For the equilibrium, `N_(2)O_(4)hArr2NO_(2)` , `(G_(N_(2)O_(4))^(@))_(298)=100kJ//mol` and `(G_(NO_(2))^(@))_(298)=50kJ//mol`.
(`a`) When `5 mol/litre` of each is taken, calculate the value of `DeltaG` for the reaction at `298 K`.
(`b`) Find the direction of reaction.

Explanation

`{:(,N_(2)O_(4),hArr,2NO_(2)),("Concentration at t=0",5,,5):}`
`[G_(N_(2)O_(4)^(@)]_(298 K)=100 kJ mol^(-1)`
`[G_(NO_(2)^(@)]_(298 K)=50 kJ mol^(-1)`
`DeltaG^(@)` for reaction `=2xxG_(NO_(2)^(@)-G_(N_(2)O_(4))^(@)`
`=2xx50-100=0`
Now `DeltaG=DeltaG^(@)+2.303RT logQ`
`DeltaG=0+2.303xx8.314xx10^(-3)xx298 log"(5^(2))/(5)`
`DeltaG=+3.99 kJ`
Since `DeltaG=+ve` and thus reaction will not proceed in forward direction. It will take place in backward direction.