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maria [59]
3 years ago
13

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mo

l . 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?
Chemistry
1 answer:
aleksandrvk [35]3 years ago
3 0

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}.

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