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Question

Two reactions A and B of the same order have equal pre-exponential factors but activation energy for A is greater than that of B by 24.9 kJ/mol. Calculate the ratio(kBkA) at 27C.
Here, kA and kB are rate constants for reactions A
and B, respectively.
(R=8.31 JK1mol1)

A
2.199×104
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B
2.199×104
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C
4.199×104
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D
4.199×104
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Solution

The correct option is A 2.199×104
We know that,
log10k=log10AE2.303RT
So, log10kB=log10AEB2.303RT
and log10kA=log10AEA2.303RT
log10(kBkA)=(EAEB)2.303RT
log10(kBkA)=24.9×10002.303×8.314×300
kBkA=2.199×104
Hence, (a) is correct.

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