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Lerok [7]
3 years ago
15

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

barrier for the reaction?\
Chemistry
1 answer:
maria [59]3 years ago
3 0

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\

\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\

\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\

\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}

E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\

E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}

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