In the reaction equilibrium
`N_(2)O_(4) hArr 2NO_(2)(g)`
When `5` mol of each is taken and the temperature is kept at `298 K`, the total pressure was found to be `20` bar.
Given : `Delta_(f)G_(n_(2)O_(4))^(ɵ)=100 kJ, Delta_(f)G_(NO_(2))^(ɵ)=50 KJ`
a. Find `DeltaG` of the reaction at `298 K`.
b. Find the direction of the reaction.

Answer

`5.0705 xx 10^(3) kJ mol^(-1)`; reverse direction

Explanation

(i) Standard Gibbs free energy change for the reaction `N_(2)O_(4) (g) hAtt 2NO_(2) (g)`
`Delta G^(@) = - 2.303 RT log K_(P) = 0, K_(P) = 1`
Initially, `P_(N_(2)O_(4)) = P_(NO_(2)) = 10` bar
Reaction quotient `= ((P_(NO_(2)))^(2))/(P_(N_(2)O_(2))) = (100)/(10) = 10`
`Delta G^(2) = 2 Delta G_(f(N_(2)O_(4)))^(@) = 100 - 100 = 0`
Intial Gibbs free energy of the above reaction,
`Delta G = Delta G^(@) + 2.303 RT log Q_(P)`
`Delta G = 0 + 2.303 xx 8.314 xx 2981 log 10`
`= 5.0705 xx 10^(3) kJ mol^(-1)`
(ii) Since initial Gibbs free energy change of the reaction is positive, so the reverse reaction will take place.