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

A, C, D

Explanation

i. Standard Gibbs free energy change for the reaction,
`N_(2)O_(4)(g) 2NO_(2)(g)`
`DeltaG^(ɵ)=DeltaG_("Product")^(ɵ)-DeltaG_("Reactant")^(ɵ)=2xx50-100=0`
`DeltaG^(ɵ)=-2.303 RT log K_(p)=0`
`K_(p)=1`
Initially,
So, `=((P_(NO_(2)))^(2))/P_(N_(2)O_(4))=10`
Initial Gibbs free energy of the above reaction,
`DeltaG=DeltaG^(ɵ)+2.303 RT log Q_(P)`
`DeltaG=0+2.303xx8.314xx298 log 10`
`=5.705xx10^(3) kJ "mol"^(-1)`
ii. Since initial Gibbs free energy change of the reaction is potive, so the reverse reaction will take place.