For the reaction takes places at certain tempreature `NH_(4) HS(s) hArrNH_(3)(g) + H_(2)S(g)`,
if eqilibrium pressure is X bar, then `Delta_(r) G^(@)` would be

Options

A. `-2RT ln X`
B. `- RT ln (X- ln2)`
C. `-2RT (ln X- ln2)`
D. None of these

Answer

C

Explanation

`DeltaG^(@) = - 2.303 RT log Keq`
` = - RT In K_(P)`
`P_(NH^(3)) = P_(H_(2)S) = (x)/(2) "bar"`
`= - RT ln (("x")/(2))^(2) = - 2RT (ln X - ln 2)`