Edexcel Jun 2014 (IAL) Paper 4 Q21 (with explained solutions) |

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Edexcel Jun 2014 (IAL) Paper 4 Q21

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21 This question concerns the reaction of hydrogen with iodine to form hydrogen iodide at 700 K. H2(g) + I2(g) (cid:85) 2HI(g) (cid:507)H = 10 kJ mol1 (a) (i) Write the expression for the equilibrium constant, Kp, for this reaction. *(ii) 1 mol of hydrogen was mixed with 1 mol of iodine in a sealed container and left to reach equilibrium at 700 K. The total pressure was 5 atm. At equilibrium, the amount of iodine remaining was 0.21 mol. Calculate the partial pressure of each gas at equilibrium. Use the partial pressures to calculate the value of Kp, stating its units, if any. (1) (5) Kp =units*P42992A02128* 21 Turn over<br />
(b) State the effect of increasing the pressure on the equilibrium position. Justify your answer by using the equation: H2(g) + I2(g) (cid:85) 2HI(g) (cid:507)H = 10 kJ mol1 (1)(c) (i) Explain how increasing the temperature affects the value of (cid:507)Stotal of this reaction. Assume that (cid:507)Ssystem does not change when the temperature increases. (2)*(ii) Use your answer to (c)(i) to explain the effect of an increase in temperature on the value of Kp and the equilibrium yield of hydrogen iodide. (2)(Total for Question 21 = 11 marks) TOTAL FOR SECTION B = 49 MARKS 22 *P42992A02228*<br />

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Acceptable Answers Question Number 21(a)(i) (Kp =) P2 HI H2 P x P I2 ALLOW 2 PHI ALLOW lower or upper case p /pp/curved brackets IGNORE state symbols even if incorrect IGNORE missing x Mark Reject square brackets expressions without p/pp/P/PP to show partial pressure WCH0401 1406<br />
Question Number 21(a)* (ii) Acceptable Answers initial mol eqm mol partial pressure Kp H2 1 I2 1 2HI 0.21 0.21 1.58 0.21 x 5 0.21 x 5 1.58 x 5 = 0.525 = 0.525 3.952 = 56.6/57 no units 0.5252 = 3.95 Reject Mark First mark eqm mol of H2 = 0.21 (1) Second mark eqm mol of HI = 1.58 (see note at end) (1) Third mark all 3 partial pressures, either working or answer (1) consequential on their equilibrium moles Fourth mark correct value (56.6/57) for Kp with or without working (1) consequential on their partial pressures and their expression for Kp in (a)(i) Fifth mark no units stated / working to show that units cancel / statement that units cancel ALLOW - /atm0 or similar indication of no units ALLOW units based on expression for Kp in (i) or the expression they have used in the calculation in (ii) (1) Note Only award 5 marks if partial pressures are stated or working to show them If 0.21 mol H2 (1) and 0.79 mol HI(0), then partial pressures H2/I2 = 0.868 and HI = 3.264 (1) Kp = 14/14.2 (1) no units (1) If 0.21 mol H2 (1) and 0.42 mol HI(0), then partial pressures H2/I2 = 1.25 and HI = 2.5 (1) Kp = 4 (1) no units (1) Alternative method for marks 2 and 3 If initial P(H2) and P(I2) = 2.5 (atm) Eqm P(H2) and P(I2) = 2.5x0.21 = 0.525(atm) (1) P(HI) = 5-(2 x 0.525) = 3.95 (1) IGNORE SF except 1SF WCH0401 1406<br />
Question Number 21(b) Acceptable Answers Reject Mark No effect because (there are) equal numbers of (gas) mole(cule)s on each side of the equation Just equal numbers of mole(cule)s Acceptable Answers Question Number 21(c)(i) First mark Reject Mark Question Number 21(c)* (ii) (H is negative/exothermic so)H/T gets less positive ALLOW decreases OR (H is negative/exothermic so) Ssurroundings gets less positive ALLOW decreases (1) Second mark (so, since Stotal =H/T + Ssystem) (1) Stotal decreases Mark independently No TE on incorrect Ssurroundings Ignore comments based on Kp Acceptable Answers Reject Mark First mark lnK = Stotal/R OR K = eStotal/R OR Stotal = RlnK OR Stotal is (directly) proportional to lnK (1) Second mark Kp decreases and yield (of HI) decreases (1) consequential on their Stotal in (c)(i) Ignore comments based on Le Chateliers principle Stotal is (directly) proportional to K Just equilibrium position moves to the left without reference to yield and kp (Total for Question 21 = 11 marks) TOTAL FOR SECTION B = 49 MARKS WCH0401 1406<br />

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