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Q.

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be (R=8.314 JK−1mol−1  and log2=0.301)

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a

48.6 kJ mol−1

b

58.5 kJ mol−1

c

60.5 kJ mol−1

d

53.6 kJ mol−1

answer is D.

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Detailed Solution

As per Arrhenius equation
In  K2K1=EaR(1T2−1T1)=2.303 log 2 

= −Ea8.314(1310−1300) 

 = Ea= 53.6 kJ/mol
 

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