Chemistry - Chemical Kinetics Question with Solution | TestHub

The problem involves calculating the rate of a second-order reaction at a different temperature, given the activation energy and the rate at a specific temperature. We can use the Arrhenius equation to solve this.

Arrhenius equation:

$k=A{e}^{-\frac{E_a}{RT}}$

where:

$k$ is the rate constant

$A$ is the pre-exponential factor

$E_a$ is the activation energy

$R$ is the gas constant ($8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ or $2calmo{l}^{-1}{K}^{-1}$)

$T$ is the temperature in Kelvin

Since we are given the activation energy in $\mathrm{kcal}\mathrm{mo}{\mathrm{l}}^{-1}$, we'll use $R=2calmo{l}^{-1}{K}^{-1}$.

We have two temperatures, $T_1=500\mathrm{K}$ and $T_2=1000\mathrm{K}$, and the rate at $T_1$ is $k_1=0.08\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$. We want to find $k_2$ at $T_2$.

$\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Plugging in the values:

$E_a=9.2\mathrm{kcal}\mathrm{mo}{\mathrm{l}}^{-1}=9200\mathrm{cal}\mathrm{mo}{\mathrm{l}}^{-1}$

$R=2calmo{l}^{-1}{K}^{-1}$

$T_1=500\mathrm{K}$

$T_2=1000\mathrm{K}$

$k_1=0.08\mathrm{mol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$

$\ln \frac{k_2}{0.08}=\frac{9200}{2}\left(\frac{1}{500}-\frac{1}{1000}\right)$

$\ln \frac{k_2}{0.08}=\frac{9200}{2}\left(\frac{2-1}{1000}\right)$

$\ln \frac{k_2}{0.08}=\frac{9200}{2}\left(\frac{1}{1000}\right)$

$\ln \frac{k_2}{0.08}=\frac{9.2}{2}=4.6$

Now, we need to find $\frac{k_2}{0.08}$.

$\frac{k_2}{0.08}=e^{4.6}$

Since $\ln x=2.3\log x$, we can rewrite $e^{4.6}$ as follows:

$e^{4.6}=10^{\frac{4.6}{2.3}}=10^2=100$

$k_2=0.08\times 100=8$

Final Answer: The final answer is $\boxed{8}$