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

The reaction is:
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
Since, number of moles of both `N_(2)O_(4)` and `NO_(2)` are same hence their partial pressure will also be same.
`P_(N_(2)O_(4))=P_(NO_(2))=(20)/(2)=10` bar
`Q_(P)=([P_(NO_(2))]^(2))/([P_(N_(2)O_(4))])=(10^(2))/(10)=10`bar
`DeltaG_("reaction")^(@)=2DeltaG_(f)^(@)NO_(2)-DeltaG_(f)^(@)N_(2)O_(4)`
`=2xx50-100=0`
We know that, `DeltaG=DeltaG^(@)-2.303RT" "log" "Q`
`=0-2.303xx8.314xx298" log "10`
`=5705J`
Since, `DeltaG` is negative hence reaction will be spontaneous in forward direction.