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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Answer:
Correct Answer: 71. $(3.26 \times 10^{-2} $ $min^{-1})$
Solution:
- (i) $A \xrightarrow{k _A}$ Product
(ii) $B \xrightarrow{k _B}$ Product
$\text { For (i) } \dfrac{E _a}{R}\left(\dfrac{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) } =\dfrac{E _a(i)}{2}=26.8 $ $kJ$
At $310 K \quad t _{1 / 2}(i)=30$ $ min$
$\because \quad$ Rate of (ii) $=2$ rate of (i)
$ \Rightarrow \quad t _{1 / 2}(ii)=15$ $ min $
Now for reaction (ii) :
$ \begin{aligned} & \ln \{\dfrac{k _B(310)}{k _B(300)} \}=\ln \{\dfrac{t _{1 / 2}(300)}{t _{1 / 2}(310)} \}=\dfrac{E _a(ii)}{R}\left(\dfrac{10}{300 \times 310}\right) \\ & \Rightarrow \quad \ln \{\dfrac{t _{1 / 2}(300)}{15} \}=\dfrac{\ln 2}{2} \Rightarrow t _{1 / 2}(300)=21.2 min \\ & \Rightarrow \quad k _B(300)=\dfrac{\ln 2}{t _{1 / 2}}=\dfrac{0.693}{21.2}=3.26 \times 10^{-2} min^{-1} \end{aligned} $
BETA
