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Question

A reaction proceeds five times faster at 57 C as it does at 27 C. The energy of activation is:
(take ln3=1.1, ln5=1.6)

A
10.56 kcalmol1
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B
10.80 kcalmol1
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C
10.64 kcalmol1
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D
10.90 kcalmol1
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Solution

The correct option is A 10.56 kcalmol1
We know,
2.303logk2k1=EaRT2T1T1T2
Here, T2=57+273=330 K
T1=27+273=300 K
R=2×103 kcal. Ea=?
As r=k[A]n (at a temperature T)
Hence r2r1=k2k1 (at a temperature T2 andT1)
But r2r1=5 (Given)
k2k1=5
ln5=Ea2×103330300330×300
Ea=10.56 Kcal/mol
(Using ln 5 = 1.6)

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