Q.
The rate of a reaction doubles, when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be R=8.314JK−1mol−1and log 2=0.301)
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a
53.6 kJ mol−1
b
48.6 kJ mol−1
c
58.5 kJ mol−1
d
60.5 kJ mol−1
answer is A.
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Detailed Solution
From Arrhenius equation, log k2k1=−Ea2.303R1T2−1T1Given, k2k1=2,T2=310 K and T1=300 KOn putting values, ⇒ log 2=−Ea2.303×8.3141310−1300 ⇒ Ea=53598.6J/mol≈53.6kJ/mol
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