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For the reaction, 2NOCl(g) hArr 2NO(g)...

For the reaction,
`2NOCl(g) hArr 2NO(g)+Cl_(2)(g)`
Calculate the standard equilibrium constant at `298 K`. Given that the value of `DeltaH^(ɵ)` and `DeltaS^(ɵ)` of the reaction at `298 K` are `77.2 kJ mol^(-1)` and `122 J K^(-1) mol^(-1)`.

Text Solution

Verified by Experts

Using the relation
`DeltaG^(ɵ)=DeltaH^(ɵ)-TDeltaS^(ɵ)`
`=77200-298xx122`
`=40844 J "mol"^(-1)`
Let the equilibrium constant be `K_(c )^(ɵ)`. We know that
`DeltaG^(ɵ)=-2.303 RT log K_(c )^(ɵ)`
or `log K_(c )^(ɵ)=-(DeltaG^(ɵ))/(2.303xx8.314xx298)`
`=-40844/(2.303xx8.314xx298)=-7.158`
`K_(c )^(ɵ)=6.95xx10^(-8)`
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