**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 = -w_{max}

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 w_{max}?

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 DGº_{f}, 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 DGº_{f} for O_{2}(g), Fe(s), any element?

For any element, DGº_{f} = 0 kJ/mol by definition.

3. From tables of DGº_{f}, calculate the spontaneity, or not, of the following processes under standard conditions:

3a. CaC_{2}(s) + 2 H_{2}O(l) Ž Ca(OH)_{2}(s) + HCÀCH

Although not very useful except at 100 kPa and 298 K, DGº_{298} = Sm_{i}DGº_{f}(i).

So: DGº_{298} = (-1)(-64.8) + (-2)(-285.8) + (+1)(-899.2) + (+1)(209) = -53.8 kJ.

A negative value indicates the process is spontaneous.

3b. SbCl_{3}(s) + Cl_{2}(g) Ž SbCl_{5}(l)

As above: DGº_{298} = (-1)(-323.9) + (-1)(0) + (+1)(-350.5) = -26.6 kJ

Another spontaneous process.

3c. BaCO_{3}(s) Ž BaO(s) + CO_{2}(g)

As above: DGº_{298} = (-1)(-1139) + (+1)(-520.4) + (+1)(-394.4) = 224.2 kJ.

A positive value indicate a non-spontaneous process.

3d. Use the reaction :

SO_{2}(g) + Cl_{2}(g) Ž SO_{2}Cl_{2}(l)

to find the absolute entropy, S_{298}, of SO_{2}Cl_{2}(l).

DHº_{f} for SO_{2}Cl_{2}(l) is -389 kJ/mol; DGº_{f} for SO_{2}Cl_{2}(l) is -314 kJ/mol.

Data for SO_{2} (g) and Cl_{2}(g) are in your text.

We know: DGº_{298} = (-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 DGº_{298} = DHº - 298*DSº

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

From which X = 208.0 J/mol.K.

Answer: Sº_{298}(SO_{2}Cl_{2}) = 208 J/mol.K.

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

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 DGº_{T} (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. CaC_{2}(s) + 2 H_{2}O(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 DGº_{500} = (-128.4 kJ) - (500 K)(.07442 kJ/K) = -165.6 kJ (spontaneous at 500 K)

At 1000K DGº_{1000} = (-128.4 kJ) - (1000 K)(.07442 kJ/K) = -202.8 kJ (spontaneous at 1000 K)

b. SbCl_{3}(s) + Cl_{2}(g) Ž SbCl_{5}(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 DGº_{500} = (-58.1 kJ) - (500 K)(-0.1057 kJ/K) = -5.25 kJ (spontaneous at 500 K)

At 1000 K DGº_{500} = (-58.1 kJ) - (1000 K)(-0.1057 kJ/K) =47.6 kJ (non-spontaneous at 1000 K)

c. BaCO_{3}(s) Ž BaO(s) + CO_{2}(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 DGº_{500} = (277.4 kJ) - (500 K)(0.1738 kJ/K) = 190.5 kJ (non-spontaneous at 500 K)

At 1000 K DGº_{500} = (277.4 kJ) - (1000 K)(0.1738 kJ/K) = 103.6 kJ (non-spontaneous at 1000 K)

7a. What is the value of DGº_{T} when the system is at equilibrium under standard conditions?

Under standard conditions, DGº_{T} = 0 at equilibrium.

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

i) 2 ClF_{3}(g) Ž Cl_{2}(g) + 3 F_{2}(g)

At equilibrium DGº_{T} = 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) MgCO_{3}(s) Ž MgO(s) + CO_{2}(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 DG_{T} vary with pressure (or concentrations)?

The relationship is DG_{T} = DGº_{T} + 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 a_{i}^{m(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: DGº_{T} = -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, K_{p} and K_{c} for a reaction?

K is the activities equilibrium constant.

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

K_{c} is the concentration constant for a reaction in concentration measurements.

If K_{c} is for gases, then it must be converted to K_{p} to relate to K.

If K_{c} 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) PCl_{3}(g) + Cl_{2}(g) ¾ PCl_{5}(g)

K may be calculated: DGº_{298} = -RT ln K.

and DGº_{298} can be calculated from DGº_{f} values:

So: DGº_{298} = (-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 10^{9}.

ii) H_{2}O(l) ¾ H_{2}O(g)

As above:

then DGº_{298} = (-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(H_{2}O_{(g)}) / a(H_{2}O_{(l)}) = P(H_{2}O)/Pº / 1

So 0.0311 = P(H_{2}O)/100 kPa

and P(H_{2}O) = 3.11 kPa. (Experimental value: 3.17 kPa)

9. For the reaction

PCl_{5}(g) Ž PCl_{3}(g) + Cl_{2}(g) at 25ºC

DHº = + 92.5 kJ and DSº = + 182 J/K

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

DGº_{298} = 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?

DGº_{300} = 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, DGº_{T} = 0, so T = DHº/DSº = 92500J/182J.K^{-1}. = 508 K.

10. Show that DG_{T} = RTln(Q/K)

We have: DG_{T} = DGº_{T} + RT ln Q

At equilibrium, DG_{T} = 0, and DGº_{T} = -RT ln K.

So DG_{T} = -RT ln K + RT ln Q = RT ln (Q/K).

11a. Calculate DG_{600} for the following process:

3 H_{2}(g) + N_{2}(g) Ž 2 NH_{3}(g) ***NOTE BALANCE***

if P(H_{2}) = 25 kPa, P(N_{2}) = 10 kPa, and P(NH_{3}) = 200 kPa.

We have: DG_{600} = DGº_{600} + R*600*ln Q.

and DGº_{600} = DHº - TDSº

So: DHº = 2*(-45.9) = -91.8 kJ.

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

Then: DGº_{600} = (-91.8 kJ) - 600K*(-0.159 kJ/K) = 3.6 kJ.

So DG_{600} = 3.6 kJ + R*600*ln Q.

Q = a(NH_{3})^{2}/( a(H_{2})^{3}.a(N_{2})) = (2.00)^{2}/((0.25)^{3}.(0.10)) = 2560

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

11b. Calculate K_{600} for the ammonia synthesis reaction.

At 600 K, ln K_{600} = -DGº_{600} / RT = 3600 J / (8.314*600) = 0.722

Thus K_{600} = 2.06.