At `27^(@)C,N_(2)O_(4)` has vapour density`=(230)/(6)`. If the equilibrium pressure is `0.96 atm`, then find `DeltaG^(@)`. ( given `log2 =0.3)`

Options

A. `1.1Kcal`
B. `193Kcal`
C. `79.3Kcal`
D. `8.041Kcal//mol`

Answer

1

Explanation

`(D)/(d)=1+(n-1)alpha`
`(92//2)/(230//6)=1+(2-1)alpha`
`alpha=0.2`
`K_(p)=(4alpha^(2))/(1-alpha^(2))P=(4(0.2)^(2))/(1-(0.2)^(2))=0.16`
Now`" "DeltaG^(@)=-2.303RTlogK_(p)`
`=-2.303xx8.314xx300log0.16`
`=+5744.1426xx0.80`
`=4595.31J=4.595kJ=1.1kcal`