Question

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?\

Answer 1

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

`\ln( (k_2 )/(k_1 )) = ((E_(a) )/(R ))(( 1)/(T_1) - (1 )/(T_2 ))\\`

`\ln( (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)}) * 1.436 *10^(-4)\\`

`\ln2 = E_(a) * 1.727 * 10^(-5) \text{ mol} \cdot \text{J}^(-1)`

`E_(a) = \frac{\ln2 }{ 1.727 *10^(-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)`