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Following reaction in equilibrium at 25^...

Following reaction in equilibrium at `25^(@)C`:
`2NO(g,1 xx 10^(-5) "atm")+Cl_2(g,1 xx 10^(-2)"atm") harr 2NOCl(g, 1 xx 10^(-2) "atm") DeltaG^@` is

A

`-45.65 KJ`

B

`-28.53 KJ`

C

`-22.82 KJ`

D

`-57.06 KJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(P)=(P_(NoCl)^(2))/(P_(NO)^(2) xx P_(Cl_(2)))=((10^(-2))^(2))/((10^(-5))^(2) xx 10^(-2))=10^(8)`
`DeltaG^(@)= -RT` 1 n k = -45.65 kJ
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