The values for the standard enthalpy of combustion of graphite and carbon monoxide are given below: H c /kj mol 1 C (graphite) 394 CO(g) 283

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1 1. (a) (i) Define the term standard enthalpy of combustion..... (3) (ii) The values for the standard enthalpy of combustion of graphite and carbon monoxide are given below: H c /kj mol 1 C (graphite) 394 CO(g) 283 Use these data to find the standard enthalpy change of formation of carbon monoxide using a Hess s law cycle. C(graphite) O2 (g) CO(g) (3) (iii) Suggest why it is not possible to find the enthalpy of formation of carbon monoxide directly... Maltby Academy 1

2 (iv) Draw an enthalpy level diagram below for the formation of carbon monoxide from graphite. (b) Natural gas consists of methane, CH 4. When methane burns completely in oxygen the reaction occurs as shown in the equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H c = 890 kj mol 1 Methane does not burn unless lit. Use this information to explain the difference between thermodynamic and kinetic stability (4) (Total 12 marks) 2. Thermochemical data, at 298 K, for the equilibrium between zinc carbonate, zinc oxide and carbon dioxide is shown below. ZnCO 3 (s) ZnO(s) + CO 2 (g) H ο = kj mol 1 S ο [ZnO(s)] = J mol 1 K 1 S ο [ZnCO 3 (s)] = J mol 1 K 1 S ο [CO 2 (g)] = J mol 1 K 1 Maltby Academy 2

3 (a) (i) Suggest reasons for the differences between the three standard entropies..... (ii) Calculate the entropy change for the system, S ο, for this reaction. Include the system sign and units in your answer. (b) Calculate the entropy change for the surroundings, S ο surroundings, at 298 K, showing your method clearly. Maltby Academy 3

4 (c) (i) Calculate the total entropy change for this reaction, S ο total, at 298 K. (ii) What does the result of your calculation in (c)(i) indicate about the natural direction of this reaction at 298 K? Justify your answer... (d) (i) Write an expression for the equilibrium constant, Kp, for this reaction. Maltby Academy 4

5 (ii) State how you would alter ONE condition to increase the yield of carbon dioxide from this equilibrium reaction. Justify your answer. (Total 11 marks) 3. The reaction between solid barium hydroxide and solid ammonium chloride can be represented by the equation below. Ba(OH) 2 (s) + 2NH 4 Cl(s) BaCl 2 (s) + 2NH 3 (g) + 2H 2 O(l) ΔH ο = kj mol 1 The standard entropies, at 298 K, for the reactants and products are: S ο [Ba(OH) 2 (s)] = J mol 1 K 1 S ο [NH 4 Cl(s)] = J mol 1 K 1 S ο [BaCl 2 (s)] = J mol 1 K 1 S ο [NH 3 (g)] = J mol 1 K 1 S ο [H 2 O(l)] = J mol 1 K 1 (a) Why is the standard entropy of ammonia more positive than the standard entropy of barium chloride? Maltby Academy 5

6 (b) Use the values given to calculate the standard entropy change, ΔS ο system, for this reaction. Include the sign and units in your answer. (c) Calculate the standard entropy change of the surroundings, ΔS ο surroundings, at 298 K for this reaction. (d) Use your answers to (b) and (c) to show that this reaction is feasible at 298 K Maltby Academy 6

7 (e) Calculate the minimum temperature, in kelvin, at which the reaction is spontaneous. (Total 8 marks) 4. 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) H 2 (g) CH 3 OH(l) (i) Suggest why methanol has the highest entropy value of the three substances..... Maltby Academy 7

8 (ii) 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 (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. Maltby Academy 8

9 (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) (i) Suggest reasons for the choice of temperature and pressure. Temperature Pressure (3) (ii) 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) Maltby Academy 9

10 (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. (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. (i) What does the shaded area on the graph represent?.. (ii) Draw a line on the graph, labelled E C, to show the activation energy of the catalysed reaction. (Total 17 marks) Maltby Academy 10

11 5. The reaction between nitrogen and hydrogen can be used to produce ammonia. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH ο = 92.2 kj mol 1 Standard entropies are given below S ο [N 2 (g)] = J mol 1 K 1 S ο [H 2 (g)] = J mol 1 K 1 S ο [NH 3 (g)] = J mol 1 K 1 (a) Calculate the entropy change of the system, ΔS ο system, for this reaction. Include a sign and units in your answer. (b) Calculate the entropy change of the surroundings, ΔS ο surroundings, at 298 K. Include a sign and units in your answer. Maltby Academy 11

12 (c) (i) Calculate the total entropy change, ΔS ο total, at 298 K. Include a sign and units in your answer. (ii) Is this reaction feasible at 298 K? Justify your answer. (d) In industry the reaction is carried out at about 700 K using an iron catalyst and high pressures. (i) The yield of ammonia produced at equilibrium is less at 700 K than at 298 K, if the pressure remains constant. In terms of entropy, explain why this happens. (ii) Higher pressures increase the yield of ammonia at equilibrium. Suggest a reason why pressures greater than 300 atmospheres are not routinely used. Maltby Academy 12

13 (iii) Iron is a heterogeneous catalyst. Explain what is meant by heterogeneous. (Total 9 marks) 6. (a) The distribution of the energy of particles in a gas at temperature T1 is shown below. (i) On the diagram above, draw the distribution of energy of particles at a lower temperature, T2. Maltby Academy 13

14 (ii) Use the diagram to explain why the rate of a reaction increases with an increase in temperature. (3) (iii) Explain fully why a catalyst increases the rate of a reaction. Maltby Academy 14

15 (b) The fermentation of glucose is an exothermic reaction and is catalysed by enzymes in yeast. C 6 H 12 O 6 (aq) 2C 2 Η 5 ΟΗ(aq) + 2CO 2 (g) The reaction is slow at room temperature. (i) Describe, with the aid of a diagram, an experiment you could do to follow the progress of this reaction at different temperatures. (4) Maltby Academy 15

16 (ii) Would you expect S system to be positive or negative for this reaction? Justify your answer with TWO pieces of evidence. (iii) Deduce the sign of S surroundings. Show your reasoning. (Total 15 marks) 7. (a) State Hess s Law. Maltby Academy 16

17 (b) Methane burns in oxygen. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) (i) Calculate the enthalpy change for this reaction, using the bond enthalpies given below. Bond enthalpy / kj mol 1 C H +435 O = O +498 C = O +805 H O +464 (3) (ii) State the name of this enthalpy change. (iii) The value of this enthalpy change, under standard conditions, is 890 kj mol 1. State the meaning of standard conditions. (iv) Suggest, with a reason, why the enthalpy change calculated in (i) is different from the standard value quoted in (iii). Maltby Academy 17

18 (c) Although the reaction between methane and oxygen is exothermic, it does not occur unless the mixture is ignited. Use these facts to explain the difference between thermodynamic and kinetic stability. (4) (Total 14 marks) ô 8. When barium nitrate is heated it decomposes as follows: Ba(NO 3 ) 2 (s) BaO(s) + 2NO 2 (g) + ½O 2 (g) ΔH = kj mol 1 (a) Use the following data when answering this part of the question. Substance Standard entropy, S ο / J mol 1 K 1 Ba(NO3) 2 (s) BaO(s) NO 2 (g) O 2 (g) Maltby Academy 18

19 (i) Explain why: S ο [NO 2 (g)] is greater than S ο [BaO(s)] S ο [Ba(NO 3 ) 2 (s)] is greater than S ο [BaO(s)]. (ii) Calculate the entropy change of the system, ΔS ο system, for this reaction. Include a sign and units in your answer. (b) Calculate the entropy change of the surroundings, ΔS ο surroundings, for the reaction at 298 K. Include a sign and units in your answer. (c) Calculate ΔS ο total, and explain the significance of the sign for this value. Maltby Academy 19

20 (d) Calculate the minimum temperature at which the decomposition of barium nitrate should occur. You can assume that ΔH and ΔS system are not affected by a change in temperature. (Total 10 marks) 9. (i) Define the term enthalpy of hydration, ΔH hyd, of an ion. Maltby Academy 20

21 (ii) The table below gives some information about the sulphates of the Group 2 elements magnesium and barium. sulphate lattice energy / kj mol 1 hydration enthalpy of cation / kj mol 1 solubility / mol dm 3 MgSO BaSO Use the lattice energy and hydration enthalpy values to explain the difference in the solubility of the two salts. (4) (Total 6 marks) 10. (a) (i) Enthalpy or heat change / released when 1 mol of substance is burned in excess oxygen / completely all substances in standard states (at a specified temp)/ at a pressure of 1 atm. 3 Maltby Academy 21

22 (ii) Suitable cycle (need not be labelled but if labelled, these must be correct) working answer e.g. C + 1 / O CO ( H CO) 2 2 form ( H Carbon) comb CO ( H Carbon monoxide) 2 comb = 394 ( 283) = 111 (kj mol 1 ) Penalise 1 mark if units incorrect 3 (iii) (some) CO 2 is always produced in the reaction 1 (iv) (Energy) C (+ O ) 1 /2 2 1 CO Consequential on (a) (ii) n.b. if no answer in(a) (ii), correct diagram can still score (b) Methane (and oxygen) / reactants thermodynamically unstable w.r.t. products Must be a comparison Since reactants are at a higher energy level (than products) or reverse argument Reactants / methane, oxygen kinetically stable Due to high activation energy 4 If no reference to methane in the answer (max 3) [12] 11. (a) (i) Gases have much higher entropies than solids as there are many more ways of arranging the entities / less ordered / more random(ness) OR reverse argument ZnCO 3 has more atoms/is more complex than ZnO 2 Maltby Academy 22

23 (ii) S ο system = (+43.6) + (+213.6) (+82.4) =+174.8/175 J mol 1 K 1 method answer, sign and units Correct answer, sign and units with no working 2 (b) As printed Amended S ο surroundings = H T S ο surroundings = H T OR = ( ) OR = ( ) = 1560 / 1559 J mol 1 K 1 answer, sign and units = 238(.3) J mol 1 K 1 answer, sign and units ONLY accept 3 or 4 SF 2 IF correct answer, sign and units with no working (c) (i) S ο total = = 1384 / 1380 J mol 1 K 1 = 63.5 / 64 / 63 / 63.2 / 63.4 J mol 1 K 1 IF = 1385(.2) = 1385 / 1390 J mol 1 K 1 IF = 1384 J mol 1 K 1 ONLY penalise incorrect units OR no units in (a)(ii), (b) and (c)(i) once 1 (ii) Natural direction is right to left /reverse as S ο total /total entropy change is negative / less than zero. 1 MUST be consistent with (i) Maltby Academy 23

24 (d) (i) K p = p co 2 ((g) eqm) 1 (ii) Increase temperature / reduce pressure Decreases S ο surroundings (negative) and hence increases S ο total / Le Chatelier s principle applied (i.e increasing temperature, reducing pressure) 2 [11] 12. (a) Many more ways of arranging / more disordered gas molecules than solid (particles) 1 (b) S ο system = (+192.3) + 2(+69.9) (+99.7) 2(+94.6) = +359(.2) J K 1 mol 1 Method 2 Sign, value, units (c) S ο 3 H ( ) surroundings = / = T 298 = / 171 J K 1 mol 1 2 (d) S ο total = S ο system + S ο surroundings So the total entropy change has a positive value / is greater than zero. OR S ο total = / +188 J K 1 mol 1 1 (e) 0 = T 3 Some recognisable correct method T = 142(.3) / 143 K 2 [8] Maltby Academy 24

25 13. (a) (i) Methanol is the biggest/ most complex molecule / greatest M R /most atoms/most electrons 1 (ii) S system = (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) = kj mol 1 K 1 / +433 J mol 1 K 1 / for wrong units/ no units / more than 4 SF 1 for wrong sign/ no sign 2 (v) S total = = / J mol 1 K 1 / +214 J mol 1 K 1 / kj mol 1 K 1 Positive so possible 2 (b) (i) Temperature Faster at 400 C even though yield is lower Pressure Higher pressure improves yield of methanol Higher pressure increases rate Maximum 3 3 (ii) 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 = = 125 atm K p = (125)/ = / atm 2 2 (c) (i) Number of molecules / fraction of molecules with energy E A /number of molecules which have enough energy to react. 1 (ii) Vertical line / mark on axis to show value to the left of line E A 1 [17] 14. Penalise units only once in this question (a) ( ) [ ( )] = 198.8/199 J mol 1 K 1 2 (b) / 92.2 / H / T = + 309(.4) J mol 1 K 1 / (4) kj mol 1 K 1 2 Maltby Academy 25

26 (c) (i) = J mol 1 K 1 (3 SF) OR = J mol 1 K 1 (3 SF) [Do not penalise missing + sign if penalised already in (b)] NOT 4SF. Penatise SF only once on paper 1 (ii) Yes, as S total is positive / total entropy change 1 (d) (i) Higher T makes S surroundings decrease (so S total is less positive) 1 (ii) Cost (of energy) to provide compression/ cost of equipment to withstand high P/ maintenance costs. NOT safety considerations alone 1 (iii) Different phase/state (to the reactants) 1 [9] 15. (a) (i) Starts at zero and approaching x-axis Maximum greater and at lower energy T 2 needs only to be just higher than T 1 T 2 curve must go below T 1 curve approaching the x-axis 2 (ii) As the temperature increases the energy of the particles increases Use the diagram shading areas OR more particles to the right hand side of E A line and so more (successful) collisions/particles have energy greater / equal or greater than the activation energy NOT equal on its own NOT mention of frequency of collisions on its own 3 Maltby Academy 26

27 (iii) A catalyst provides an alternative route with a lower activation energy/ which requires less energy so more collisions / particles have energy greater than the activation energy 2 (b) (i) e.g. Measure the volume of gas given off in a given time / count bubbles / obscuring cross using limewater and then repeat over a range of temperatures No diagram max 3 If method shown cannot possibly work max 1 ie waterbath or sensible range of temperatures BUT NOT different temperatures Penalty 1 for poor diagram 4 (ii) Positive 1 mol goes to 4 moles/particles (so more disorder) /increase in number of moles/particles products include a gas (and so more disorder) NOT 1 mole of compound/element goes to 4 moles of compound/element If negative 0 (out of 2) 2 (iii) Positive with some explanation e.g. S surroundings = H/T OR because reaction is exothermic H is therefore negative and so S surroundings must be positive If negative given in (ii) allow TE here 2 [15] Maltby Academy 27

28 16. (a) Heat / enthalpy / energy change (for a reaction) / H is independent of the pathway / route (between reactants and products) OR depends only on its initial and final state Both marks can score from a diagram and equation 2 (b) (i) H = {(4x + 435) + (2x + 498)} + {(2x 805) + (4x 464)} IGNORE signs for first two marks, ie marks for total enthalpies of bonds broken and made. = 730 (kjmol 1 ) 3 rd mark is consequential on their values for first two marks (kjmol 1 ) (max 2) 3 (ii) (Enthalpy of) combustion DO NOT penalise standard 1 (iii) At 1 atm pressure OR 101 / 100 kpa OR 1 bar stated temperature ACCEPT 298 K / 25 C 2 (iv) Reaction has H 2 O(g) (rather than H 2 O(l)) So not standard conditions 2 nd mark is conditional on the 1 st Average bond enthalpies used (so not specific) (1 max) 2 QWC (c) (Exothermic so) products are at lower energy than reactants Reactants are therefore thermodynamically unstable (with respect to products) Consequential on 1 st mark NOT reaction or system is thermodynamically unstable Can argue from point of view of products. E a is high (for noticeable reaction at room temperature) NOT E a high on its own So reactants are kinetically stable (with respect to products) Consequential on 3 rd mark If reaction instead of reactants is used (3 max) 4 [14] Maltby Academy 28

29 17. (a) (i) NO 2 is a gas (whereas BaO is a solid) Ba(NO 3 ) 2 has a more complicated structure than BaO Allow 2 nd mark if a correct statement is combined with a neutral wrong statement 2 Accept Ba(NO 3 ) 2 molecule has more electrons / is larger than BaO molecule Accept more atoms/ions/particles Accept more complicated/complex compound Reject Ba(NO 3 ) 2 has a larger molar mass than BaO Reject more molecules/elements (ii) S ο system = ( ) + (½ 205.0) = J mol 1 K 1 1 per error 2 Accept +439 J mol 1 K 1 Accept J/ mol /K (b) S ο surroundings = H = T = 1700 J mol 1 K 1 (3 s.f.) Penalise wrong units in (a)(ii) and (b) once only 2 Accept 1690 J mol 1 K Accept 1695 J mol 1 K 1 Answers in kj mol 1 K 1 Reject 1694 J mol 1 K 1 Reject J mol 1 K 1 Reject J mol 1 K 1 (c) S ο total = = 1260 (J mol 1 K 1 ) Allow TE [follow through working from (a)(ii) and (b)] Mark consistently with (a)(ii) and (b) The reaction isn t spontaneous / doesn t go (at 298K) Must be consistent with sign in calculation 2 Accept 1256 J mol 1 K 1 Accept 1261 J mol 1 K 1 Accept J mol 1 K 1 Maltby Academy 29

30 (d) When just spontaneous, S ο total = 0 505OR or implied by calculation i.e a(ii) S ο surroundings = J mol 1 K 1 T = = 1150 (K) Accept K Accept 877 C Accept 1151K with no working (1 max) Reject 1151K for 2 nd mark Reject any negative value for T (in K): no 2 nd mark Reject 1150 C ignore 0 K Allow full marks for an answer without working 2 [10] 18. (i) Enthalpy change when 1 mol of gaseous ions Accept energy or heat Reject any implication of an endothermic process e.g. energy required Reject.1 mol of gaseous atoms is dissolved such that further dilution causes no further heat change Accept added to water / reacts with water instead of dissolved Reject just hydrated Reject just completely hydrated IGNORE standard conditions Accept is dissolved to form an infinitely dilute solution OR Is dissolved in a large/excess/infinite amount of water Mark each aspect independently 2 Maltby Academy 30

31 (ii) EITHER H SOLN = ( H LE + H HYD ) Expression quoted or correctly used in at least one of the calculations below Accept answer only with no working H SOLN MgSO 4 = ( 2874) + ( 1920) = +954 (kj mol 1 ) H SOLN = BaSO 4 = ( 2374) + ( 1360) = (kj mol 1 ) Accept answer only with no working Enthalpy of solution of MgSO4 less endothermic/more exothermic/more negative than for BaSO 4, so MgSO 4 more soluble than BaSO4 (or reverse argument) Reject just solubility/ H soln depends on a balance between lattice and hydration energies OR (both) lattice energies and hydration enthalpies decrease from MgSO 4 to BaSO 4 (or down group) Accept The hydration energies decrease faster.. Reject ( )500 and ( )560 stated without further explanation (but) lattice energies change less H SOLN = ( H LE + H HYD ) stated in words or symbols Reject just solubility/ H solution depends on a balance between lattice and hydration energies so H soln less exothermic/more endothermic/more positive for BaSO 4 so less soluble OR so H soln more exothermic/more negative/less endothermic for MgSO 4 so MgSO 4 more soluble 4 [6] Maltby Academy 31