The decomposition of dinitrogen pentaoxide to nitrogen dioxide and oxygen follows first-order kinetics and has an activation energy of 102 kJ/mol. By what factor will the fraction of collisions with energy greater than or equal to the activation energy increase if the reaction temperature goes from 30°C to 60°C?

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Quizplus · Accepted Answer

The factor by which the fraction of collisions with energy greater than or equal to the activation energy increases can be calculated using the Arrhenius equation, which relates the rate constant of a reaction to the temperature and activation energy. The equation for the ratio of two rate constants at two different temperatures (T1 and T2) is given by:$$ \frac{k_2}{k_1} = e^{\frac{-E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})} $$where $E_a$ is the activation energy, $R$ is the gas constant (8.314 J/mol·K), $T_1$ and $T_2$ are the initial and final temperatures in Kelvin, respectively.Given:- $E_a = 102 \times 10^3 \, J/mol$ (since 1 kJ = 1000 J)- Initial temperature $T_1 = 30°C = 303.15 K$- Final temperature $T_2 = 60°C = 333.15 K$Plugging these values into the equation:$$ \frac{k_2}{k_1} = e^{\frac{-102 \times 10^3}{8.314}(\frac{1}{333.15} - \frac{1}{303.15})} $$$$ \frac{k_2}{k_1} \approx e^{(-12293.7)(-0.000903)} $$$$ \frac{k_2}{k_1} \approx e^{11.1} $$$$ \frac{k_2}{k_1} \approx 38.4 $$Therefore, the fraction of collisions with energy greater than or equal to the activation energy increases by a factor of approximately 38.4 when the temperature is increased from 30°C to 60°C.

The Decomposition of Dinitrogen Pentaoxide to Nitrogen Dioxide and Oxygen