NEET Chemistry Chemical Kinetics Questions Solved

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The rate of  a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R=8.314 J K-1 mol -1, log =0.6021]

We know, log K2K1=Ea2.303RT2-T1T1T2

      Given : T1 = 293 K,     T2 = 313 K,

      Let K1 = K(consider)  K2 = 4K,

          R = 8.314,     E= ?

As,    log4KK=Ea2.303×8.314313-293293×313 

  or  log 4 =Ea19.1471×2091709

  or  0.6020 =Ea19.1471×2091709

 Ea0.6020×19.1471×9170920

        = 52854.43 = 52.8 K J mol-1

Difficulty Level:

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