Chemical Kinetics - Result Question 68
####67. At $380^{\circ} C$, the half-life period for the first order decomposition of $H _2 O _2$ is $360 min$. The energy of activation of the reaction is $200 kJ mol^{-1}$. Calculate the time required for $75 %$ decomposition at $450^{\circ} C$.
(1995, 4M)
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Answer:
Correct Answer: 67. $(1.386 min)$
Solution:
- For 1 st order reaction :
$$ \begin{gathered} k \propto \frac{1}{t _{1 / 2}} \ \Rightarrow \quad \ln {\frac{k\left(450^{\circ} C\right)}{k\left(380^{\circ} C\right)} }=\ln {\frac{t _{1 / 2}\left(380^{\circ} C\right)}{t _{1 / 2}\left(450^{\circ} C\right)} }=\frac{E _a}{R}\left(\frac{450-380}{727 \times 653}\right) \ \Rightarrow \quad \ln {\frac{360}{t _{1 / 2}\left(450^{\circ} C\right)} }=\frac{200 \times 10^{3}}{8.314} \times \frac{70}{727 \times 653}=3.54 \ \Rightarrow \quad t _{1 / 2}\left(450^{\circ} C\right)=10.37 min \ \Rightarrow \quad \text { Time for } 75 % \text { reaction at } 450^{\circ} C \ \quad=2 \times t _{1 / 2}=2 \times 10.37=20.74 min \end{gathered} $$
