Reactants A and B are combined to form product C as shown by the equation below:

Question

Reactants A and B are combined to form product C as shown by the equation below:

2A(g) + B(g) ⟷3C(g)

0.258 mol of A is combined with 0.163 mol of B in a 2.14 L rigid container at standard conditions. When the reaction reaches equilibrium, it is found that 0.067 mol of C are present in the container. What is the standard free energy of this reaction (in kJ/mol)?

Answer 1

Standard free energy = ∆Gº = -RT ln K

Since we know R (gas constant) and T(298K Standard temp), we need to first find K before solving for ∆Gº

2A(g) + B(g) <==> 3C(g)

0.258......0.163.......0........Initial

-2x..........-x............+3x.....Change

0.258-2x..0.163-x...3x......Equilibrium

We know that @ equilibrium 3x = 0.067 moles. This is in 2.14 L so that translates to 0.0313 M

Final concentrations @ equilibrium are as follows:

[C] = 0.0313 M

[A] = (0.258 - 2*0.067) / 2.14 = 0.124 / 2.14 = 0.0579 M

[B] = (0.163 - 0.067) / 2.14 = 0.0444 M

K = [C] / [A]²[B] = (0.0313) / (0.0579)²(0.0444)

K = 210

(Be sure to check the math)

Solving for ∆Gº, we have...

∆Gº = - RT ln K = -(8.314 J/Kmol)(298K) ln 210 = -2478 J/mol x 5.35

∆Gº = -13,250 J/mol = -13.3 kJ/mol