CHEM J-9 June The equilibrium constant expressions for reactions (1) and (2) are given by:

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1 CHEM J-9 June 014 Use the following equilibria: CH 4 (g) C H 6 (g) + H (g) K 1 = CH 4 (g) + H O(g) CH 3 OH(g) + H (g) K = to calculate the equilibrium constant, K 3, for the following reaction: Show all working. CH 3 OH(g) + H (g) C H 6 (g) + H O(g). The equilibrium constant expressions for reactions (1) and () are given by: K 1 = C H 6 g [H g ] [CH 4 g ] and K = CH 3OH g [H g ] [CH 4 g ][H O g ] The equilibrium constant for reaction (3) is: K 3 = C H 6 g [H O g ] [CH 3 OH g ] [H g ] K 3 can be obtained by multiplying K 1 by 1 / K : K 1 / K = C H 6 g [H g ] / CH 3OH g [H g ] [CH 4 g ] [CH 4 g ] [H O g ] = C H 6 g [H O g ] [CH 3 OH g ] [H g ] = K 3 K 3 = ( ) / ( ) = Answer:

2 CHEM J-10 June 014 Consider the following reaction. N O 4 (g) NO (g) K c = at 5 C 4 A mol sample of NO is allowed to come to equilibrium with N O 4 in a 0.50 L container at 5 C. Calculate the amount (in mol) of NO and N O 4 present at equilibrium. Show all working. The initial concentration of NO (g) is: [NO (g)] = number of moles / volume = mol / 0.50 L = 0.17 mol L -1 A reaction table can be used to calculate the equilibrium concentrations. N O 4 (g) NO (g) initial change +x -x equilibrium x 0.17 x Hence, K c = [NO g ] [N O 4 ] = (0.17!x) x = So, (0.17 x) = x x + 4x = x 4x 0.693x = 0 Solving the quadratic gives x = and x = The second route would give a negative value for [NO (g)]. Using x = : [N O 4 (g)] = M and [NO (g)] = ( ) M = M As these concentrations are in a 0.50 L container: number of moles of N O 4 (g) = concentration volume = mol L L = mol number of moles of NO (g) = mol L L = mol Amount of NO : mol Amount of N O 4 : mol

3 CHEM N-10 November 014 Use the following equilibria: CH 4 (g)! C H 6 (g) + H (g) K 1 = CH 4 (g) + H O(g)! CH 3 OH(g) + H (g) K = to calculate the equilibrium constant, K 3, for the following reaction. CH 3 OH(g) + H (g)! C H 6 (g) + H O(g) Show all working. The equilibrium constant expressions for the three reactions are: K 1 = C H 6 g [H g ] [CH 4 g ] K = CH 3OH g [H g ] CH 4 g [H O g ] Hence, K 3 = K 1 / K : K 3 = C H 6 g [H O g ] CH 3 OH g [H g ] K 1 / K = C H 6 g [H g ] [CH 4 g ] / CH 3OH g [H g ] CH 4 g [H O g ] = C H 6 g [H O g ] CH 3 OH g [H g ] = ( ) / ( ) = Answer:

4 CHEM N-1 November 014 The standard Gibbs free energy of the following reaction is kj mol 1. COCl (g)! CO(g) + Cl (g) What is the expression for the equilibrium constant, K p, for this reaction? 5 K p = P COP Cl P COCl Calculate the value of the equilibrium constant at 98 K. Using ΔG o = -RTlnK p : J = -(8.314 J K -1 mol -1 ) (98 K) lnk p lnk p = K p = K p = In which direction will this reaction proceed if a mixture of gases is made with: P COCl = 1.00 atm; P Cl = 0.01 atm; P CO = 0.50 atm? Show working. Using Q p = P COP Cl P COCl : Q p = (0.50)(0.01) (1.00) = As Q p > K p, the reaction will proceed towards reactants: P COCl will increase P CO and P Cl will decrease.

5 CHEM J-9 June 013 Consider the following reaction. N O 4 (g) NO (g) 4 An equilibrium mixture in a 1.00 L container is found to contain [N O 4 ] = 1.00 M and [NO ] = 0.46 M. The vessel is then compressed to half its original volume while the temperature is kept constant. Calculate the concentration [N O 4 ] when the compressed system has come to equilibrium. Show all working. For this reaction, the equilibrium constant expression is given by: K c = [NO g ] [N O 4 ] As mixture is at equilibrium when [N O 4 ] = 1.00 M and [NO ] = 0.46 M: K c = (0.46) (1.00) = 0.1 If the volume of the vessel is halved, the initial concentrations will double: [N O 4 ] =.00 M and [NO ] = 0.9 M. The reaction is no longer at equilibrium and Le Chatelier s principle predicts it will shift towards the side with fewer moles: it will shift towards reactants. A reaction table needs to be used to calculate the new equilibrium concentrations. N O 4 (g) NO (g) initial change +x -x equilibrium.00 + x 0.9 x Hence, K c = [NO g ] [N O 4 ] = (0.9!x) (.00!x) = 0.1 So, (0.9 x) = 0.1 (.00 + x) x + 4x = x 4x 3.89x = 0 ANSWER CONTINUES ON THE NEXT PAGE

6 CHEM J-9 June 013 With a = 4, b = and c = 0.43, this quadratic equation has roots: x =!b ± b!4ac a = 3.89 ± (!3.89)! This gives x = 0.13 or The latter makes no chemical sense as it gives a negative concentration for NO. Hence using x = 0.13: [N O 4 ] = (.00 + x) M = ( ) M =.13 M [NO ] = (0.9 - x) M = ( ) M = 0.66 M Answer: [N O 4 ] =.13 M

7 CHEM N-1 November 013 The standard Gibbs free energy of the following reaction is kj mol 1. COCl (g) CO(g) + Cl (g) 5 What is the expression for the equilibrium constant, K p, for this reaction? K p = p(cl )p(co) p(cocl ) Calculate the value of the equilibrium constant at 98 K. As G = RT lnk, = -( ) lnk K = K p = In which direction will this reaction proceed if a mixture of gases is made with: P COCl = 1.00 atm; P Cl = 0.01 atm; P CO = 0.50 atm? Show working. The reaction quotient, Q p, is: Q p = p(cl )p(co) p(cocl ) = (0.01)(0.50) (1.00) = As Q p > K p, the reaction proceeds to decrease Q p. It moves to reduce the amount of produce and increase the amount of reactant: it shifts to the left (towards reactants). THIS QUESTION CONTINUES ON THE NEXT PAGE.

8 CHEM N-13 November 013 This reaction mixture is now allowed to come to equilibrium at 98 K in a fixed volume container. Calculate the equilibrium pressure of Cl. 4 As shown in 013-N-1, the reaction will shift towards reactants to reach equilibrium. The equilibrium constant for the forward reaction is very small: K = With P Cl = 0.01 atm initially, it is easiest to assume that (i) all of the Cl reacts and (ii) a little of the COCl then reacts to reach equilibrium. (i) All of the Cl reacts with CO to make COCl : P COCl = ( ) atm = 1.01 atm P Cl = ( ) atm = 0.00 atm P CO = ( ) atm = 0.49 atm (iii) This reaction will then shift to equilibrium: COCl (g) CO(g) + Cl (g) initial change -x +x +x equilibrium 1.01 x x x K p = p(cl )p(co) p(cocl ) = x( x) (1.01 x) As the equilibrium constant is so small, x will be tiny. In this case, x ~ 0.49 and 1.01 x ~ Hence: K p ~ x(0.49) (1.01) = x = = Answer:

9 CHEM J-10 June 01 Consider the following reaction. SO (g) + NO (g) SO 3 (g) + NO(g) An equilibrium mixture in a 1.00 L vessel was found to contain [SO (g)] = M, [NO (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 (g) of M? 4 From the chemical equation, K eq = [SO 3 g ][NO g ] [SO g ][NO g ] As the original mixture is at equilibrium: K eq = [SO 3 g ][NO g ] [SO g ][NO 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 (g) NO (g) SO 3 (g) NO(g) initial x change +y +y -y -y equilibrium y y y x - y As [NO (g)] = M at the new equilibrium, y = ( ) M = 0.00 M. Hence, the new equilibrium concentrations are: [SO (g)] = ( ) M = M [NO (g)] = M [SO 3 (g)] = ( ) M = M [NO(g)] = ( x 0.00) M = ( x) M As the system is at equilibrium, K eq = [SO 3 g ][NO g ] [SO g ][NO g ] = (0.400)(0.00!x) = 3.00 (1.000)(0.300) Solving this gives x =.05 M. As the reaction is carried out in a 1.00 L container, this is also the number of moles required. Answer:.05 mol THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY

10 CHEM J-1 June 01 The diagram below represents the Gibbs Free energy change associated with the formation of 4 different oxides. 4 Temperature ( C) ΔG (kj mol 1 ) Using the free energy data above, write down the equation and indicate with an arrow the direction of the expected spontaneous reaction under the following conditions. If you think no reaction would occur, write no reaction. a) C and SnO are mixed at 400 C At 400 C, the Sn / SnO line is below the C / CO line. Hence, there is no reaction. b) C and SnO are mixed at 900 C SnO + C à Sn + CO c) SnO, Sn, Zn and ZnO are mixed at 900 C Zn + SnO à ZnO + Sn Of the 4 oxide formation reactions, write down one for which the entropy change is negative. Provide a brief explanation for your choice. The entropy change is likely to be negative in three of them: Sn + O à SnO Zn + O à ZnO 4/3Al+ O à /3AlO Each of these involves a decrease in the number of moles of gas. Gases have far higher entropy than solids. The fourth reaction involves an increase in the number of moles of gas so is likely to produce an increase in entropy.

11 CHEM N-11 November 01 The standard Gibbs free energy of formation for ammonia, NH 3 (g), is 16.4 kj mol 1. Consider the following reaction at 98 K. 5 N (g) + 3H (g) NH 3 (g) What is the expression for the equilibrium constant, K p, for this reaction? K p = p (NH 3 ) p(n ) p 3 (H ) Calculate the value of the equilibrium constant at 98 K. The reaction as written produces mol of NH 3, hence: Δ r G o = Δ f G o (NH 3 (g)) = kj mol -1 = -3.8 kj mol -1 As ΔG o = -RTlnK p : J mol -1 = -(8.314 J K -1 mol -1 ) (98 K) lnk p lnk p = 13.4 K p = K p = In which direction will this reaction proceed if a mixture of gases is made with: P NH3 = 1.00 atm; P H = 1.00 atm; P N = 0.50 atm? The reaction quotient with this mixture is: p (NH Q p = 3 ) p(n ) p 3 (H ) = (1.00) (0.50)(1.00) =.00 3 As Q p < K p, the reaction will proceed to products (i.e. to the right).

12 CHEM N-13 November 01 Consider the following reaction. SO (g) + NO (g) SO 3 (g) + NO(g) An equilibrium mixture in a 1.00 L vessel was found to contain [SO (g)] = M, [NO (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 (g) of M? 4 From the chemical equation, K eq = [SO 3 g ][NO g ] [SO g ][NO g ] As the original mixture is at equilibrium: K eq = [SO 3 g ][NO g ] [SO g ][NO 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 (g) NO (g) SO 3 (g) NO(g) initial x change +y +y -y -y equilibrium y y y x - y As [NO (g)] = M at the new equilibrium, y = ( ) M = 0.00 M. Hence, the new equilibrium concentrations are: [SO (g)] = ( ) M = M [NO (g)] = M [SO 3 (g)] = ( ) M = M [NO(g)] = ( x 0.00) M = ( x) M As the system is at equilibrium, K eq = [SO 3 g ][NO g ] [SO g ][NO g ] = (0.400)(0.00!x) = 3.00 (1.000)(0.300) Solving this gives x =.05 M. As the reaction is carried out in a 1.00 L container, this is also the number of moles required. Answer:.05 mol

13 CHEM J-10 June 010 Ammonia, NH 3 (g), has a standard Gibbs free energy of formation equal to 16.4 kj mol 1. Consider the following reaction at 98 K. 6 N (g) + 3H (g) NH 3 (g) In which direction will this reaction proceed if a mixture of gases is made with: P NH3 = 1.00 atm P H = 0.50 atm P N = 0.50 atm The reaction as written forms moles of NH 3 (g). The standard Gibbs free energy of the reaction is therefore ( kj mol -1 ) = -3.8 kj mol -1. Using G = -RTlnK p, the value of K p can be determined: lnk p = - G / RT = -( J mol -1 ) / (8.314 J K -1 mol -1 )(98 K) = 13.4 K p = The direction of the reaction can be determined by calculating the reaction quotient: Q = =... = 16 As Q < K p, the reaction will proceed to the right. This will increase the partial pressure of NH 3 (g) and decrease the partial pressures of N (g) and H (g). Answer: to the right What pressure of hydrogen gas should be added to a mixture already containing 0.0 atm NH 3 and 0.50 atm N so that the amounts of NH 3 and N will not change? If the partial pressures of NH 3 (g) and N (g) do not change, these values must be the values at equilibrium. Hence: K p = = = atm 3 = atm.. = Answer: atm

14 CHEM J-11 June 010 Determine the value of the equilibrium constant (at 98 K) for the following reaction. CO (g) + H O(l) H CO 3 (aq) 3 Substance f Hº / kj mol 1 Sº / J K 1 mol 1 H CO 3 (aq) H O(l) CO (g) Using rxn H = Σm f H (products) - Σn f H (reactants), the enthalpy of this reaction is: H = ( f H (H CO 3 (aq)) - ( f H (CO (g) + f H (H O(l)) = [(-700.) ( )] kj mol -1 = -0 kj mol -1 Using rxn S = ΣmS (products) - ΣnS (reactants), the entropy change of this reaction is: S = (S (H CO 3 (aq)) - (S (CO (g) + S (H O(l)) = [(187) ( ))] kj mol -1 = -97 J K -1 mol -1 Using G = H - T S : G = ( J mol -1 ) (98 K)(-97 J K -1 mol -1 ) = J mol -1 Using G = -RTlnK p, the value of K p can be determined: lnk p = - G / RT = -(+8906 J mol -1 ) / (8.314 J K -1 mol -1 )(98 K) = K p = 0.07 Answer: 0.07 Consider the following equilibrium. CO(g) + H O(g) CO (g) + H (g) K c = 31.4 at 588 K If a L vessel contains.50 mol CO(g),.50 mol H O(g), 5.00 mol CO (g) and 5.00 mol H (g) at 588 K, what are the concentrations of all species at equilibrium? The initial concentrations are: [CO(g)] = number of moles / volume =.50 mol / L = 0.5 M [H O(g)] =.50 mol / L = 0.50 M [CO (g)] = 5.00 mol / L = M [H (g)] = 5.00 mol / L = M ANSWER CONTINUES ON THE NEXT PAGE

15 CHEM J-11 June 010 The reaction table is therefore: CO(g) H O(g) CO (g) H (g) initial change -x -x +x +x equilibrium x x x x The equilibrium constant is therefore: Hence, K c = =.... = =.. = = (31.4)1/ = 5.60 Finally: x = 5.60 (0.50 x) x = x x = M [CO(g)] = ( ) M = M [H O(g)] = ( ) M = M [CO (g)] = ( ) M = M [H (g)] = = ( ) M = M [CO] = M [H O] = M [CO ] = M [H ] = M

16 CHEM N-8 November 010 The reaction below is endothermic. N O 3 (g) NO(g) + NO (g) Indicate whether the equilibrium will shift right, shift left, or remain unchanged when disturbed in the following ways. 3 adding more NO(g) increasing the pressure at constant temperature removing NO (g) increasing the volume at constant temperature adding some Ar(g) increasing the temperature at constant pressure left left right right no change right

17 CHEM N-9 November 010 At 1000 K, a reaction mixture containing SO (g), O (g) and SO 3 (g) was allowed to come to equilibrium in a reaction vessel. The reaction is: 3 SO (g) + O (g) SO 3 (g) At equilibrium, the system was found to contain the following concentrations: [SO ] = M, [O ] = M and [SO 3 ] = M. Calculate K c for this reaction. The equilibrium constant in terms of concentrations is given by: K c = =... = 56 K c = 56 If a mixture containing [SO ] = M, [O ] = M, and [SO 3 ] = M is placed in the vessel, is the reaction at equilibrium? If not, which way will it shift in order to achieve equilibrium, right or left? With these initial concentrations, Q = =... = 0.05 As Q < K c, the reaction will proceed to the right to increase the concentration of the product and reduce the concentrations of the reactants.

18 CHEM N-10 November 010 What does it mean to say that a reaction has reached equilibrium? 1 A reaction is at equilibrium when the rate of the forward reaction equals the rate of the backward reaction. Consider the following equilibrium. Br (g) + Cl (g) BrCl(g) K c = 7.0 at 400 K 4 If 0.44 mol of Br and 0.44 mol of Cl are introduced into a.0 L container at 400 K, what are the equilibrium concentrations of Br (g), Cl (g) and BrCl(g)? The initial concentrations are: [Br (g)] = number of moles / volume = (0.44 mol) / (.0 L) = 0. M [Cl (g)] = (0.44 mol) / (.0 L) = 0. M [BrCl(g)] = 0 M A reaction table can then be used: Br (g) Cl (g) BrCl(g) initial change -x -x +x equilibrium 0. x 0. x x The equilibrium constant in terms of concentrations is then: or Hence, K c = = =... = 7.0. = (7.0)1/ x = 0.15 [Br (g)] = ( ) M = M [Cl (g)] = ( ) M = M [BrCl(g)] = 0.5 M [Br (g)] = M [Cl (g)] = M [BrCl(g)] = 0.5 M

19 CHEM J-9 June 009 Consider the following reaction at equilibrium. CH 3 OH(g) CO(g) + H (g) K c = What is the concentration of CO(g) when [CH 3 OH(g)] = M and [H (g)] = M? The equilibrium constant for the reaction is given by: K eq = CO g [H g ] [CH 3 OH g ] = When [CH 3 OH(g)] = M and [H (g)] = M, [CO(g)] = K eq[ch 3 OH g ] [H g ] = ( ) ( ) M = M Answer: M Explain briefly the chemical principles behind a) froth flotation, or b) electrorefining. Froth flotation is a technique to separate a mineral from unwanted dirt and rocks. The crude ore is crushed to a fine powder and then treated with water to produce a slurry. A surfactant that selectively coats the mineral, thus making it more hydrophobic, is added and the mixture agitated and aerated. The mineral attaches to the air bubbles and floats to the surface (as a froth) where it is collected before undergoing further refining. Electrorefining is a technique for purifying a metal, e.g. copper. An electrolytic cell consisting of a pure copper cathode and an impure copper anode is constructed. A voltage is selectively applied so that noble metals (less electropositive than Cu) do not dissolve. When operating, the current causes the impure copper anode to dissolve, including metal impurities more electropositive than copper. The noble metals do not dissolve and form a sludge. Only pure copper is deposited at the cathode - the more electropositive metals stay in solution as cations.

20 CHEM J-10 June 009 Solid NH 4 HS in placed in an evacuated container at 5 ºC and the following equilibrium is established. 3 NH 4 HS(s) NH 3 (g) + H S(g) Hº = +93 kj mol 1 At equilibrium, some solid NH 4 HS remains in the container. Predict and explain each of the following. (a) The effect on the equilibrium partial pressure of NH 3 gas when additional solid NH 4 HS is introduced into the container. The equilibrium constant is given by K p = [NH 3 (g)][h S(g)]. It does not include the concentration of the solid as this is a constant during the reaction. If extra solid is introduced, it does not change the concentration of the solid and there is no effect on the equilibrium. The solid is not included in the equilibrium constant so changing the amount present has no effect on the equilibrium. (b) The effect on the amount of solid NH 4 HS present when the volume of the container is decreased. Decreasing the volume of the container will increase the partial pressure of both gases. The reaction will move to the left to reduce the overall pressure. The amount on solid will therefore increase. (c) The effect on the amount of solid NH 4 HS present when the temperature is increased. The reaction is endothermic, as Hº is positive. If the temperature is increased, the reaction will move to the right to absorb the extra heat. The amount of solid will therefore decrease.

21 CHEM J-13 June 009 Fe O 3 can be reduced by carbon monoxide according to the following equation. Fe O 3 (s) + 3CO(g) Fe(s) + 3CO (g) K p =19.9 at 1000 K At 1000 K, what are the equilibrium partial pressures of CO and CO if the only gas initially present is CO at a partial pressure of atm? The reaction table is: Fe O 3 (s) 3CO(g) Fe(s) 3CO (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 = p(co ) 3 p(co) 3 = (3x) 3 ( x) 3 = 19.9 (3x) ( x) = (19.9)1/3 =.71 3x = (.71)( x) = x or 11.1x =.65 or x = 0.38 From the reaction table, p(co) = ( x) atm = 0.64 atm p(co ) = 3x atm = atm p(co) = 0.64 atm p(co ) = atm

22 CHEM N-8 November 009 The value of the equilibrium constant, K c, for the following reaction is mol L 1. CO (g) + N (g) CO(g) + NO(g) What is the equilibrium concentration of CO(g) if the equilibrium concentration of [CO (g)] = 0.49 M, [N (g)] = M and [NO(g)] = M? The equilibrium constant for the reaction is given by: K eq = CO g [NO g ] [CO g ] [N g ] = When [CO (g)] = 0.49 M, [N (g)] = M and [NO(g)] = M, [CO(g)] = K eq[co g ] [N g ] [NO(g)] = (0.49) (0.319) (0.350) M = M [CO(g)] = 0.73 M Answer: 0.73 M

23 CHEM J-9 June 008 Explain the difference between an equilibrium constant and a reaction quotient. Both the equilibrium constant (K) and the reaction quotient (Q) show the relationship between the amounts of product and reactant. K refers to equilibrium concentrations and so refers to a system at equilibrium. It is a constant for any system, depending only on the temperature. Q refers to concentrations that are not necessarily at equilibrium. The following reactions have been demonstrated in mammalian liver at 37 C and ph 7.5. aspartate + citrulline argininosuccinate + H O K eq = 13 (1) argininosuccinate arginine + fumarate K eq = 4.5 () fumarate + NH 4 + aspartate K eq = Calculate the equilibrium constant at 37 C and ph 7.5 for the following reaction. arginine + H O citrulline + NH 4 + (4) If the first three reactions are combined by reversing each one and adding the reactions together, reaction (4) results: -(1) argininosuccinate + H O aspartate + citrulline K eq = 1/13 -() arginine + fumarate argininosuccinate K eq = 1/4.5 -(3) aspartate fumarate + NH 4 + (4) arginine + H O citrulline + NH 4 + K eq = 1/0.17 K eq = (1/13) (1/4.5) (1/0.17) When a reaction is reversed, the new equilibrium constant is the reciprocal of the original value. This has been used in the final column to obtain the equilibrium constants for the reverse of reactions (1), () and (3). When reactions are added together, the new equilibrium constant is the product of the equilibrium constants for the individual reactions. Overall, this gives for reaction (4): K eq = (1/13) (1/4.5) (1/0.17) = 0.10 Answer: K eq = 0.10

24 CHEM J-10 June 008 Methanol, CH 3 OH, is produced commercially by the catalysed reaction of carbon monoxide and hydrogen gas. K p for this reaction at 600 K is CO(g) + H (g) CH 3 OH(g) The reaction is exothermic, yet the equilibrium favours the reactants. Explain why this is the case. For a process to be spontaneous, the entropy of the universe must increase. Exothermic reactions lead to heat being given to the surroundings leading to an increase in the entropy of the surroundings, surroundings S. Formation of reactants is endothermic so surroundings S is negative. However, as formation of reactants leads to the formation of 3 molecules from each CH 3 OH, it leads to an increase in the entropy of the system, system S is postivie.. s For reactants to be favoured requires system S to be more positive than surroundings S is negative. As surroundings S = q / T, it becomes less negative as T increases and so formation of reactants becomes more favourable at high T. The reaction vessel at 600 K is filled with 0.0 atm of CO(g) and 0.0 atm H (g). What is the final pressure of CH 3 OH(g) at equilibrium? The equilibrium constant in terms of partial pressures, K p, is given by: K p = ( pch ) 3OH = ( pco )( ph ) CO(g) + H (g) CH 3 OH(g) initial change -x -x +x equilibrium x 0.0 x x Thus, at equilibrium: K p = ( pch ) 3OH ( pco )( ph ) = x (0.0 - x)(0.0 - x) = As K p is very small, x will be tiny compared to 0.0 atm and so (0.0 x) ~ 0.0. This approximation simplifies the expression for K p : x K p ~ = so x = ( ) (0.0) (0.0) (0.0)(0.0) x = p CH3OH = atm Answer: p CH3OH = atm ANSWER CONTINUES ON THE NEXT PAGE

25 CHEM J-10 June 008 Under what conditions of temperature and pressure do you think an industrial reactor would function to optimise the production of methanol? Explain. As formation of methanol in this reaction is exothermic, its formation is favoured by low temperature. However, as a low temperature would also lead to a slow reaction, the temperature used would be chosen as a compromise between maximizing the yield whilst keeping the rate of its formation high. As the reaction leads to a decrease in the number of moles of gas (3 mol of gas 1 mol of gas), high pressure will favour product formation. Thus, the industrial process uses high pressures.

26 CHEM N-9 November 008 Consider the following reaction. N O 4 (g) NO (g) 4 An experiment was conducted in which mol of N O 4 (g) was introduced into a L flask. After equilibrium had been established at a particular temperature, the concentration of N O 4 (g) was found to be M. Calculate the equilibrium constant, K c, for the reaction as written at that temperature. As initially mol of N O 4 (g) is present in L, its concentration is M. The reaction table is: N O 4 (g) NO (g) initial change -x +x equilibrium x x As [N O 4 (g)] eq = x M = M, x = M. Hence, [NO (g)] eq = x = ( ) M = 0.10 M. The equilibrium constant in terms of concentrations, K c, is: K c = [NO g ] [N O 4 g ] = = 0.11 Answer: 0.11 THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY

27 CHEM N-11 November 008 Hydrogen cyanide, HCN(g), is prepared commercially by the reaction of methane, CH 4 (g), ammonia, NH 3 (g), and oxygen, O (g), at high temperature. The other product is gaseous water. Write a chemical equation for the reaction. CH 4 (g) + NH 3 (g) + 3 /O (g) HCN(g) + 3H O(l) What volume of HCN(g) can be obtained from 0.0 L of CH 4 (g), 0.0 L of NH 3 (g) and 0.0 L of O (g)? The volumes of all gases are measured at the same temperature and pressure. From the ideal gas law, pv = nrt, the number of moles is directly proportional to the volume of the substance, if the same temperature and pressure are used. As the volumes of CH 4 (g), NH 3 (g) and O (g) are the same, the number of moles of each is also the same. From the chemical equation, 1.5 mol of O (g) is required for every 1 mol of CH 4 (g) and every 1 mol of NH 3 (g). As the number of moles of O (g) is equal to the number of moles of CH 4 (g) and NH 3 (g), the limiting reagent is O (g). From the chemical equation, 1 mol of HCN(g) is made from every 1.5 mol of O (g). Each mole of O (g) will lead to /3 mol of HCN(g). Hence, the volume of HCN(g) = /3 0.0 L = 13.3 L. Answer: 13.3 L The reaction of carbon disulfide with chlorine is as follows. 3 CS (g) + 3Cl (g) CCl 4 (g) + S Cl (g) H 98 = 38 kj mol 1 In which direction will the reaction move when the following changes are made to the system initially at equilibrium? (a) The pressure on the system is doubled by halving the volume. As the reaction involves 4 mol of gas mol of gas, increasing the pressure favours products (Le Chatelier s principle). The reaction will shift to the right. (b) CCl 4 is removed. Removing product will lead to the reaction shifting to produce more product (Le Chatelier s principle). The reaction will shift to the right. (c) The system is heated. The reaction is exothermic. Increasing the temperature will cause the reaction to reduce the amount of the exothermic reaction (Le Chatelier s principle). The reaction will shift to the left.

28 CHEM J-8 June 007 A major disadvantage of hydrogen as a fuel is that it is a gas, and therefore hard to store. There is an enormous world-wide effort, including research performed in the University of Sydney, to develop novel chemical structures in which H can be stored much more efficiently. One of the structures being tested in the School of Chemistry is shown below. 5 H.39 Å Cu + O C CH ~ O Cu + H What type of intermolecular force (or forces) are responsible for the binding between the Cu + and the H? There are ion-induced dipole forces between Cu + and the H molecules as well as weak dispersion (London or induced dipole-induced dipole) forces. In order that such a material be useful for fuel storage, the binding of the H must be reversible: cage(s) + H (g) cage H (s) One simple way to reverse the binding is to increase the temperature, so that at low temperature the equilibrium lies to the right and at high temperature to the left. Use this information, plus any chemical knowledge or intuition to infer the sign of G, H and S at low and high temperatures. (You may assume that H and S do not change greatly with temperature.) At low temperature, the equilibrium lies to the right favouring products: G < 0. At high temperature, it lies to the left favouring reactants: G > 0. The reaction involves formation of a solid from a solid and a gas. There is therefore a decrease in the entropy: S < 0. This is true at all temperatures. The forward reaction becomes less favourable as the temperature is increased. Le Chatelier s principle therefore suggests that the reaction is exothermic: H < 0. This is true at all temperatures. (If the temperature is increased, the equilibrium shifts to remove heat by increasing the backward reaction.) Temperature G S H low < 0 < 0 < 0 high > 0 < 0 < 0

29 CHEM J-9 June 007 N (g) and O (g) react with each other to a small extent if a catalyst is present to form nitric oxide, NO(g), according to the following equation. 5 N (g) + O (g) NO(g) The equilibrium constant, K p, for this reaction is at 98 K and at 500 K. Is the reaction exothermic or endothermic? Give reasons for your answer. The reaction is favoured at higher temperatures (i.e. increasing the temperature increases the amount of product). Le Chatelier s principle therefore suggests that the reaction is endothermic: H > 0 The partial pressures of O (g) and N (g) in air are 0.10 atm and atm respectively. If air at atmospheric pressure is sealed in a 1.00 L container containing the catalyst at 98 K, what will be the partial pressure of NO(g) and the total pressure inside the container at equilibrium? Two moles of NO are produced for every mole of O and N that is lost. partial pressures O N NO start change -x -x +x equilibrium 0.1-x 0.78-x x The equilibrium constant in terms of partial pressures is then: ( p NO ) (x) 35 K p = = = ( p )( p ) (0.1 x)(0.78 x) O N As K p is very small, x will be tiny so that (0.1-x) ~ 0.1 and (0.78-x) ~ 0.78 to a very good approximation. Substituting these approximations in gives: 4x = ( ) (0.1) (0.78) x = The partial pressure of NO is twice this value = The overall number of moles of gas does not change during the reaction so neither does the total pressure. The pressure at equilibrium = 1.00 atm 18 atm Pressure of NO(g): atm Total pressure: 1.00 atm THIS QUESTION CONTINUES ON THE NEXT PAGE.

30 CHEM J-10 June 007 Oxidation of NO(g) to produce the pollutant NO (g) is favoured at higher temperatures, such as those in a car exhaust: 6 NO(g) + O (g) NO (g) The equilibrium constant, K p, for this reaction is at 500 K. What is the value of K c at 500 K? K p and K c are related through the equation K p = K c (RT) n where n is the change in the number of moles of gas during the reaction. As three moles go to two moles in the reaction, n = 1 and hence: K p c = K (RT) c p 1 K = K (RT) = ( ) ( ) = K c = Using this value and the equilibrium constant for the formation of NO(g) from the previous page, calculate the value of K c for the formation of NO (g) from N (g) and O (g) at 500 K according to the following equation. N (g) + O (g) NO (g) From 007-J-9, K p = for the reaction N (g) + O (g) NO(g). For this reaction, n = 0 and hence K p = K c (RT) 0 or K c = K p. This reaction and the one above can be combined: N (g) + O (g) NO(g) K c = NO(g) + O (g) NO (g) K c = N (g) + O (g) NO (g) K c = ( ) ( ) = K c =

31 CHEM N-10 November 007 Water (1.00 mol) was placed in an otherwise empty container with a fixed volume of 100 L. The container was heated to 1705 K, at which temperature the following equilibrium was established. 5 H O(g) Calculate K c for this reaction at 1705 K. H (g) + O (g) K p = atm For a reaction involving a change in the number of moles of gas of n, K p and K c are related by: K p = K c (RT) n In this reaction, mol of gas 3 mol of gas and hence n = 1. Thus, K p = K c (RT) 1 or K c = K 9 p = = RT (0.0806) (1705) K c = Determine the amount of O (in mol) in the container at equilibrium at 1705 K. As 1.00 mol of H O is initially present in a container with volume 100 L, the initial concentration of H O is: concentration = number of moles = 1.00mol = M volume 100 L The reaction table, with [O (g)] equilibrium = x, is: H O(g) H (g) + O (g) initial change -x +x +x equilibrium x x x [H (g)] [O (g)] (x) (x) -11 Hence, K c = = = [HO(g)] ( x) Although the amount of water initially present is quite small, the value of K c is so very small that x ~ This approximation simplifies the expression so that: 3 4x -11 K c = ~ = (0.0100) or 4x 3 = (0.0100) ( ) Hence, x = [O (g)] equilibrium = M

32 CHEM N-10 November 007 ANSWER CONTINUES ON THE NEXT PAGE As this concentration of O is present in a 100 L container, the number of moles of O present is: number of moles = concentration volume = ( M) (100 L) = mol Answer: mol

33 CHEM J-9 June 006 The thermal decomposition of lithium peroxide produces oxygen. Li O (s) Li O(s) + O (g) A 1.0 g sample of Li O was placed in a closed container and heated to a temperature, where some, but not all, of the Li O decomposes. The experiment is then repeated using a.0 g sample, heated to the same temperature in an identical container. How does the pressure of O (g) produced vary between these two experiments? Explain. The equilibrium constant for this reaction is given by: K p = p O (g) as all other reactants and products are solids. As solid is left at the equilibrium point in both reactions, it does not enter the equilibrium expression and has no effect on the equilibrium constant. The pressure of O (g) will be a constant, equal to K p.

34 CHEM N-10 November 006 Consider the following reaction. N O 4 (g) NO (g) An experiment was conducted in which mol of N O 4 (g) was introduced into a 1.00 L flask. After equilibrium had been established at 100 C, the concentration of N O 4 (g) was found to be M. Calculate the equilibrium constant, K c, for the reaction as written at 100 C. 5 The initial concentration of N O 4 (g) is: [N O 4 (g)] 0 = number of moles = = M volume 1.00 At equilibrium, [N O 4 (g)] = M. The change in the concentration is ( ) = M. From the chemical equation, each mole of N O 4 that reacts leads to two moles of NO. Hence, [NO (g)] = = 0.10 M. The equilibrium constant in terms of concentrations, K c, is therefore: K c = [NO (g)] (0.10) = = 0.1 [N O (g)] (0.0491) 4 K c = 0.1 Use your calculated value for K c to calculate whether a mixture of N O 4 (g) (0.10 M) and NO (g) (0.550 M) is at equilibrium at 100 C? If not, in which direction will it move? Using these concentrations, the reaction quotient Q is: Q = [NO (g)] (0.550) = =.5 [N O (g)] (0.10) 4 As Q > K p, the reaction will proceed to reduce Q by increasing the amount of reactant (N O 4 (g)) and decreasing the amount of product (NO (g)). It will proceed towards the left.

35 CHEM J-9 June 005 The value of the equilibrium constant, K c, for the following reaction is mol L 1. CO (g) + N (g) CO(g) + NO(g) What is the equilibrium concentration of CO(g) if the equilibrium concentration of [CO (g)] = 0.39 M, [N (g)] = M and [NO(g)] = 0.46 M? The equilibrium constant in terms of concentrations, K c, is: K c = [CO(g)] [NO(g)] [CO(g) ](0.46) [CO (g)] [N (g)] (0.39) (0.419) Therefore, [CO(g)] = 0.16 M or [CO(g)] = M Answer: M When hydrogen cyanide (HCN) is dissolved in water it dissociates into ions according to the following equation. 3 HCN(aq) H + (aq) + CN (aq) The equilibrium constant for this reaction is K c = mol L 1. If 1.00 mol of HCN is dissolved to make 1.00 L of solution, calculate the percentage of HCN that will be dissociated. The initial concentration of HCN is 1.00 M. The reaction table is: HCN(aq) H + (aq) CN - (aq) [initial] change -x +x +x [equilibrium] 1.00-x x x K c = [H (aq)][cn (aq)] (x)(x) x [HCN(aq)] (1.00 x) (1.00 x) = As K c is very small, the amount of dissociation will be tiny and 1.00-x ~ Hence, x ~ x, so x = = [H + (aq)] = [CN - (aq)]. As [HCN(aq)] = 1.00 x, [HCN(aq)] = 1.00 and the percentage dissociation is: percentage dissociation = % Answer: %

36 CHEM J-1 June 004 A saturated solution of iodine in water contains g I per litre, but more than this amount can dissolve in a potassium iodide solution because of the following equilibrium. I (aq) + I (aq) I 3 (aq) A M KI solution dissolves 1.5 g of I per litre, most of which is converted to I 3 (aq). Assuming that the concentration of I (aq) in all saturated solutions is the same, calculate the equilibrium constant for the above reaction. 4 The molar mass of I is ( 16.90) g mol -1 = 53.8 g mol -1. As g of I dissolves in a litre of water, the concentration of I in the saturated solution of iodine in water is therefore: [I (aq)] = M g of I corresponds to a litre of KI solution, the concentration is M. For the equilibrium, the reaction table is therefore: 1.5g mol and when this dissolves in gmol I - (aq) I (aq) I - 3 (aq) [initial] change -x -x +x [equilibrium] x x x Assuming that [I (aq)] is the same as in the saturated solution (as stated in the question), x = so x = giving: [I - (aq)] = = 0.05 M, [I (aq)] = M and [I 3 - (aq)] = M. The equilibrium constant is therefore: K c = 3 [I (aq)] (0.048) 710 [I (aq)][i (aq)] (0.05)(0.0013) Answer: 710

37 CHEM N-9 November 004 When 1.0 mol of acetic acid and 1.0 mol of ethanol are mixed they react according to the following equation. 3 C H 5 OH(l) + CH 3 COOH(l) CH 3 COOC H 5 (l) + H O(l) At equilibrium the mixture contains 0.67 mol of the ester (CH 3 COOC H 5 ). What is the equilibrium constant for the reaction? At equilibrium, there are ( ) mol of C H 5 OH and CH 3 COOH together with 0.67 mol of CH 3 COOC H 5 and of H O(l). The equilibrium constant is therefore: K c = [CH3COOCH 5(l)][HO(l)] (0.67)(0.67) = = 4.1 [C H OH(l)][CH COOH(l)] ( )( ) 5 3 Note that although amounts (mol) rather than concentrations are known, the volume is the same for each species and cancels through in the equation. The reaction involves mixtures of liquids and not aqueous solution so the concentration of water is included in the expression. ANSWER: 4.1 Bromine-containing compounds are even more ozone depleting than the analogous chlorine-containing compounds. In the stratosphere, an equilibrium exists between bromine and the NO x species. One of these equilibrium reactions is: 3 NOBr(g) NO(g) + Br (g) To study this reaction, an atmospheric chemist places a known amount of NOBr in a sealed container at 5 C to a pressure of 0.50 atm and observes that 34% of it decomposes into NO and Br. What is K p for this reaction? The initial partial pressure of NOBr is 0.50 atm. At equilibrium, 34% of it has decomposed so its partial pressure at equilibrium is = atm. The reaction stoichiometry means that moles of NO(g) are produced for every moles of NOBr(g) that decompose. The equilibrium partial pressure of NO is therefore = atm. 1 mole of Br (g) is produced for every moles of NOBr(g) that decompose. The equilibrium partial pressure of Br (g) is therefore ½ = atm. The equilibrium constant in terms of partial pressures, K p, is therefore: K p =... = ANSWER:

38 CHEM N-11 November 004 The bacterium Azotobacter chroococcum, growing aerobically in a medium free of nitrogen containing compounds, obtains all of its nitrogen by the "fixation" of atmospheric N. The solubility of N in water is governed by the following equilibrium: 4 N (aq) N (g) K = atm L mol 1 What is the concentration of dissolved N available to the bacterium at 1.0 atm and 30 C? (Air is 78% N.) The equilibrium constant is given by K = pressure of N (g). ( p ) N (g) [N (aq)] where p N (g) is the partial The atmospheric pressure is the sum of the partial pressures of the constituent gases in the air. As N represents 78% of the air and the total pressure is 1.0 atm, the partial pressure of N is given by p = atm = 0.78 atm Hence, the concentration of dissolved N is: N (g) [N (aq)] = P N (g) K = 0.78 atm atm L mol 1 = M Answer: M A culture of these bacteria (1.0 L) grows to a density of 0.84 mg dry weight per ml of culture and has a nitrogen content of 7.0% of the dry weight. What volume of air at 1.0 atm and 30 C would supply this nitrogen requirement? As the density of the culture is 0.84 mg ml -1, the mass of 1.0 L (1000 ml) is 1000 ml ( g ml -1 ) = 0.86 g. The nitrogen content is 7.0% so the mass of nitrogen in the culture is g = g. As nitrogen has a atomic mass of g mol -1, this mass corresponds to: moles of nitrogen atoms = mass of nitrogen g = atomic mass g mol = mol Nitrogen is present in the air as N so the number of moles of N required is ½ mol = mol. The partial pressure due to N is 0.78 atm if the atmospheric pressure is 1.0 atm. Using PV = nrt, the volume of air containing mol of N (g) is: V -1-1 nrt (0.01mol) (0.0806L atm K mol ) ((30+73)K) = = P (0.78atm) = L = 68 ml Answer: L or 68 ml