Ethyl chloride vapor decomposes by the first-order reaction: C2H5Cl -> C2H4 + HCl. The activation energy is 249 kJ/mol, and the frequency factor is 1.6 * 10^14 s^-1. Find the temperature at which the rate of the reaction would be twice as fast.

Verified step by step guidance

1. Identify the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$, where $k$ is the rate constant, $A$ is the frequency factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
2. Recognize that the problem asks for the temperature at which the rate of the reaction is twice as fast, meaning $k_2 = 2k_1$.
3. Use the Arrhenius equation to set up the relationship between the two rate constants: $\frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} = 2$.
4. Simplify the equation: $e^{-\frac{E_a}{RT_2}} = 2 e^{-\frac{E_a}{RT_1}}$ and take the natural logarithm of both sides to solve for $T_2$: $-\frac{E_a}{RT_2} = \ln(2) - \frac{E_a}{RT_1}$.
5. Rearrange the equation to solve for $T_2$: $T_2 = \frac{E_a}{R(\ln(2) + \frac{E_a}{RT_1})}$, where $E_a = 249 \text{ kJ/mol}$, $R = 8.314 \text{ J/mol K}$, and $T_1$ is the initial temperature.