1. What is the definition of equilibrium:

a. In thermodynamic terms.

A reaction at equilibrium has DG = 0.

b. In reaction kinetic terms.

In kinetic terms a reaction is at equilibrium when the forward and backward overall reaction rates are equal.

2. Define K, K_p and K_c (not forgetting the units where they are necessary) for the following processes.

Example only for the gas reactions:

a. 2 NH₃(g) ¾ N₂(g) + 3 H₂(g)

K = a(N₂)*a(H₂)³ / a(NH₃)²

K_p = P(N₂)*P(H₂)³ / P(NH₃)² kPa²

K_c = [N₂]*[H₂]³ / [NH₃]² M²

b. PCl₅(g) ¾ PCl₃(g) + Cl₂(g)

c. 2 HI(g) ¾ H₂(g) + I₂(g)

d. CaCO₃(s) ¾ CaO(s) + CO₂(g)

K = a(CO₂)

K_p = P(CO₂) kPa

K_c = [CO₂] M

e. CH₃COOH(aq) + H₂O(l) ¾ CH₃COO^-(aq) + H₃O⁺(aq)

K = a(CH₃COO^-)*a(H₃O⁺) / a(CH₃COOH)

K_p is not defined for solutions.

K_c = [CH₃COO^-][H₃O⁺] / [CH₃COOH] M

e. AgCl(s) ¾ Ag⁺(aq) + Cl^-(aq)

K = a(Ag⁺)*a(Cl^-)

K_p is not defined for solutions.

K_c = [Ag⁺][Cl^-]

3. From thermodynamic tables, find values for these equilibrium constants at standard temperature and at 95º C.

Example only:

(a.) 2 NH₃(g) ¾ N₂(g) + 3 H₂(g)

Working backwards from what we want:

Required is K₃₆₈ and this can be calculated from DG₂₆₈ which in turn can be calculated fron DHº and DSº which in their turn can be calculated from tables.

So, from tables: DHº = (-2)(-46.11) + (+1)(0) + (+3)(0) = 92.22 kJ

And: DSº = (-2)(192.3) + (+1)(191.5) + (+3)(130.6) = 198.7 J / K.

From which: DG₃₆₈º = 92.22 - 368*(198.7 / 1000) = 19.10 kJ.

And finally: ln K₃₆₈ = -DGº₃₆₈ /RT = -19100 / (8.314 * 368) = -6.24.

And K₃₆₈ = 1.95 x 10^-3.

3b. In the case of 2b, above, find the equilibrium pressures in the system at 95º C if 100 kPa of PCl₅ is placed in a 10.0-L pot.

For 2b, K₃₆₈ can be worked out:

PCl₅(g) ¾ PCl₃(g) + Cl₂(g)

First DHº = 92.5 kJ; DSº = 181.7 J / K.

Then DG₃₆₈º = 25.6 kJ

And finally, K₃₆₈ = 2.30 x 10^-4.

K = a(PCl₃) * a(Cl₂) / a(PCl₅)

2.30*10^-4 = (x/100)(x/100) / ((100 - x) / 100)

From which x = 1.5 or -113 (impossible)

So P(PCl₅) = 98.5 kPa; P(PCl₃) = 1.5 kPa; P(Cl₂) = 1.5 kPa.

3c. In the case of 2c (above), find the pressures of all gases at equilibrium at 450º C if 50 kPa of each of the three gases are placed in a 20.0-L container at 450º C.

For 2c, K₇₂₃ can be worked out:

2 HI(g) ¾ H₂(g) + I₂(g)

First DHº = -53.0 kJ.

Then DSº = -21.8 J / K.

Then DG₇₂₃º = -37.2 kJ

From which K₇₂₃ = 490.

K = a(H₂)*a(I₂) / a(H₂)²

490 = ((50 + x)/100) * ((50 + x)/100) / ((50 - x)/100)²

Required are the pressures of the gases at equilibrium, these are known when we find x:

Simplifying: 490 = (50 + x)² / (50 - x)²

Take square root of each side (since RHS is a perfect square): 22.14 = (50 + x) / (50 - x)

From which x = 45.7

So the required answer is:

At equilibrium: P(HI) = 4.3 kPa; P(H₂) = 95.7 kPa; P(I₂) = 95.7 kPa.

3d. In the case of 2e (above) find the solubility of silver chloride in water at 298 K, and at 368 K.

From 2e, AgCl(s) ¾ Ag⁺(aq) + Cl^-(aq)

First DHº = (-1)(-127.0) + (+1)(105.9) + (+1)(-167.5) = 65.4 kJ

Then DSº = (-1)(96.11) + (+1)(73.93) + (+1)(55.10) = 32.92 J / K.

From which DG₂₉₈º = 54.7 kJ

and DG₃₆₈º = 52.4 kJ.

So K₂₉₈ = 2.58 x 10^-10 and K₃₆₈ = 3.65 x 10^-8.

The solubility of silver chloride is the moles/L which dissolve, which (by stoichiometry) is the mol/L of Ag⁺:

AgCl(s) ¾ Ag⁺(aq) + Cl^-(aq)

some x mol/L x mol/L

So K = x², and x = 1.61 x 10^-5 at 298 K and 1.91 x 10^-4 at 368 K.

Thus solubility of silver chloride in water is 1.61 x 10^-5 mol/L at 298 K and 1.91 x 10^-4 mol/L at 368 K.

4. Consider the process SO₂(g) + NO₂(g) ¾ SO₃(g) + NO(g)

a) Calculate K and K_p at 25ºC, and at 100ºC. Use thermodynamic tables for data.

K and K_p are the same in this case (as Dn = 0).

First: DHº = (-1)(-298.6) + (-1)(33.2) + (+1)(-396) + (+1)(90.29) = -42.11 kJ

Then: DSº = (-1)(248.1) + (-1)(239.9) + (+1)(256.7) + (+1)(210.7) = -20.6 J / K.

Finally: DGº₂₉₈ = -36.0 kJ and K₂₉₈ = 2.04 x 10⁶,

and: DGº₃₇₃ = -34.4 kJ and K₃₇₃ = 6.57 x 10⁴.

b) A mixture comprises l.00 g of each gas at l00ºC in a 2.00 L container. Calculate Q and decide which direction the reaction will have to move to reach equilibrium.

Q₃₇₃ = ( a(SO₃) * a(NO) ) / ( a (SO₂) * a(NO₂) ) and P_i = n_i * (8.314 * 373 / 2.00)

Thus P(SO₃) = 19.38 kPa; P(SO₂) = 24.23 kPa; P(NO) = 51.69 kPa; P(NO₂) = 33.71 kPa.

From which Q₃₇₃ = 1.23.

Since Q₃₇₃ is smaller than K₃₇₃, the reaction will move to the right, producing more products.

5. The normal boiling point of acetone is 56.2º C and its standard enthalpy of vapourization is 30.3 kJ/mol. Calculate the vapour pressure of acetone at 21ºC.

The equation for the process is:

Acetone(l) ¾ Acetone(g)

K = a(Acetone(g)) / 1. = P_Acetone / Pº. This finding K will give the pressure.

K can be found from: ln K = DSº/R - DHº/RT.

We are given DHº and the normal boiling point, at which DSº = DHº/T_b.

So ln K = DHº/RT_b - DHº/RT = 30300/8.314(329.2^-1 - 294^-1) = -1.33.

and K = 0.264.

From which P_Acetone = 26.4 kPa.

6. At 500 K, the equilibrium constant, K, = 49.9 for the reaction

2 NO(g) + Cl₂(g) ¾ 2 NOCl(g).

At 600 K, the equilibrium constant, K, = 2.28.

Calculate DHº, DSº, DGº, and K at 298 K assuming that DHº and DSº are independent of temperature changes.

First: DHº can be calculated using: ln (K₁ / K₂) = DHº/R(T₂^-1 - T₁^-1).

From which DHº = -76.97 kJ.

Then DSº can be calculated from: ln K = DSº/R - DHº/RT

From which DSº = -121.5 J / K.

Then DGº₂₉₈ can be obtained from: DGº₂₉₈ = DHº - 298*DSº = -40.8 kJ.

Finally K₂₉₈ is obtainable from DGº_T = -RTln K. K = 1.42 x 10⁷.

7. For the equilibrium H⁺(aq) + OH^-(aq) ¾ H₂O(l),

DHº = 55.9 kJ and DSº = -80.4 J/K at 25ºC.

a) Calculate K at 25ºC.

K can be calculated from: ln K = DSº/R - DHº/RT.

K₂₉₈ = 1.00 x 10^-14.

b) Assuming that DHº and DSº do not change with increasing temperature, calculate K at 100ºC.

K₃₇₃ = 9.4 x 10^-13.

c) The pH of pure water is given by pH = -½log₁₀ K. Does the pH of pure water increase or decrease with temperature? Is pure water acidic, basic or still neutral at 100ºC ? Comment.

The pH of pure water decreases (from 7 at 25 ºC to ca. 6 at 100 ºC).

Pure water remains neutral.

8. For a reaction NOT at equilibrium, Q can be defined. How might you use a value of Q and of K to find which way the reaction will proceed to reach equilibrium.

The value of Q changes towards the value of K as the reaction moves to equilibrium.

Thus if Q < K, Q must increase, more products are produced.

And if Q > K, Q must decrease and more reactants are produced.

9a. Show that DG_T = RTln(Q/K)

We have shown that: DG_T = DGº_T + RT ln Q.

At equilibrium, DG_T = 0 and Q = K, so DGº_T = -RT ln K.

Substituting: DG_T = -RT ln K + RT ln Q = RT ln (Q / K).

9b. For reaction 2b, calculate which way the reaction must move to reach equilibrium at 95º C if P(PCl₅) = 100 kPa, P(PCl₃) = 10 kPa, and P(Cl₂) = 5 kPa.

Known:

PCl₅(g) ¾ PCl₃(g) + Cl₂(g); K₃₆₈ = 2.30 x 10^-4.

Q = Pa^m(i) = 0.1 * 0.05 / 1 = 5.00 x 10^-3.

Since Q is larger than K, the reaction will move left to give more reactants.

9c. For the reaction in 2d, which direction will the process move if the pressure of CO₂(g) is 5.0 kPa at 95º C?

CaCO₃(s) ¾ CaO(s) + CO₂(g)

K = a(CO₂) = P(CO₂)/Pº

K₃₆₈ must be calculated from DHº and DSº which are calculated from tables.

First DHº = (-1)(-1207.9) + (+1)(-635.1) + (+1)(-393.5) = 179.3 kJ.

Then: DSº = (-1)(92.9) + (+1)(39.75) + (+1)(213.6) = 160.5 J / K.

And so K₃₆₈ = exp(DSº/R - DHº/RT) = 8.56 x 10^-18.

Thus P(CO₂) = 8.56 x 10^-16 kPa.

10. What is meant by the "Le Châtelier Principle"?

The "Le Châtelier Principle"is a rule which gives a qualitative view of what will happen to the equilibrium position of a reaction when changes are made. Stated it is:

When a disturbance is imposed on a reaction initially at equilibrium, that reaction will respond by moving in the direction which will minimize that disturbance.

A disturbance is any change that changes at least one of the reactions activities.

11a. What will happen to the position of equilibrium for the following processes if the amount of the first reactant is doubled (without changing a volume)?

(i) 2 NH₃(g) ¾ N₂(g) + 3 H₂(g)

(ii) CaCO₃(s) ¾ CaO(s) + CO₂(g)

(iii) CH₃COOH(aq) + H₂O(l) ¾ H₃O⁺(aq) + CH₃COO^-(aq)

(iv) AgCl(s) ¾ Ag⁺(aq) + Cl^-(aq)

(i) Reaction moves R

(ii) a(CaCO₃) cannot be changed: no disturbance, no movement.

(iii) Reaction moves R.

(iv) a(AgCl) cannot change: no disturbance, no movement.

11b. For the same reactions what will happen to the position of equilibrium if the amount of the first product is doubled without changing the volume?

(i) Reaction moves L.

(ii) a(CaO) cannot be changed: no disturbance, no movement.

(iii) Reaction moves L.

(iv) Reaction moves L.

11c. Again for the reactions in 11a, what would happen to the position of equilibrium if the total pressure were doubled?

(i) Product side more affected, reaction moves L.

(ii) Product side affected, reaction moves L.

(iii) P does not affect activities in solution, no disturbance, no movement.

(iv) As in (iii)

11d. Finally with respect to the same three reactions, what would happen to the position of equilibrium if the temperature were increased? (You need thermodynamic tables to answer this!)

Reaction will move in the endothermic direction:

(i) DHº = 91.8 kJ. Reaction moves R

(ii) DHº = 178.3 kJ. Reaction moves R.

(iii) DHº = -0.42 kJ. Reaction moves L.

(iv) DHº = 65.4. Reaction moves R.

12. The reaction C₂H₄(g) + H₂(g) ¾ C₂H₆(g) is exothermic as written. Assume that an equilibrium is established.

How would the equilibrium amount (moles) of C₂H₆(g) change with:

13. At 250ºC, 0.110 mol of PCl₅(g) was introduced into a 1.00 L container. Equilibrium was established: PCl₅(g) ¾ PCl₃(g) + Cl₂(g). [PCl₃] = 0.050 mol/L. Find the value of K₅₂₃ and of DGº₅₂₃ for this reaction.

***For gases it is better to work in pressures***

All pressures are calculated from the given concentrations.

K₅₂₃ = ( a(PCl₃) * a(Cl₂) ) / a(PCl₅) = (x/100)² / (478.3-x)/100

DGº_T = -RT ln K.

To find DGº₅₂₃ we need to find K₅₂₃ which requires x in the above quadratic.

At equilibrium, P(PCl₃) = 217.4 kPa (from the concentration) = x

Thus K₅₂₃ = 0.1812 and DGº₅₂₃ = 7.43 kJ.

14. For the equilibrium Br₂(g) + Cl₂(g) ¾ 2 BrCl(g) at 400 K, K = 7.0. If 50 kPa each of Br₂(g) and Cl₂(g) are introduced into a container at 400 K what is the partial pressure of BrCl(g) at equilibrium.

K = (P(BrCl)/100)² / {(P(Br₂)/100) * (P(Cl₂)/100)}

K = (2x)² / (50 - x)² = 7.

Since the "x" side is a perfect square, take the square root of both sides to solve quickly:

x = 28.5.

Thus at equilibrium: P(BrCl) = 57.0 kPa, P(Cl₂) = 21.5 kPa, P(Br₂) = 21.5 kPa.

15. For the reaction H₂O(g) + CO(g) ¾ H₂(g) + CO₂(g) K₂₉₈ = 9.78 x 10⁴ and Hº₂₉₈ = -1.2 kJ. What will be the equilibrium partial pressures of all gases if 100 kPa each of H₂O(g) and CO(g) are introduced into a container and allowed to equilibrate at 600 K.

K₆₀₀ is needed. It can be calculated from the given data using:

ln (K₂₉₈ / K₆₀₀) = DHº/R(600^-1 - 298^-1)

from which K₆₀₀ = 7.67 x 10⁴.

K = { a(H₂) * a (CO₂) } / { a(H₂O) * a (CO) }

7.67 x 10⁴ = (x/100)² / ((100 - x)/100)².

Solve quickly by taking the square root of both sides to give x = 99.7.

Thus at equilibrium, P(H₂O) = P(CO) = 0.3 kPa and P(H₂) = P(CO₂) = 99.7 kPa.

16a. What is the "common ion" effect, and how does it relate to the Le Châtelier Principle?

The "common ion" effect is the movement of an equilibrium when a solution containing one of the ions in the equilibrium is added. The Le Châtelier Principle predicts that the equilibrium will be disturbed by the change and (usually) the equilibrium position will move away from the added ion.

For example if the equilibrium: AgCl(s) ¾ Ag⁺(aq) + Cl^-(aq) is established and then a source of Ag⁺ ions is added (say a solution of AgNO₃), then the added Ag⁺ will disturb the equilibrium such that it will move away from the Ag⁺ and more solid AgCl will form.

16b. Show qualitatively how the pH of a buffer solution containing a weak acid differs from the pH of the weak acid itself. (say CH₃COOH containing CH₃COO^- ions)

In this example, the common ion in the buffer is the CH₃COO^- ion.

The equilibrium: CH₃COOH(aq) + H₂O(l) ¾ CH₃COO^-(aq) + H₃O⁺(aq) gives the pH of the solution. If the common ion CH₃COO^- is added, the equilibrium shifts to remove added ion, that is to the left. This causes the [H₃O⁺] to drop as well which is reflected in an increase in the pH of the solution. Thus the pH of a buffer solution is higher than the pH of the acid on its own.

16c. Show how the solubility of a poorly soluble salt drops on addition of another source of one of its ions (say AgCl in a solution of NaCl).

This was answered in part (a), except that Cl^- is the common ion instead of Ag⁺.