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} $ $s^{-1}$
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Answer:
Correct Answer: 9. (c)
Solution:
- The temperature dependence of a chemical reaction is expressed by Arrhenius equation,
$ k=A e^{-E _a / R T} $$\quad$ …….. (i)
Taking natural logarithm on both sides, the Arrhenius equation becomes,
$ \ln k=\ln A-\dfrac{E _a}{R T} $
where, $-\dfrac{E _a}{R}$ is the slope of the plot and $\ln A$ gives the intercept.
Eq. (i) at two different temperatures for a reaction becomes,
$ \ln \dfrac{k _2}{k _1}=\dfrac{E _a}{R}\left(\dfrac{1}{T _1}-\dfrac{1}{T _2}\right) $ $\quad$ …….. (ii)
$\Rightarrow$ In the given problem,
$ \begin{gathered} T _1=400 K, T _2=500 K \\ k _1=10^{-5} s^{-1}, k _2=? \end{gathered} $
$-\dfrac{E _a}{R}$ (Slope) $=-4606$
On substituting all the given values in Eq. (ii), we get
$ \begin{gathered} \ln \dfrac{k _2}{10^{-5}}=4606\left(\dfrac{1}{400}-\dfrac{1}{500}\right) \\ \ln \dfrac{k _2}{10^{-5}}=2.303 \\ \dfrac{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} $ $S^{-1}$.
BETA
