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For the reaction NH(4)HS(g) hArr NH(3)...

For the reaction
`NH_(4)HS(g) hArr NH_(3)(g)+H_(2)S(g)`
in a closed flask, the equilibrium pressure is `P` atm. The standard free energy of the reaction would be:

A

`-RT ln p`

B

`-RT (ln p-ln 2)`

C

`-2 RT ln p`

D

`-2 RT (ln p-ln 2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`NH_(4)HS(g) hArr NH_(3)(g)+H_(2)S(g)`
`:. K_(p)=p_(NH_(3)).p_(H_(2)S)=P/2.P/2=P^(2)/4`
`DeltaG=-RT ln K_(p)=-RT ln (P^(2)/4)`
`=-RT ln (P/2)^(2)=-2 RT (ln p-ln 2)`
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