`K_(c)` for the reaction `N_(2)O_(4) hArr 2NO_(2)` in chloroform at `291 K` is `1.14`. Calculate the free energy change of the reaction when the concentration of the two gases are `0.5` mol `dm^(-3)` each at the same temperature. `(R=0.082 L atm K^(-1) mol^(-1))`

Explanation

From the given data
`T=291 K, R=0.082 L "atm" K^(-1) "mol"^(-1)`
`K_(c )=1.14, C_(NO_(2))=C_(N_(2)O_(4))=0.5 "mol" dm^(-3)`
The reaction quotient `Q_(c )` for the reaction `N_(2)O_(4) hArr 2NO_(2)`,
`Q_(c )=([NO_(2)]^(2))/([N_(2)O_(4)])=(0.5xx0.5)/0.5=0.5`
Since `Q_(p)=Q_(c ) (RT)^(Deltan)` and `Deltan=2-1=1` in this case
`:. Q_(p)=0.5(0.082xx291)=11.93`
`K_(p)=K_(c )(RT)^(Deltan)=1.14 (0.082xx291)=27.1`
Substituting these values in the following equation, we get
`DeltaG=DeltaG^(ɵ)+RT ln Q_(p)`
`=-RT ln K_(p)+RT 1n Q_(p)`
`=-2.303 RT (log K_(p)-log Q_(p))`
`DeltaG=-(0.082xx291xx2.303)(log 27.2-log 11.93)`
`-54.95 (1.4346-1.0766)=-19.67 L "atm"`