The standard Gibbs energy change for the reaction
`N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g)`
is `-33.2 kJ "mol"^(-1)` at 298 K.
(a) Calculate the equilibrium constant for the above reaction.
(b) What would be the equilibrium constant if the reaction is written as
`1/2N_(2)(g) + 3/2H_(2)(g) iff NH_(3)(g)`
(c) What will be the equilibrium constant if the reaction is
`NH_(3)(g) iff 1/2N_(2)(g) + 3/2 H_(2)(g)`.

Explanation

(a) For the reaction :
`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
`DeltaG^(@) = -33.2 kJ "mol"^(-1)`
`log K = -(DeltaG^(@))/(2.303 RT)`
`R = 8.314 J K^(-1) "mol"^(-1),T = 298 K`
`therefore log K = -(-33.2xx 10^(3)J "mol"^(-1))/(2.3030 xx (8.314 J K^(-1) "mol"^(-1)) xx(298 K))`
` = 5.82`
or K = `6.6 xx 10^(5)`.
(b) For the reaction :
`1/2N_(2)(g) + 3/2H_(2)(g) to NH_(3)(g)`
`DeltaG^(@) = 1/2 xx (-33.2) = -16.6 kJ "mol"^(-1)`
or `logK = (-16.6 xx 10^(3)J mol^(-1))/(2.303 xx (8.314 J K^(-1) "mol"^(-1)) xx (298 K)) = 2.91`
or `K = 8.1 xx 10^(2)`.
(c) For the reaction :
`NH_(3)(g) to 1/2N_(2)(g) + 2/3 H_(2)(g)`
`DeltaG^(@) = -[-16.6 kJ "mol"^(-1)] = 16.6 kJ mol^(-1)`
`logK = -(16.6 xx 10^(3)J "mol"^(-1))/(2.303 xx (8.314 J K^(-1) "mol"^(-1))xx (298 K)) = -2.91`
or `K = 1.23 xx 10^(-3)`.