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^(ɵ)` of the reaction at `298 K`.
b. Find the direction of the reaction.

Explanation

`N_(2)O_(4) rarr 2NO_(2)`
`DeltaG^(@) = 2DeltaG_(f)^(@)(NO_(2))- DeltaG_(f)^(@)(N_(2)O_(4)) = 0`
Also `DeltaG^(@) = -RT ln K = 0, K = 1`
Let the reaction shifts in forward direction.
`N_(2)O_(4) rarr 2NO_(2)`
`implies K = (5+2x)^(2)/(10+x)^(2) xx 10-x/5-x xx 20 = 1 implies 81x^(2)+405x+450 = 0 implies x = -1.66 and -3.33`
Both values of x indicates that reaction actually proceeds in backward direction.