Question

Given that the initial rate constant is 0.0191 s-1 at an initial temperature of 24°C, what would the rate constant be at a temperature of 120°C for a reaction with an activation energy of 43.4 kJ/mol?

Answer 1

Explanation:

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

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

Since, it is given that $K_{1}$ is 0.0191 $s^{-1}$, $T_{1}$ is 24 degree celsius equals 297 K, $T_{2}$ is 120 degree celsius equals 393 K, and $E_{a}$ is 43.4 kJ/mol or 43400 J/mol.

Therefore, putting the value in the above formula as follows.

               ln $\frac{K_{2}}{K_{1}}$ = $\frac{-43400 J/mol}{8.314 j/mol K}[\frac{1}{393} - \frac{1}{297}]$

                ln $\frac{K_{2}}{0.0191 s^{-1}}$ = $\frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

                      $K_{2}$ = 1.40 $s^{-1}$

Thus, we can conclude that rate constant be at a temperature of 120°C for a reaction with an activation energy of 43.4 kJ/mol is 1.40 $s^{-1}$.