Question

Consider the system:

A (aq) → B (aq)

at 279 K where ΔG^o_form A = -17.1 kJ/mol and ΔG^o_form B = -12.9 kJ/mol. Calculate the concentration of B at equilibrium when 2.19 mol of A and 1.47 mol of B are mixed in water to form exactly one liter of solution.

Answer 1

0.515M

Explanation

A(aq) <-------> B(aq)

∆G°rxn = ∆G°f(products) - ∆G°(reactants)

= -12.9kJ/mol - (-17.1kJ/mol)

= 4.2kJ/mol

∆G° = - RTlnK

- lnK = ∆G°/RT

= 4200(J/mol)/(8.314(J/K mol) × 279K)

= 1.81

- logK = 0.786

K = 0.164

Consider the reverse reaction

B(aq)< --------> A(aq)

K_-1 = [A]/[B] = 6.10

Initial [ A] = 2.19M

initial [ B]= 1.47M

Change in [ B] = -x

Change in [A] = +x

Equillibrium [A] = 2.19 + x

Equillibrium [ B] = 1.47 - x

2.19 + x = 8.967 - 6.10x

7.10x = 6.777

x = 0.955

Therefore,

[ A] = 2.19 + x = 2.19 + 0.955 = 3.145M

[ B] = 1.47 - x = 1.47 - 0.955 = 0.515M