1}{20} \ln \dfrac{4}{3}=55.66 \times 10^{-3} min^{-1} \\ & \text { Now } \quad k\left(40^{\circ} C\right) \times 20=\ln \dfrac{100}{100-x} \\ & \Rightarrow \quad 55.66 \times 10^{-3} \times 20=\ln \dfrac{100}{100-x} \Rightarrow x=67 \% \end{aligned} $
BETA
