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.

Explanation

We need to calculate `Delta_(r)G`.
Use: `Delta_(r)G = Delta_(r)G^(Theta) +RT InQ`
So first calculated `Delta_(r)G^(Theta)` using:
`Delta_(r)G^(Theta) = sum G^(Theta) ("Products") - sum G^(Theta) ("Reactants")`
`= sum (Delta_(f)G^(Theta)) ("Product") -sum (Delta_(f)G^(Theta)) ("Reactants")`
` [ :'` At standard conditions: `G^(Theta)underset("combound") =Delta_(f)G^(Theta)underset("compound").]`
`= 2 xx Delta_(r)G^(Theta) (NO_(2),g) -Delta_(r)G^(Theta)(N_(2)O_(4),g)`
`=2 xx 50 - 100 = 0`
and `Delta_(r)G = Delta_(r)G^(Theta) +RT In Q`
`= 0 +8.314 xx 298 In ((10^(2))/(10)) = 5.70 kJ mol^(-1)`
`[ :' Q=(p_(NO_(2))^(2))/(p_(N_(2)O_(4))^(2))=(((5)/(10)xx20)^(2))/(((5)/(10)xx20))]`