Question

Ba(OH)2.8H2O(s) + 2NH4Cl(s) -----> BaCl2. 2H2O (s) + 2NH3(g) + 8H2O(l)

Substance	DHfo (kJ/mol)	DGfo (kJ/mol)
Ba(OH)2.8H2O(s)	-3342	-2793
NH4Cl(s)	-314.4	-203.0
BaCl2. 2H2O (s)	-1460.1	-1296.5
H2O(l)	-285.83	-237.2
NH3(g)	-80.29	-26.6

I have the answers, but I dont understand how to get to it.

'change'Grxn= -48.3 kJ, 'change'Hrxn= 63.5 kJ, K=2.93 x 108

Answer 1

To calculate del G_rxn, we have to take the sum of change in G_f of all products and the sum of delta G_f of all reactants, and then take their difference. That gives delta G_reaction.

              = [-1296.5 + 2(-26.6) + 8(237.2)] - [ -2793 + 2(-203)] kJ

             = -48.3 kJ

similarly,

              = [ -1460.1+2(-80.29) + 8(-285.83)] - [ -3342.0 +2(-314.4)]

            = - 63.58 kJ

equilibrium constant , K (assuming T = 298 K):

-48.3 x 10³ J = -(8.314 J/K.mol)(298 K)lnK

lnK = 19.49

Taking antilog of 19.49, we get

K = 2.91 x 10⁸   (you can do K = e^19.49 to take antilog)