Question

Thermodynamics

CALCULATIONS Question 1. Write the complete expression for the free energy change of the L = V equilibrium reaction for pure water (equation 2) at 100 bars and 300-C using the following data: Gy = -210330 joules/mole GL = -210330 joules/mole Recall that the right side (water vapor) of the reaction is the product side and left side (water liquid) of the reaction is the reactant side, Recall that we always calculate the AG, by subtracting the sum of the free energies of the reactant phases from the sum of the free energies of the product phases: AG, = 2G -G. Question 2. Next, calculate the free energy of an aqueous solution of 10 wt.% NaCl at these same P-T conditions (100 bars, 300°C) using equation (I). For a 10 wt. NaCl solution, the mole fraction of H0 in this solution will be 0.9466. Assume that this solution is an ideal solution, which means that mole fraction of H,O= activity of H.O, or X(H,0) = a(H,O): Question 3. Now calculate the free energy change (AG) of the L = V equilibrium reaction for the aqueous liquid at 100 bars and 300°C using your result for AG from question 2 above: Question 4. Is the AG, for the aqueous liquid-water vapor phase boundary still zero at 100 bars and 300°C? What is the sign of the AG, for this aqueous fluid-H,Oy boundary now?

example information to help solve each question

Answer 1

Solution:

1) For the conversion of water liquid (L) and water vapor (V).

H2O (L) = H2O (V)

Suppose, free energy for H2O(L) is GL and for H2O(V) is GV.

Then, free energy change ( ΔG) is calculated as,

ΔG = GV -  GL

ΔG = - 210330 J/mol - (-210330 J/mol)

ΔG = -210330 J/mol + 210330 J/mol

ΔG = 0

This, indicates both phases exist in equilibrium.