Question

A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 degree​

Answer 1

Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree​

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

where $k_2$ = rate constant at temp $T_2$

$k_1$ = rate constant at temp $T_1$

$E_a$= activation energy

R= gas constant

$T_1$= temperature = $20^0C=(20+273)K=293K$

$T_2$= temperature = $30^0C=(30+273)K=303K$

$ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]$

$ln \frac{k_{2}}{k_{1}}=1.15$

$\frac{k_{2}}{k_{1}}=3$

Thus rate increases 3 times on raising the temperature from 20degree to 30 degree​