At 320 K, a gas `A_(2)` is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in `J mol^(-1)` is approximately : `(R =8.314 Jk^(-1)"mol"^(-1), "In" 2=0.693, "In" 3 = 1.098)`

Options

A. 1844
B. 2069
C. 4281
D. 4763

Answer

C

Explanation

In the reactionn `A_(2)hArr2A`
Intitially Let `[A_(2)]=1M` and `[A]=0M`
After 20% dissociation, 80% of `A_(2)` remains.
`[A_(2)]=1xx80/100=0.8M`
20% of 1M is `1xx20/100=0.2[A]=2xx0.2=0.4M`
The equilirium constant
`K=([A]^(2))/([A_(2)]), K=([0.4]^(2))/([0.8])=0.2`
`DeltaG^(@)=-RT"In"K=-8.314JK^(-1)"mol"^(-1)xx320Kxx"In"0.2`
`=4281` J/mol