Question
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 ___
Answer: Option B
:
B
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
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:
B
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
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