Question

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer 1

Explanation:

Relation between temperature and activation energy according to Arrhenius equation is as follows.

            k = $A exp^{\frac{-E_{a}}{RT}}$

where,   k = rate constant

              A = pre-exponential factor

           $E_{a}$ = activation energy

             R = gas constant

              T = temperature in kelvin

Also,  

         $ln (\frac{k_{2}}{k_{1}}) = (\frac{-E_{a}}{R}) \times (\frac{1}{T_{2}} - \frac{1}{T_{2}})$

      $T_{1}$ = $244^{o}C$ = (244 + 273) K = 517.15 K

      $T_{2}$ = $324^{o}C$ = (597.15 + 273) K = 597.15 K

     $k_{1}$ = 6.7 $M^{-1} s^{-1}$,     $k_{2}$ = ?

         R = 8.314 J/mol K

      $E_{a}$ = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows.

   $ln (\frac{k_{2}}{6.7}) = (\frac{-71000}{8.314} \times (\frac{1}{597.15} - \frac{1}{517.15})$

                   = 2.2123

        $\frac{k_{2}}{6.7} = exp(2.2123)$

                    = 9.1364

                $k_{2} = 61 M^{-1}s^{-1}$

Thus, we can conclude that rate constant of this reaction is $61 M^{-1}s^{-1}$.