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
10.56kcalmol−1
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B
10.80kcalmol−1
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C
10.64kcalmol−1
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D
10.90kcalmol−1
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Solution
The correct option is A10.56kcalmol−1 We know, 2.303logk2k1=EaR⌊T2−T1T1T2⌋ Here, T2=57+273=330K T1=27+273=300K R=2×10−3kcal.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=Ea2×10−3⌊330−300330×300⌋ ∴Ea=10.56Kcal/mol (Using ln 5 = 1.6)