Discussion Forum

The rate constant is doubled when the temperature increases from 27°C to 37°C. Activation energy in kJ is

Answer:

Explanation:

$log\frac{K_{2}}{K_{1}}$= $\frac{E_{a}}{2.303R}\left[\frac{1}{T_{1}}- \frac{1}{T_{2}}\right]$

If $log\frac{K_{2}}{K_{1}}$ = 2

log 2 = $\frac{E_{a}}{2.303\times8.314}\left[\frac{1}{300}- \frac{1}{310}\right]$

$E_{a} = 0.3010\times2.303\times8.314\left(\frac{300\times310}{10}\right)$

= 53598.59 J/mol = 54kJ