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Question

The rate of a reaction A doubles on increasing the temperature from 300 to 310K. By how much, the temperature of reaction B should be increased from 300K so that rate doubles if activation energy of the reaction B is twice to that of reaction A?

A
4.92K
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B
19.67K
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C
2.45K
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D
9.84K
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Solution

The correct option is A 4.92K
For the reaction A
logkk=E2.303R[TTTT]
k=2k
T=300KT=310K
Substitute values in the above equation.
log2kk=E2.303R[310300300×310]
log2=E2.303R[10300×310] ......(1)
For the reaction B
logkk=E2.303R[TTTT]
k=2k
Ea(B)=2Ea(A)
T=300KT=?
Substitute values in the above equation.
log2kk=2E2.303R[T300300×T]
log2=2E2.303R[T300300×T] ......(2)
Divide equation (2) with (1)
log2log2=2E2.303RE2.303R×[T300300×T][10300×310]
1=2×[T300T][131]
162=[T300T]
T300=0.01613T
0.9838T=300
T=304.92K
Hence, the temperature of reaction B should be increased from 300K by 304.92300=4.92K.

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