Calculate as many thermodynamic quantities as you can for a given reaction.

## Introduction

Teachers who have not used the problems before should read the section Using the problems before starting.

## Prior knowledge

Algebraic definition of pH, and Kw; and the relationships between equilibrium constant, standard free energy change, standard enthalpy change, standard entropy change and standard electrode potential. A detailed knowledge is unnecessary as students are encouraged to consult textbooks and data books during the exercise. NB To get the students started, you may have to tell them the value of Kw, if necessary by explaining that the pH of pure water is 7.

## Resources

Scientific calculators, and perhaps data books for thermodynamic constants should be available. Students are not allowed to use thermodynamic data such as Eo, so they will have to know the difference between thermodynamic constants and thermodynamic data.

## Possible method

The following can be calculated starting from the pH of water: the dissociation constant of water, ΔGθ and Eθ, then using ΔHθ, ΔSθ.

1. For pure water, pH = 7 at 25°C

[H+(aq)] = 107 mol dm3

Since water is neutral,

[H+(aq)] = [OH (aq)] = 107 mol dm3

H2 O(l) H+(aq) + OH (aq)

K = [H+(aq)][OH (aq)] = 1 x 1014 1 (see Background information)

ΔGθ = –RT lnK

= – 8.31 J K1 mol1 x 298 K x – 32.2

= 79.8 kJ mol1,

or – 79.8 KJ mol1 for the reverse reaction, where ΔHθ is given.

H+(aq) + OH (aq) → H2 O(l) ΔHθ = –57.1 kJ mol1

ΔGθ = –79.8 kJ mol1

1. 2. ΔGθ = –nFEθ

E° = –ΔGθ/nF = 79800/96500 V

= 0.83 V

NB Any Eθ is the emf for a galvanic cell. In this case the cell could be constructed from the following two half-reactions:

2H+ (aq) + 2e → H2 (g) Eθ = 0.00 V

H2 (g) + 2OH (aq) → 2H2 O(l) + 2e Eθ = 0.83 V

Overall this gives

H2 (g) + 2H+ (aq) + 2OH (aq) → 2H2 O(l) + H2 (g) Eθ cell = 0.83 V

1.  ΔGθ = ΔHθ – TΔSθ

TΔSθ = ΔHθ – ΔGθ = – 57.1 – (– 79.8) kJ mol1

= 22.7 kJ mol–1

22700

ΔSθ = ————

298

= 76 J mol1 K1

Because this value is positive the H+ (aq) and OH (aq) must be more ordered than H2 O(l). This should provoke discussion about the ordering of water molecules round the H+ (aq) and the OH (aq).

Students may calculate the temperature at which the dissociation of water becomes feasible, ie when K ≥ 1. If so, you could point out that this figure is meaningless because none of the reactants or products can exist at the calculated temperature.

## Suggested approach

During trialling the following instructions were given to students and proved to be extremely effective:

1. Working as a group discuss the problem and decide what you can calculate. You should divide the work amongst yourselves but keep one another informed of your progress. Such discussion can play a vital part in working out a solution to an open-ended problem like this where divergent thinking is required. Several minds working on a problem together can stimulate ideas that one on its own could not manage. About 10 minutes should be spent on this problem initially with further discussion as required.
2. Write a brief account of your findings.
3. Working as a group, prepare a short (ca 5-minute maximum) presentation to give to the rest of the class. If possible all group members should take part: any method of presentation (such as a blackboard, overhead projector, etc) can be used.

Outline the problem, describe and explain what you did; do not go through the detailed calculation. If possible try and think back to how you decided on what you could calculate. After the presentation, be prepared to accept and answer questions and to discuss what you did with the rest of the class.