Chlorofluorocarbons released into the atmosphere from refrigerants and aerosol sprays were found to catalyze the decomposition of ozone, leading to the formation of an ozone hole over the South Pole. The catalyzed reaction was found to have an activation energy of 2.2 kJ/mol compared with an activation energy of 17.7 kJ/mol for the uncatalyzed reaction. By what factor is the decomposition rate constant increased in the catalyzed reaction at 300 K?

Question

Quizplus · Accepted Answer

The rate constant $k$ for a reaction can be calculated using the Arrhenius equation: $k = A \exp\left(-\frac{E_a}{RT}\right)$, where $A$ is the pre-exponential factor (assumed to be the same for both the catalyzed and uncatalyzed reactions in this context), $E_a$ is the activation energy, $R$ is the gas constant (8.314 J/mol·K), and $T$ is the temperature in Kelvin. The factor by which the rate constant increases due to catalysis is given by the ratio of the rate constants for the catalyzed ($k_{\text{cat}}$) and uncatalyzed ($k_{\text{uncat}}$) reactions:$$\text{Factor} = \frac{k_{\text{cat}}}{k_{\text{uncat}}} = \frac{\exp\left(-\frac{E_{a,\text{cat}}}{RT}\right)}{\exp\left(-\frac{E_{a,\text{uncat}}}{RT}\right)} = \exp\left(-\frac{E_{a,\text{cat}} - E_{a,\text{uncat}}}{RT}\right)$$Given $E_{a,\text{cat}} = 2.2 \, \text{kJ/mol} = 2200 \, \text{J/mol}$ and $E_{a,\text{uncat}} = 17.7 \, \text{kJ/mol} = 17700 \, \text{J/mol}$, and $T = 300 \, \text{K}$, we can calculate the factor:$$\text{Factor} = \exp\left(-\frac{2200 \, \text{J/mol} - 17700 \, \text{J/mol}}{8.314 \, \text{J/mol·K} \cdot 300 \, \text{K}}\right) = \exp\left(-\frac{-15500}{2494.2}\right) \approx \exp(6.21) \approx 500$$Therefore, the decomposition rate constant is increased by a factor of approximately 500 in the catalyzed reaction at 300 K.

Chlorofluorocarbons Released into the Atmosphere from Refrigerants and Aerosol Sprays