`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 interpret the difference.

Explanation

`N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
`Delta n= 2-4 = -2`
`K_(p) = K_(c) (RT)^(Delta n)`
`1*64 xx 10^(-4) " atm"^(-2) = K_(c) (0*0821) L " atm "K^(-1) mol^(-1) xx 673 K ) ^ (-2) " "(T=400 + 273 K = 673 K)`
or `K_(c) = (1*64 xx 10^(-4) "atm"^(-2))/((0*821 xx 673 " atm"^(-1) mol^(-1))^(2))=0*5372 " mol"^(2)L^(-2)`
Now, `Delta G^(@) = - 2*303" RT " log K`
If `K=K_(p)`
` Delta G^(@) = - 2*303 xx (8*314 " JK"^(-1) mol^(-1)) (673 K) xx log (1*64 xx 10^(-4))`
`=-2*303 xx 8*314 xx 673 xx (-3* 7852) " J "mol^(-1) = +48*78 " kj "mol^(-1)`
If `K=K_(c)` ,
`Delta G^(@)= - 2*303 xx (8*314 xx 673 xx (-0*27) " J "mol^(-1) = + 3479 " J "mol^(-1)`
Interpreting the difference .`DeltaG^(@)` is the free energy change when all the reactants and products are in their standard state. In case of `K_(p)`, standard state pressure is used which is 1 atm (or now 1 bar) wheras in case of `K_(c)` , standard concentration is used which is 1 mol `L^(-1)` (as already explained on page 7/22).