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Question

A catalyst lowers the activation energy for a certain reaction from 75 kJ to 25 kJ mol1. The effect on the rate of reaction at 25C, other things being equal is :
  1. rate of reaction decreases 5.81×108 times
  2. rate of reaction decreases 6.34×108 times
  3. rate of reaction increases 6.34×108 times
  4. rate of reaction increases 5.81×108 times

A
rate of reaction increases 5.81×108 times
B
rate of reaction increases 6.34×108 times
C
rate of reaction decreases 5.81×108 times
D
rate of reaction decreases 6.34×108 times
Solution
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The relationship between the rate of the reaction and the activation energy is logk=Ea2.303RT+logA

This equation in presence and in absence of catalyst can be written as log(k2k1)=12.303RT(Ea2Ea1)

Here, Ea2 and Ea1 represents the activation energies in presence and in absence of catalyst and k2 and k1 represents the rate constants in presence and in absence of catalyst.

Hence, log(k2k1)=12.303×8.314×298(2500075000)=8.763

(k2k1)=5.81×108

Hence, the rate of the reaction increase 5.81×108 times.

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