The activation energy for a reaction at the temperature TK was found to be 2.303 RTJmol−1. The ratio of the rate constant to Arrhenius factor is ___
A
10−2
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B
10−1
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C
2×10−2
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D
2×10−3
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Solution
The correct option is B10−1 k=Ae−EaRT Ea=2.303RT ∴kA=e−(2.303RTRT);kA=e−2.303 loge(kA)=logee−2.303 loge(kA)=−2.303 or 2.303log10(kA)=−2.303 log(kA)=−1;log(AK)=1 ∴Ak=antilog1=10 or kA=110=10−1