The activation energy of
`H_(2)+I_(2) hArr 2HI(g)` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence of catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reactions are made at `27^(@)C` and the frequency factor for forwatd and backward reactions are `4xx10^(-4)` and `2xx10^(-3)` respectively, calculate `K_(c)`.

Explanation

A catalyst lowers the activation energy for forward reaction as well as for backward reaction by equal amount.
Thus, in presence of catalyst,
Energy of activation for forward reaction
`(E_(a1))=167-80=87 kJ "mol"^(-1)`
Energy of activation for backward reaction
`(E_(a2))=180-80=100 kJ "mol"^(-1)`
`:.` For forward reaction, `K_(1)=A_(1)e^(-Ea//RT)`
For backward reaction, `K_(2)=A_(2)e^(-Ea//RT)`
where `A_(1)` and `A_(2)` are frequency factors and `E_(a1)` and `E_(a2)` are energies of activation after addition of catalyst.
`:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2) e^([(-E_(a1)//RT)+(-E_(a2)//RT)])`
`=A_(1)/A_(2)e^(([E_(a2)-E_(a1)])/(RT))=(4xx10^(-4))/(2xx10^(-3)) e^(((100-87))/((8.314xx10^(-3)xx300)))`
`:. K_(c )=2xx10^(-1) e^([13//8.314xx10^(-3)xx300])`
`K_(c )=36.8`