Two reactions `R_(2) and R_(2)` have identical pre - exponential factors. Activations enery of `R_(1)` exceeds that of `R_(2)` by 10 kJ `mol_(-1)` . If `k_(1) and k_(2)` are rate constants for rate constants for reactions `R_(1) and R_(2)` respectively at 300k , then In `(k_(2)/k_(1))`is equal to `(R=8.314 J mol^(-1)K^(-1))`A. 8B. 12C. 6D. 4

Two reactions `R_(2) and R_(2)` have identical pre - exponential facto

1.	Two reactions `R_(2) and R_(2)` have identical pre - exponential factors. Activations enery of `R_(1)` exceeds that of `R_(2)` by 10 kJ `mol_(-1)` . If `k_(1) and k_(2)` are rate constants for rate constants for reactions `R_(1) and R_(2)` respectively at 300k , then In `(k_(2)/k_(1))`is equal to `(R=8.314 J mol^(-1)K^(-1))`A. 8B. 12C. 6D. 4
Answer» Correct Answer - D We know `k_(2)=Ae^(-E_(a)//RT)` `(k_(2))/(k_(1))=e^((E_(a)^(1)-E_(b)^(2))//RT)` `ln.(k_(2))/(k_(1))=(E_(a)^(1)-E_(b)^(2))/(RT)` `=(10xx1000)/(8.314xx300)` `~~4`

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