Equilibrium exam questions with solutions 1

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Equilibrium exam questions with solutions 1

  1. 1. Equilibrium / Reversible Reactions 2002 Question 10 (c) Question State Le Chatelier’s principle. (7) When 30 g of ethanoic acid and 23 g of ethanol were placed in a conical flask and a few drops of concentrated sulfuric acid added, an equilibrium was set up with the formation of ethylethanoate and water. The equilibrium is represented by the following equation. CH3COOH + C2H5OH = CH3COOC2H5 + H2O When the equilibrium mixture was analysed it was found to contain 10 g of ethanoic acid. (i) Write the equilibrium constant expression, K c, for this reaction. (6) (ii) Calculate the value of the equilibrium constant, Kc. (12) 2002 Question 10 Solution (c) (i) (ii) 2003 STATE: a system at equilibrium (3) reactions oppose applied stresses (4) [CH3COOC2H5][H2O] [CH3COOH][C2H5OH] (6) Kc = 4 (12) Accept 3.7 – 4.3 30 ÷ 60 = 0.5 mol; 23 ÷ 46 = 0.5 mol; 10 ÷ 60 = 0.16• mol CH3COOH + C2H5OH = CH3COOC2H5 + H2O 0.5 mol 0.5 mol 0 0 (3) 0.16• mol 0.16• mol (3) 0.3• mol 0.3• mol (3) 0.16• may be given as 0.16, 0.17, 0.166 0.3• x 0.3• 0.3• as 0.33 etc • • 0.16 x 0.16 = 4 (3) [NB < 3.7/>4.3 (– 1)] (7) (6) (12) Question 11 (a) State Le Chatelier’s principle. (7) A gaseous mixture of hydrogen, iodine and hydrogen iodide form an equilibrium according to the following equation. H2(g) + I2(g) = 2HI(g) (i) Write an expression for the equilibrium constant, Kc, for this system. (6) (ii) The value of the equilibrium constant, Kc, for this reaction is 50 at 721 K. If 2 moles of hydrogen iodide gas were introduced into a sealed vessel at this temperature calculate the amount of hydrogen iodide gas present when equilibrium is reached. (12) 2003 Question 11 (a) (i) (ii) State: reactions at equilibrium (3) oppose applied stresses* (4) [HI]2 [H2][I2] 1.56 mol (12) H2 + I2 = 2HI 0 mol 0 mol 2 mol ½x ½x 2–x (3) (2 – x)2 = 50 [Square root both sides] (3) (½ x)2 2 – x = 7.07 (or 7.1) (3) ½x X = 0.44 => [HI] = 1.56 (3) (7) (6) (12)
  2. 2. 2004 Question 9 (a) What is meant by chemical equilibrium? Why is it described as a dynamic state? (8) Consider the following reversible chemical reaction: N2(g) + 3H2(g) = 2NH3(g) ΔH = − 92.4 kJ (b) Use Le Chatelier’s principle to predict the levels (high or low) of temperature and pressure needed to maximise the yield of ammonia when equilibrium is established. Give a reason (i) for the temperature level you have predicted, (ii) for the level of pressure you have predicted. (12) (c) Are the temperature levels predicted using Le Chatelier’s principle actually used to maximise ammonia yield in industry? Explain your answer. (6) (d) What is the effect of a catalyst on a reversible reaction? (6) (e) In an experiment 6.0 moles of nitrogen and 18.0 moles of hydrogen were mixed and allowed to come to equilibrium in a sealed 5.0 litre vessel at a certain temperature. It was found that there were 6.0 moles of ammonia in the equilibrium mixture. Write the equilibrium constant expression for the reaction and calculate the value of the equilibrium constant (Kc) at this temperature. (18) 2004 Question 9 (a) (b) (i) (ii) (c) (d) (e) EQUIL: state in which rate of forward reaction = rate of reverse reaction (5) DYNAMIC: reaction has not stopped / product forming and returning to reactants constantly (3) PREDICT: low temperature (3) high pressure (3) REASON: forward reaction is exothermic (produces heat, raises temp.) (3) forward reaction goes to fewer molecules (smaller volume, lowers pressure) (3) [Marks for (i) and (ii) can be given independently of PREDICT] ARE: No / high temperature is used [compromise temperature / specified temp] (3) EXPLAIN: at low temperatures rate is too slow) / activation energy too high (3) EFFECT: lowers activation energy / increases rate for both / catalyses both / reaches equilibrium faster / equilibrium position unaffected (6) [Allow 3 marks for ‘increases rate’] (5) (3) (6) (3) (3) Kc: = [NH3]2 [N2][H2]3 (6) CALC: 0.4 l2 mol–2 (M–2) [or answer that rounds off to 0.4] (12) N2 + 3H2 = 2NH3 6 mol 18 mol 0 mol 3 mol 9 mol 6 mol (3) ÷5 0.6 M 1.8 M 1.2 M (3) Kc = (1.2)2 (3) = 0.4 (3) (0.6)(1.8)3 (6) (3) (3) (6) (12)
  3. 3. 2005 Question 9 (a) State Le Chatelier’s principle. (5) (b) A student is provided with glassware and other laboratory apparatus as well as the following chemicals: potassium dichromate(VI) (K2Cr2O7.2H2O), hydrochloric acid (HCl(aq)), sodium hydroxide (NaOH), cobalt(II) chloride crystals (CoCl2.6H2O) and deionised water (H2O). (i) Describe clearly how the student could use a selection of the chemicals listed above to establish a chemical equilibrium. Write a balanced equation for the equilibrium. (12) (ii) Describe how the student could then demonstrate the effect of concentration on that chemical equilibrium. State the observations made during the demonstration. (9) (c) The value of Kc for the following equilibrium reaction is 4.0 at a temperature of 373 K. CH3COOH + C2H5OH = CH3COOC2H5 + H2O (i) Write the equilibrium constant (Kc) expression for this reaction. (6) (ii) What mass of ethyl ethanoate (CH3COOC2H5) would be present in the equilibrium mixture if 15 g of ethanoic acid and 11.5 g of ethanol were mixed and equilibrium was established at this temperature? (18) 2005 Question 9 (a) (b) (i) (ii) STATE: reactions at equilibrium (2) oppose (minimise, relieve) applied stress(es) (3) *If the word stress is replaced by particular examples (e.g. pressure), all three (temp., pressure & conc.) must be given. DESCRIBE: choice of methods Method 1 make solution of (dissolve, put) potassium dichromate / in water Cr2O72- + H2O = CrO42- + 2 H+ FORMULAS: (3) BALANCING: (3) Method 2A Method 2B Method 2C make soln. of cobalt chloride make soln. of cobalt chloride in make soln. of cobalt in hydrochloric acid water chloride in water and add HCl to inter- mediate colour (violet, lilac) CoCl42- + 6H2O = Co(H2O)62+ + 4Cl─ FORMULAS: (3) BALANCING: (3) [Accept make solution of hydrochloric acid (3) // in water (3) // HCl + H2O = H3O+ + Cl¯ ] (6) Note: the equations may be given in reverse. DESCRIBE: choice of methods (i) (ii) Method 1 (alternative) Add HCl (no marks) solution changes from yellow to orange Method 2A add water solution from blue to pink (purple) (c) Method 2C(a) add water soln from intermediate to pink (purple) Question 11 (b) (3) (3) (6) (3) (3) (3) Method 2C(b) add HCl (3) soln from intermediate. (3) to blue (3) (Accept ‘red’ in place of ‘pink’ for Method 2) [Accept add sodium hydroxide (3) // monitor pH using a pH meter or pH probe (3) // note increase in pH ] (3) WRITE: (6) [CH3COOC2H5OH][H2O] [CH3COOH][C2H5OH] CALC: 14.08 – 14.96 g (18) CH3COOH + C2H5OH = CH3COOC2H5 + H2O 0.25 mol* 0.25 mol* 0 mol 0 mol [* addition must be shown for error to be treated as slip] 0.25 – x 0.25 – x x x ___x2__ =4 [square root both sides] (0.25 – x)2 x =2 0.25 – x x = 0.16• [allow 0.16 / 0.17] 0.16• x 88 = 14.6• (14.08 – 14.96) 2006 (3) (3) (6) (9) Method 1 add sodium hydroxide the solution changes from orange to yellow Method 2B add HCl solution from pink to blue (3) (3) (3) (3) (3) (3) (3) (6) (18)
  4. 4. State Le Châtelier’s principle. The following equilibrium is set up in solution by dissolving cobalt (II) chloride crystals in water to form the pink species Co(H2O)6 and then adding concentrated hydrochloric acid until the solution becomes blue. Co(H2O)62+ + 4Cl ¯ = CoCl42- + 6H2O pink blue (i) When the solution becomes blue, has reaction ceased? Explain. (ii) The forward reaction is endothermic. State and explain the colour change observed on cooling the reaction mixture. (iii) Other than heating, mention one way of reversing the change caused by cooling the reaction mixture. 2006 (6) (6) (6) Question 11 (b) (i) (ii) (iii) 2007 (7) STATE: reactions at equilibrium // oppose (minimise, relieve) applied stress(es)* [*If the word stress(es) is replaced by particular examples (e.g. pressure), all three (temperature, pressure & concentration) must be given.] (4 + 3) WHEN: no EXPLAIN: forward and reverse reactions continue at same rate STATE: becomes pink EXPLAIN: equilibrium shifts to left / backwards / in reverse / in the exothermic direction / to produce heat / to oppose cooling / to minimise (oppose) stress HOW: add conc. hydrochloric acid (HCl) / add source of chloride ions (Cl¯) e.g NaCl / remove water (7) (3) (3) (3) (3) (6) Question 10 (a) (a) (i) Write the equilibrium constant (Kc) expression for the reaction (7) (ii) Three moles of nitrogen gas and nine moles of hydrogen gas were mixed in a 1 litre vessel at a temperature T. There were two moles of ammonia in the vessel at equilibrium. Calculate the value of Kc for this reaction at this temperature. (12) (iii) Henri Le Chatelier, pictured on the right, studied equilibrium reactions in industry in the late 19th century. According to Le Chatelier’s principle, what effect would an increase in pressure have on the yield of ammonia at equilibrium? Explain. (6) 2007 (a) Question 10 (i) (ii) (iii) (b) 2008 (i) [NH3]2 [Square brackets essential] [N2][H2]3 CALC: 0.009 (1/108) M–2 (12) [or answers that give 0.009 correct to one significant figure] N2 + 3H2 = 2NH3 start: 3 mol/M 9 mol/M equil: 2 mol/M 6 mol/M 2 mol/M (6) Kc = 22 2 x 63 (3) = 0.009 ( 1/108 ) (3) WHAT: it would increase the yield of ammonia (3) [The increase in yield of ammonia must be mentioned] EXPL: reaction shifts in direction (to side) of fewer molecules (moles) (smaller volume) to decrease the pressure (3) STATE: equal volumes of gases contain equal numbers of molecules (particles) (4) under same conditions* of temperature and pressure (3) [* Do not accept “under all conditions”.] [Do not accept “at s.t.p.”] WRITE: Question 7. (7) (12) (6) (7)
  5. 5. A chemical equilibrium is established when eleven moles of hydrogen and eleven moles of iodine are mixed at a temperature of 764 K. Initially the colour of the mixture is deep purple due to the high concentration of iodine vapour. The purple colour fades and when equilibrium is established the colour of the mixture is pale pink and there are seventeen moles of hydrogen iodide present. The equilibrium is represented by the equation H2 (g) + I2 (g) = 2HI (g) ΔH = 51.8 kJ colourless purple colourless (b) What is meant by chemical equilibrium? When the colour of the mixture has become pale pink, has reaction ceased? Explain. (c) Write an expression for the equilibrium constant (Kc) for the reaction. Calculate the value of the equilibrium constant (Kc) at 764 K. (11) (6) (d) State Le Châtelier’s principle. Use Le Châtelier’s principle to predict and explain the effect of a decrease in temperature on (i) the yield of hydrogen iodide, (ii) the intensity of colour of the equilibrium mixture. (12) (6) (9) What change, if any, will an increase in the pressure on the equilibrium mixture have on the yield of hydrogen iodide? Explain. 2008 (6) Question 7 (a) (b) (i) (ii ) WHAT: state in which rate of forward reaction = rate of reverse (backward) reaction (5) [Accept “rates equal (the same) in both directions”.] WHEN: no (continuing) (3) EXPLAIN: chemical equilibrium a dynamic state / concentrations of reactants and products unchanged as rates equal / both forward and reverse reactions still occur (3) WRITE: [HI]2 [H2][I2] [Square brackets essential] CALC: 46.24 [46.2 or 46 ( – 1)] (12) H2 + I2 = 2 HI 11 mol 11 mol 0 mol 11 – x 11 – x 2x (3) 2.5 mol 2.5 mol 17 mol (3) Kc = 172 2.52 (3) = 46.24 (3) STATE: reactions at equilibrium (3) / oppose (minimise, relieve) applied stress(es)*} (3) [*If the word ‘stress(es)’ is replaced by particular examples (e.g. pressure), all three (temperature, pressure and concentration) must be given.] PREDICT: lower (decreased) yield of hydrogen iodide darker pink (purple) (3) EXPLAIN: reaction shifts backward (reverse) which is exothermic (heat producing, to raise temperature) / disfavours forward endothermic [Accept “reaction is endothermic so reverse occurs”.] (3) [Answers need not be linked] CHANGE: none (no change) (3) EXPLAIN: equal numbers of molecules (moles) on both sides of equation / pressure does not change ( has no affect on) the number of molecules (moles) for this reaction (3) [Answers must be linked] (11) (6) (12) (6) (3) (12)
  6. 6. 2009 Question 11 (a) Ammonia is formed in the Haber process according to the following balanced equation. 3H2(g) + N2 (g) = 2NH3(g) The table shows the percentages of ammonia present at equilibrium under different conditions of temperature T and pressure P when hydrogen and nitrogen gases were mixed in a 3:1 molar ratio. (i) Find from the table the conditions of temperature and pressure at which the highest yield of ammonia is obtained. (4) (ii) Deduce from the data whether this reaction is exothermic or endothermic. Explain your reasoning. (6) (iii) Identify one industrial problem associated with the use of high pressures. (3) (iv) Write an equilibrium constant (Kc) expression for this reaction. (6) (v) State the effect on the value of Kc of using a catalyst. Justify your answer. (6) 2009 (a) Question 11 (i) (ii) (iii ) (iv ) (v) FIND: T = 573 K P = 1000 atm [No need to designate numbers as T & P] (4) DEDUCE: forward reaction is exothermic / backward (reverse) reaction is endothermic (3) EXPLAIN: more NH3 (product) at lower temp / less N2 (H2, reactants) at lower temp or opposite at higher temp / forward reaction favoured at lower temp / backward (reverse) at higher temp higher yield at low temperature / lower yield at high temperature (3) IDENTIFY: high (building, maintenance) costs / danger of leaks / danger of explosions (3) (4) (6) [NH3]2 [Square brackets essential] (6) [N2][H2]3 STATE: no effect / none (3) JUSTIFY: forward and backward (reverse) rates equally affected by catalyst / rates increased but position of equilibrium unchanged / reaches equilibrium faster but yield unchanged (3) (6) WRITE: (3) (6)
  7. 7. 2009 Question 11 (a) Ammonia is formed in the Haber process according to the following balanced equation. 3H2(g) + N2 (g) = 2NH3(g) The table shows the percentages of ammonia present at equilibrium under different conditions of temperature T and pressure P when hydrogen and nitrogen gases were mixed in a 3:1 molar ratio. (i) Find from the table the conditions of temperature and pressure at which the highest yield of ammonia is obtained. (4) (ii) Deduce from the data whether this reaction is exothermic or endothermic. Explain your reasoning. (6) (iii) Identify one industrial problem associated with the use of high pressures. (3) (iv) Write an equilibrium constant (Kc) expression for this reaction. (6) (v) State the effect on the value of Kc of using a catalyst. Justify your answer. (6) 2009 (a) Question 11 (i) (ii) (iii ) (iv ) (v) FIND: T = 573 K P = 1000 atm [No need to designate numbers as T & P] (4) DEDUCE: forward reaction is exothermic / backward (reverse) reaction is endothermic (3) EXPLAIN: more NH3 (product) at lower temp / less N2 (H2, reactants) at lower temp or opposite at higher temp / forward reaction favoured at lower temp / backward (reverse) at higher temp higher yield at low temperature / lower yield at high temperature (3) IDENTIFY: high (building, maintenance) costs / danger of leaks / danger of explosions (3) (4) (6) [NH3]2 [Square brackets essential] (6) [N2][H2]3 STATE: no effect / none (3) JUSTIFY: forward and backward (reverse) rates equally affected by catalyst / rates increased but position of equilibrium unchanged / reaches equilibrium faster but yield unchanged (3) (6) WRITE: (3) (6)

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