Lattice energy, Born–Haber & entropy: A Level Chemistry questions with answers
Free Lattice energy, Born–Haber & entropy questions for A Level Chemistry (9701), part of Physical chemistry. Each question has the answer and the marking points an examiner would look for.
Lattice energy, Born–Haber & entropy practice questions
Question 1Structured · 4 marksA* challenge
Entropy and Gibbs free energy determine feasibility.
(a)
Explain why the entropy increases when a solid dissolves.
[2]
(b)
For a reaction ΔH = +178 kJ mol−1 and ΔS = +161 J K−1 mol−1. Calculate the minimum temperature for the reaction to be feasible.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
more disorder, more ways to arrange
particles become more dispersed/disordered B1 more ways of arranging energy B1
2
(b)
1106 K
T = ΔH/ΔS = 178 000/161 C1 1106 K A1
2
Question 2Structured · 11 marksA* challenge
Data for the formation of CaCl2 (kJ mol−1): ΔHat(Ca) = +178; first IE = +590; second IE = +1145; E(Cl–Cl) = +242; electron affinity of Cl = −349; ΔHf(CaCl2) = −795.
(a)
Define lattice energy.
[2]
(b)
Construct a Born–Haber cycle and calculate the lattice energy of CaCl2.
[4]
(c)
Explain how the lattice energy of MgCl2 compares with that of CaCl2.
[2]
(d)
A reaction has ΔH = −90 kJ mol−1 and ΔS = −120 J K−1 mol−1. Determine the temperature range over which the reaction is feasible.
[3]
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Part
Answer and marking points
Marks
(a)
energy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions
one mole of ionic solid from gaseous ions B1 standard conditions B1
2
(b)
−2252 kJ mol−1
correct terms in cycle M1 2 × electron affinity and full E(Cl–Cl) used M1 correct arithmetic M1 −2252 A1
4
(c)
MgCl2 more exothermic: Mg2+ smaller, higher charge density, stronger attraction
MgCl2 more exothermic/negative B1 smaller ion radius, same charge B1
For the reaction 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g), ΔH = +129 kJ mol−1 and ΔS = +335 J K−1 mol−1.
(a)
Explain why ΔS is positive for this reaction.
[2]
(b)
Calculate ΔG at 298 K and state whether the reaction is feasible at this temperature.
[3]
(c)
Calculate the minimum temperature at which the reaction becomes feasible.
[2]
(d)
Many reactions with a negative ΔG at 298 K do not take place noticeably at room temperature. Explain why.
[1]
(e)
Sketch how ΔG varies with temperature for this reaction, labelling the temperature at which it becomes feasible.
Fig. 3.1
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
more moles of gas produced
the number of moles of gas increases B1 so the disorder / number of possible arrangements increases B1
2
(b)
29.2 kJ mol−1; not feasible
ΔG = ΔH − TΔS with ΔS in kJ M1 29.2 kJ mol−1A1 not feasible as ΔG > 0 A1
3
(c)
385 K
T = ΔH/ΔS M1 385 K A1
2
(d)
high activation energy, kinetically slow
the activation energy may be high, so the rate is negligible (kinetic stability) B1
1
(e)
straight line with gradient −ΔS, crossing zero at T = ΔH/ΔS
straight line with negative gradient (−ΔS) and intercept ΔH B1 crossing ΔG = 0 at 385 K B1
2
Question 4Structured · 14 marks
This question is about the lattice energy of MgO.
enthalpy change
value / kJ mol−1
atomisation of Mg
+148
first + second ionisation energies of Mg
+2186
atomisation of O
+249
first + second electron affinities of O
+657
formation of MgO
−602
Table 4.1
(a)
Define lattice energy.
[2]
(b)
Construct a Born–Haber cycle and use it to calculate the lattice energy.
[4]
(c)
Explain why the lattice energy of MgO is much more exothermic than that of NaCl.
[2]
(d)
Explain why the second electron affinity of oxygen is endothermic.
[2]
(e)
For the decomposition CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol−1 and ΔS = +161 J K−1 mol−1. Calculate the minimum temperature at which the reaction is feasible.
[3]
(f)
Explain why the entropy change for this decomposition is positive.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
enthalpy change when 1 mol of ionic solid formed from gaseous ions under standard conditions
enthalpy change when one mole of an ionic compound is formed B1 from its gaseous ions (under standard conditions) B1
2
(b)
−3842 kJ mol−1
cycle with correct species and state symbols M1 ΔHf = sum of steps + LE M1 LE = −602 − (148 + 2186 + 249 + 657) M1 −3842 kJ mol−1A1
Mg²⁺ and O²⁻ have higher charges than Na⁺ and Cl⁻ B1 and smaller ionic radii, so stronger electrostatic attraction B1
2
(d)
adding electron to negative ion; repulsion must be overcome
the electron is added to an O⁻ ion B1 energy is needed to overcome repulsion between the negative ion and the electron B1
2
(e)
1106 K
ΔG = ΔH − TΔS ≤ 0 M1 T = 178 000/161 (units converted) M1 1106 K A1
3
(f)
gas produced from solid; more disorder
a gas is formed from a solid, so the number of ways of arranging particles/energy increases B1
1
Question 5Structured · 11 marks
Table 5.1 shows data for a Born–Haber cycle for calcium oxide, CaO.
process
ΔH / kJ mol⁻¹
enthalpy change of formation of CaO(s)
−635
enthalpy change of atomisation of Ca(s)
+178
first ionisation energy of Ca
+590
second ionisation energy of Ca
+1145
enthalpy change of atomisation of oxygen, ½O₂(g) → O(g)
+249
first electron affinity of oxygen
−141
second electron affinity of oxygen
+798
Table 5.1
(a)
Define the term lattice energy.
[2]
(b)
Calculate the lattice energy of CaO.
[3]
(c)
Explain why the second electron affinity of oxygen is endothermic.
[2]
(d)
Explain, in terms of ionic charge and ionic radius, why the lattice energy of CaO is much more exothermic than that of KCl.
[2]
(e)
The ionic radius of Mg²⁺ is 0.072 nm and that of Ca²⁺ is 0.100 nm. Suggest how the lattice energy of MgO compares with that of CaO. Explain your answer.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
energy change when 1 mol of ionic solid formed from gaseous ions
the enthalpy change when one mole of an ionic compound is formed B1 from its gaseous ions (under standard conditions) B1
2
(b)
−3454 kJ mol−1
ΔHf = sum of other steps + LE (cycle) M1 LE = −635 − (178 + 590 + 1145 + 249 −141 + 798) M1 −3454 kJ mol−1A1
3
(c)
adding electron to negative ion; repulsion must be overcome
the electron is added to a negative ion (O⁻) B1 energy is needed to overcome the repulsion between the ion and the electron B1
2
(d)
double charges; smaller ions; stronger attraction
Ca²⁺ and O²⁻ have twice the charges of K⁺ and Cl⁻ B1 Ca²⁺ and O²⁻ are smaller than K⁺ and Cl⁻, so the ions are closer and the electrostatic attraction is stronger B1
2
(e)
MgO more exothermic
magnesium oxide is more exothermic B1 Mg²⁺ is smaller, so the attraction between ions is stronger B1
2
Question 6Structured · 11 marks
This question is about the enthalpy change of solution of MgCl₂. Lattice energy = −2526 kJ mol−1; enthalpy change of hydration of Mg²⁺ = −1920 kJ mol−1; of Cl⁻ = −378 kJ mol−1.
(a)
Define enthalpy change of hydration.
[2]
(b)
Construct an energy cycle and calculate the enthalpy change of solution.
[3]
(c)
Explain why the enthalpy change of hydration of Mg²⁺ is more exothermic than that of Ca²⁺.
[2]
(d)
Define entropy and predict the sign of the entropy change when a solid ionic compound dissolves.
[2]
(e)
Explain how a salt with an endothermic enthalpy of solution can still be soluble at room temperature.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
enthalpy change when 1 mol of gaseous ions dissolves in water to give infinitely dilute solution
enthalpy change when one mole of gaseous ions B1 is dissolved in a large excess of water (infinitely dilute solution) B1
Mg²⁺ smaller, higher charge density; stronger ion–dipole attraction to water
Mg²⁺ has a smaller ionic radius (same charge) so a higher charge density B1 it attracts the δ− oxygen of water molecules more strongly B1
2
(d)
measure of disorder / number of ways of arranging; usually positive
entropy is a measure of the number of ways particles and energy can be arranged (disorder) B1 usually positive because ions are freed from the lattice B1
2
(e)
ΔS positive enough that ΔG = ΔH − TΔS negative
dissolving can be feasible if ΔG = ΔH − TΔS < 0 B1 a large positive ΔS makes −TΔS outweigh the positive ΔH B1
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