Chemistry · Chemical thermodynamics · chapter 5 of 9

Class 11 Chemistry Chapter 5: Thermodynamics case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

A student measures the enthalpy of solution of a salt MX in a calorimeter made of a foamed polystyrene cup with a lid and a thermometer, open to the atmosphere. She dissolves 0.100 mol of MX in 100 mL of water. The temperature of the solution falls from 25.0 °C to 21.8 °C. The heat capacity of the cup and its contents, found earlier with a heater, is 440 J K⁻¹. Heat lost to the room during the short experiment can be neglected. The lattice enthalpy of MX is +720 kJ mol⁻¹.

Part 1 · 1 mark

The quantity this experiment measures directly is

(a) ΔU, because the cup is closed with a lid(b) ΔH, because the process occurs at constant (atmospheric) pressure(c) ΔS, because heat flows at constant temperature(d) ΔG, because the dissolution is spontaneous

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Answer: (b) ΔH, because the process occurs at constant (atmospheric) pressure

The cup is at atmospheric pressure, so the heat measured is q_p = ΔH, the enthalpy of solution.

Part 2 · 2 marks

Calculate Δ_solH of MX in kJ mol⁻¹ and state its sign with a reason.

Answer and marking scheme

Heat drawn from the solution = C × ΔT = 440 J K⁻¹ × 3.2 K = 1408 J. This heat is absorbed by the dissolving salt, so Δ_solH = +1408 J / 0.100 mol = +14080 J mol⁻¹ = +14.1 kJ mol⁻¹. Positive, since the temperature fell: dissolution is endothermic.

q = 440 × 3.2 = 1408 J (1)
Δ_solH = +14.1 kJ mol⁻¹, endothermic as temperature fell (1)

Would the value found change noticeably if the same dissolution were carried out in a sealed vessel at constant volume? Explain with the relation between ΔH and ΔU.

Answer and marking scheme

ΔH = ΔU + pΔV. Only a solid and a liquid take part and no gas is formed, so the volume change is negligible and pΔV is negligible. ΔH and ΔU are therefore practically equal, and the value would not change noticeably.

ΔH = ΔU + pΔV (1)
no gas, negligible ΔV, so ΔH ≈ ΔU: no noticeable change (1)

Part 3 · 1 mark

Using Δ_solH = +14.1 kJ mol⁻¹, the enthalpy of hydration of the ions of MX is

(a) +734.1 kJ mol⁻¹(b) −734.1 kJ mol⁻¹(c) +705.9 kJ mol⁻¹(d) −705.9 kJ mol⁻¹

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Answer: (d) −705.9 kJ mol⁻¹

Δ_solH = Δ_latticeH + Δ_hydH, so Δ_hydH = 14.1 − 720 = −705.9 kJ mol⁻¹.

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