The initial reaction free energy is given by the expression derived above for reaction free energy:

To get the standard reaction free energy, recall that when we set the reaction free energy equal to zero above we found

Since we know the equilibrium constant we can now find the standard reaction free energy which is -5705 J/mole. Plugging this value of the standard reaction free energy into the first equation for the reaction free energy and plugging in the values for the molalities gives -15.0 kJ/mole.

Now to calculate the value at equilibrium, we need to use the equilibrium relationship

and some stochiometry and mass action equations. First stochiometry tells us that if a is the molality of A that is converted in the reaction as it goes to equilibrium (the change in the # moles of A per kg solvent required to meet equilibrium) then the following will be true:

Remember that we know the initial values of each so this means that we can put the final values of each in terms of a.

Now we can plug in the numbers for KEQ and the initial concentrations:

Now just solve for a using the quadratic equations and then back calculate the final values of each component. Remember that for the equation:

The solution for x is just

In our case we want to solve for the variable a and can see that

Thus, there are two solutions. However, one of them is not right. The sum gives an a of 4.11 and if we look back up at the expressions for the final values of the molality of A and B, we can see that if we subtract 4.11 from the initial molality of A (0.09) we obtain a negative molality. This is physically impossible. Thus the second solution, 0.69, must be correct. This gives the following equilibrium values of the molalities of A, B and C: