InterviewSolution
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InterviewSolution
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`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. |
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Answer» 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. |
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