Question

`K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` at `400^(@)C` is `1.64xx10^(-4)`. Find `K_(c)`. Also find `DeltaG^(ɵ)` using `K_(p)` and `K_(c)` values and interprest the difference.

Answer 1

Correct Answer - A::C::D
`N_(2)+3H_(2) hArr 2NH_(3)`
`Deltan=Sigman` products - `Sigman` reactions `=2-4=-2`
`K_(p)=K_(c)(RT)^(Deltan)`
`1.64xx10^(-4)=K_(c)(0.0821xx673)^(-2)`
`:. K_(c)=0.5006`
Now since `DeltaG^(ɵ)=-2.303 RT log K`
If `K=K_(p)`
then `DeltaG^(ɵ)=-2.303xx2xx673 log (1.64xx10^(-4))`
`=+11.733 kcal`
If `K=K_(c)`
then `DeltaG^(ɵ)=-2.303xx2xx673 log (0.5006)=+931 cal`
where `DeltaG^(ɵ)` is the free energy change when all the reactants and products are in the standard state. For `DeltaG^(ɵ)=-2.303 RT log K_(p)`, standard state means the partial pressure of each as `1` atm.
When `DeltaG^(ɵ)=-2.303 RT log K_(C)` standard state means that concentration of each is `1` mole `L^(-1)`. So the value are different.