1. The reaction between sulphur dioxide and oxygen is a dynamic equilibrium (2)

Transcription

1 1. The reaction between sulphur dioxide and oxygen is a dynamic equilibrium. SO + O SO 3 H = 196 kjmol 1 (a) Explain what is meant by dynamic equilibrium () (b) In the table below state the effect on this reaction of increasing the temperature and of increasing the pressure. Effect on the rate of the reaction Effect on the position of equilibrium Increasing the temperature Increases Increasing the pressure (3) (c) This reaction is one of the steps in the industrial production of sulphuric acid. The normal operating conditions are a temperature of 450 C, a pressure of atmospheres and the use of a catalyst. Justify the use of these conditions. (i) A temperature of 450 C: (3) Maltby Academy 1

2 A pressure of atmospheres: () (iii) A catalyst: (d) Give the name of the catalyst used.... (e) Give one large scale use of sulphuric acid.... (Total 13 marks). In the vapour phase sulphur trioxide dissociates: SO 3 (g) SO (g) + O (g) (a) (i) Write an expression for K p for this dissociation. Maltby Academy

3 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 O 5, is an effective catalyst for this reaction. State the effect of using double the mass of catalyst on: (i) the position of the equilibrium;.. the value of K p... (Total 8 marks) Maltby Academy 3

4 3. Consider the following equation: SO + O SO 3.0 moles of SO and 1.0 mole of O were allowed to react in a vessel of volume 60 dm 3. At equilibrium 1.8 moles of SO 3 had formed and the pressure in the flask was atm. (a) (i) Write the expression for K c for this reaction between SO and O. Calculate the value of K c, with units. (3) (b) The reaction between SO and O is exothermic. State the effect on the following, if the experiment is repeated at a higher temperature: (i) K c.. the equilibrium position... (c) State the effect of a catalyst on: (i) K c.. Maltby Academy 4

5 the equilibrium position... (d) (i) Write the expression for K p for the reaction between SO and O. Calculate the mole fractions of SO, O and SO 3 at equilibrium. () (iii) Calculate the partial pressures of SO, O and SO 3 at equilibrium. (iv) Calculate the value of K p, with units. () (Total 14 marks) Maltby Academy 5

6 4. The equation below shows a possible reaction for producing methanol. CO(g) + H (g) CH 3 OH(l) ΔH ο = 19 kj mol 1 (a) The entropy of one mole of each substance in the equation, measured at 98 K, is shown below. Substance S ο /J mol 1 K 1 CO(g) H (g) CH 3 OH(l) 39.7 (i) 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 6

7 (iv) Calculate the entropy change of the surroundings ΔS ο surroundings, at 98 K. () (v) Show, by calculation, whether it is possible for this reaction to occur spontaneously at 98 K. () (b) When methanol is produced in industry, this reaction is carried out at 400 ºC and 00 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) + H (g) CH 3 OH(g) (i) Suggest reasons for the choice of temperature and pressure. Temperature Pressure (3) Maltby Academy 7

8 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) + H (g) CH 3 OH(g) (iv) In the equilibrium mixture at 00 atmospheres pressure, the partial pressure of carbon monoxide is 55 atmospheres and the partial pressure of hydrogen is 0 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 8

9 (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?.. Draw a line on the graph, labelled E C, to show the activation energy of the catalysed reaction. (Total 17 marks) 5. The reaction between nitrogen and hydrogen can be used to produce ammonia. N (g) + 3H (g) NH 3 (g) ΔH ο = 9. kj mol 1 Standard entropies are given below S ο [N (g)] = J mol 1 K 1 S ο [H (g)] = J mol 1 K 1 S ο [NH 3 (g)] = J mol 1 K 1 Maltby Academy 9

10 (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 98 K. Include a sign and units in your answer. () (c) (i) Calculate the total entropy change, ΔS ο total, at 98 K. Include a sign and units in your answer. Maltby Academy 10

11 Is this reaction feasible at 98 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 98 K, if the pressure remains constant. In terms of entropy, explain why this happens Higher pressures increase the yield of ammonia at equilibrium. Suggest a reason why pressures greater than 300 atmospheres are not routinely used (iii) Iron is a heterogeneous catalyst. Explain what is meant by heterogeneous (Total 9 marks) Maltby Academy 11

12 6. (a) Define the term standard enthalpy of formation (3) (b) In the Haber process, ammonia is manufactured from nitrogen and hydrogen as shown in the equation. N (g) + 3H (g) NH 3 (g) (i) Use the bond enthalpies below to calculate the standard enthalpy of formation of ammonia. Bond Bond enthalpy / kj mol 1 N N in N +945 H H in H +436 N H in NH (4) Maltby Academy 1

13 Draw a labelled enthalpy level diagram for the formation of ammonia in the Haber process. Enthalpy () (iii) State the temperature used in the Haber process and explain in terms of the rate of reaction and position of equilibrium, why this temperature is chosen. Temperature... (3) Maltby Academy 13

14 (iv) Identify the catalyst used in the Haber process and state what effect, if any, it has on the equilibrium yield of ammonia. Catalyst... Effect on yield... () (v) Explain why it is necessary to use a catalyst in this process. (c) The pressure used in the Haber process is 50 atmospheres. (i) State and explain an advantage of increasing the pressure to 1000 atmospheres. () Suggest a disadvantage of using a pressure of 1000 atmospheres. (Total 18 marks) Maltby Academy 14

15 7. One stage in the manufacture of sulphuric acid is SO (g) + O (g) SO 3 (g) The equilibrium constant K p = p p SO SO 3 p O (a) 10.0 mol of SO and 5.00 mol of O were allowed to react. At equilibrium, 90.0% of the SO was converted into SO 3. (i) Calculate the number of moles of SO, O and SO 3 present in the equilibrium mixture. () Calculate the mole fractions of SO, O and SO 3 at equilibrium. Maltby Academy 15

16 (iii) Assuming that the total pressure of the equilibrium mixture was.00 atm, calculate the partial pressures of SO, O and SO 3 at equilibrium. (iv) Calculate the value of K p. () (b) The reaction between sulphur dioxide and oxygen is exothermic. (i) State the effect, if any, on K p of increasing the temperature at constant pressure. Maltby Academy 16

17 Use your answer to (i), and the expression K p = p p SO SO 3 p O to explain the effect on the position of equilibrium of increasing the temperature at constant pressure. () (c) The reaction was repeated at a higher pressure whilst maintaining a constant temperature. (i) State the effect, if any, of an increase in the total pressure on the value of K p.... State the effect, if any, of this increase in pressure on the amount of sulphur trioxide in the equilibrium mixture.... (d) State the effect, if any, of a catalyst on: (i) K p... the equilibrium position.... (Total 13 marks) Maltby Academy 17

18 8. Methanoic acid and ethanol react together to form ethyl methanoate, HCOOC H 5, and water. This reaction is reversible and can be allowed to reach equilibrium. HCOOH(l) + C H 5 OH(l) HCOOC H 5 (l) + H O(l) ΔH = +45 kj mol 1 (a) Draw the full structural formula of ethyl methanoate, showing all bonds. (b) What type of organic compound is ethyl methanoate?... (c) In an experiment, 3.00 mol methanoic acid, HCOOH, and 6.5 mol ethanol, C H 5 OH, were mixed together. A small quantity of catalyst was added. The mixture was left for several days in a water bath to reach equilibrium at constant temperature. (i) Complete the table. at start of experiment Number of moles in the reaction mixture HCOOH C H 5 OH HCOOC H 5 H O at equilibrium 0.50 () Maltby Academy 18

19 Write an expression for the equilibrium constant, K c, for the reaction. (iii) Calculate K c for the reaction at the temperature of the experiment. The total volume of the equilibrium mixture was 485 cm 3. () (iv) State and explain whether K c for this reaction has units. Maltby Academy 19

20 (d) (i) The temperature of this equilibrium mixture is lowered. Explain the effect of this on the value of the equilibrium constant and hence on the yield of ethyl methanoate. (4) A student added more catalyst to the mixture. State, giving a reason, what would happen to the composition of the equilibrium mixture. (Total 13 marks) 9. (a) Still reacting / rate of forward reaction and backward reaction equal / implication that forward and backward reactions are still taking place But concentrations constant / no macroscopic changes (b) Temp (Increases) Left / to SO / to endothermic / lower yield Press Increases/faster Right to SO 3 / to smaller number of molecules 3 Maltby Academy 0

21 (c) (i) Increases rate / or suitable comment on rate Moves position of equilibrium to endothermic side / or suitable comment on equilibrium such as reasonable yield / less SO 3 Either compromise in which the rate is more important than the position of equilibrium or optimum temperature for catalyst to operate or valid economic argument 3 Increases rate / more SO 3 / only needs small pressure to ensure gas passes through plant / high or reasonable yield obtained at 1 atms or at low pressure anyway and references to economic cost against yield benefit e.g increase in pressure would increase yield of product but the increase in yield would not offset the cost of increasing the pressure (iii) Catalyst speeds up reaction 1 (d) Vanadium (V) oxide / vanadium pentoxide / V O 5 1 (e) Any one use production of fertilizers, detergents, dyes, paints, pharmaceuticals (in) car batteries, pickling metal 1 [13] PSO P 10. (a) (i) Kp = P SO3 O [ ] no mark ( ) OK 1 SO 3 SO + O Mols at start 0 0 mols at equ 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 1

22 i.e. P SO = P O = P SO3 = = = = 1.83 n.b. could show mole fraction for all 3 and then 10 later to give partial pressure. Kp = (5.46) (.73) / (1.83) = 4.5 atm 5 (b) (i) No effect 1 No effect 1 [8] 11. (a) (i) K c = [SO 3 ] / [SO ] [O ] = (0.03) K c = 3 3 ( ) = 4860or mol 1 dm 3 3 (b) (i) K c decreases 1 shifts to left / in reverse 1 (c) (i) no effect 1 no effect 1 (d) (i) K p = pso 3 / pso po penalise square brackets 1 Total number of moles consequential on a SO = 0.095(4); O = (); SO 3 = Maltby Academy

23 (iii) Partial pressures: SO = (5) atm; O = (4) atm; 1 SO 3 = 1.71(4) atm i.e. multiply answer in by (iv) / = 850 atm 1 [14] 1. (a) (i) Methanol is the biggest/ most complex molecule / greatest M R /most atoms/most electrons 1 S system = (130.6) = 19.1/ 19 J mol l K 1 Method answer + units (iii) yes as 3 molecules 1 OR yes as () gases a liquid 1 (iv) S surr = H/T (stated or used) = ( 19/ 98) = 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 (v) S total = = / J mol 1 K 1 / +14 J mol 1 K 1 / kj mol 1 K 1 Positive so possible (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 Not in same phase as reactants. ALLOW state instead of phase 1 (iii) K p = p(ch 3 OH)/p(CO) p(h ) 1 (iv) Partial pressure of methanol = = 15 atm K p = (15)/55 0 = / atm (c) (i) 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 3

24 13. Penalise units only once in this question (a) ( 19.3) [ ( )] = 198.8/199 J mol 1 K 1 (b) / 9. / H / T = + 309(.4) J mol 1 K 1 / (4) kj mol 1 K 1 (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 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 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] 14. (a) Enthalpy/heat/energy change for one mole of a compound/substance/ a product NOT solid/molecule/species/element Reject heat released or heat required unless both mentioned to be formed from its elements in their standard states ALLOW normal physical state if linked to standard conditions Reject natural state / most stable state standard conditions of 1 atm pressure and a stated temperature (98 K) 3 Reject room temperature and pressure Reject under standard conditions Maltby Academy 4

25 (b) (i) Bonds broken Bonds made N N (+)945 6N H ( )346 and ( )1308 3H H ( )53 H = = 93 sign and value H ο = 93 = 46.5 (kj mol 1 ) sign and value q on 3 rd mark 4 Accept 46.5 (kj mol 1 ) with working (4) Accept with working max (3) Accept +93 with working max () ( Enthalpy ) N + (3)H H OR 93 ()NH 3 Accept 46.5 Correct labelled levels Reject Reactants and Products as labels H labelled direction of arrow must agree with thermicity Accept double headed arrow Diagram marks cq on sign and value of H in (b)(i) IGNORE activation energy humps Maltby Academy 5

26 (iii) C Accept any temperature or range within this range higher temperature gives higher rate but a lower yield because reaction is exothermic Accept favours endothermic reaction more than exothermic so lower yield OR Lower temperature give higher yield because reaction is exothermic but rate is slower 3 Accept cq on sign of H f in (b)(i) or levels in Reject lower temp favours exothermic reaction (iv) Iron / Fe IGNORE any promoters no effect on yield (v) temp would have to be much higher for a reasonable rate then yield would be too low lower activation energy implies reasonable rate OR Allows reaction at a lower temp at a reasonable/fast rate giving a reasonable yield. 1 Accept rate too slow without catalyst at a temp giving a reasonable yield Reject to lower activation energy of reaction (c) (i) advantage higher (equilibrium) yield/more NH3 in equilibrium mixture/equilibrium shifts to right because smaller number of (gaseous) moles/molecules on rhs IGNORE any reference to change in rate Reject just more ammonia Maltby Academy 6

27 disadvantage (plant more) expensive because thicker pipes would be needed OR cost (of energy) for compressing the gases/cost of pump OR Cost of equipment/pressure not justified by higher yield 1 Accept stronger or withstand high pressure for thicker Accept vessel/container/plant /equipment/reaction vessels for pipes Reject just more expensive Reject just thicker pipes etc Reject apparatus [18] 15. (a) IGNORE s.f. throughout this question (i) moles SO ( ) = 1.00 (mol) moles O ( ) = (mol) moles SO (mol) all 3 correct () correct Reject multiples of the stated moles All three total number of moles i.e. X X X SO O SO ( 0.095) ( ) or ( 0.857) or or or 6 7 Reject rounding to 1 sig fig Mark consequential on (a)(i) 1 Maltby Academy 7

28 (iii) All three total pressure i.e pso =.00or = (atm) po =.00or = (atm) 9.00 pso 3 =.00or or = 1.71 (atm) Mark consequential on (a) 1 (iv) K p (1.71) (0.190) (0.095) K p = 851 atm 1 Mark consequential on (a)(iii) and (a)(iv) Accept answer with units and no working () Accept correct answers between 845 and 855 as this covers rounding up etc Reject wrong units e.g. mol 1 dm 3 (b) (i) (K p ) decreases 1 (K p decreases so) Reject any Le Chatelier argument (this prevents access to 1 st mark) p SO3 fraction/quotient p SO po has to decrease (to equal new k p ) so shifts to left hand side this mark only available if (b)(i) answer was k p decreases. Reject shifts to right, even if answer to (b)(i) was k p increases (as p SO 3 decreases whereas p SO and p O increase) (c) (i) No effect/none/zero (effect) 1 Maltby Academy 8

29 Increases OR more SO 3 /more sulphur trioxide OR increases amount of SO 3 /sulphur trioxide 1 (d) (i) No effect/none/zero (effect) 1 No effect/none/zero (effect) 1 [13] 16. (a) O H C H H O C C H H H 1 (b) ester 1 (c) (i) Moles: C H 5 OH: 3.75 Moles: HCOOC H 5 :.50 and moles H O :.50 for both [HCOOC H 5 ][H O] K c 1 [HCOOH][C H OH] 5 Reject obviously round brackets ( ) Maltby Academy 9

30 (iii) K c Must have clearly divided moles of each component by for 1 st mark e.g. [HCOOC H 5 ] = [H O] = 5.16 (mol dm 3 ) and [HCOOH] = 1.03 (mol dm 3 ) and [C H 5 OH] = 7.73 (mol dm 3 ) = 3.33 stand alone mark IGNORE sig.figs. (.50) Accept K c = 3.33 only scores () if it is stated that V cancels either here or in (iv) If [H O] omitted in, then answer K c = mol 1 dm 3 () but this will give K c = 1.33 mol 1 dm 3 with V omitted from calculation Reject 1 st mark if 485 used as V in expression (iv) No, (as) equal numbers of moles on both sides OR volumes cancel OR mol dm 3 cancel OR units cancel OR crossing out units to show they cancel 1 Accept equal powers/moles on both sides OR powers cancel Mark CQ on K c expression in Reject concentrations cancel (d) (i) (as reaction) endothermic Accept exothermic in backward direction (or words to that effect) K c decreases If state exothermic in forward direction, 1 mark only (out of 4) for CQ increase in K c numerator in quotient (has to) decrease OR denominator in quotient (has to) increase OR fraction (has to) decrease yield of HCOOC H 5 decreases 4 Maltby Academy 30

31 no effect as catalysts do not affect (the value of) K OR no effect as catalysts do not affect the position of equilibrium OR no effect as catalysts do not affect the yield OR No effect as catalysts increase the rate of the forward and backward reactions equally/to the same extent OR no effect as catalysts only increase the rate OR no effect as catalysts only alter the rate no effect can be stated or implied IGNORE any references to activation energy 1 Reject just catalysts increase rate [13] Maltby Academy 31