The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature.
`"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)`
Arrhenius gave an equation which describes aret constant `k` as a function of temperature
`k = Ae^(-E_(a)//RT)`
where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant.
Equation when expressed in logarithmic form becomes
`log k = log A - (E_(a))/(2.303 RT)`
Activation energies of two reaction are `E_(a)` and `E_(a)'` with `E_(a) gt E'_(a)`. If the temperature of the reacting systems is increased form `T_(1)` to `T_(2)` (`k'` is rate constant at higher temperature).

Options

A. `(k_(1)')/(k_(1)) = (k_(2)')/(k_(2))`
B. `(k_(1)')/(k_(1)) lt (k_(2)')/(k_(2))`
C. `(k_(1)')/(k_(1)) gt (k_(2)')/(k_(2))`
D. `(k_(1)')/(k_(1)) gt (2k_(2)')/(k_(2))`

Answer

C

Explanation

For first reaction:
`log k_(1) = log A-E_(a)//RT_(1)`
`log k_(2) = log A - E_(a)//RT_(2)`
`:. log.(k_(2))/(k_(1)) = (E_(a))/(R )((T_(2)-T_(1))/(T_(1)T_(2)))` …(i)
ismilarly, for second reaction,
`log.(k'_(2))/(k'_(1)) = (E'_(a))/(R )((T_(2)-T_(1))/(T_(1)T_(2)))` ....(ii)
form Eqs. (i) and (ii),
`k_(2)//k_(1) prop E_(a)` and `k'_(2)//k'_(1) prop E'_(a)`
Since `E_(a) gt E'_(a)`
`:. (k_(2))/(k_(1)) gt (k_(2)')/(k_(1)') rArr (k_(1)')/(k_(1)) gt (k_(2)')/(k_(1))`