Activation energy `(E_(a))` and rate constants (`k_(1)` and `k_(2)`) of a chemical reaction at two different temperatures (`T_(1)` and `T_(2)`) are related by

Options

A. `In (k_(2))/(k_(1)) = (E_(a))/(R ) ((1)/(T_(1)) - (1)/(T_(2)))`
B. `In (k_(2))/(k_(1)) = -(E_(a))/(R ) ((1)/(T_(1)) - (1)/(T_(2)))`
C. `In (k_(2))/(k_(1))= - (E_(a))/(2.303R) ((1)/(T_(2)) - (1)/(T_(1)))`
D. `In (k_(2))/(k_(1))= -(E_(a))/(R ) ((1)/(T_(2)) + (1)/(T_(1)))`

Answer

A

Explanation

Rate constants for most chemical reactions closely follow Arrhenius equation:
`k=Ae^(-E_(a)//RT)`
Taking the natural logarithm of both sides of the Arrhenius equation gives
`Ink=InA-E_(a)//RT`
Now we write equation for two different absolute temperatures `T_(1)` and `T_(2)` . We write `K_(1)` for the rate constant at temperature `T_(1)` and `K_(2)` for the constant at temperature `T_(2)` .
`InK_(2)=InA-(E_(a))/(RT_(2))`
`InK_(2)=InA-(E_(a))/(RT_(1))`
Here we assume that the activation energy `E_(a)` and the frequency factor `A` do not vary with temperature.
We eliminate In `A` by subtracting the equations.
`Ink_(2)-InK_(1)=-(E_(a))/(RT_(2))+(E_(a))/(RT_(11))`
or `In(k_(2))/(k_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`