Chemical Kinetics - Result Question 72
####71. Two reactions (i) $A \rightarrow$ products (ii) $B \rightarrow$ products, follow first order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from $300 K$ to $310 K$. The half-life for this reaction at $310 K$ is $30 min$. At the same temperature $B$ decomposes twice as fast as $A$. If the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at $300 K$.
(1992, 3M)
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Solution:
- (i) $A \xrightarrow{k _A}$ Product
(ii) $B \xrightarrow{k _B}$ Product
$$ \begin{array}{ll} \text { For (i) } & \frac{E _a}{R}\left(\frac{10}{300 \times 310}\right)=\ln 2 \ \Rightarrow \quad E _a(i) & =9300 R \ln 2=53.6 kJ \ \Rightarrow \quad E _a \text { (ii) } & =\frac{E _a(i)}{2}=26.8 kJ \end{array} $$
At $310 K \quad t _{1 / 2}(i)=30 min$
$\because$ Rate of (ii) $=2$ rate of (i)
$$ \Rightarrow \quad t _{1 / 2}(ii)=15 min $$
Now for reaction (ii) :
$$ \begin{aligned} & \ln {\frac{k _B(310)}{k _B(300)} }=\ln {\frac{t _{1 / 2}(300)}{t _{1 / 2}(310)} }=\frac{E _a(ii)}{R}\left(\frac{10}{300 \times 310}\right) \ & \Rightarrow \quad \ln {\frac{t _{1 / 2}(300)}{15} }=\frac{\ln 2}{2} \Rightarrow t _{1 / 2}(300)=21.2 min \ & \Rightarrow \quad k _B(300)=\frac{\ln 2}{t _{1 / 2}}=\frac{0.693}{21.2}=3.26 \times 10^{-2} min^{-1} \end{aligned} $$
