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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Solution:
- $\ln {\frac{k\left(40^{\circ} C\right)}{k\left(25^{\circ} C\right)} }=\frac{E _a}{R}\left(\frac{15}{298 \times 313}\right)$
$$ \begin{aligned} & =\frac{70 \times 1000}{8.314} \times \frac{15}{298 \times 313}=1.35 \ & \Rightarrow \quad \frac{k\left(40^{\circ} C\right)}{k\left(25^{\circ} C\right)}=3.87 \ & \text { Also } k\left(25^{\circ} C\right)=\frac{1}{20} \ln \frac{100}{75}=\frac{1}{20} \ln \frac{4}{3} \ & \Rightarrow \quad k\left(40^{\circ} C\right)=3.87 \times k\left(25^{\circ} C\right) \ & =3.87 \times \frac{1}{20} \ln \frac{4}{3}=55.66 \times 10^{-3} min^{-1} \ & \text { Now } \quad k\left(40^{\circ} C\right) \times 20=\ln \frac{100}{100-x} \ & \Rightarrow \quad 55.66 \times 10^{-3} \times 20=\ln \frac{100}{100-x} \Rightarrow x=67 % \end{aligned} $$
