Entropy and EnergeticsEdexcel International A Level Chemistry: Topic test

20 questions, 54 marks

Edexcel International A Level Chemistry

Entropy and Energetics topic test

Total 54 marks

Name

Class

Date

1
Solid carbon dioxide ('dry ice') sublimes in a warm room: CO₂(s) → CO₂(g). The process is endothermic.
(a)
What is the sign of the entropy change of the system, ΔS system, for this change, and why?
[1 mark]
APositive, because the gas has many more ways of arranging its molecules and energy than the solid
BNegative, because the process absorbs heat
CZero, because the number of molecules does not change
DNegative, because the molecules become more ordered
(b)
Which reaction has the most positive entropy change of the system?
[1 mark]
AN₂(g) + 3H₂(g) → 2NH₃(g)
BH₂O(g) → H₂O(l)
C2H₂O₂(l) → 2H₂O(l) + O₂(g)
DCa²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
(c)
Explain, in terms of the dispersal of molecules and energy, why the entropy of carbon dioxide increases when it sublimes.
[2 marks]
Total for question 1: 4 marks
2
Sodium hydrogencarbonate decomposes on heating: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g). For the equation as written, ΔH = +136 kJ mol⁻¹ and ΔS system = +334 J K⁻¹ mol⁻¹.
(a)
What is the entropy change of the surroundings, ΔS surroundings, at 298 K?
[1 mark]
A+456 J K⁻¹ mol⁻¹
B−456 J K⁻¹ mol⁻¹
C−0.456 J K⁻¹ mol⁻¹
D−122 J K⁻¹ mol⁻¹
(b)
What happens to the total entropy change as the temperature is raised?
[1 mark]
AΔS system becomes more negative, so ΔS total decreases
BΔS surroundings does not change because ΔH is constant
CΔS total is unchanged because ΔS system is fixed
DΔS surroundings becomes less negative, so ΔS total increases and the reaction becomes feasible above a certain temperature
(c)
Calculate the minimum temperature, in K, at which the decomposition becomes feasible.
[2 marks]
Total for question 2: 4 marks
3
Potassium oxide, K₂O, is an ionic solid. Data in kJ mol⁻¹: standard enthalpy change of formation of K₂O −361; standard enthalpy change of atomisation of potassium +89; first ionisation energy of potassium +419; standard enthalpy change of atomisation of oxygen, per mole of O atoms, +249; first electron affinity of oxygen −141; second electron affinity of oxygen +798.
(a)
Define the term lattice energy, and explain why the second electron affinity of oxygen is endothermic.
[3 marks]
(b)
Use the data to calculate the lattice energy of potassium oxide.
[4 marks]
Total for question 3: 7 marks
4
Lithium fluoride is only sparingly soluble in water but lithium chloride is very soluble. Data in kJ mol⁻¹: lattice energy (for formation of the solid from gaseous ions) of LiF −1049 and of LiCl −846; enthalpy change of hydration of Li⁺ −519, of F⁻ −506 and of Cl⁻ −364. The entropy change of the system on dissolving at 298 K is −37 J K⁻¹ mol⁻¹ for LiF and +9 J K⁻¹ mol⁻¹ for LiCl.
(a)
Calculate the enthalpy change of solution and the total entropy change at 298 K for each compound, and use the results to explain the difference in their solubilities.
[6 marks]
(b)
Explain why the lattice energy of LiF is more exothermic than that of LiCl, why the hydration enthalpy of the fluoride ion is more exothermic than that of the chloride ion, and why the entropy change of the system on dissolving is negative for LiF.
[6 marks]
Total for question 4: 12 marks
5
A data book lists enthalpy changes of hydration for gaseous ions and enthalpy changes of solution for ionic compounds, which a student uses to explain how ionic solids dissolve in water.
(a)
Which equation represents the enthalpy change of hydration of the chloride ion?
[1 mark]
ACl(g) + aq → Cl⁻(aq)
BCl⁻(aq) → Cl⁻(g) + aq
C½Cl₂(g) + e⁻ → Cl⁻(g)
DCl⁻(g) + aq → Cl⁻(aq)
(b)
Which gaseous ion has the most exothermic enthalpy change of hydration?
[1 mark]
AMg²⁺
BNa⁺
CK⁺
DCa²⁺
(c)
State what is meant by the standard enthalpy change of solution of an ionic compound.
[2 marks]
Total for question 5: 4 marks
6
Diamond is thermodynamically unstable relative to graphite at 298 K: C(diamond) → C(graphite), ΔH = −1.9 kJ mol⁻¹ and ΔS system = +3.4 J K⁻¹ mol⁻¹. However, diamond jewellery does not turn into graphite.
(a)
What is the total entropy change for the conversion of diamond to graphite at 298 K?
[1 mark]
A+3.4 J K⁻¹ mol⁻¹
B−3.0 J K⁻¹ mol⁻¹
C+9.8 J K⁻¹ mol⁻¹
D+3400 J K⁻¹ mol⁻¹
(b)
Why does diamond not turn into graphite at room temperature?
[1 mark]
AThe total entropy change is negative
BThe activation energy is very high, so the rate of change is negligible
CThe conversion is endothermic
DDiamond has zero entropy at 298 K
(c)
Distinguish between thermodynamic stability and kinetic stability, using diamond as an example.
[2 marks]
Total for question 6: 4 marks
7
Hydrogen burns in oxygen to form liquid water: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ mol⁻¹ for the equation as written. Standard molar entropies in J K⁻¹ mol⁻¹: H₂(g) 131, O₂(g) 205, H₂O(l) 70.0.
(a)
Calculate the entropy change of the system for the reaction and explain its sign.
[3 marks]
(b)
Calculate the entropy change of the surroundings and the total entropy change at 298 K. Explain why the reaction is feasible even though the entropy change of the system is negative.
[4 marks]
Total for question 7: 7 marks
8
A chemist compares potassium iodide and silver iodide. Lattice energies in kJ mol⁻¹: KI experimental (from a Born–Haber cycle) −649 and theoretical (ionic model) −636; AgI experimental −889 and theoretical −778. Enthalpy changes of hydration in kJ mol⁻¹: K⁺ −322, Ag⁺ −464, I⁻ −305. The entropy change of the system on dissolving at 298 K is +107 J K⁻¹ mol⁻¹ for KI and +69 J K⁻¹ mol⁻¹ for AgI.
(a)
Explain what the differences between the experimental and theoretical lattice energies show about the bonding in each compound, and why silver iodide is affected more than potassium iodide.
[6 marks]
(b)
Use the experimental lattice energies and the other data to calculate the enthalpy change of solution and the total entropy change at 298 K for each compound, and hence predict which is soluble in water.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).