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.

In the reaction equilibrium `N_(2)O_(4) hArr 2NO_(2)(g)` When `5` mol

1.	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» 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.

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