Copyright reserved
DEPARTMENT OF CHEMISTRY
CMY 282
Surname
Initials
Student
Number
Class Test 2 2022
Signature
The Second & Third Laws of Thermodynamics, and
Physical Transformations & Mixtures
Date: 2 June 2022
Examiners:
Mr H van der Poll
Dr C Slabbert
External Examiner:
Prof L van der Merwe
Pages: 7
Time: 50 Minutes
Marks: 25
INSTRUCTIONS
1. This paper consists of 5 Questions (7 pages), please ensure that you have them all.
2. To ensure accurate marking, students are required to write neatly and clearly.
3. All calculations must be shown in full (show all calculation steps).
4. When deriving an equation, all steps should be shown in a logical manner, with explanations where
necessary. Do not only write the equations.
5. The allocation of marks to each question is a guide to the depth of answer required.
6. The use of non-programmable scientific calculators is permitted.
7. A data sheet is provided separately.
8. Answer ALL questions.
9. Answer all questions in the spaces provided. Specifically for Question 4 answer blocks are provided,
please answer this question in these blocks. Use the back of the page if you need more space to
answer a question. Clearly indicate this action in your answer.
Question
Marks
Marker/Examiner
1
4
2
6
3
5
4
4
5
6
TOTAL:
25
Page 1
CMY 282 Class Test 2 2022
Page 2
Question 1
[4]
Predict whether the entropy change will be near zero, positive, or negative for the following reactions.
Explain your reasoning in each case.
(a) Ag+ (aq) + Cl– (aq) → AgCl (s)
(2)
(b) Fe (s) +Zn2+ (aq) → Fe2+ (aq) + Zn (s)
(2)
(a) The entropy change is negative (ΔSsys < 0), since the reaction progresses as written from a less
ordered (aqueous solution of ions) to a more ordered form (solid crystal lattice).
(b) The entropy change is near zero (ΔSsys ~ 0) it is not zero!, since on each side of the reaction
equation a solid metal and solvated (2+)-ion is present. Therefore, the system does not undergo
a major change in order and so the entropy change is close to zero.
CMY 282 Class Test 2 2022
Page 3
Question 2
[6]
A quantity of 0.50 mol of an ideal gas at 20 °C expands isothermally against a constant pressure of
2.0 atm from 1.0 L to 5.0 L. Calculate the values of ΔSsys, ΔSsurr and ΔStot. Is this process spontaneous?
Explain why it is or why it is not.
See Theme 3 Tutorial Q1.
CMY 282 Class Test 2 2022
Page 4
Question 3
[5]
Starting with the criterion for spontaneity in terms of entropy (i.e. ΔStot ≥ 0), derive the so-called
Gibbs Energy expression ΔG = ΔH – TΔS. Clearly write out all arguments symbolically and logically
explain each step in the derivation. Finally, deduce the specific criterion for spontaneity for the Gibbs
Energy.
Derivation was done in class.
CMY 282 Class Test 2 2022
Page 5
Question 4
[4]
Give a short definition/explanation/concepts for each of the following terms:
(a) A phase
(1)
A form of matter that is uniform throughout a in chemical composition and physical state
(b) A phase transition
(1)
Spontaneous conversion of one phase into another
(c) The phase rule
(1)
Expression that relates the number of variables that may be changed while the phases of a system
remains in mutual equilibrium
(d) The triple point
(1)
Set of conditions under which three different phases of a substance (typically solid, liquid and vapour)
all simultaneously coexist in equilibrium
CMY 282 Class Test 2 2022
Page 6
Question 5
[6]
A container is divided into two equal compartments as indicated in the figure below. One contains
3.0 mol H2 (g) at 25 °C and a pressure of 3p; the other contains 1.0 mol N2 (g) at the same temperature,
but at a pressure of 1p. You are asked to calculate the Gibbs energy of mixing when the partition is
removed. Assume that the gases are perfect.
Answer the following questions:
(a) Why can the equation for to calculate the Gibbs energy of mixing, ∆mix G = nRT(xA lnxA +
xB lnxB ) , not be used directly?
(1)
(b) Calculate the Gibbs energy of mixing
(4)
(c) Expanding on part (a) of this question, what does the value of ΔmixG imply?
(1)
(a) The equation cannot be used directly because the two gases are initially at different pressures. The initial
Gibbs energy from the chemical potentials must first be calculated to proceed.
(b)
(c) In this case the pressure change, and ΔG < 0 is a signpost of spontaneous change only at constant
temperature and pressure. In this case, the value of ΔmixG is the sum of two contributing components:
the mixing itself and the changes and in pressure of the two gases to their final total pressure, 2p.