Thermodynamics. IV. Free Energy and Equilibrium.

Questions for Quiz #7.

1a. What is meant by the term "Free Energy"? What is the relationship between DS(univ) and DG.

Free energy is a measure of the maximum work that a system can do: DG = -wmax

The relationship is: DS(universe) = - DG / T.

1b. In what sense is Free Energy "free"?

It is the work that a spontaneous process can produce.

1c. What is meant by the term maximum work, or wmax?

The maximum work is the work that a process can produce whilst staying at equilibrium.

1d. What is the difference between Gibbs free energy, G, and Helmoltz free energy, A?

G is the constan pressure term, A is the constant volume change.

2a. Free energy is tabulated as Df, in the same way as enthalpy is tabulated, but not the same way as entropy is tabulated. Why?

Free energy is an energy change, and there is no known zero of energy, so energy differences must be tabulated.

2b. What is Df for O2(g), Fe(s), any element?

For any element, Df = 0 kJ/mol by definition.

3. From tables of Df, calculate the spontaneity, or not, of the following processes under standard conditions:

3a. CaC2(s) + 2 H2O(l) Ž Ca(OH)2(s) + HCÀCH

Although not very useful except at 100 kPa and 298 K, D298 = SmiDf(i).

So: D298 = (-1)(-64.8) + (-2)(-285.8) + (+1)(-899.2) + (+1)(209) = -53.8 kJ.

A negative value indicates the process is spontaneous.

3b. SbCl3(s) + Cl2(g) Ž SbCl5(l)

As above: D298 = (-1)(-323.9) + (-1)(0) + (+1)(-350.5) = -26.6 kJ

Another spontaneous process.

3c. BaCO3(s) Ž BaO(s) + CO2(g)

As above: D298 = (-1)(-1139) + (+1)(-520.4) + (+1)(-394.4) = 224.2 kJ.

A positive value indicate a non-spontaneous process.

3d. Use the reaction :

SO2(g) + Cl2(g) Ž SO2Cl2(l)

to find the absolute entropy, S298, of SO2Cl2(l).

Df for SO2Cl2(l) is -389 kJ/mol; Df for SO2Cl2(l) is -314 kJ/mol.

Data for SO2 (g) and Cl2(g) are in your text.

We know: D298 = (-1)(-300.2) + (-1)(0) + (+1)(-314) = -13.8 kJ.

and DHº = (-1)(-296.8) + (-1)(0) + (+1)(-389) = -92.2 kJ.

and DSº = (-1)(248.1) + (-1)(223.0) + (+1)(X) where X is the required entropy.

Also D298 = DHº - 298*DSº

so DSº = (-92200J/298K) - (-13800J/298K) = -263.1 J/K

From which X = 208.0 J/mol.K.

4. DHº and DSº are almost independent of temperature, show that this is NOT the case for D

The relationship between these three terms is: DGº = DHº - TDSº. The dependence of DG on T is shown in this relationship.

5. Given that DHº and DSº are almost temperature independent (so we will treat them as if they are totally temperature independent), how can DT (where T is ANY temperature) be calculated.

By using the relationship : DGº = DHº - TDSº.

6. Repeat the questions in 3a, b, c (above) but find the spontaneity, or not, at a) 500 K, and b) 1000 K (assume that all phases given remain the same - which is not true, but, oh well, this is only a theoretical exercise anyway!).

a. CaC2(s) + 2 H2O(l) Ž Ca(OH)2(s) + HCÀCH

First: DHº = (-1)(-59.9) + (-2)(-285.8) + (+1)(-986.9) + (+1)(227) = -128.4 kJ.

Next: DSº = (-1)(70.0) + (-2)(69.94) + (+1)(83.4) + (+1)(200.9) = 74.42 J/K.

At 500 K D500 = (-128.4 kJ) - (500 K)(.07442 kJ/K) = -165.6 kJ (spontaneous at 500 K)

At 1000K D1000 = (-128.4 kJ) - (1000 K)(.07442 kJ/K) = -202.8 kJ (spontaneous at 1000 K)

b. SbCl3(s) + Cl2(g) Ž SbCl5(l)

As above: DHº = (-1)(-382.4) + (-1)(0) + (+1)(-440.5) = -58.1 kJ

Then: DSº = (-1)(184.2) + (-1)(223.0) + (+1)(301.5) = -105.7 J/K.

At 500 K D500 = (-58.1 kJ) - (500 K)(-0.1057 kJ/K) = -5.25 kJ (spontaneous at 500 K)

At 1000 K D500 = (-58.1 kJ) - (1000 K)(-0.1057 kJ/K) =47.6 kJ (non-spontaneous at 1000 K)

c. BaCO3(s) Ž BaO(s) + CO2(g)

As above: DHº = (-1)(-1219) + (+1)(-548.1) + (+1)(-393.5) = 277.4 kJ

Then: DSº = (-1)(112) + (+1)(72.1) + (+1)(213.7) = 173.8 J/K

At 500 K D500 = (277.4 kJ) - (500 K)(0.1738 kJ/K) = 190.5 kJ (non-spontaneous at 500 K)

At 1000 K D500 = (277.4 kJ) - (1000 K)(0.1738 kJ/K) = 103.6 kJ (non-spontaneous at 1000 K)

7a. What is the value of DT when the system is at equilibrium under standard conditions?

Under standard conditions, DT = 0 at equilibrium.

7b. Calculate the temperature at which the following systems will be at equilibrium under standard conditions:

i) 2 ClF3(g) Ž Cl2(g) + 3 F2(g)

At equilibrium DT = 0 = DHº - TDSº.

So T = DHº / DSº.

From tables: DHº = (-2)(-159.1) + (+1)(0) + (+3)(0) = 318.2 kJ.

And: DSº = (-2)(281.7) + (+1)(223.0) + (+3)(202.7) = 267.7 J/K.

So equilibrium under standard conditions occurs when T = 318200J / 267.7 J/K = 1189 K.

ii) MgCO3(s) Ž MgO(s) + CO2(g)

As above:

From tables: DHº = (-1)(-1112) + (+1)(-601.2) + (+1)(-393.5) = 117.3 kJ

and: DHº = (-1)(65.86) + (+1)(26.9) + (+1)(213.7) = 174.7 J/K.

So equilibrium under standard conditions occurs when T = 117300J / 174.7 J/K = 671 K.

8a. How does DGT vary with pressure (or concentrations)?

The relationship is DGT = DT + RT ln Q.

8b. Define Q, the reaction quotient for a process. How is it related to K, the equilibrium constant for the same process?

Q = P aim(i).

Where a is the activity of component i in a reaction.

8c. What is the relationship between free energy and K, the equilibrium constant? What are the units for K obtained from this relationship.

The relationship is: DT = -RT ln K.

K must be in terms of activities, and so has no units.

8d. Define the term "activity" for a solid, a liquid, an ideal solution, and an ideal gas.

Ideal gas: a = P/Pº.

Pure liquid: a = 1.

Pure solid: a = 1.

Solvent of ideal solution: a = 1.

Solute of ideal solution: a = [A] mol.L-1 / 1 mol.L-1.

8e. What is the relationship between K, Kp and Kc for a reaction?

K is the activities equilibrium constant.

Kp is the pressure equilibrium constant for a gas reaction (no solution). If each pressure is divided by Pº, Kp is converted to K.

Kc is the concentration constant for a reaction in concentration measurements.

If Kc is for gases, then it must be converted to Kp to relate to K.

If Kc is for solutions, the each concentration is divided by 1 M to convert to activities and to give K.

8f. Calculate K, the equilibrium constant for the following processes at 298 K (standard temperature).

i) PCl3(g) + Cl2(g) ¾ PCl5(g)

K may be calculated: D298 = -RT ln K.

and D298 can be calculated from Df values:

So: D298 = (-1)(-258) + (-1)(0) + (+1)(-313) = -55 kJ.

Then ln K = -(-55000J) /( 8.314 J/mol.K * 298 K) = 22.2

and K = 4.38 x 109.

ii) H2O(l) ¾ H2O(g)

As above:

then D298 = (-1)(-237.2) + (+1)(-228.6) = 8.6 kJ.

So ln K = -(8600 J)/(8.314 * 298) = -3.47.

and K = 0.0311.

8g. In this latter case, calculate the vapour pressure of water at 298 K from this equilibrium constant.

By definition: K = a(H2O(g)) / a(H2O(l)) = P(H2O)/Pº / 1

So 0.0311 = P(H2O)/100 kPa

and P(H2O) = 3.11 kPa. (Experimental value: 3.17 kPa)

9. For the reaction

PCl5(g) Ž PCl3(g) + Cl2(g) at 25ºC

DHº = + 92.5 kJ and DSº = + 182 J/K

a) Is the reaction spontaneous under standard conditions at 25ºC?

D298 = 92.5 kJ - 298 K * 0.182 kJ/K = 38.3 kJ.

A positive value indicates a non-spontaneous process.

b) Assuming that the values of DHº and DSº do not change with a change in temperature

i) Is the reaction spontaneous at 300ºC?

D300 = 92.5 kJ - 573 K * 0.182 kJ/K = -11.79 kJ.

A negative value indicates that the process is spontaneous at this temperature.

ii) At what temperature (if one exists) does equilibrium occur with all three gases at standard pressure?

For this, DT = 0, so T = DHº/DSº = 92500J/182J.K-1. = 508 K.

10. Show that DGT = RTln(Q/K)

We have: DGT = DT + RT ln Q

At equilibrium, DGT = 0, and DGºT = -RT ln K.

So DGT = -RT ln K + RT ln Q = RT ln (Q/K).

11a. Calculate DG600 for the following process:

3 H2(g) + N2(g) Ž 2 NH3(g) ***NOTE BALANCE***

if P(H2) = 25 kPa, P(N2) = 10 kPa, and P(NH3) = 200 kPa.

We have: DG600 = D600 + R*600*ln Q.

and D600 = DHº - TD

So: DHº = 2*(-45.9) = -91.8 kJ.

and DSº = (-3)(130.6) + (-1)(153.2) + (+2)(193) = -159 J/K.

Then: D600 = (-91.8 kJ) - 600K*(-0.159 kJ/K) = 3.6 kJ.

So DG600 = 3.6 kJ + R*600*ln Q.

Q = a(NH3)2/( a(H2)3.a(N2)) = (2.00)2/((0.25)3.(0.10)) = 2560

and DG600 = 3.6 kJ + [(8.314*600*7.85)/1000] kJ = 42.7 kJ

11b. Calculate K600 for the ammonia synthesis reaction.

At 600 K, ln K600 = -D600 / RT = 3600 J / (8.314*600) = 0.722

Thus K600 = 2.06.