Bond C=O C H C O O H. Use the enthalpy change for the reaction and data from the table to calculate a value for the H H bond enthalpy.

Transcription

1 Many chemical processes release waste products into the atmosphere. Scientists are developing new solid catalysts to convert more efficiently these emissions into useful products, such as fuels. One example is a catalyst to convert these emissions into methanol. The catalyst is thought to work by breaking a H H bond. An equation for this formation of methanol is given below. CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) H = 49 kj mol Some mean bond enthalpies are shown in the following table. Bond C=O C H C O O H Mean bond enthalpy / kj mol (a) Use the enthalpy change for the reaction and data from the table to calculate a value for the H H bond enthalpy. H H bond enthalpy =... kj mol (3) (b) A data book value for the H H bond enthalpy is 436 kj mol. Suggest one reason why this value is different from your answer to part (a). () (c) Suggest one environmental advantage of manufacturing methanol fuel by this reaction. () Page of 35

2 (d) Use Le Chatelier's principle to justify why the reaction is carried out at a high pressure rather than at atmospheric pressure. (3) (e) Suggest why the catalyst used in this process may become less efficient if the carbon dioxide and hydrogen contain impurities. () Page 2 of 35

3 (f) In a laboratory experiment to investigate the reaction shown in the equation below,.0 mol of carbon dioxide and 3.0 mol of hydrogen were sealed into a container. After the mixture had reached equilibrium, at a pressure of 500 kpa, the yield of methanol was 0.86 mol. CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) Calculate a value for K p Give your answer to the appropriate number of significant figures. Give units with your answer. K p =... Units =... (7) (Total 6 marks) 2 Which change would alter the value of the equilibrium constant (K p ) for this reaction? 2SO 2 (g) + O 2 (g) 2SO 3 (g) A Increasing the total pressure of the system. B Increasing the concentration of sulfur trioxide. C Increasing the concentration of sulfur dioxide. D Increasing the temperature. (Total mark) Page 3 of 35

4 3 An experiment was carried out to determine the equilibrium constant, K c, for the following reaction. CH 3 CH 2 COOH + CH 3 CH 2 CH 2 OH CH 3 CH 2 COOCH 2 CH 2 CH 3 + H 2 O A student added measured volumes of propan--ol and propanoic acid to a conical flask. A measured volume of concentrated hydrochloric acid was added to the flask, which was then sealed. After week, the contents of the flask were poured into water and the solution was made up to a known volume. This solution was titrated with standard sodium hydroxide solution. (a) Explain how the student could determine the amount, in moles, of propan--ol added to the flask. (2) (b) The titration described above gives the total amount of acid in the equilibrium mixture. Explain how, by carrying out a further experiment, the student could determine the amount of propanoic acid in the equilibrium mixture. (2) (c) In a repeat experiment, the student failed to seal the flask that contained the equilibrium mixture. Explain why this error would lead to the student obtaining an incorrect value for the equilibrium constant K c (2) (Total 6 marks) Page 4 of 35

5 4 A sealed flask containing gases X and Y in the mole ratio :3 was maintained at 600 K until the following equilibrium was established. X(g) + 3Y(g) 2Z(g) The partial pressure of Z in the equilibrium mixture was 6.0 MPa when the total pressure was 22.0 MPa. (a) (i) Write an expression for the equilibrium constant, K p, for this reaction. (ii) Calculate the partial pressure of X and the partial pressure of Y in the equilibrium mixture. Partial pressure of X... Partial pressure of Y... (iii) Calculate the value of K p for this reaction under these conditions and state its units. Value of K p... Units of K p... (6) (b) When this reaction is carried out at 300 K and a high pressure of 00 MPa, rather than at 600 K and 22.0 MPa, a higher equilibrium yield of gas Z is obtained. Give two reasons why an industrialist is unlikely to choose these reaction conditions. Reason... Reason 2... (2) (Total 8 marks) Page 5 of 35

6 5 The manufacture of methanol can be achieved in two stages. (a) In the first stage, methane and steam react according to the following equation. CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) H ο = +20 kj mol Discuss, with reasons, the effects of increasing separately the temperature and the pressure on the yield of the products and on the rate of this reaction. (6) (b) In the second stage, carbon monoxide and hydrogen react according to the following equation. CO(g) + 2H 2 (g) CH 3 OH(g) A 62.8 mol sample of carbon monoxide was added to 46 mol of hydrogen. When equilibrium was reached at a given temperature, the mixture contained 26.2 mol of methanol at a total pressure of 9.50 MPa. Write an expression for the equilibrium constant, K p, for this reaction. Calculate a value for K p at this temperature and give its units. (8) (Total 4 marks) 6 This question is about the reaction given below. CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) Enthalpy data for the reacting species are given in the table below. Substance CO(g) H 2 O(g) CO 2 (g) H 2 (g) ΔH / kj mol Which one of the following statements is not correct? A B C D The value of K p changes when the temperature changes. The activation energy decreases when the temperature is increased. The entropy change is more positive when the water is liquid rather than gaseous. The enthalpy change is more positive when the water is liquid rather than gaseous. (Total mark) Page 6 of 35

7 7 Sulphur dioxide and oxygen were mixed in a 2: mol ratio and sealed in a flask with a catalyst. The following equilibrium was established at temperature T 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔH = 96 kj mol The partial pressure of sulphur dioxide in the equilibrium mixture was 24 kpa and the total pressure in the flask was 04 kpa. (a) Deduce the partial pressure of oxygen and hence calculate the mole fraction of oxygen in the equilibrium mixture. Partial pressure of oxygen... Mole fraction of oxygen (3) (b) Calculate the partial pressure of sulphur trioxide in the equilibrium mixture.... () (c) Write an expression for the equilibrium constant, K p, for this reaction. Use this expression to calculate the value of K p at temperature T and state its units. Expression for Kp Calculation Units (4) (d) When equilibrium was established at a different temperature, T 2, the value of K p was found to have increased. State which of T and T 2 is the lower temperature and explain your answer. Lower temperature... Explanation (3) Page 7 of 35

8 (e) In a further experiment, the amounts of sulphur dioxide and oxygen used, the catalyst and the temperature, T, were all unchanged, but a flask of smaller volume was used. Deduce the effect of this change on the yield of sulphur trioxide and on the value of K p. Effect on yield of SO Effect on K p (2) (Total 3 marks) 8 (a) The gaseous reactants W and X were sealed in a flask and the mixture left until the following equilibrium had been established. 2W(g) + X(g) 3Y(g) + 2Z(g) ΔH = 200 kj mol Write an expression for the equilibrium constant, K p, for this reaction. State one change in the conditions which would both increase the rate of reaction and decrease the value of K p. Explain your answers. (7) (b) Ethyl ethanoate can be prepared by the reactions shown below. Reaction CH 3 COOH(l) + C 2 H 5 OH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) H = 2.0 kj mol Reaction 2 CH 3 COCl(l) + C 2 H 5 OH(l) CH 3 COOC 2 H 5 (l) + HCl(g) H = 2.6 kj mol (i) (ii) Give one advantage and one disadvantage of preparing ethyl ethanoate by Reaction rather than by Reaction 2. Use the information given above and the data below to calculate values for the standard entropy change, S, and the standard free-energy change, G, for Reaction 2 at 298 K. CH 3 COCl(l) C 2 H 5 OH(l) CH 3 COOC 2 H 5 (l) HCl(g) S /JK mol (8) (Total 5 marks) Page 8 of 35

9 9 The equation for the combustion of butane in oxygen is C 4 H O 2 4CO 2 + 5H 2 O The mole fraction of butane in a mixture of butane and oxygen with the minimum amount of oxygen required for complete combustion is A 0.33 B 0.53 C 0.67 C (Total mark) 0 At high temperatures, SO 2 Cl 2 dissociates according to the following equation. SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) ΔH = +93 kj mol When.00 mol of SO 2 Cl 2 dissociates, the equilibrium mixture contains 0.75 mol of Cl 2 at 673 K and a total pressure of 25 kpa. (a) Write an expression for the equilibrium constant, K p, for this reaction () (b) Calculate the total number of moles of gas present in the equilibrium mixture.... (2) (c) (i) Write a general expression for the partial pressure of a gas in a mixture of gases in terms of the total pressure. (ii) Calculate the partial pressure of SO 2 Cl 2 and the partial pressure of Cl 2 in the equilibrium mixture. Partial pressure of SO 2 Cl 2... Partial pressure of Cl 2... (5) Page 9 of 35

10 (d) Calculate a value for the equilibrium constant, K p, for this reaction and give its units (3) (e) State the effect, if any, of an increase in temperature on the value of K p for this reaction. Explain your answer. Effect on K p... Explanation (2) (f) State the effect, if any, of an increase in the total pressure on the value of K p for this reaction.... () (Total 4 marks) When a mixture of mol of PCl 3 and mol of Cl 2 was heated in a vessel of fixed volume to a constant temperature, the following reaction reached equilibrium. PCl 3 (g) + Cl 2 (g) PCl 5 (g) DH = 93 kj mol At equilibrium, 0.66 mol of PCl 5 had been formed and the total pressure was 225 kpa. (a) (i) Calculate the number of moles of PCl 3 and of Cl 2 in the equilibrium mixture. Moles of PCl 3... Moles of Cl 2... (ii) Calculate the total number of moles of gas in the equilibrium mixture. (3) Page 0 of 35

11 (b) Calculate the mole fraction and the partial pressure of PCl 3 in the equilibrium mixture. Mole fraction of PCl Partial pressure of PCl (3) (c) (i) Write an expression for the equilibrium constant, K p, for this equilibrium. (ii) The partial pressures of Cl 2 and PCl 5 in the equilibrium mixture were 5.3 kpa and 83.6 kpa, respectively, and the total pressure remained at 225 kpa. Calculate the value of K p at this temperature and state its units. (4) (d) State the effect on the mole fraction of PCl 3 in the equilibrium mixture if (i) the volume of the vessel were to be increased at a constant temperature, (ii) the temperature were to be increased at constant volume. (2) (Total 2 marks) Page of 35

12 2 This question relates to the equilibrium gas-phase synthesis of sulphur trioxide: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Thermodynamic data for the components of this equilibrium are: Substance ΔH / kj mol S / J K - mol - SO 3 (g) SO 2 (g) O 2 (g) This equilibrium, at a temperature of 585 K and a total pressure of 540 kpa, occurs in a vessel of volume.80 dm 3. At equilibrium, the vessel contains mol of SO 2 (g), mol of O 2 (g) and mol of SO 3 (g). The mole fraction of SO 3 in the equilibrium mixture is A B C D (Total mark) Page 2 of 35

13 3 This question relates to the equilibrium gas-phase synthesis of sulphur trioxide: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Thermodynamic data for the components of this equilibrium are: Substance ΔH / kj mol S / J K - mol - SO 3 (g) SO 2 (g) O 2 (g) This equilibrium, at a temperature of 585 K and a total pressure of 540 kpa, occurs in a vessel of volume.80 dm 3. At equilibrium, the vessel contains mol of SO 2 (g), mol of O 2 (g) and mol of SO 3 (g). With pressures expressed in MPa units, the value of the equilibrium constant, K p, is A 4.90 B 6.48 C 9.07 D 6.8 (Total mark) Page 3 of 35

14 4 This question relates to the equilibrium gas-phase synthesis of sulphur trioxide: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Thermodynamic data for the components of this equilibrium are: Substance ΔH / kj mol S / J K - mol - SO 3 (g) SO 2 (g) O 2 (g) This equilibrium, at a temperature of 585 K and a total pressure of 540 kpa, occurs in a vessel of volume.80 dm 3. At equilibrium, the vessel contains mol of SO 2 (g), mol of O 2 (g) and mol of SO 3 (g). Possible units for the equilibrium constant K p include A B C D no units kpa Mpa kpa 2 (Total mark) Page 4 of 35

15 5 This question relates to the equilibrium gas-phase synthesis of sulphur trioxide: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Thermodynamic data for the components of this equilibrium are: Substance ΔH / kj mol S / J K - mol - SO 3 (g) SO 2 (g) O 2 (g) This equilibrium, at a temperature of 585 K and a total pressure of 540 kpa, occurs in a vessel of volume.80 dm 3. At equilibrium, the vessel contains mol of SO 2 (g), mol of O 2 (g) and mol of SO 3 (g). At equilibrium in the same vessel of volume.80 dm 3 under altered conditions, the reaction mixture contains mol of SO 3 (g), mol of SO 2 (g) and mol of O 2 (g) at a total pressure of 623 kpa. The temperature in the equilibrium vessel is A 307 C B 596 K C 337 C D 642 K (Total mark) Page 5 of 35

16 6 The following information concerns the equilibrium gas-phase synthesis of methanol. CO(g) + 2H 2 (g) CH 3 OH(g) At equilibrium, when the temperature is 68 C, the total pressure is.70 MPa. The number of moles of CO, H 2 and CH 3 OH present are 0.60, and 0.80, respectively. Thermodynamic data are given below. Substance ΔH / kj mol S / J K - mol - CO(g) 0 98 H 2 (g) 0 3 CH 3 OH(g) Possible units for the equilibrium constant, K p, for this reaction are A B C D no units kpa MPa kpa 2 (Total mark) Page 6 of 35

17 7 The following information concerns the equilibrium gas-phase synthesis of methanol. CO(g) + 2H 2 (g) CH 3 OH(g) At equilibrium, when the temperature is 68 C, the total pressure is.70 MPa. The number of moles of CO, H 2 and CH 3 OH present are 0.60, and 0.80, respectively. Thermodynamic data are given below. Substance ΔH / kj mol S / J K - mol - CO(g) 0 98 H 2 (g) 0 3 CH 3 OH(g) The mole fraction of hydrogen in the equilibrium mixture is A B C D (Total mark) Page 7 of 35

18 8 The following information concerns the equilibrium gas-phase synthesis of methanol. CO(g) + 2H 2 (g) CH 3 OH(g) At equilibrium, when the temperature is 68 C, the total pressure is.70 MPa. The number of moles of CO, H 2 and CH 3 OH present are 0.60, and 0.80, respectively. Thermodynamic data are given below. Substance ΔH / kj mol S / J K - mol - CO(g) 0 98 H 2 (g) 0 3 CH 3 OH(g) With pressures expressed in MPa units, the value of the equilibrium constant, K p, under these conditions is A.37 B.66 C 2.82 D 4.80 (Total mark) Page 8 of 35

19 9 The following information concerns the equilibrium gas-phase synthesis of methanol. CO(g) + 2H 2 (g) CH3OH(g) At equilibrium, when the temperature is 68 C, the total pressure is.70 MPa. The number of moles of CO, H 2 and CH 3 OH present are 0.60, and 0.80, respectively. Thermodynamic data are given below. Substance ΔH / kj mol S / J K - mol - CO(g) 0 98 H 2 (g) 0 3 CH 3 OH(g) Which one of the following statements applies to this equilibrium? A B C D The value of K p increases if the temperature is raised. The value of K p increases if the pressure is raised. The yield of methanol decreases if the temperature is lowered. The yield of methanol decreases if the pressure is lowered. (Total mark) 20 Hydrogen and carbon monoxide were mixed in a 2: mole ratio. The mixture was allowed to reach equilibrium according to the following equation at a fixed temperature and a total pressure of kpa. 2H 2 (g) + CO(g) CH 3 OH(g) (a) The equilibrium mixture contained mol of carbon monoxide and mol of methanol. (i) Calculate the number of moles of hydrogen present in the equilibrium mixture. (ii) Hence calculate the mole fraction of hydrogen in the equilibrium mixture. Page 9 of 35

20 (iii) Calculate the partial pressure of hydrogen in the equilibrium mixture. (5) (b) In a different mixture of the three gases at equilibrium, the partial pressure of carbon monoxide was 7550 kpa, the partial pressure of hydrogen was 2300 kpa and the partial pressure of methanol was 270 kpa. (i) Write an expression for the equilibrium constant, K p, for this reaction. (ii) Calculate the value of the equilibrium constant, K p, for the reaction under these conditions and state its units. K p... Units... (3) Page 20 of 35

21 (c) Two isomeric esters E and F formed from methanol have the molecular formula C 6 H 2 O 2 Isomer E has only 2 singlet peaks in its proton n.m.r. spectrum. Isomer F is optically active. Draw the structures of these two isomers. Isomer E Isomer F (2) (Total 0 marks) Page 2 of 35

22 2 Summarised directions for recording responses to multiple completion questions A (i), (ii) and (iii) only B (i) and (iii) only C (ii) and (iv) only D (iv) alone Which of the following statements about a catalyst is / are true? (i) (iii) (iii) (iv) It speeds up the forward reaction and slows down the reversere action. It increases the proportion of molecules with higher energies. A homogeneous catalyst usually acts in the solid state. It does not alter the value of the equilibrium constant. (Total mark) Page 22 of 35

23 Mark schemes (a) Bonds broken = 2(C=O) + 3(H H) = H H Bonds formed = 3(C H) +(C O) + 3(O H) = Both required 49 = [ (H H)] [ ] 3(H H) = [ ] = 450 Both required H H = 483 (kj mol ) Allow 483.3(3) (b) Mean bond enthalpies are not the same as the actual bond enthalpies in CO 2 (and / or methanol and / or water) (c) (d) (e) (f) The carbon dioxide (produced on burning methanol) is used up in this reaction 4 mol of gas form 2 mol At high pressure the position of equilibrium moves to the right to lower the pressure / oppose the high pressure This increases the yield of methanol Impurities (or sulfur compounds) block the active sites Allow catalyst poisoned Stage : moles of components in the equilibrium mixture Extended response question CO 2 (g) + 3H 2 (g) CH 3 OH(g) + H 2 O(g) Initial moles Eqm moles ( 0.86) = 0.4 ( ) = Page 23 of 35

24 Stage 2: Partial pressure calculations Total moles of gas = 2.28 Partial pressures = mol fraction p total p CO2 = mol fraction p total = / 2.28 = 30.7 kpa p H2 = mol fraction p total = / 2.28 = 92. kpa M3 is for partial pressures of both reactants Alternative M3 = pp CO2 = pp H2 = p CH3OH = mol fraction p total = / 2.28 = 88.6 kpa p H2O = mol fraction p total = / 2.28 = 88.6 kpa M4 is for partial pressures of both products Alternative M4 = pp CH3OH = pp H2O = Stage 3: Equilibrium constant calculation K p = p CH3OH p H2O / p CO2 (p H2 ) 3 Hence K p = / 30.7 (92.) 3 = = Answer must be to 2 significant figures Units = kpa 2 2 D [6] [] 3 (a) Multiply volume of propan ol by density Allow measure the mass of the volume added Any reference to concentration of propan ol CE = 0 Divide the mass by the M r of propan ol Page 24 of 35

25 (b) (c) Titrate a measured volume of the concentrated HCl added initially to determine moles of HCl used in the experiment Allow addition of AgNO 3 to form AgCl precipitate. Use mass of precipitate to calculate initial moles of HCl added. Subtract this number of moles of HCl from the total moles of acid at equilibrium M ester will evaporate / escape Allow reactants / products will evaporate M2 incorrect values used (to determine K c ) Allow the system will no longer be at equilibrium Do not allow references to equilibrium position shifting alone [6] 4 (a) (i) (K p ) = (p z ) 2 /(p x )(p y ) 3 (penalise use of square brackets, allow ()) (ii) X (22 6)/4 = 4 (MPa) (mark is for value 4 only, ignore units) Y obtained by multiplying value for X by 3 (allow conseq on wrong value for X) Y = 2 (MPa) (mark is for value 2 only) (iii) K p = / = (allow conseq on wrong values for X and Y e.g.6 2 /3 9 3 = 0.65) (if K p wrong in (a)(i) CE) MPa 2 (allow any unit of P 2 provided ties to P used for K p value) Page 25 of 35

26 (b) high pressure expensive (due to energy or plant costs) (Rate is) slow (at lower temperatures) [8] 5 (a) (must state correct effect on yield or rate to score the reason mark) T effect: higher temp: yield greater or shifts equilibrium to right; effect: higher temp: rate increased; reason: endothermic OR more particles have E>E a OR P more successful/productive collisions; effect: higher pressure: yield less or shifts equilibrium to left; effect: higher pressure: rate increased; reason: increase in gas moles L to R OR greater collision frequency; (Q of L mark) Page 26 of 35

27 (b) M equilibrium moles of CO = = 36.6 M2 equilibrium moles of H 2 = 46 2(26.2) = 93.6 M3 total no moles = = 56.4 M4 partial pressure = mole fraction x total pressure M5 M6 Page 27 of 35

28 M () 2.2(l) (l) 0 4 M8 MPa 2 kpa 2 Pa 2 If no subtraction lose M, M2 and M3) (If 2 missed in M2, lose both M2 and M3) (If M gained but moles of H 2 = 73.2 (i.e. double CO), M2 and M3 lost) (If M gained but mol H 2 = 2( ), M2 and M3 lost) (If M and M2 correct but M3 lost for CE, penalise M6 also) (M4 can be gained from the numbers in the expression for M6 even if these numbers are wrong) (If K p contains [ ] lose M5 but then mark on) (If chemically wrong expression for K p, lose M5, M6 and M7 (allow M8 conseq on their K p )) (If divided by 9.5, or not used 9.5 at all, lose M6 and M7 (and M4)) (If tried to convert to kpa and is factor(s) of 0 out, penalise in M6 and allow M8 for kpa 2 ) [4] B 6 [] 7 (a) 2 (kpa) pp = mole fraction total pressure or mole fraction = 2/04 = 0.5 (allow 0.2) (b) 68 (kpa) Page 28 of 35

29 (d) T 2 (Must be correct to score any marks in this section) (c) K p = (If K p wrong, allow consequential units only) (penalise square brackets in expression but then mark on) = = (Allow 0.67) (Allow full marks in calculation consequential on their values in (a) and (b)) kpa (e) Exothermic Reduce T to shift equilibrium to the right or forward reaction favoured by low T or K p increases for low T or low T favours exothermic reaction Increase None [3] Page 29 of 35

30 8 (a) M K p = ( P Y) 3. ( P Z) 2 / ( P W) 2.( PX) NB [ ] wrong M2 M3 M4 M5 M6 M7 temperature increase particles have more energy or greater velocity/speed more collisions with E > E a or more successful collisions Reaction exothermic or converse Equilibrium moves in the left Marks for other answers Increase in pressure or concentration allow M, M5, M6 Max 3 Addition of a catalyst; allow M, M5, M6 Max 3 Decrease in temperature; allow M, M2, M6 Max 3 Two or more changes made; allow M, M6 Max 2 (b) (i) Advantage; reaction goes to completion, not reversible or faster Disadvantage; reaction vigorous/dangerous (exothermic must be qualified) or HCl(g) evolved/toxic or CH 3 COCl expensive NB Allow converse answers Do not allow reactions with other reagents e.g. water or ease of separation Page 30 of 35

31 (ii) ΔS = ΣS products ΣS reactants ΔS = ( ) (20 + 6) ΔS = 84 (JK mol ) (Ignore units) Allow 84 to score () mark ΔG = ΔH TΔS = /000 = 46.6 kj mol or J mol Allow (2) for 46.6 without units (Mark ΔG consequentially to incorrect ΔS) (e.g. ΔS = 84 gives ΔG = +3.4 kj mol ) [5] A 9 [] 0 (a) K p = () (b) =.75 () () 2 (c) (i) p = Total pressure mol fraction () (ii) Partial of SO 2 Cl 2 : 25 = 7.9 kpa () Partial pressure of Cl 2 : 25 = 53.6 kpa () () 5 (d) K p = () = 6 () kpa () 3 Page 3 of 35

32 (e) Effect on K p : increase () Explanation: increase T sends equilibrium in endothermic direction () (f) no effect () 2 Notes (a) If K p has [ ] lose mark in (a) but allow full marks in (d) If K p wrong/upside down etc, allow max 2 in (d) for substitution of numbers () and consequential units () (b) Mark for moles of SO 2 Cl 2 can be scored in part (c) (ii) if not gained in (b).75 get (2) If moles of SO 2 Cl 2 =, this is a Chemical Error, hence a 2 mark penalty If total moles given in (b) =.75, this scores [2] in (b); but if the no moles of SO 2 Cl 2 = in (c)(ii), lose both marks in (c)(ii) for pp of SO 2 Cl 2 = (/.75) 25, i.e. the 2 mark penalty is in (c)(ii). If total moles given in (b) = 2.5, score zero in (b), but can gain full marks in (c)(ii) consequentially, i.e. the 2 mark penalty is in (b). If moles of SO 2 Cl 2 = and total in (b) does not equal 2.5, still lose both in (b) but can get all 4 conseq in (c)(ii) for /x etc and 0.75/x etc (c) (i) Allow Total pressure = sum of partial pressures for () or p A = x A p tot (ii) First mark is for mole fraction. If either number in either mole fraction is not consequential on (b), then lose both marks for that partial p. (d) If pcl 2 is not equal to pso 2 or any number used in K p is not conseq on (c)(ii), allow units only (e) SIG FIGS; must be 3 sig figs in (b) but then allow 2 sig figs in (c) and (d); (ignore extra figs) but penalise incorrect rounding If effect wrong, no marks for explanation. If effect missing, e.g. answer states equm shifts to right, mark on. In the explanation, the word endothermic (or its equivalent) is essential. [4] Page 32 of 35

33 (a) (i) Moles of PCl 3 : = 0.79 () Moles of Cl 2 : = 0.02 () 3 sig figs (ii) () allow 2 sig figs conseq on (i) 3 (b) Mole fraction of PCl 3 : 0.79/0.447 () = 0.4(00) Partial pressure of PCl 3 : pp = mol f n total P () = = 90 () kpa () 3 (c) (i) K p = () ignore brackets except [ ] must show P (ii) K p = () =.8() 0 2 () Kpa () (or Pa ) If 83.6 and 5.3 wrong way round, AE, answer = If K p in (i) allow max 2 for substitution of numbers and conseq units 4 (d) (i) increased () (ii) increased () Organic points () Curly arrows: must show movement of a pair of electrons, i.e. from bond to atom or from lp to atom / space e.g. 2 [2] Page 33 of 35

34 (2) Structures penalise sticks (i.e. ) once per paper Penalise once per paper allow CH 3 or CH 3 or or CH 3 or H 3 C B 2 C 3 C 4 D 5 D 6 B 7 B 8 D 9 [] [] [] [] [] [] [] [] 20 (a) (i) 0.86 () Page 34 of 35

35 (ii) total moles = =.375 () mole fraction of H 2 = = () Conseq on (i) ( ) (iii) pp = mole fract n total P () = = (kpa) () or.(0) Ignore units Conseq on (ii) 5 (b) (i) K p = () Penalise [ ] (ii) K p = = 2.37 (2.4) 0 9 () OR Units: kpa 2 () (c) Isomer E: or Pa 2 not conseq to wrong K p expression 3 Isomer F: 2 [0] D 2 [] Page 35 of 35