CHEM J-8 June /01(a) With 3 C-O bonds and no lone pairs on the C atom, the geometry is trigonal planar.

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1 CHEM J-8 June /01(a) What is the molecular geometry of the formate ion? Marks 7 With 3 C-O bonds and no lone pairs on the C atom, the geometry is trigonal planar. Write the equilibrium constant expression for Equation 1. K = [H+ (aq)][hcoo (aq)] [HCOOH(aq)] At equilibrium at 25 C, the amount of formate ion formed from a M solution of formic acid is 4.2 %. Calculate the concentration of H + (aq) in this solution. From equation 1, the amount of formate ion is equal to the amount of H +. Hence: [H + (aq)] = M = M Answer: M Calculate the value of the equilibrium constant, K, for Equation 1 at this temperature. From above, [HCOO - (aq)] = [H + (aq)] = M. The remaining formic acid has is: [HCOOH(aq)] = ( ) M Hence: K = [H+ (aq)][hcoo (aq)] = (0.0042)(0.0042) = M [HCOOH(aq)] ( ) Answer: M Hence calculate the concentration of formate ion in a M solution of formic acid. The reaction table for this equilibrium is: HCOOH(aq) H + (aq) HCOO - (aq) initial (M) change (M) -x +x +x equilibrium x x x ANSWER CONTIUNES IN THE NEXGT PAGE

2 CHEM J-8 June /01(a) The equilibrium constant in terms of x is therefore: K = [H+ (aq)][hcoo (aq)] = (x)(x) [HCOOH(aq)] ( x) = x 2 = ( x) As K is very small, x ~ Using this approximation: x 2 ( x) ~ x 2 = x 2 = x = [HCOO - (aq)] = M Answer: M

3 CHEM J-10 June /01(a) Nitrogen and acetylene gases react to form hydrogen cyanide according to the reaction N 2 (g) + C 2 H 2 (g) 2HCN(g) K c = at 300 C Marks 8 Write out the equilibrium constant expression for K c for this reaction as shown above. K c = [HCN g ] 2 N 2 g [C 2 H 2 g ] The value of K p for this reaction at 300 C is also Why are the values of K p and K c the same for this reaction? The number of moles of gas remains constant during the reaction. 2 mol of gas react to give 2 mol of product gas. Write a balanced equation and calculate the value of the equilibrium constant K c ' for the formation of 1.0 mol of hydrogen cyanide gas from nitrogen and acetylene gases. ½ N 2 (g) + ½ C 2 H 2 (g) HCN(g) For this reaction, K c = N 2 g [HCN g ] 1/2 [C 2 H 2 g ] 1/2 = K c 1/2 = Answer: What is the equilibrium concentration of HCN(g) if nitrogen and acetylene are mixed so that both are at starting concentrations of 1.0 mol L 1? The reaction table for this is: ½ N 2 (g) ½ C 2 H 2 (g) HCN(g) initial change -x -x +2x equilibrium x 1.0 x 2x Hence, the equilibrium constant expression in terms of x is: K c = [HCN g ] N 2 g 1/2 [C 2 H 2 g ] 1/2 = (2x) (1.0! x) 1/2 (1.0! x) 1/2 = 2x (1.0! x) = ANSWER CONTINUES ON THE NEXT PAGE

4 CHEM J-10 June /01(a) So, 2x = x 2.015x = ˆ x = Hence, [HCN(g) = 2x N = M (The small x approximation can be used but as no quadratic needs to be solved, this is unnecessary.) Answer: M

5 CHEM J-10 June /01(a) Consider the following reaction. SO 2 (g) + NO 2 (g) SO 3 (g) + NO(g) An equilibrium mixture in a 1.00 L vessel was found to contain [SO 2 (g)] = M, [NO 2 (g)] = M, [SO 3 (g)] = M and [NO(g)] = M. If the volume and temperature are kept constant, what amount of NO(g) needs to be added to the reaction vessel to give an equilibrium concentration of NO 2 (g) of M? Marks 4 From the chemical equation, K eq = [SO 3 g ][NO g ] [SO 2 g ][NO 2 g ] As the original mixture is at equilibrium: K eq = [SO 3 g ][NO g ] [SO 2 g ][NO 2 g ] = (0.600)(0.400) (0.800)(0.100) = 3.00 This equilibrium is now disturbed by the addition of x M of NO(g). To reestablish equilibrium, the reaction will shift to the left by an unknown amount y. The reaction table for this is: SO 2 (g) NO 2 (g) SO 3 (g) NO(g) initial x change +y +y -y -y equilibrium y y y x - y As [NO 2 (g)] = M at the new equilibrium, y = ( ) M = M. Hence, the new equilibrium concentrations are: [SO 2 (g)] = ( ) M = M [NO 2 (g)] = M [SO 3 (g)] = ( ) M = M [NO(g)] = ( x 0.200) M = ( x) M As the system is at equilibrium, K eq = [SO 3 g ][NO g ] [SO 2 g ][NO 2 g ] = (0.400)(0.200!x) = 3.00 (1.000)(0.300) Solving this gives x = 2.05 M. As the reaction is carried out in a 1.00 L container, this is also the number of moles required. Answer: 2.05 mol THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY

6 CHEM J-8 June /01(a) Consider the following equilibrium reaction. 4HCl(g) + O 2 (g) 2H 2 O(g) + 2Cl 2 (g) K c = 885 at 500 o C If mol HCl, mol O 2, mol H 2 O and mol Cl 2 are mixed in a 1.0 L container at 500 o C, in what direction will the reaction proceed? Marks 3 As the gases are present in a 1.0 L container, their concentrations are: concentration = number of moles / volume [HCl] = mol / 1.0 L = M [O 2 ] = mol / 1.0 L = M [H 2 O] = mol / 1.0 L = M [Cl 2 ] = mol / 1.0 L = M The reaction quotient, Q, is: Q = [H 2O(g)] 2 [Cl 2 (g)] 2 [HCl(g)] 4 [O 2 g ] = (0.090)2 [0.085] 2 [0.030] 4 [0.020] = 3600 As Q > K c, the reaction will proceed towards the reactants. What is the value of K p for the reaction at 500 o C? Answer: towards reactants The reaction involves 5 mol of gaseous reactants going to 4 mol of gaseous products. The number of moles of gas decreases in the reaction with Δn = -1. K p and K c are related by: K p = K c (RT) Δn Hence, K p = (885) [ ( )] -1 = 14 Answer: 14

7 CHEM J-9 June /01(a) Write a balanced equation for the following reaction: WO 3 (s) + H 2 (g) W(s) + H 2 O(g) Marks 3 WO 3 (s) + 3H 2 (g) W(s) + 3H 2 O(g) What is the equilibrium constant expression, K p, for the above reaction? The equilibrium constant only involves the gaseous reactants and products. The equilibrium constant in terms of partial pressures, K p, is therefore: K p = What is the equilibrium constant, K c, for the above reaction, in terms of K p? The reaction involves no change in the moles of gas during the reaction: n = 0. As K p = K c (RT) n, for this reaction K p = K c. Fe 2 O 3 can be reduced by carbon monoxide according to the following equation. 3 Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) K p =19.9 at 1000 K At 1000 K, what are the equilibrium partial pressures of CO and CO 2 if the only gas initially present is CO at a partial pressure of atm? The reaction table is: Fe 2 O 3 (s) 3CO(g) 2Fe(s) 3CO 2 (g) initial / atm change / atm -3x +3x equilibrium / atm x 3x The solids do not appear in the equilibrium constant expression and do not need to be considered. The equilibrium constant in terms of partial pressures, K p, is given by: Hence, K p = =. = = (19.9)1/3 = x = (2.71)( x) = x or 11.1x = 2.65 or x = ANSWER CONTINUES ON THE NEXT PAGE

8 CHEM J-9 June /01(a) From the reaction table, p(co) = ( x) atm = atm p(co 2 ) = 3x atm = atm p(co) = atm p(co 2 ) = atm

9 CHEM J-10 June /01(a) Calculate the standard-free energy change for the oxidation of ammonia to nitric oxide and water, according to the following equation. Data: f G (NO(g)) = 87.6 kj mol 1 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(l) f G (NH 3 (g)) = 16.5 kj mol 1 f G (H 2 O(l)) = kj mol 1 Marks 3 Using rxn G = Σm f G (products) - Σn f G (reactants), rxn G = (( ) (4-16.5)) kj mol -1 as f G (O 2 (g)) = 0. = kj mol -1 Answer: kj mol -1 Is the reaction spontaneous under standard conditions? Give a reason for your answer. The reaction is spontaneous as rxn G < 0. This is the condition for a reaction to be spontaneous.

10 CHEM J-11 June /01(a) Heating SbCl 5 causes it to decompose according to the following equation. 4 SbCl 5 (g) SbCl 3 (g) + Cl 2 (g) A sample of 0.50 mol of SbCl 5 is placed in a 1.0 L flask and heated to 450 ºC. When the system reaches equilibrium there is 0.10 mol of Cl 2 present. Calculate the value of the equilibrium constant, K c, at 450 C. From the chemical equation, 1 mol of Cl 2 (g) is produced from every 1 mol of SbCl 5 (g) that reacts. As 0.10 mol of Cl 2 (g) is present at equilibrium, 0.10 mol of SbCl 5 (g) has reacted. Initially, 0.50 mol of SbCl 5 (g) was present so the amount left at equilibrium is ( ) mol = 0.40 mol. From the chemical equation, 1 mol of SbCl 3 (g) is produced from every 1 mol of SbCl 5 (g) that reacts. As 0.10 mol of SbCl 5 (g) has reacted, 0.10 mol of SbCl 3 (g) is present at equilibrium. [SbCl 5 (g)] eq =. = 0.40 mol L -1 = 0.40 M.. [SbCl 3 (g)] eq = [Cl 2 (g)] eq = = 0.10 mol L -1 = 0.10 M. The equilibrium constant in terms of concentrations, K c is given by: K c = =... = Answer: K c = 0.025

11 CHEM J-8 June /01(a) Ammonia, NH 3, is produced from nitrogen and hydrogen gas at high temperatures using the Haber process. At a temperature of 670 K and 50.0 MPa pressure, an equilibrium mixture was found to contain mol nitrogen, mol hydrogen and 1.50 mol ammonia. Write a balanced equation for the Haber process. Marks 6 N 2 (g) + 3H 2 (g) 2NH 3 (g) Calculate the mole fraction of each gas in the mixture. Total number of moles = (0.925 (N 2 ) (H 2 ) (NH 3 )) mol = 5.20 mol As mole fraction of A = X A =, = = = = = = Calculate the partial pressure of each gas. The mole fraction, X A, and the partial pressure, p A, are related by: p A = X A P total As P total = 50.0 MPa, = MPa = 8.90 MPa = MPa = 26.7 MPa = MPa = 14.4 MPa Calculate the value for K p for the reaction at this temperature. The equilibrium constant in terms of partial pressures is given by: The partial pressures need to be in atmospheres to be used in this equation. ANSWER CONTINUES ON THE NEXT PAGE

12 CHEM J-8 June /01(a) The equilibrium constant in terms of partial pressures is given by: The partial pressures need to be in atmospheres to be used in this equation. As 1 atm = kpa = Pa, = 8.90 MPa = atm = 88.0 atm = 26.7 MPa = atm = 263 atm = 14.4 MPa = atm = 142 atm Hence, = = (dimensionless) Answer:

13 CHEM J-9 June /01(a) A sample of 0.62 mol CCl 4 was placed in a 2.0 L container and heated to a certain temperature. At equilibrium, [Cl 2 ] = M. What is the value of the equilibrium constant K c for the following reaction at that temperature? 3 CCl 4 (g) C(s) + 2Cl 2 (g) The initial concentration of CCl 4 is: number of moles concentration = volume = 0.62 mol 2.0 L = 0.31 mol L 1 = 0.31 M The reaction table is: concentration CCl 4 (g) C(s) 2Cl 2 (g) start change -x - +2x equilibrium 0.31-x - 2x As C(s) is a solid, its concentration is essentially constant and is not included in the equilibrium calculations. At equilibrium, [Cl 2 (g)] = M and hence x = M. Hence at equilibrium, [CCl 4 (g)] = (0.31 x) M = 0.28 M. The equilibrium constant in terms of the concentrations, K c, is thus: K c = [Cl 2 g ] 2 = (0.060)2 [CCl 4 g ] (0.28) = Answer: 0.013

14 CHEM J-6 June /01(a) Determine K c for the reaction ½O 2 (g) + Na 2 O(s) Na 2 O 2 (s) at 25 C. 2 Data: Na 2 O(s) 2Na(s) + ½O 2 (g) K c = at 25 C. Na 2 O 2 (s) 2Na(s) + O 2 (g) K c = at 25 C. The reaction involves the formation of Na 2 O 2 (s) from Na 2 O(s) and thus involves the first reaction and the reverse of the second reaction. The reactions can be combined: Na 2 O(s) 2Na(s) + ½ O 2 (s) K c = Na(s) + O 2 (g) Na 2 O 2 (s) K c = 29 (5 10 ) = Na 2 O(s) + ½ O 2 (g) Na 2 O 2 (g) K c = ( ) 29 (5 10 ) = Answer:

15 CHEM J-9 June /01(a) K p = 7.0 for the reaction Br 2 (g) + Cl 2 (g) 2BrCl(g) at 400 K. Suppose a 1.0 L flask is filled with 0.30 atm Br 2 (g) and 0.30 atm Cl 2 (g) at 400 K. Find the pressures of all three gases at equilibrium. Marks 4 The equilibrium constant in terms of partial pressures, K p, is given by: K p = (P ) BrCl 2 (P )(P ) Br2 Cl2 The reaction table is: pressure Br 2 (g) Cl 2 (g) 2BrCl(g) start change -x -x +2x equilibrium 0.30-x 0.30-x 2x The equilibrium constant in terms of partial pressures, K p, is given by: K p = (P ) (2x) (2x) = = (0.30 x)(0.30 x) (0.30 x) BrCl (P Br )(P 2 Cl ) = 2 As K p is not small, the assumption that 0.30 x ~ x cannot be made. Hence, (2x) = (0.30 x) 7.0 2x = (0.30 x) 7.0 = (0.30 7) x 7 (0.30 7) x = = 0.17 (2 + 7) The partial pressures at equilibrium are therefore: P Br 2 = P Cl 2 = 0.30 x = 0.13 atm, PBrCl = 2x = 0.34 atm p(br 2 ): 0.13 atm p(cl 2 ): 0.13 atm p(brcl): 0.34 atm

16 CHEM J-7 June /01(a) Heating SbCl 5 causes it to decompose according to the following equation. SbCl 5 (g) SbCl 3 (g) + Cl 2 (g) A sample of 0.50 mol of SbCl 5 is placed in a 1.0 L flask and heated to 450 ºC. When the system reaches equilibrium there is 0.10 mol of Cl 2 present. Calculate the value of the equilibrium constant, K c, at 450 C. 4 One mole of Cl 2 is generated by the decomposition of one mole of SbCl 5. As 0.10 mol of Cl 2 is present at equilibrium, ( ) = 0.40 mol of SbCl 5 must be left. One mole of SbCl 3 is generated alongside the production of one mole of Cl 2 so the number of moles of SbCl 3 = number of moles of Cl 2 = 0.10 mol. number of moles The volume of the flask is 1.0 L so the concentration =. The volume concentrations are therefore: [SbCl 5 (g)] = 0.40 M, [Cl 2 (g)] = [SbCl 3 (g)] = 0.10 M. The equilibrium constant in terms of concentrations, K c, is therefore: K c = [Cl 2(g)][[SbCl 3(g)] (0.10) (0.10) = = [SbCl (g)] (0.40) 5 Answer: K c = 0.025

17 CHEM J-7 June /01(a) Aluminium acts as a reducing agent in the thermite reaction where Fe 2 O 3 is reduced to metallic iron. Write a balanced equation for the thermite reaction. Marks 4 2Al(s) + Fe 2 O 3 (s) à Al 2 O 3 (s) + 2Fe(s) What is the maximum theoretical mass of Fe that can be produced when 270 g of Al reacts with excess Fe 2 O 3 in the thermite reaction? 270 g of aluminium corresponds to: number of moles = mass of Al atomic mass of Al = 270 =10. mol From the chemical equation, 2 mol of Fe is produced for every 2 mol of Al consumed. Hence, mol of Fe is the maximum yield. This corresponds to: mass of Fe = number of moles of Fe molar mass of Fe = 10. g g mol -1 = 560 g Answer: 560 g What does the superscript o mean in the symbol ΔH f? 1 All reactants and products are in their standard states: the most stable form of the substance at a pressure of 1 atm and a temperature of 298 K. Briefly describe what is meant by Dynamic Equilibrium? 1 A reaction at equilibrium has not stopped - the rate of the forward reaction is equal to the rate of the backward reaction - a dynamic situation.

18 CHEM J-6 June 2004 Briefly describe the following ideas or phenomena. Dynamic equilibrium 3 A reaction at equilibrium has not stopped - the rate of the forward reaction is equal to the rate of the backward reaction - a dynamic situation. The difference between Q and K K is the equilibrium constant. For a particular reaction it is dependent on the temperature only. Q is the reaction quotient. It can be calculated from the concentrations and pressures of the reactants and products. Q = K when the reaction is at equilibrium. If Q < K, the reaction will proceed to the right to produce more product(s). If Q > K, the reaction will proceed to the left to produce more reactant(s). Effect of a catalyst on equilibrium A catalyst lowers the activation energy of a reaction by providing an alternative reaction pathway or mechanism. It has no effect on the equilibrium position, but it does allow equilibrium to be established faster.

19 CHEM J-9 June 2003 A saturated solution of iodine, I 2, in water contains g I 2 per litre, but more than this amount can dissolve in a KI solution because of the following equilibrium. 4 I (aq) + I 2 (aq) I 3 (aq) A solution of KI (0.100 M) dissolves 12.5 g of iodine per litre, most of which is converted to I 3 (aq). Assuming that the concentration of I 2 in all saturated solutions is the same, calculate the equilibrium constant for the above reaction. The molar mass of I 2 is ( ) g mol -1 = g mol -1. Hence, g number of moles = g mol 1 = mol As this amount dissolves in a litre of water, the concentration of I 2 in the saturated solution of iodine in water is M g Similarly 12.5 g of I 2 corresponds to g mol 1 = mol and in a litre of KI solution, this has a concentration is M. For the equilibrium, the reaction table is therefore: I - (aq) I 2 (aq) I - 3 (aq) [initial] change -x -x +x [equilibrium] x x x Assuming that [I 2 (aq)] is the same as in the saturated solution (as stated in the question), x = so x = giving: [I - (aq)] = = M, [I 2 (aq)] = M and [I 3 - (aq)] = M. The equilibrium constant is therefore: K c = [I 3 aq ] I aq [I 2 aq ] = (0.048) (0.0013) = 710 Answer: 710

20 CHEM1001/CHEM1101 combined 2003-N-8 November 2003 You are a member of a research team of industrial chemists who are discussing the operation of an ammonia plant. Ammonia is formed from nitrogen and hydrogen according to the following equilibrium reaction. Marks 3 N 2 (g) + 3H 2 (g) 2NH 3 (g) The plant operates close to 700 K, at which K p is atm 2 and employs the stoichiometric ratio 1:3 of N 2 :H 2. At equilibrium the partial pressure of NH 3 is 50 atm. Calculate the partial pressures of each reactant and hence the total pressure under these conditions. The equilibrium constant in terms of partial pressures, K p, is given by: (P NH3 ) 2 K p = (P N2 )(P H2 ) 3 If the partial pressure of N 2 (g) = x, the partial pressure of H 2 (g) = 3x: (50) 2 K p = (x)(3x) 3 = or 27x 4 = (50) 2 = Hence, x = 31 atm. Therefore, P N2 = 31 atm, P H2 = 3 31 atm = 93 atm P total = P X2 = ( ) atm = 174 atm p(n 2 ) = 31 atm p(h 2 ) = 93 atm p(total) = 174 atm Ammonium carbamate (NH 2 COONH 4 ) is a salt of carbamic acid that is found in the blood and urine of mammals. At 250 ºC, K c = M 3 for the following equilibrium: NH 2 COONH 4 (s) 2NH 3 (g) + CO 2 (g) If 7.81 g of NH 2 COONH 4 is introduced into a L evacuated container, what is the total pressure inside the container at equilibrium at 250 ºC? 3 If x mol of ammonium carbamate decomposes, it will produce 2x mol of NH 3 (g) and x mol of CO 2 (g). As ammonium carbamate is a solid, it does not appear in the expression for K c. Therefore: K c = [NH 3 (g)] 2 [CO 2 (g)] = (2x) 2 (x) = 4x 3 = Hence, x = M or [NH 3 (g)] = = M and [CO 2 (g)] = M. ANSWER CONTINUES ON THE NEXT PAGE

21 The concentration is given by c = n nrt. Hence for each component P = V V = crt. Hence, P total = ( M) RT + ( M) RT = (( ) M) ( L atm K -1 mol -1 ) (( ) K) = atm