Thermodynamics questions
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Explain why the enthalpy of lattice dissociation for sodium oxide is greater than the enthalpy of lattice dissociation for sodium chloride.
Oxide ions M1 have higher (negative) charge M2 stronger attraction between (O2- and Na+/oppositely charged) ions Ignore electronegativity
Give a reason why data books do not contain a value for the enthalpy of solution of sodium oxide.
It reacts with water OR It reacts to form (a solution of) NaOH
Table 1 shows some enthalpy change data. Table 1 Enthalpy change / kJ mol–1 Ca2+(g) → Ca2+(aq) –1650 Cl–(g) → Cl–(aq) –364 Ca2+(g) + 2 Cl– (g) → CaCl2(s) –2237 Use the data in Table 1 to calculate the molar enthalpy change when calcium chloride dissolves in water.
(Enthalpy change = lattice dissociation energy + hydration energies of ions) M1 Enthalpy change = +2237 − (2 x 364) − 1650 M2 = − 141 kJ mol−1
Use your answer to part (a) to deduce how the temperature changes when calcium chloride dissolves in water.
Temperature goes up/increases
Explain why the enthalpy of hydration of fluoride ions is more negative than the enthalpy of hydration of chloride ions.
M1 fluoride ions/F− (ions) are smaller OR M1 F− has a higher charge density Do not accept fluorine atoms/ions are smaller M2 stronger attraction (of fluoride ion) to δ+ on H/ electron deficient H (in water)
Explain why the second ionisation energy of calcium is greater than the first ionisation energy of calcium.
Electron removed from a positive charge/ion OR Electron removed from smaller ion/electron removed closer to nucleus OR Stronger attraction between same number of protons and fewer electrons
Table 3 shows lattice enthalpies based on a perfect ionic model and lattice enthalpies from Born–Haber cycles for three metal chlorides. Table 3 Lattice enthalpy of dissociation / kJ mol–1 Perfect ionic model Born–Haber cycle Calcium chloride 2223 2237 Potassium chloride 690 701 Silver chloride 770 905 Discuss the values in Table 3. In your answer you should * compare the three values based on a perfect ionic model * compare the values based on a perfect ionic model to the values from a Born–Haber cycle for each compound.
1a Value for CaCl2 is larger OR Values for KCl and AgCl are similar OR Values for CaCl2 > AgCl > KCl 1b Ca2+ has a larger charge/ is a smaller ion OR Ag+ and K+ have smaller charge or larger ions 1c CaCl2 has stronger ionic bonds or stronger attraction between + and - ions (Ca2+and Cl- ) OR AgCl and KCl have weaker ionic bonds or weaker attraction between + and - ions (Ag+/ K+ and Cl- ) Stage 2 similarities in the perfect ionic model and Born-Haber cycle values 2a CaCl2 has similar values (between the perfect ionic model and Born-Haber cycle) 2b KCl has similar values (between the perfect ionic model and Born-Haber cycle) 2c CaCl2 and KCl have (almost) perfect ionic bonding or + ions are point charges/(perfectly) spherical Stage 3 difference in the perfect ionic model and Born-Haber cycle values 3a AgCl has larger difference in values (between the perfect ionic model and Born-Haber cycle) 3b AgCl contains (some) covalent character 3c Ag+ more polarising/distorts electron cloud more
Another student obtained a value of +15 kJ mol−1 using the same method. Suggest the main reason for the difference between this experimental value for the enthalpy of solution and the correct value of +26 kJ mol−1
Heat gain from surroundings
Table 3 shows some entropy data at 298 K Table 3 Entropy S / J K–1 mol–1 NH4NO3(s) 151 NH4 +(aq) 113 NO3 −(aq) 146 Calculate a value for the Gibbs free-energy change (∆G), at 298 K, for the reaction when ammonium nitrate dissolves in water. NH4NO3(s) + aq → NH4 +(aq) + NO3 −(aq) ∆H = +26 kJ mol−1 Use data from Table 3 and the value of ∆H from the equation. Assume for the solvent, water, that the entropy change, ∆S = 0
ΔS = (113 + 146) − 151 = +108 (J K−1 mol−1) 1 ΔG = ΔH − TΔS OR 26 − (298 × 108 × 10−3) ΔG = −6.184 / −6.18 / −6.2
Explain what the calculated value of ∆G indicates about the feasibility of this reaction at 298 K
negative value for ΔG indicates reaction is feasible/spontaneous
State why there is a difference between theoritical and experimental values
Has covalent character
Explain why enthalpy of hydration becomes less exothermic from Li+ to K+
From Li+ to K+ the size of ion increases and electrostatoc attarction betwen metal ion and Odeltanegative of water decreases
State the meaning of the term enthalpy change.
Heat (energy) change at constant pressure
Magnesium chloride dissolves in water. Give an equation, including state symbols, to represent the process that occurs when the enthalpy of solution of magnesium chloride is measured.
MgCl2(s) → Mg2+(aq) + 2 Cl‒ (aq)
The enthalpy of hydration of Ca2+(g) is –1650 kJ mol–1 Suggest why this value is less exothermic than that of Mg2+(g)
1 Ca2+ (ion) bigger/lower charge to size ratio (than Mg2+) M2 weaker attraction/bond to (Oδin) water
Use the equation and the data in the table above to calculate the minimum temperature, in K, at which this reaction becomes feasible.
M1 ΔH = (3 ͯ– 394) –(–1669 x 2) M1 correct expression 1 M2 = 2156 (kJ mol–1 ) M2 if –2156 seen allow 1 mark out M1 and M2 1 M3 ΔS = (28 x 4 + 214 x 3) – (51 x 2 + 6 x 3) M3 correct expression 1 M4 = 634 (J K–1 mol –1 ) M4 if – 634 allow 1 mark from M4 and M4 1 M5 ΔG= ΔH – T ΔS or ΔH =T ΔS or T = ΔH ÷ΔS M5 expression or rearranged expression or with numbers 1 M6 ΔS = 0.634 kJ K–1mol –1 M6 ΔS = M4 ÷ 1000 1 M7 T = = 3400 to 3401 (K)