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For a reaction A(g) hArr B(g) at equilib...

For a reaction `A(g) hArr B(g)` at equilibrium . The partial pressure of B is found to be one fourth of the partial pressure of A . The value of `DeltaG^(@)` of the reaction `A rarrB` is

A

RT ln `4`

B

`-"RT" ln 4`

C

`"RT" log 4`

D

`-"RT" log 4`

Text Solution

Verified by Experts

The correct Answer is:
A
`A(g) hArr B(g) " " P_(B) = (1)/(4)P_(A) " "P_(A)=4P_(B)`
`K_(P) = (P_(6))/(P_(A)) = (P_(A)//4)/(P_(A)) = (1)/(4)`
At equilibrium , `DeltaG=O`.
`DeltaG^(@) =-RT l nK_(P) =- RTInK_(P) =- RTIn (1)/(4) = RT In 4`
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