Chemical Kinetics - Result Question 9
####9. For a reaction, consider the plot of $\ln k$ versus $1 / T$ given in the figure. If the rate constant of this reaction at $400 K$ is $10^{-5} s^{-1}$, then the rate constant at $500 K$ is
(2019 Main, 12 Jan II)
(a) $4 \times 10^{-4} s^{-1}$
(b) $10^{-6} s^{-1}$
(c) $10^{-4} s^{-1}$
(d) $2 \times 10^{-4} s^{-1}$
10 Decomposition of $X$ exhibits a rate constant of $0.05 \mu g /$ year. How many years are required for the decomposition of $5 \mu g$ of $X$ into $2.5 \mu g$ ?
(2019 Main, 12 Jan I)
(a) 20
(b) 25
(c) 40
(d) 50
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Answer:
Correct Answer: 9. (d)
Solution:
- The temperature dependence of a chemical reaction is expressed by Arrhenius equation,
$$ k=A e^{-E _a / R T} $$
Taking natural logarithm on both sides, the Arrhenius equation becomes,
$$ \ln k=\ln A-\frac{E _a}{R T} $$
where, $-\frac{E _a}{R}$ is the slope of the plot and $\ln A$ gives the intercept.
Eq. (i) at two different temperatures for a reaction becomes,
$$ \ln \frac{k _2}{k _1}=\frac{E _a}{R}\left(\frac{1}{T _1}-\frac{1}{T _2}\right) $$
$\Rightarrow$ In the given problem,
$$ \begin{gathered} T _1=400 K, T _2=500 K \ k _1=10^{-5} s^{-1}, k _2=? \end{gathered} $$
$-\frac{E _a}{R}$ (Slope) $=-4606$
On substituting all the given values in Eq. (ii), we get
$$ \begin{gathered} \ln \frac{k _2}{10^{-5}}=4606\left(\frac{1}{400}-\frac{1}{500}\right) \ \ln \frac{k _2}{10^{-5}}=2.303 \ \frac{k _2}{10^{-5}}=10 \Rightarrow k _2=10^{-4} s^{-1} \end{gathered} $$
Therefore, rate constant for the reaction at $500 K$ is $10^{-4} S^{-1}$.
