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. $(20.74$ $ min)$
Solution:
- For 1 st order reaction :
$k \propto \dfrac{1}{t _{1 / 2}} $
$\Rightarrow \quad \ln \{\dfrac{k\left(450^{\circ} C\right)}{k\left(380^{\circ} C\right)} \}=\ln \{\dfrac{t _{1 / 2}\left(380^{\circ} C\right)}{t _{1 / 2}\left(450^{\circ} C\right)} \}=\dfrac{E _a}{R}\left(\dfrac{450-380}{727 \times 653}\right) $
$\Rightarrow \quad \ln \{\dfrac{360}{t _{1 / 2}\left(450^{\circ} C\right)} \}=\dfrac{200 \times 10^{3}}{8.314} \times \dfrac{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 $ $min$
BETA
