A level chemistry AQA Thermodynamics exam questions and mark scheme. these are high quality exam questions which will test your understanding of the topic and will increase your ability to acheive an A* in the exams
,Question 1. The energy level diagram (Born-Haber cycle) for caesium chloride is shown below.
–
C s+ (g ) + C l(g ) + e
–1
H 4 = + 1 2 1 k J m o l
–1
H 5 = – 3 6 4 k J m o l
–
C s +(g ) + 12 C l 2(g ) + e
+ –
H 3 = + 3 7 6 k J m o l
–1
C s (g ) + C l (g )
C s(g ) + 12 C l 2(g )
H 2 = + 7 9 k J m o l – 1
H 6
1
C s(s) + 2 C l 2 (g )
–1
H 1 = – 4 3 3 kJ m o l
C sC l(s)
(a) Give the names of the enthalpy changes represented by H1, H2 and H5. (3)
(b) Calculate the value of the lattice energy H6. (2)
(c) Explain why the enthalpy change represented by H3 has a lower magnitude for caesium than for sodium.
(3) (Total 8 marks)
Question 2. Below is the energy level diagram (Born-Haber cycle) for the formation of rubidium iodide from its
elements.
–1 H 1
H I(R b ) = + 4 02 k J m o l
–
R b + (g ) + I (g )
R b (g ) + I(g )
–1
H at ( 12 I 2 ) = + 10 7 k J m ol
R b (g ) + 1
2 I 2 (s)
–1
H at (R b ) = + 8 5.8 k J m o l
–1
H 2 = – 6 0 9 k J m o l
–1
H f(R b I) = – 3 28 k J m ol
R b I(s)
(i) Complete the diagram giving the identities of the missing species (2)
(ii) Give the names of the enthalpy changes represented by H1 and H2. (2)
(iii) Calculate the value of the enthalpy change represented by H1. (2)
(Total 6 marks)
, Question 3. The following data relate to lithium chloride.
Standard molar enthalpy change of solution is –37.0 kJ mol–1.
Lattice enthalpy is –846 kJ mol–1.
+ –
L i (g ) + C l (g )
L i(s) + ½ C l 2 (g )
B
E n th alp y C
A
L iC l(s)
+ –
L i (aq ) + C l (aq )
(i) Give the name of each of the changes A and B. (2)
(ii) Calculate the value of the enthalpy change represented by C and suggest the name(s) of the
enthalpy change(s). (3)
(Total 5 marks)
Question 4. The following reaction occurs in the high-temperature preparation of titanium(IV) chloride.
TiO2(s) + C(s) + 2Cl2(g) TiCl4(l) + CO2(g)
(i) Use the data given below to calculate the standard enthalpy change and the standard entropy change for this
reaction.
Substance TiO2(s) C(s) Cl2(g) TiCl4(l) CO2(g)
Hf / kJ mol –1 –940 0 0 –804 –394
S / J K–1mol–1 49.9 5.7 223 252 214
(ii) Calculate the temperature at which this reaction ceases to be feasible. (9)
(Total 9 marks)
Question 5. Sulphuric acid is manufactured by the Contact Process. One stage of the process is the oxidation of
sulphur dioxide according to the equation:
–1
2SO2(g) + O2(g) 2SO3(g) H = –196 kJ mol
Typical operating conditions are a temperature of about 450 °C, close to atmospheric pressure and the presence
of a catalyst. The use of a converter containing four catalyst beds results in a 99.5% conversion.
–1 –1
The value of the entropy change, S, for the reaction between sulphur dioxide and oxygen –188 J mol K .
(i) Explain why there is a decrease in entropy of the system. (2)
(ii) Use the equation G = H – TS to calculate the value of G for the reaction between sulphur dioxide and
oxygen at 25 °C. Explain the significance of the sign of the value obtained. (4)
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