1. (a) When solid calcium nitrate is heated, brown fumes of nitrogen dioxide, NO 2, are seen and the solid remaining after decomposition is calcium oxide. Write a balanced equation for the thermal decomposition of calcium nitrate.... Describe the changes you would see when cold water is added drop by drop to cold calcium oxide and give the chemical equation for the reaction............. (3) (iii) State whether barium nitrate will decompose more easily or less easily than calcium nitrate on heating with a Bunsen burner.... (iv) Account for the trend in the thermal stability of the nitrates of the elements in group 2............. (3) (b) The brown fumes in part (a) are not pure NO 2 but a mixture of N 2 O 4 and NO 2. N 2 O 4 (g) 2NO 2 (g) Pale yellow dark brown A transparent glass syringe was filled with the gaseous mixture of N 2 O 4 and NO 2 and its tip sealed. When the piston of the syringe was rapidly pushed well into the body of the syringe, thereby compressing the gas mixture considerably, the colour of the gas became momentarily darker but them became lighter again. Maltby Academy 1
Suggest why compressing the gases causes the mixture to darken....... Explain why the mixture turns lighter on standing.......... (iii) Write an expression for the equilibrium constant, K p, for this equilibrium. (iv) 1.0 mole of N 2 O 4 was allowed to reach equilibrium at 400K. At equilibrium the partial pressure of N 2 O 4 was found to be 0.15 atm. Given that the equilibrium constant K p for this reaction is 48 atm, calculate the partial pressure of NO 2 in the equilibrium mixture. (3) (Total 16 marks) Maltby Academy 2
2. (a) Calculate the concentration, in mol dm 3, of a solution of hydrochloric acid, HCl, which has a ph of 1.13. Calculate the concentration, in mol dm 3, of a solution of chloric(l) acid, HOCl, which has a ph of 4.23. Chloric(l) acid is a weak acid with K a = 3.72 10 8 mol dm 3. (4) (b) The ph of 0.100 mol dm 3 sulphuric acid is 0.98. Calculate the concentration of hydrogen ions, H +, in this solution. Write equations to show the two successive ionisations of sulphuric acid, H 2 SO 4, in water. Maltby Academy 3
(iii) Suggest why the concentration of hydrogen ions is not 0.20 mol dm 3 in 0.100 mol dm 3 sulphuric acid. (c) Many industrial organic reactions produce hydrogen chloride as an additional product. This can be oxidised to chlorine by the Deacon process: 4HCl(g) + O 2 (g) 2Cl 2 (g) + 2H 2 O(g) H = 115 kj mol 1. 0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of volume 10.0 dm 3 in the presence of a copper(i) chloride catalyst at 400 ºC. At equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride. Write an expression for the equilibrium constant K c. Calculate the value of K c at 400 ºC. (4) Maltby Academy 4
(d) State and explain the effect, if any, on the position of equilibrium in (c) of: decreasing the temperature; decreasing the volume; (iii) removing the catalyst. (Total 20 marks) 3. In the vapour phase sulphur trioxide dissociates: 2SO 3 (g) 2SO 2 (g) + O 2 (g) (a) Write an expression for K p for this dissociation. Maltby Academy 5
At a particular temperature, 75% of the sulphur trioxide is dissociated, producing a pressure of 10 atm. Calculate the value of K p at this temperature paying, attention to its units. (5) (b) Solid vanadium(v) oxide, V 2 O 5, is an effective catalyst for this reaction. State the effect of using double the mass of catalyst on: the position of the equilibrium; the value of K p. (Total 8 marks) 4. (a) Methane reacts with steam in a reversible reaction. In industry this reaction, carried out at a pressure of 30 atm, is used to produce hydrogen for the manufacture of ammonia CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) H = +210 kj mol 1 Maltby Academy 6
Define the term partial pressure as applied to a gas mixture. Write an expression for the equilibrium constant, K p, for this reaction. (iii) State and explain the effect of increasing the total pressure on the position of this equilibrium; (b) State the effect on the value of K p for this equilibrium of the following. Increasing the total pressure. Increasing the temperature. Maltby Academy 7
(iii) Adding a catalyst. (c) There is a theory that methane, CH 4, constantly leaks from the earth s crust. This is not noticeable on land but at the bottom of a cold sea, such as off the Canadian coast, the methane is trapped in a solid cage of water molecules. CH 4 (g) + 6H 2 O(s) [CH 4 (H 2 O) 6 ](s) methane hydrate At 29 C the equilibrium pressure of the methane is 101.3 kpa. Write an expression for K p for this equilibrium. Deduce the value of K p at 29 C, stating its units. Maltby Academy 8
(iii) At 0 C the equilibrium pressure of methane rises to 2600 kpa. What does this tell you about the effect of temperature change on the position of equilibrium and about the enthalpy change for this reaction? (iv) Some people have suggested collecting the methane hydrate from the bottom of the sea and allowing it to warm up to 0 C on board a ship. Comment on whether this would be a useful method for collecting methane. (Total 12 marks) 5. (a) Define the term partial pressure.......... (b) If phosphorus pentachloride is heated in a sealed tube the following equilibrium is set up PCl 5 (g) PCl 3 (g) + Cl 2 (g) Maltby Academy 9
Write an expression for the equilibrium constant, K p, for the above reaction. At a given temperature T a sample of phosphorus pentachloride is 40.0% dissociated, the total equilibrium pressure being 2.00 atm. Calculate the partial pressures of each of the components of the equilibrium mixture, to three significant figures. Hence calculate the value of K p under these conditions. (5) (c) At a higher temperature, T + t, the equilibrium mixture in (b) contains a greater proportion of phosphorus trichloride and chlorine. State, with a reason, whether the dissociation of phosphorus pentachloride is exothermic or endothermic. Maltby Academy 10
How does K p change when the temperature is increased? (d) If calcium carbonate is heated in a sealed vessel in the absence of air the following equilibrium is set up: CaCO 3 (s) CaO(s) + CO 2 (g) Write the expression for K p for this equilibrium. At 1030 C the total pressure in the vessel is 16 atm. What is the value of K p? (Total 12 marks) 6. When dinitrogen tetroxide, N 2 O 4, dissociates, the following equilibrium is established. N 2 O 4 (g) 2NO 2 (g) (a) State a property which could be measured to follow the progress of this reversible reaction.... (b) Write an expression for the equilibrium constant, K c, for this reaction. Maltby Academy 11
(c) When a sample of 0.0370 moles of gaseous dinitrogen tetroxide is allowed to dissociate at 25 C in a container of volume 1 dm 3, 0.0310 moles of N 2 O 4 (g) remain in the equilibrium mixture. Complete the table below, and use the data to calculate K c for the reaction. Include a unit in your answer. N 2 O 4 NO 2 Number of moles at start 0.0370 0 Number of moles in 1 dm 3 at equilibrium 0.0310 K c calculation: (3) (d) The reaction was repeated at a higher pressure, maintaining the temperature at 25 C. How does this increase in pressure affect the amount of nitrogen dioxide, NO 2 (g), in the equilibrium mixture?... How does this increase in pressure affect the value of K c?... Maltby Academy 12
(e) The reaction was repeated at the original pressure, but the temperature was increased to 75 C. The value of K c was approximately twenty times greater. How does this information show that the reaction is endothermic?......... (f) Predict the sign of ΔS system for the reaction, giving a reason for your answer.......... (g) Write the equation for the relationship between ΔS surroundings and ΔH for the reaction. (h) The magnitude of ΔS system for the reaction is greater than the magnitude of ΔS surroundings. Explain why this must be the case............. (Total 13 marks) Maltby Academy 13
7. This question concerns the equilibrium 2NO(g) N 2 (g) + O 2 (g) H = 180 kj mol 1 (a) Define the term partial pressure....... (b) Write the expression for K p for the above reaction. At 1600 C and 1.5 atm pressure NO is 99 % dissociated at equilibrium. Calculate the value of K p under these conditions. (4) Maltby Academy 14
(c) State and explain the effect on K p and hence on the position of equilibrium of decreasing the temperature at constant pressure................ (3) (d) The reaction Ni(s) + 4CO(g) Ni(CO) 4 (g) is used to purify nickel. Write the expression for K p for this system. In order to achieve a high equilibrium yield of Ni(CO) 4 should a low or a high partial pressure of carbon monoxide be used? Explain your answer in terms of K p.......... (Total 12 marks) Maltby Academy 15
8. The equation below shows a possible reaction for producing methanol. CO(g) + 2H 2 (g) CH 3 OH(l) ΔH ο = 129 kj mol 1 (a) The entropy of one mole of each substance in the equation, measured at 298 K, is shown below. Substance S ο /J mol 1 K 1 CO(g) 197.6 H 2 (g) 130.6 CH 3 OH(l) 239.7 Suggest why methanol has the highest entropy value of the three substances... Calculate the entropy change of the system, ΔS ο system, for this reaction. (iii) Is the sign of ΔS ο system as expected? Give a reason for your answer.... Maltby Academy 16
(iv) Calculate the entropy change of the surroundings ΔS ο surroundings, at 298 K. (v) Show, by calculation, whether it is possible for this reaction to occur spontaneously at 298 K. (b) When methanol is produced in industry, this reaction is carried out at 400 ºC and 200 atmospheres pressure, in the presence of a catalyst of chromium oxide mixed with zinc oxide. Under these conditions methanol vapour forms and the reaction reaches equilibrium. Assume that the reaction is still exothermic under these conditions. CO(g) + 2H 2 (g) CH 3 OH(g) Suggest reasons for the choice of temperature and pressure. Temperature...... Pressure..... (3) Maltby Academy 17
The catalyst used in this reaction is heterogeneous. Explain this term... (iii) Write an expression for the equilibrium constant in terms of pressure, K p, for this reaction. CO(g) + 2H 2 (g) CH 3 OH(g) (iv) In the equilibrium mixture at 200 atmospheres pressure, the partial pressure of carbon monoxide is 55 atmospheres and the partial pressure of hydrogen is 20 atmospheres. Calculate the partial pressure of methanol in the mixture and hence the value of the equilibrium constant, K p. Include a unit in your answer. Maltby Academy 18
(c) The diagram below shows the distribution of energy in a sample of gas molecules in a reaction when no catalyst is present. The activation energy for the reaction is E A. What does the shaded area on the graph represent?. Draw a line on the graph, labelled E C, to show the activation energy of the catalysed reaction. (Total 17 marks) 9. (a) 2Ca(NO 3 ) 2 2CaO + 4NO 2 + O 2 formulae correct balance. Ignore any state symbols. The balance mark is not stand alone. 2 steam / fizzing sound / crumbles solid swells up / milky liquid produced / comment about sparingly soluble substance CaO + H 2 O Ca(OH) 2 ignore any state symbols 3 (iii) less 1 Maltby Academy 19
(iv) (Cat)ion size increases down the Group / charge density decreases (not atom size) The polarizing power of the cation decreases down the Group. The less polarized the anion is by the cation the more difficult the nitrate is to decompose. Polarisation mark could come from the less the electron cloud is distorted or trend in cation size comparison of the lattice energies of the nitrate and the oxide balance in favour of oxide at top of group and the nitrate at the bottom 3 (b) same number of particles in a smaller volume / gas density increased 1 comment related to the number of molecules on each side to explain a shift to l.h.s. (not just due to Le Chatelier ) so at higher pressure equilibrium moves to favour N 2 O 4 2 (iii) K p = p(no 2 ) 2 p(n 2 O 4 ) There must be some symbolism for pressure, and no [ ] 1 (iv) (K p = p(no 2 ) 2 = 48) p(n 2 O 4 ) p(no 2 ) 2 = 48 0.15 = 7.2 p(no 2 ) = 2.7 atm accept 2.683 / 2.68 / 2.7 Answer and units conditional on (iii). 3 [16] 10. (a) HCl: ph = 1.13 [H + ] = 0.074 mol dm 3 [HCl] = 0.074 mol dm 3 [0.074 to 0.07413] 1 HOCl: ph = 4.23 [H + ] = 5.89 10 5 mol dm 3 K a = [H + ] [OCl ] [HOCl] [H + ] = [OCl ] or implied later in calculation [HOCl] = [H + ] 2 / K a = 0.0932 mol dm 3 4 Maltby Academy 20
(b) [H + ] = 0.10 / 0.1047 / 0.105 H 2 SO 4 H + + HSO 4 or H 2 SO 4 + H 2 O H 3 + O + HSO 4 ignore state symbols HSO 4 H + + SO 4 2 Must be H 2 SO 4 + H 2 O H 3 O + + SO 4 2 ignore state symbols (iii) second ionisation suppressed by the first ionisation 4 (c) K c 2 [Cl 2 ] [H2O] 4 [HCl] [O ] 1 2 2 4HCl + O 2 2Cl 2 + 2H 2 O equilibrium mols 0.20 0.050 0.30 and 0.30 [ ] eq 10 0.020 0.0050 0.030 0.030 K c = [0.030] 2 x [0.030] 2 = 1010 or 1012 or 1013 or 1012.5 [0.020] 4 x [0.005] (mol 1 dm 3 ) 4 (d) As reaction (left to right) is exothermic Decrease in temperature drives equilibrium to from left to right 2 As more (gas) molecules on the left, equilibrium is driven from left to right (iii) A catalyst has no effect As it only alters the rate of the reaction not the position of equilibrium / it alters the rate of the forward and reverse reactions equally 2 2 [20] PSO2 P 11. (a) Kp = 2 P 2 SO3 O2 [ ] no mark ( ) OK 1 2SO 3 2SO 2 + O 2 Mols at start 2 0 0 mols at equ 0.5 1.5 0.75 Mark by process 1 mark for working out mole fraction 1 mark for 10 1 mark for correct substitution in K p and answer 1 mark for unit Maltby Academy 21
i.e. P SO2 = P O2 = P SO3 = 1.5 2.75 0.7 2.75 0.5 2.75 10 = 5.46 10 = 2.73 10 = 1.83 n.b. could show mole fraction for all 3 and then 10 later to give partial pressure. Kp = (5.46) 2 (2.73) / (1.83) 2 = 24.5 atm 5 (b) No effect 1 No effect 1 [8] 12. (a) fraction of the total pressure generated by a gas or or pressure gas would generate if it alone occupied the volume or P total mol fraction 1 K p p(co) p(h2 ) not [ ] 1 p(ch ) p(h O) 4 2 3 (iii) Increase in total pressure will result in less product molecules in the equilibrium mixture / equilibrium moves to left because more molecules on product side of the equilibrium than on left 2 (b) No change 1 K P increase 1 (iii) No change 1 (c) K 1 p p(ch ) 1 4 Maltby Academy 22
9.87 10 3 kpa 1 / 9.87 10 6 Pa 1 consequential on 1 Allow 3 5 sig fig (iii) equilibrium has moved left in favour of gas exothermic going left to right/in the forward direction / as written Stand alone 2 (iv) Answer yes or no with some sensible justification e.g. No the costs would not justify the amount produced 1 [12] 13. (a) The pressure which the gas exerts if it alone occupies the same volume at the same temperature or partial pressure of a gas in a mixture = mole fraction of gas total pressure 1 (b) K p = p(pcl 3 )p(cl 2 )/p(pcl 5 ) no square brackets Or K p = P P PCl3 P PCl5 Cl2 1 mark consequentially on PCl 5 PCl 3 + Cl 2 mols at start 1 0 0 mols at eqm 0.60 0.40 0.40 mol fraction 0.60 0.40 0.40 1.40 1.40 1.40 partial pressures 0.60 2 0.40 2 0.40 2 1.40 1.40 1.40 = 0.857 = 0.571 = 0.571 Kp = (0.571) 2 /0.857 = 0.38 or 0.381 atm but 2-4 sf acceptable i.e. mols at eqm mols fraction partial pressures (ie. multiply x2) substitute in Kp + answer units 5 (c) Endothermic conditional on increase of temperature moves equilibrium in Maltby Academy 23
direction that absorbs heat 2 K p increases 1 (d) K p = P(CO 2 ) or K p = PCO 2 1 16 (atm) ignore units. Consequential on (d) 1 [12] 14. (a) Pressure NOT partial pressure ) intensity or change of colour ) Any one volume ) 1 (b) K c = [NO 2 (g)] 2 / [N 2 O 4 (g)] State symbols required 1 (c) Mol NO 2 at equilibrium = 0.0120 / 1.20 10 2 K c = (0.0120) 2 (0.0310) = 4.6 / 4.65 10 3 mol dm 3 3 (d) Amount of NO 2 reduced 1 No effect 1 (e) As K c is bigger, more NO 2 is produced so heat helps forward reaction / by Le Chatelier s principle reaction goes forward to use up heat / as temperature increases S total must be more positive so S surroundings (= H/T must be less negative 1 (f) Positive / + with some attempt at explanation 1 mol / molecule gas 2 mol / molecule gas / products more disordered than reactants 2 (g) H S surroundings = T OR H 1000 T 1 (h) S total is positive as reaction occurs So S system must be more positive than S surroundings is negative 2 [13] 15. (a) Pressure exerted by the gas if it alone occupied the same volume at the same temperature/mole fraction total pressure 1 Maltby Academy 24
(b) K p = p(n ) p(o ) 2 p(no) 2 2 Correct number of moles Correct mole fractions Correct partial pressures 2.45 10 3 ACCEPT 2 4 SF 4 1 (c) K p increases Equilibrium moves to r.h.s. which is the exothermic direction 3 (d) K p = p (Ni(CO) 4 ) / p(co) 4 1 High partial pressure with some reason so the pressure Ni(CO) 4 increases to keep Kp constant. 2 [12] 16. (a) Methanol is the biggest/ most complex molecule / greatest M R /most atoms/most electrons 1 S system = 239.7 197.6 2(130.6) = 219.1/ 219 J mol l K 1 Method answer + units 2 (iii) yes as 3 molecules 1 OR yes as gases a liquid 1 (iv) S surr = H/T (stated or used) = ( 129/ 298) = +0.433 kj mol 1 K 1 / +433 J mol 1 K 1 /+ 432.9 1 for wrong units/ no units / more than 4 SF 1 for wrong sign/ no sign 2 (v) S total = 219.1 + 433 = +213.9 / +213.8 J mol 1 K 1 / +214 J mol 1 K 1 / +0.214 kj mol 1 K 1 Positive so possible 2 Maltby Academy 25
(b) Temperature Faster at 400 C even though yield is lower Pressure Higher pressure improves yield of methanol Higher pressure increases rate Maximum 3 3 Not in same phase as reactants. ALLOW state instead of phase 1 (iii) K p = p(ch 3 OH)/p(CO) p(h 2 ) 2 1 (iv) Partial pressure of methanol = 200 55 20 = 125 atm K p = (125)/55 20 2 = 5.68 10 3 / 5.7 10 3 atm 2 2 (c) Number of molecules / fraction of molecules with energy E A /number of molecules which have enough energy to react. 1 Vertical line / mark on axis to show value to the left of line E A 1 [17] Maltby Academy 26