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Zadanie 4a

7 pktobliczeniowetrudne (szac.)Paper 2, listopad 2023, TZ2

Wzorowane na: Chemistry HL Paper 2, November 2023, TZ2, pytanie 4(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 4a-4b.

Carbonyl sulfide, COS, is an impurity in some industrial gas streams. It can be removed by gas-phase hydrolysis according to the equilibrium

COS(g) + H2O(g) ⇌ CO2(g) + H2S(g)

Polecenie

(i) Calculate the enthalpy change for this reaction, using section 13 of the data booklet and the values given below.

QuantityCOS(g)H2S(g)
ΔHf⊖ / kJ mol−1−142−20.6

(ii) Outline why the entropy change of this reaction is expected to be small.

(iii) Ignore the entropy change. Use your answer to (i) together with sections 1 and 2 of the data booklet to estimate Kc at 500 K. (If you did not obtain an answer to (i), use −25.0 kJ mol−1, but this is not the correct value.)

(iv) The concentrations of the species involved at equilibrium at 500 K are:

COS(g)H2O(g)CO2(g)H2S(g)
4.00 × 10−4 mol dm−31.00 × 10−3 mol dm−3x mol dm−3x mol dm−3

Using your answer to (iii), calculate x, the equilibrium concentration of carbon dioxide. (If you did not obtain an answer to (iii), use 4.10 × 102, but this is not the correct value.)

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Odpowiedź z klucza

(i) «∑ΔHf⊖\sum \Delta H_\mathrm{f}^\ominus(reactants) = −142 + (−242) =» −384 «kJ mol−1» AND «∑ΔHf⊖\sum \Delta H_\mathrm{f}^\ominus(products) = −394 + (−20.6) =» −414.6 «kJ mol−1» ✔

«ΔH⊖ = −414.6 − (−384) =» −30.6 «kJ mol−1» ✔

(ii) the same number of moles of gas on each side of the equation «2 mol to 2 mol» ✔

(iii) «ΔG⊖ = ΔH⊖ − TΔS⊖ ≈ ΔH⊖ = −30.6 kJ mol−1»

«ln⁡Kc=−ΔG⊖RT=30 600 J mol−18.31 J K−1 mol−1×500 K=7.36\ln K_\mathrm{c} = -\dfrac{\Delta G^\ominus}{RT} = \dfrac{30\,600\ \mathrm{J\,mol^{-1}}}{8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \times 500\ \mathrm{K}} = 7.36» ✔

Kc = 1.58 × 103 ✔

(iv) «Kc=[COX2][HX2S][COS][HX2O]K_\mathrm{c} = \dfrac{[\ce{CO2}][\ce{H2S}]}{[\ce{COS}][\ce{H2O}]}» 1.58×103=x×x4.00×10−4×1.00×10−31.58 \times 10^{3} = \dfrac{x \times x}{4.00 \times 10^{-4} \times 1.00 \times 10^{-3}} ✔

«x=1.58×103×4.00×10−7=x = \sqrt{1.58 \times 10^{3} \times 4.00 \times 10^{-7}} =» 0.0251 «mol dm−3» ✔

Schemat punktowania

(i) [2]: M1 for the two sums, M2 for the enthalpy change with its sign. Award [2] for −30.6 «kJ mol⁻¹». Award [1] for +13.4 «kJ mol⁻¹», obtained with −286 for liquid water. (ii) [1]. (iii) [2]: M1 for ln Kc = −ΔG⊖/RT with ΔG⊖ in joules, M2 for Kc. Award [2] for a correct final answer. Using −25.0 gives Kc = 4.10 × 102; ECF from +13.4 gives Kc = 3.98 × 10−2. (iv) [2]: M1 for the expression with the data substituted, M2 for x. Award [2] for a correct final answer. Using 4.10 × 102 gives 0.0128; ECF from 3.98 × 10−2 gives 1.26 × 10−4.

Komentarz

The estimate ignores the term TΔS, which is justified here: ΔS is only about −1 J K⁻¹ mol⁻¹, so TΔS is about −0.4 kJ mol⁻¹ at 500 K, far smaller than ΔH.

Umiejętności w zadaniu