The activation energy of `H_(2)+I_(2) hArr 2HI` in equilibrium for the forward reaction is `167 kJ mol^(-1)` whereas for the reverse reaction is `180 kJ mol^(-1)`. The presence if catalyst lowers the activation energy by `80 kJ mol^(-1)`. Assuming that the reaction are made at `27^(@)C` and the frequency factorr for the forward and backward reactions are `6xx10^(-4)` and `3xx10^(-3)`, respectively, calculate `K_(c )`.

Explanation

The lowering of activation energy by a catalyst occurs for forward reaction as well as for backward reaction.
Thus, in presence of catalyst,
Energy of activation for forward reaction `(DeltaH_(1))`
`=167-80=37 kJ "mole"^(-1)`
Energy of activation for backward reaction `(DeltaH_(2))`
`=180-80=100 kJ "mole"^(-1)`
For forward reaction, `K_(1)=A_(1) e^(-DeltaH_(1)//RT)`
For backward reaction `K_(2)=A_(2) e^(-DeltaH_(2)//RT)`
where `A_(1)` and `A_(2)` are frequency factors and `DeltaH_(1)` and `DeltaH_(2)` are energies of activation.
`:. K_(c )=K_(1)/K_(2)=A_(1)/A_(2). e^([(-DeltaH_(1)//RT)+(DeltaH_(2)//RT)])`
`=(6xx10^(-4))/(3xx10^(-3))xxe^([(-87+100)//(8.314xx10^(-3)xx300)])`
`K_(c )=2xx10^(-1) e^(+13//(8.314xx300xx10^(3)))=36.8`