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:

  1. 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 $10^{-4} S^{-1}$.



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