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^(ɵ)` of the reaction at `298 K`.
b. Find the direction of the reaction.

Options

A. `-4.68 KJ`
B. `-6.04 KJ`
C. `-5.705 KJ`
D. `0.4 KJ`

Answer

C

Explanation

Reaction quotient = `p[NO_(2)]^(2)/p[N_(2)O_(4)] = 100/10 = 10`
`DeltaG^(@) = 2DeltaG^(@)_(f)(NO_(2))-DeltaG^(@)_(f)(N_(2)O_(4)) = 2 xx 50 - 100 = 0`
`DeltaG = DeltaG^(@) - 2.303RT log_(10)Q_(p) = 0 - 2.303 xx 8.314 xx 298 log_(10)10 = -5705.8 J = -5.705 KJ`