A-LevelChemistryEquilibria9701/22 · May/June 2022 · Q312 Marks

0.025 mol of HI(g) is added to a closed vessel and left to reach dynamic equilibrium. The total pressure of the vessel is 100 kPa. equation 1 2HI(g) ⇌ H₂(g) + I₂(g)

(a)(i)Explain what is meant by dynamic equilibrium.[2]
(a)(ii)Describe one difference in the initial appearance of the reaction mixture compared to the mixture at equilibrium.[1]
(a)(iii)Write an expression for Kp for the reaction described in equation 1.[1]
(a)(iv)At equilibrium the partial pressure of HI(g) is 86.4kPa. Calculate the amount of HI(g) present in the mixture at equilibrium. Show your working.[2]
(b)Use equation 1 and the bond energy values in Table 3.1 to calculate the change in enthalpy, ∆H, for the thermal decomposition of 1 mole of HI(g). Show your working. Table 3.1 bond bond energy/kJmol⁻¹ H-H 436 I-I 151 H-I 299[2]
(c)Describe the effect of increasing pressure on the value of Kp for the decomposition of HI(g).[1]
(d)(i)HCl(g) is prepared by adding NaCl(s) to concentrated H2SO4. HI(g) is not prepared by adding NaI(s) to concentrated H₂SO₄ because the HI(g) produced also reacts with concentrated H2SO4. Identify the type of reaction that occurs when NaI(s) reacts with concentrated H2SO4 to form HI(g).[1]
(d)(ii)Write an equation for the reaction of HI(g) and concentrated H2SO4.[1]
(d)(iii)Explain why HI(g) reacts with concentrated H2SO4 whereas HCl does not.[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[2] M1 rate(s) of forward and reverse / backward reactions are equal / are the same M2 no change in measurable properties OR concentration of reactants and products remain constant
[1] (colourless gas) becomes purple (gas)
[1] Kp = (pH²)(pI²) / (PHI)2
[2] M1 use correct expression for finding amount of HI used up in reaction 86.4 = 100 × (0.025 – x) / 0.025 − x + x/ 2 + x/2) where x = HI mol used up OR 86.4 = 100 × (0.025 – 2x) / (0.025 – 2x + x + x) where 2x = HI mol used up M2 correct calculation to find amount of HI(mol) at equilibrium based on HI(mol) at eq'm = 0.025 – HI(mol) used = 0.025 – 0.0034 = 0.0216 OR 0.022 (mol) OR alternative method M1 use the pp H2 and pp I2 at equilibrium (= 13.6) to calculate no mol H2 & I2 at equilibrium (either 0.0034 in total OR 0.0017 each) M2 use value for no mol H2 + I2 in expression to find no mol HI 0.025 - (no mol H2 + I2) = 0.0216
[2] evaluation, to give a value, based on calculation using: all three correct bond energies [HI = 299, H2 = 436, I2 = 151] correct use of stoichiometry to calculate ΔΗ for 1 mol HI x = 1 and y = ½ calculate a value for ∆H correctly i.e. x(299) – y(436 + 151) ΔΗ = (+)5.5 kJ mol-1
[1] none / no change
[1] acid-base
[1] 8HI + H2SO4 → 4I2 + H2S + 4H2O OR 8H+ + 8I- + H2SO4 → 4I2 + H2S + 4H2O
[1] explanation in terms of comparison of reducing nature of HI to HCl i.e. HI / I is stronger reducing agent compared to Cl-/HCl OR explanation in terms of additional behaviour of HI (compared to HCl) i.e. HI / I- reduces S (in sulfuric acid) / (concentrated) sulfuric acid