The activation energies of two reaction are `E_1 and E_2`with E_1 greater than E_2. IF the temperature of the system is increased from `T_1` to `T_2` the rate constant of the reactions changes from `k_1` to `k_1^'` in the first reaction and `k_2` to `k_2^'` in the second reaction. Predict which of the following expression is correct?

Options

A. `(k_1^')/k_1=(k_2^')/k_2`
B. `(k_1^')/k_1gt(k_2^')/k_2`
C. `(k_1^')/k_1lt(k_2^')/k_2`
D. `(k_1^')/k_1=(k_2^')/k_2=`

Answer

B

Explanation

For first reaction,
`E_1=(2.303 RT_1T_2)/((T_1-T_2)) log""(k_1.)/(k_1)`.....(i)
For second reaction,
`E_2= (2.303 RT_1T_2)/((T_1-T_2))log""(k_2.)/k_2.....(ii)`
Given `E_1 gt E_2`
`implies (2.303 RT_1T_2)/((T_1-T_2))log""(k_1.)/k_1gt (2.303 RT_1T_2)/((T_1-T_2)) log""(k_2.)/k_2`
`therefore (k_1.)/k_1 gt (k_2.)/k_2`