Answers to problems set for Quiz 8

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, Kp and Kc (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

Kp = P(N2)*P(H2)3 / P(NH3)2 kPa2

Kc = [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)

Kp = P(CO2) kPa

Kc = [CO2] M

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

K = a(CH3COO-)*a(H3O+) / a(CH3COOH)

Kp is not defined for solutions.

Kc = [CH3COO-][H3O+] / [CH3COOH] M

e. AgCl(s) ¾ Ag+(aq) + Cl-(aq)

K = a(Ag+)*a(Cl-)

Kp is not defined for solutions.

Kc = [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 = -D368 /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.

 PCl5(g) ¾ PCl3(g) + Cl2(g) I (kPa) 100 0 0 C (kPa) (-1)x (+1)x (+1)x E (kPa) 100 - x x x

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.

 2 HI(g) ¾ H2(g) + I2(g) I (kPa) 50 50 50 C (kPa) (-1)x (+1)x (+1)x E(kPa) 50 - x 50 + x 50 + x

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

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 Kp at 25ºC, and at 100ºC. Use thermodynamic tables for data.

K and Kp 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: D298 = -36.0 kJ and K298 = 2.04 x 106,

and: D373 = -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 Pi = ni * (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. = PAcetone / 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/Tb.

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

and K = 0.264.

From which PAcetone = 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 D298 can be obtained from: D298 = DHº - 298*DSº = -40.8 kJ.

Finally K298 is obtainable from DT = -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 DGT = RTln(Q/K)

We have shown that: DGT = DT + RT ln Q.

At equilibrium, DGT = 0 and Q = K, so DT = -RT ln K.

Substituting: DGT = -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 = Pam(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:

 C2H6(g) change K change an increase in temperature. down down an increase in total pressure up none the addition of C2H4(g). up none the addition of Pt(s) as a catalyst. none none the removal of H2(g). down none

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 D523 for this reaction.

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

All pressures are calculated from the given concentrations.

 PCl5(g) ¾ PCl3 + Cl2(g) Initial (kPa) 478.3 0 0 Change (-1)x (+1)x (+1)x Equilibrium 478.3-x x x

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

DT = -RT ln K.

To find D523 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 D523 = 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.

 Br2(g) + Cl2(g) ¾ 2 BrCl(g) Initial (kPa) 50 50 0 Change (-1)x (-1)x (+2)x Equilibrium 50 - x 50 - x 2x

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.

 H2O(g) + CO(g) ¾ H2(g) + CO2(g) Initial (kPa) 100 100 0 0 Change (-1)x (-1)x (+1)x (+1)x Equilibrium 100 - x 100 - x x x

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+.