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Question

The rate of a particular reaction doubles when the temperature changes from 300K to 310K. Calculate the energy of activation of the reaction. [Given R=8.314JK−1mol−1 ]


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Solution

Arrhenius equation

  • The Arrhenius equation can be used to determine the influence of temperature change on the rate constant and the rate of the reaction.
  • The Arrhenius equation is as follows:

logk2k1=Ea2.303RT2-T1T2T1

  • k2 and k1 are the rate constants.
  • T2 and T1 are the temperatures.
  • Ea is the activation energy.

Determine activation energy

  • Initial temperature, T1=300K
  • Final temperature, T2=310K
  • Initial rate constant is k1.
  • Final rate constant, k2=2k1
  • The activation energy is as follows:

logk2k1=Ea2.303RT2-T1T2T1Ea=logk2k1×2.303R×T2T1T2-T1=log2k1k1×2.303×8.314JK-1mol-1×310K×300K310K-300K=0.3010×19.15JK-1mol-1×9300K=53606.60Jmol-1=53.60kJmol-1

  • The activation energy of the reaction is 53.60kJmol-1.

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