Question

Scoring scheme: 3-3-2-1 for many reactions near room temperature, the rate and the rate constant approximately double for a 10 °c rise in temperature. what is the value of activation energy in kj/mol for such a reaction

Answer 1

Answer:

\boxed{\text{52.9 kJ/mol}}

Explanation:

To solve this problem, we must use the Arrhenius equation:

$\ln \dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{R}\left(\dfrac{1}{T_{2}} - \dfrac{1}{T_{1}}\right)$

The activation energy depends on the starting temperature, so, let's assume that

T₁ = 25 °C = 298.15 K

T₂ = 35 °C = 308.15 K

k₂/k₁ = 2

This gives

$\ln \dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{R}\left(\dfrac{1}{T_{2}} - \dfrac{1}{T_{1}}\right)\\\\\ln \dfrac{2}{1} = \dfrac{E_{a}}{8.314}\left(\dfrac{1}{308.15} - \dfrac{1}{298.15}\right)\\\\\ln 2 = \dfrac{E_{a}}{8.314}\left(3.3540 \times 10^{-3} - 3.2452\times 10^{-3}\right)\\\\8.314 \ln 2 = E_{a}\left(1.088 \times 10^{-4}\right)\\\\E_{a} = \dfrac{8.314 \ln 2}{1.088 \times 10^{-4}}\\\\E_{a} = 5.29 \times 10^{4}\text{ J/mol}\\\\E_{a} = \boxed{\textbf{52.9 kJ/mol}}$