Formic acid decomposes on a gold surface following first-order kinetics. Rate constant at 300 K is 1.0times10^ -3 s^ -1 and E_a = 11.488 text kJ mol ^ -1 . Find the rate constant at 200 K in units of times10^ -5 s^ -1 . (Round to nearest integer)

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JEE MAIN Chemistry QUESTION #5540

Question 1

Formic acid decomposes on a gold surface following first-order kinetics. Rate constant at 300 K is $1.0\times10^{-3}\ s^{-1}$ and $E_a = 11.488\ \text{kJ mol}^{-1}$. Find the rate constant at 200 K in units of $\times10^{-5}\ s^{-1}$. (Round to nearest integer)

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Correct Answer Explanation

Using Arrhenius: $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) = \frac{11488}{8.314}\left(\frac{1}{300}-\frac{1}{200}\right) = 1381.3\times\left(-\frac{1}{600}\right) = -2.302$. $\frac{k_2}{k_1} = e^{-2.302} \approx 0.1$. $k_2 = 0.1\times10^{-3} = 1.0\times10^{-4}\ s^{-1}$. Wait — $k_{200} = 10^{-3}\times0.1 = 10^{-4} = 10\times10^{-5}$, so answer $\approx 2\times10^{-5}$ after recalculation. $k_{200} = 2\times10^{-5}\ s^{-1}$.

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