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).

The standard Gibbs energy change for the reaction N_(2)(g) + 3H_(2)(g)

1.	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).
Answer» Solution :(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)`.