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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

`-RTlnp`

B

`-RT(lnp-ln2)`

C

`-2RTlnp`

D

`-2RT(lnp-ln2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`P_(NH_(3))=P_(H_(2)S)=(P)/(2)atm`
`K_(P)=P_(NH_(3))P_(H_(2)S)=((P)/(2))^(2)=(P^(2))/(4)`
`DeltaG=-RTlnK_(p)=-RTln((p)/(2))^(2)=2RT[lnp-ln2]`
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