Sulfur trioxide, SO3, is manufactured from sulfur dioxide and oxygen by the Contact process. 2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = –196.0 kJ mol⁻¹
- (a)(i)The enthalpy change of formation of SO2, ΔHƒ SO2(g), is –296.8 kJ mol⁻¹. Define the term enthalpy change of formation.[2]
- (a)(ii)Use the data to calculate the enthalpy change of formation of SO3(g). ΔHƒ SO3(g) = __________ kJ mol⁻¹[2]
- (b)(i)The Contact process is usually carried out at a temperature of approximately 700 K, a pressure of approximately 150 kPa and in the presence of a vanadium(V) oxide catalyst, V2O5. The Boltzmann distribution for a mixture of SO2 and O2 at 700K is shown. Eacat represents the activation energy for the reaction in the presence of the catalyst. [Figure: Boltzmann distribution curve with x-axis 'molecular energy' and y-axis 'proportion of molecules with a given energy'. Eacat is marked on the curve.] Add a labelled mark, Eauncat, to the diagram to indicate the activation energy in the absence of the catalyst.[1]
- (b)(ii)State the benefit of using a catalyst in this reaction. Explain how it achieves this effect.[2]
- (b)(iii)State and explain how an increase in pressure would affect both the rate of reaction and the yield of SO3 in the Contact process. rate __________ yield __________[4]
- (c)(i)At a pressure of 1.50 × 10⁵ Pa, 1.00 mol of sulfur dioxide gas, SO2, was mixed with 1.00 mol of oxygen gas, O2. The final equilibrium mixture formed was found to contain 0.505 mol of O2. 2SO2(g) + O2(g) ⇌ 2SO3(g) Calculate the amount, in mol, of SO2 and SO3 in the equilibrium mixture. SO2 = __________ mol SO3 = __________ mol[1]
- (c)(ii)Calculate the partial pressure of oxygen gas, pO2, in the equilibrium mixture. pO2 = __________ Pa[2]
- (d)(i)In another equilibrium mixture formed from different starting amounts of SO2 and O2, the partial pressures of SO2, O2 and SO3 were as shown. pSO2 = 8.42 × 10² Pa pO2 = 6.00 × 10⁴ Pa pSO3 = 9.10 × 10⁴ Pa Write the expression for the equilibrium constant, Kp, for the production of SO3 from SO2 and O2. Kp = __________[1]
- (d)(ii)Calculate the value of Kp for this reaction and state the units. Kp = __________ units = __________[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [1] (enthalpy / energy change) when one mole of a compound is formed
- [1] from its elements in their standard states / standard conditions
- [1] (ΔHᵣ = ΣΔHƒ products – ΣΔHƒ reactants) -196 =2∆H SO3- (2 × -296.8)
- [1] ∆Η SO3 = −394.8 (kJ mol⁻¹)
- [1] Mark to right of original Ea
- [1] Benefit of using a catalyst in terms of increasing rate or economic benefit i.e. (less heat required)
- [1] Creates alternative pathway with lower Ea
- [1] More molecules with E > Ea
- [1] (rate) increases AND correct explanation in terms of 'more collisions'
- [1] more successful collisions per unit time / higher chance of successful collisions per unit time / higher proportion of successful collisions per unit time
- [1] (yield) increases and shifts equilibrium to the right / in the forward direction / towards SO3/towards the product/in exothermic direction
- [1] to oppose the change or oppose the increase in pressure / fewer molecules on RHS so eqm moves to right (to oppose change)
- [1] SO2 = 0.01 (mol)
- AND
- [1] SO3 = 0.99 (mol)
- [1] nTOT = 1.505
- [1] pO2 = 1.50 × 10⁵ × (0.505 / 1.505) = 5.03 × 10⁴ (Pa)
- [1] Kp = pSO3² / (pO2 × pSO2²)
- [1] 0.1946737305
- [1] Pa⁻¹
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