CHEM 102 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA CHAPTER 14, Chemical Equilibrium Chapter

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1 CHEM 10 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 717 CHAPTER 14, Chemical Equilibrium Chapter 14. Chemical Equilibrium 14.1 Characteristics of Chemical Equilibrium 14. The Equilibrium Constant 14.3 Determining Equilibrium Constants 14.5 The Meaning of Equilibrium Constant 14.6 Using Equilibrium Constants 14.7 Shifting a Chemical Equilibrium: Le Chatelier's Principle 14.8 Equilibrium at the Nanoscale 14.9 Controlling Chemical Reactions: The Haber-Bosch Process Objectives are as follows: Basic Skills Students should be able to: 1. Recognize a system at equilibrium and describe the properties of equilibrium systems (Section 14.1).. Describe the dynamic nature of equilibrium and the changes in concentrations reactants and products that occur as a system approaches equilibrium sections 14.1 and 14.). 3. Write equilibrium constant expressions, given balanced chemical equations (Section 14.). 4. Obtain equilibrium constant expressions for related reactions from the Expression for one or more known reactions (Section 14.). 5. Calculate Kp from K, or Kc from Kp for the same equilibrium (Section 14.). 6. Calculate a value of K, for an equilibrium system, given information about initial concentrations and equilibrium concentrations (Section 14 3). 7. Make qualitative predictions about the extent of reaction based upon equilibrium constant values; that is, be able to predict whether a reaction product favored or reactant favored based on the size of the equilibrium constant (Section 14.4). 8. Calculate concentrations of reactants and products in an equilibrium system Kc and initial concentrations are known (Section 14 5). 9. Use the reaction quotient Q to predict in which direction a reaction will go reach equilibrium (Section 14 5). 10. Show by using Le Chatelier's principle how changes in concentrations, pressure or volume, and temperature shift chemical equilibria (Section 14.6). 11. Use the change in enthalpy and the change in entropy qualitatively to pre whether products are favored over reactants (Section 14.7). 1. List the factors affecting chemical reactivity, and apply them to predict' optimal conditions for producing products (Section 14.8).

2 CHEM 10 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 717 CHAPTER 14, Chemical Equilibrium KEY CONCEPTS dynamic nature of equilibrium: Law of mass action equilibrium constant expression K p and K c forward and reverse reaction initial concentration equilibrium concentration calculate equilibrium constant calculate K p from K c and vice versa heterogeneous equilibria homogeneous equilibria reaction quotient (Q) Le Châtelier's Principle predicting equilibrium shifts calculate equilibrium concentrations from K p and K c equilibrium constants from experiments Characteristics of Chemical Equilibrium In doing stoichiometry calculations we assumed that reactions proceed to completion, that is, until one of the reactants runs out. Many reactions do proceed essentially to completion: complete reactions are indicated by. For such reactions it can be assumed that the reactants are quantitatively converted to products and that the amount of limiting reactant that remains is negligible. Irreversible or complete reactions: Chemical reactions can be considered to have forward and backward reactions. Forward reaction is when reactants combine to form products whereas products are converted back to reactants in the backward reaction. In most chemical reactions, the rate of backward reaction is so small all reactants are completely converted to products. This condition is usually represented in a chemical equation by an arrow pointing to right E.g. H + O H O Equilibrium Chemical Reactions On the other hand, there are many chemical reactions that stop far short of completion. An example is the dimerization of nitrogen dioxide: NO (g) + NO (g) N O 4 (g) The reactant, NO, is a dark brown gas, and the product, N O 4, is a colorless gas. When NO is placed in an evacuated, sealed glass vessel at 5 C, the initial dark brown color decreases in intensity as it is converted to colorless N O 4. However, even over a

3 long period of time, the contents of the reaction vessel do not become colorless. Instead, the intensity of the brown color eventually becomes constant, which means that the concentration of NO is no longer changing. Making the container colder makes equilibrium to shift to left and warming shift the equilibrium to right. Other examples: 3H (g) + N (g) NH 3 (g) Any chemical reaction could be considered as a forward and backward reactions occurring at the same time( ) as described previously. If the rates of backward and forward reactions chemical reactions are comparable both reactants and products can coexist leading to a condition called chemical equilibrium reaction. Dynamic Equilibrium: Equilibrium chemical reaction could be considered as a forward and backward reaction occurring at the same time( ). The reactants and products will interchange constantly, however maintaining same concentrations of reactants and products. Dynamic equilibrium means a constantly changing system. In an equilibrium reaction reactants and products change continuously among each other (dynamic) as opposed to a static equilibrium. This change maintaining a constant concentration of reactants and products is called a dynamic equilibrium in equilibrium chemical reactions. Double arrows are used in the equation to indicate this. 3H (g) + N (g) NH 3 (g) The Equilibrium Constant Equilibrium constant comes from a law that applies to chemical equilibrium called Law of Mass action. Law of Mass Action:

4 Law of mass action describes an equilibrium process by quantifying the equilibrium concentration of reactants and products. It uses the ratio of backward and forward reactions and express it terms of an equilibrium constant (K). For example consider a hypothetical equation: j A + k B l C + m D [C] l [D] m etc. K = [A] j [B] k [A] are equilibrium concentration of A, B, C, D j, k, l, m are stoichiometric coefficients K = equilibrium constant Equilibrium Expression: Equilibrium expression is the Law of mass action equation with the equilibrium concentration of reactants and products and the equilibrium constant (K). Example: Equilibrium expression for the formation of NH 3 gas: N (g) + 3H (g) NH 3 (g) [NH 3 ] K = = 6.0 x 10 - L /mol at 17 o C [N ][H ] 3 Equilibrium Constant (K): The constant in the equilibrium expression is called equilibrium constant (K). [H O] [NH 3 ] K = and K = are [H ] [O ] [N ][H ] 3 Examples of equilibrium expressions containing K. From the numerical value of equilibrium constant following could be said about the reactions. K = α (infinity) - Irreversible reactions K = 0 - No reaction K = between 0 and 1 - Equilibrium reactions Write the equilibrium constant for the following reactions: a) 3H (g) + N (g) NH 3 (g) b) CO(g) + Cl (g) COCl (g) c) N O 4 (g) NO (g) d) MgCO 3 (s) MgO(s) + CO (g) e) NaCl(s) Na + (aq) + Cl - (aq)

5 f) 3/H (g) + 1/N (g) NH 3 (g) g) C(s) + CO (g) CO(g) h) NH 4 Cl(s) NH 3 (g) + HCl(g) i) N O(g) + N H 4 (g) 3N (g) + H O(g) j) NOBr(g) NO(g) + Br (g) k) N (g) + O (g) NO(g) l) CaCl (s) + H O(g) CaCl.H O(s) m) H (g) + CO (g) H O(g) + CO(g) a) 3H (g) + N (g) NH 3 (g) [NH 3 ] K = [H ] 3 [N ] b) CO(g) + Cl (g) COCl (g) [COCl ] K = [CO][Cl ] c) N O 4 (g) NO (g) [NO ] K = [N O 4 ] d) MgCO 3 (s) MgO(s) + CO (g) K = [CO ] Pure liquid or solid concentrations are not written in the expression. e) NaCl(s) Na+(aq) + Cl-(aq) K = [Na+(aq)] [Cl-(aq)] f) 3/ H (g) + 1/ N (g) NH 3 (g) [NH 3 ] K = [H ] 3/ [N ] 1/ g) C(s) + CO (g) CO(g) [CO] K = [CO ] h) NH 4 Cl(s) NH 3 (g) + HCl(g) K = [NH 3 ][HCl] i) N O(g) + N H 4 (g) N (g) + H O(g) [N ] [H O] K = [N O] [N H 4 ] j) NOBr(g) NO(g) + Br (g)

6 [NO] [Br ] K = [NOBr] k) N (g) + O (g) NO(g) [NO] K = [N ][O ] Which of the above chemical reactions, a-k, are examples of homogeneous equilibrium and which ones are examples of heterogeneous equilibrium? Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase. Usually homogenous equilibrium is where all reactants and products are gases. E.g. a, b, c, f, i, j, k in problem the problem above. Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest. E.g. d, e, g, h. The Meaning of Equilibrium Constant j A + k B l C + m D rate constant forward reaction k + K = rate constant backward reaction k - Equilibrium occurs when k + = k -. [C] l [D] m etc. K = [A] j [B] k [A]. are equilibrium concentration of A, B, C, D j, k, l, m are stoichiometric coefficients K = equilibrium constant E.g. Formation of water H + O H O [H O] since the backward reaction rate is almost 0 or very small K = = a equal to 0. K is equal to (infinity- α) i.e. only [H ] [O ] H O is found after equilibrium is reached. rate constant of forward Reaction k + K = = --- = α (very large) rate constant of backward Reaction 0 K eq >> 1 K eq ~ 1 reaction will go mainly to products reaction will produce roughly equal amounts of product and reactant

7 K eq << 1 reaction will go mainly to reactants Consider, for example, the equilibrium between N O 4 (g) and NO (g): N O 4 (g) NO (g) K eq = [ ] NO [ NO] 4 Listed below is experimental data giving initial concentrations for N O 4 (g) and NO (g). After some time the reaction reaches equilibrium and the concentrations listed. Equilibrium N O 4 NO N O 4 NO K eq E.g. Formation of NH 3 gas. N (g) + 3H (g) NH 3 (g) [NH 3 ] K = = 6.0 x 10 - L /mole at 17 o C [N ][H ] 3 Forward reaction is 100 times slower than the backward reaction at this temperature. Taking reaction a in the above list, the formation of ammonia (3 billion pounds per year produced in US alone) describes the importance of studying the equilibrium in a chemical reaction. Anhydrous ammonia is produced using process called Haber Process: 3H (g) + N (g) NH 3 (g) ; K = 6.0 x 10 - L /mol At 5 o K = small, no or very little product is formed. However, at 500 o C,

8 K = 6.0 x 10 - L /mol some ammonia is formed NH 3 is removed continuously from the mixture using a "cold trap" to obtain liquid ammonia. Removing ammonia shift the equilibrium to right (As discussed under Le Chatelier's Principle later in detail). Determining Equilibrium Constants Following definitions are needed to do an equilibrium constant calculation. These ICE (Initial-Change-Equilibrium) calculations are based on the idea that initial, change and equilibrium concentrations along with equilibrium constant are needed. Initial concentration: The concentrations in moles per liter (M) or partial pressure (P) of reactants and products before the equilibrium is reached is called initial concentration. E.g. 3H (g) + N (g) NH 3 (g) [H ] or P H = Concentration of H before the equilibrium is reached. [ N ] or P N = Concentration of N before the equilibrium is reached. [NH 3 ] or P NH3 = Concentration of NH 3 before the equilibrium is reached. Equilibrium concentration: The concentrations in moles per liter (M) of reactants and products after the equilibrium is reached is called equilibrium concentration. E.g. 3H (g) + N (g) NH 3 (g) [H ] or P H = Concentration of H after the equilibrium is reached. [ N ] or P N = Concentration of N after the equilibrium is reached. [NH 3 ] or P NH3 = Concentration of NH 3 after the equilibrium is reached. K (K c ) K (K c ) is when the mole/l (molarity which normally indicated by [A] ) units are used for equilibrium concentration of reactants and products in the equilibrium expression K p K p is the equlibrium constant of gaseous reactions when the concentrations are expressed in terms of individual partial pressures (p) of a mixture of gases at equilibrium. Consider the following reactions: 3H (g) + N (g) NH 3 (g)

9 CO(g) + Cl (g) COCl (g) NaCl(s) Na + (aq) + Cl - (aq) N O(g) + N H 4 (g) 3N (g) + H O(g) K (K c ) in the equilibrium expression is usually the constant calculated based on moles/liters concentration of reactants and products. This is also called K c - equilibrium constant based on M, [A] or moles/l -Concentrations. However, many chemical reactions occur in gas phase and concentration of products and reactants is easier to give K in partial pressures (P). The equilibrium constant calculated based on partial pressure is called K p - P - partial pressure. Relationship between K (K c ) and K p K p = K(RT) n K p and K (K c ) are related by the following equation: E.g. K p = K(RT) n K = constant based on concentration in mole/liters. K p = constant based on partial pressures. R = universal gas constant T = Kelvin Temperature, n = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants) N (g) + 3H (g) CO(g) + Cl (g) NH 3 (g) n = - 4 = - P NH3 K p = K p = K(RT) - P N P 3 NH3 K p K = = K p (RT) (RT) - COCl (g) n = 1 - = -1 P COCl K p = P CO P Cl K K p = K(RT) -1 ; K p = ---- ; K = K p RT RT NaCl(s) Na + (aq) + Cl - (aq)

10 K p = 1; K p = K(RT) n ; n = 0 K p = K N O(g) + N H 4 (g) 3N (g) + H O(g) n = 5-3 = P 3 N P HO K p K p = ; K p = K(RT) ; K = P NO P NH4 (RT) For example in the equilibrium system: H (g) + I (g) = HI(g); K c = 49.7 at 458 C What would be the K p of this system if R = liter-atm/mole K? 1. Check to make sure the equation is balanced. Calculate n n = Total gas moles on the right - Total gas moles on the left n = - = 0 3. Calculate T in Kelvin from given Celsius temperature T = = 731 K 4. Using the relationship plug values in and solve for K p K p = K c (RT) n K p = 49.7 [(0.081)(731)] 0 = 49.7 Equilibrium Constant K p Calculation At 500 o C, Consider the following reaction (assume an IDEAL gas mixture): PCl 5 (g) PCl 3 (g) + Cl (g) A 1.0-liter vessel was initially filled with pure PCl 5, at a pressure of.0 atm, at 5 o C. After equilibrium was established, the partial pressure of PCl 3 was 3.16 x 10 - atm. What is K p for the reaction? P PCl3 P Cl Kp = P PCl5 PCl 5 PCl 3 Cl Initial partial pressure:.0 atm

11 Change: -x x x if x amount of PCl 5 decomposed (.0 - x)atm x atm x atm to reach equilibrium ( x 10 - )atm 3.16 x 10 - atm 3.16 x10 - atm At equilibrium x = 3.16 x 10 - atm 3.16 x 10 - atm x 3.16 x 10 - atm 9.99 x 10-4 K p = = = 5.07 x 10-4 atm ( x 10 - )atm A 5.0-g sample of solid NH 4 Cl is heated in a.5-l container to 900 o C. At equilibrium the pressure of NH 3 (g) (reaction h) is 0.60 atm. Calculate the equilibrium constant, K p, for this reaction. NH 4 Cl(s) A 5.0-g sample of solid NH 4 Cl is heated in a.5-l container to 900 o C. At equilibrium the pressure of NH 3 (g) (reaction h) is 0.60 atm. Calculate the equilibrium constant, K p, for this reaction. NH 4 Cl(s) NH 3 (g) + HCl(g) Kp = P NH3 P HCl expression since it's a solid. P NH3 ;note that NH 4 Cl(s) doesn't appear in the P HCl Initial concentration Change x x Equilibrium concentration 0.6 atm x atm 0.6 atm 0.6 atm P HCl should be equal to that of P NH3 since for each NH 3 formed a HCl is formed. Therefore, K p = 0.6 atm x 0.6 atm = 0.36 atm NH 3 (g) + HCl(g) Kp = P NH3 P HCl expression since it's a solid. P NH3 ;note that NH 4 Cl(s) doesn't appear in the P HCl Initial concentration Change x x Equilibrium concentration 0.6 atm x atm 0.6 atm 0.6 atm P HCl should be equal to that of P NH3 since for each NH 3 formed a HCl is formed. Therefore, K p = 0.6 atm x 0.6 atm = 0.36 atm Using Equilibrium Constants Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction: the tendency of the reaction to occur (but not the speed of the reaction which is the area of chemical kinetics), whether or not a given set of concentrations represents an equilibrium condition, and the equilibrium position that will

12 be achieved from a given set of initial concentrations. The reaction quotient is used for predicting the net direction of equilibrium reactions. Reaction Quotient (Q) When the reactants and products of a given chemical reaction are mixed, it is useful to know whether the mixture is at equilibrium or, if not, the direction in which the system must shift to reach equilibrium. If the concentration of one of the reactants or products is zero, the system will shift in the direction that produces the missing component. However, if all the initial concentrations are nonzero, it is more difficult to determine the direction of the move toward equilibrium. To determine the shift in such cases, we use the reaction quotient Q. The reaction quotient is obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations. j A + k B l C + m D [C] l [D] m Q = [A] j [B] k Q > K eq Q = K eq Q < K eq [A]. are initial concentration of A, B, C, D etc. j, k, l, m are stoichiometric coefficients Q = reaction quotient reverse reaction will be spontaneous equilibrium forward reaction will be spontaneous Consider the following reaction: SO (g) + NO (g) NO(g) + SO 3 (g) (K c = 85.0 at 460 o C) Given mole of SO (g), mole of NO (g), 0.30 mole of NO(g),and 0.00 mole of SO 3 (g) are mixed in a 5.00 L flask, determine: a) The net the reaction quotient, Q. b) Direction to achieve equilibrium. The net the reaction quotient, Q. The reaction quotient (Q) is constant in the equilibrium expression when initial concentration of reactants and products are used. [NO][SO 3 ] Q = Q = equilibrium constant calculated based on initial concentrations. [SO ][NO ] mole mole 0.30 mole 0.00 mole [SO ] = ; [NO ] = ; [NO] = ; [SO 3 ] = L 5.00L 5.00L 5.00 L

13 [SO ] = 8 x 10-3 mole/l ; [NO ] =0.1mole/L; [NO] = 0.06 mole/l; [SO 3 ] = 4 x 10-3 mole/l 0.06 (4 x 10-3 ) Q = = x 10-3 x 0.1 The net direction to achieve equilibrium. Since the equilibrium constant K = 85.0, at 460 o C and Q = 0.3, to reach equilibrium Q should increase to To do that top term or product concentrations should be increased. Therefore the reaction goes to right (towards products) to achieve equilibrium. Calculation of unknown concentration of reactants or products in an equilibrium mixture At 100 o C the equilibrium constant (K) for the reaction: H (g) + I (g) HI(g) is 1.15 x 10. If moles of H and moles of I are placed into a 1.0-liter container and allowed to react at this temperature, what is the HI concentration (moles/liter) at equilibrium? [HI] K = ; K =1.5 x 10. [H ][I ] Initial moles: [H ]= mol; [I ] = mol ; [HI] = 0.00 mol Change in moles: [H ]= -x mol [I ] = -x mol [HI] = x mol Equilibrium concentration in M (mole/l): If x of H and x of I moles reacts to achieve equilibrium x mol x mol x mole [H ]= ; [I ] = ; [HI] = ; 1.0 L 1.0L 1.0L [HI] K = =1.5 x 10. [H ][I ] (x/1) x K = = (0.400-x/1)(0.400-x/1) (0.400-x) 4x x K =1.5 x 10 = ; Sq.Rt (1.5 x 10 ) = 10.7 = (0.4-x) (0.4-x) x = 10.7(0.4 -x) = x

14 x x = x = 4.89 x = 4.89/1.7 = mole [HI] = x/1 = mole/l At a certain temperature the value of the equilibrium constant is 3.4 for the reaction: H (g) + CO (g) H O(g) + CO(g) If mol H and mol CO are placed in a 1.00 L vessel, what is the concentration of CO at equilibrium? K = 3.4 Here we assume mole/l concentration is same as moles since volume is 1L. [H ] [ CO ] [H O ] [CO ] Initial concentration: Change in concentration: -x -x x x Equilibrium concentration: 0.40-x 0.40-x x x If x of H and x of CO reacted to produce x moles of H O and CO [ H O][CO] x K = = [H ][CO ] (0.40-x)(0.40-x) x x K = 3.4 = = (0.40-X) (0.40-X) x sq.rt. 3.4 =1.8 = (0.40-x) x = 1.8 (0.400-x) = x x + 1.8x = 0.7.8x = 0.7 x = 0.7/.8 = 0.57 mol/l [H ] [CO ] [H O ] [CO ] Equilibrium: 0.40-x 0.40-x x x Concentration mol/l 0.57 mol/l Shifting a Chemical Equilibrium: Le Chatelier's Principle Le Chatelier's principle is used to predict the shift of an equilibrium. Le Chatelier's Principle is one of the general principles applicable to any equilibrium process. It simply states that: If a change is imposed on a system at Equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

15 Listed below are how various "changes" that affect equilibria: 1) Adding products (unless one of the products is a solid!) to a reaction will cause the equilibrium to shift back to produce more reactants. ) Adding reactants (unless one of the reactants is a solid!) to a reaction will cause the equilibrium to shift forward to produce more products. 3) Removing reactants (unless one of the reactants is a solid and as long as there is some left) will cause the equilibrium to shift back to produce more reactants. 4) Removing products (unless one of the products is a solid and as long as there is some left) will cause the equilibrium to shift forward to produce more products. 5) The effect of temperature on a reaction is dependent on whether the reaction is exothermic ( H rxn = negative) or endothermic ( H rxn = positive). 6) Pressure changes could affect a reaction if there is net change in gaseous reactants and products Temperature changes could be either heating or cooling and also due to heat absorbed/released in the reaction. Pressure changes could be either by increasing or decreasing the pressure of the vessel. In addition changes in number of total particles in gaseous reaction mixtures going from reactant to products can create pressure changes E.g. Increasing T of the equilibrium should shift equilibrium to left and vice versa. Increasing P will shift equilibrium to right and vice versa. Note: volume changes can be considered as pressure changes. Increased volume have the same effect as a decrease in pressure. For the following equilibrium reactions: H (g) + CO (g) H O(g) + CO(g)?H = 40 kj Predict the equilibrium shift if: a) The temperature is increased b) The pressure is decreased a) Temperature increased for the above equilibrium: Reaction is endothermic ( H = + value). Equilibrium should shift to a direction to absorb heat. Forward reaction absorbs heat. Backward reaction should release heat. Therefore, forward reaction occurs or equilibrium shift to right to absorb heat.

16 b) Pressure is decreased: In this equilibrium there is no change in number of particles in going from reactant to products. Equilibrium cannot respond to pressure changes directly. There is no change in the position of equilibrium. Equilibrium at the Nano-scale At the nano or sub-microscopic level (i.e. looking at the molecular species), the state of equilibrium does not mean that there is no change. However, the equilibrium constant K remain essentially constant. The following point will be illustrated. Equilibrium is a dynamic process at the molecular level. There is no net change, but two opposing processes are taking place. The equilibrium constant fluctuates slightly due to unequal reaction rates in opposite directions. A system moves spontaneously toward a state of equilibrium. The driving forces for equilibrium are: (a) molecules assume the state of lowest energy, (b) molecules tend to reach a maximum disorder or entropy. Controlling Chemical Reactions: The Haber-Bosch Process The Haber-Bosch Process (aka Haber process) is the reaction of nitrogen and hydrogen to produce ammonia. The nitrogen and hydrogen are reacted over an iron catalyst under conditions of 00 atmospheres, 450 C: N (g) + 3H (g) <--> NH 3(g) The process was developed by Fritz Haber and Carl Bosch in 1909 and patented in It was first used on an industrial scale by the Germans during World War I: Germany had previously imported nitrates from Chile, but the demand for munitions and the uncertainty of this supply in the war prompted the adoption of the process. The ammonia produced was oxidised for the production of nitric acid in the Ostwald process, and the nitric acid for the production of various explosive nitro compounds used in munitions. The nitrogen is obtained from the air, and the hydrogen is obtained from water natural gas in the reaction: CH 4(g) + H O (g) CO (g) + 3H (g) Equilibrium and the Haber Process The reaction of nitrogen and hydrogen is reversible, meaning the reaction can proceed in either the forward or the reverse direction depending on conditions. The forward reaction is exothermic, meaning it produces heat and is favored at low temperatures. Increasing the temperature tends to drive the reaction in the reverse direction, which is undesirable if the goal is to produce ammonia. However, reducing the temperature reduces the rate of

17 the reaction, which is also undesirable. Therefore, an intermediate temperature high enough to allow the reaction to proceed at a reasonable rate, yet not so high as to drive the reaction in the reverse direction, is required. The forward reaction favors high pressures because there are fewer molecules on the right side. So the only compromise in pressure is the economical situation trying to increase the pressure as much as possible. The iron catalyst has no effect on the position of equilibrium, however it does increase the reaction rate. This allows the process to be operated at lower temperatures, which as mentioned before favors the forward reaction. Other catalysts are also active for this reaction, in fact the first Haber-Bosch reaction chambers, used uranium catalysts. The ammonia is formed as a gas but on cooling in the condenser liquefies at the high pressures used, and so is removed as a liquid. Unreacted nitrogen and hydrogen is fed back in to the reaction. Notwithstanding its original adoption as a military necessity, the Haber process now produces about half of all the nitrogen used in agriculture: billions of people are alive and fed from its use.

18 10 HOMEWORK HOMEWORK FOR CHAPTER Consider the reaction Ag 3 PO 4 (s) <===> 3Ag + (aq) + PO 4 3- (aq) The equilibrium constant expression for this reaction is a. [Ag 3PO 4 ] K = c + 3- [Ag ][PO 4 ] b. [Ag 3PO 4 ] K = c [Ag ] [PO 4 ] c [Ag ][PO 4 ] K = c [Ag 3PO 4 ] d. [Ag ] [PO 4 ] K = c [Ag PO ] e K = [ Ag ] [ PO ] c Chemical equilibrium exists in a reaction mixture when a. reactants are completely changed to products. b. there are equal concentrations of reactants and products. c. the rate at which reactants form products becomes zero. d. the concentrations of both products and reactants do not change. e. the concentrations of both products and reactants do not change. 3. Consider the reactions NO + O <===> NO K = a NO <===> N O 4 K = b The value of the equilibrium constant for the reaction 4NO + O <===> N O 4 is a. a + b b. ab c. (a/b) d. (ab) e. ab/ 4. Consider the reaction COCl (g) <===> CO(g) + Cl (g), with K c = In a system where the original concentration of COCl was M, calculate the equilibrium concentrations of CO and Cl. a M b M c M d M e M 5. Consider the reaction system: N (g) + O (g) N O(g) K c = At equilibrium, the concentration of N was measured as M and the concentration of O as M. What is the concentration of N O in this system? a b c

19 d e When 1.50 moles of A and.00 moles of B are placed in a 5.00 L flask and allowed to come to equilibrium, there are 0.90 moles of C in the mixture. Calculate the concentration of B at equilibrium. A + 3 B <==> C a. 0.3 M b M c M d M 7. One way to increase the production of CO gas in the equilibrium at 00 ºC shown below is to C(coal) + H O(g) <==> CO(g) + H (g) a. decrease the coal supply. b. decrease the water present. c. remove H. d. decrease the pressure.. 8. The homogeneous equilibrium in this list is a. Ag + (aq) + Cl - (aq) <===> AgCl(s). b. P 4 O 10 (s) <===> P 4 (g) + 5O (g). c. HCl(g) + NH 3 (g) <===> NH 4 Cl(s). d. HC H 3 O (aq) - <===> H+(aq) + C H 3 O (aq) For the reaction SO (g) + O (g) <===> SO 3 (g) at 87 o C, the value of K c = The K p for this reaction at 87 o C is (R= atm-l/mol K) a b c d. impossible to calculate from the data available. 10. For the gaseous reaction, NO(g) + O(g) <===> NO(g), the constant Kp is equal to: a b. Kc(RT) -. c. Kc(RT). d. Kc(RT) -1. e. Kc(RT). 11. For the reaction C(s) + CO (g) <====> CO(g) K c = 168, determine whether the system is at equilibrium when (CO) = 0.50 M and (CO ) = 0.75 M. The system at equilibrium, because. a. is; the value of Q is 0.33 b. is not; the value of Q is 0.33 c. is; the value of Q is 0.67 d. is not; the value of Q is 0.67 e. More information is needed to answer this question. 1. The value of Kc = 1.6 x 10-5 for the reaction NOCl(g) <==> NO(g) + Cl (g) at 35 C. If 1.0 mol of each gas were added to a 1.0 L flask, then a. the pressure in the flask would increase. b. the concentration of NOCl would increase. c. mostly product would be present at equilibrium. d. the value of Kc would become For the reaction CS (g) + 3 Cl (g) <===> S Cl (g) + CCl4(g), H = kj. The change that would move the equilibrium position to the left is a. increasing the temperature. b. adding Cl to the system. c. decreasing the size of the container. d. removing some CCl 4 from the system. 14. A student made a reaction mixture as directed in the lab manual, except he forgot to

20 add the catalyst. If everyone's experiment reached equilibrium, how was that student's experiment different? a. The one without a catalyst had less product. b. The one without a catalyst had more product. c. The one without a catalyst took longer to reach equilibrium, but the amount of product was the same as in the other experiments. d. The one without a catalyst reached equilibrium faster, but the amount of product was the same as in the other experiments. e. The one without a catalyst took longer to reach equilibrium and had much less product. was the same as in the other experiments. 15. Which statement concerning product-favored reactions is not correct? a. The value of the equilibrium constant is greater than 1. b. An endothermic reaction is product-favored. c. If the entropy of the products is greater than the entropy of the reactants, the reaction is productfavored. d. If a reaction is product-favored at high temperature, the entropy of the products is probably greater than the entropy of the reactants. e. If a reaction is product-favored at low temperature, the enthalpy of the products is probably less than the enthalpy of the reactants.

21 10 Sample Test 1 CHAPTER 14 Chapter 14. Chemical Equilibrium 1. The equilibrium expression for Kc for the system CO (g) + CaO <====> CaCO3 (s) is a. [CaCO3] [CO][CaO] b. [CaCO3] [CO] c. [CaCO3] [CO][CaO] d. 1 [CO] e. CO. The equilibrium constant expression, Kc, for the reaction C3H8(g) + 5O(g) <==> 3CO(g) + 4HO(g). [C3H8][O] 5 a. Kc= [CO] 3 [HO] 4 b. Kc= [CO] 3 [HO] 4 [CO] 3 [HO] 4 c. Kc= [C3H8][O] 5 d. Kc= [C3H8][O] 5 [C3H8][O] 3 e. Kc= [CO] 5 [HO] 4 3. Chemical equilibrium a. is a dynamic equilibrium. b. describes opposing chemical reactions. c. involves changes that occur at equal rates. d. is described by each of these statements. 4. Chemical equilibrium exists when a. reactants are completely changed to products. b. there are equal amounts of reactants and products. c. the rate at which reactants form products becomes zero. d. the rate at which reactants form products is the same as the rate at which products form reactants. 5. Homogeneous equilibria are those that involve a. only chemical changes. b. only one phase. c. static, not dynamic, equilibria. d. a single reactant. 6. The term initial concentration refers to the a. first concentration listed in the problem. b. concentration of the first reactant in the equation. c. concentrations at the start of the experiment. d. amount of product formed by the first reaction of a series. 7. Which of the following correctly describes the equilibrium constant for the gas-phase reaction between H and O to form gaseous HO? H(g) + O(g) <====> HO(g) [HO] a. Kc = [H][O] [HO] b. Kc = [H][O] [HO]

22 c. Kc = [H][O] [H][O] d. Kc = [HO] e. Kc = [HO] 8. For the reaction, 5CO(g) + IO5(s) <----> I(g) + 5CO(g) the expression for Kc is a. [CO] 5 [IO5] [I][CO] 5 b. [I][CO] 5 [CO] 5 c. [I][CO] 5 [CO] 5 [IO5] d. [I][CO] [CO] 9. The homogeneous equilibrium in this list is a. Ag+(aq) + Cl-(aq) <==> AgCl(s). b. P4O10(s) <==> P4(g) + 5O(g). c. HCl(g) + NH3(g) <==> NH4Cl(s). d. PCl5(g) <==> PCl3(g) + Cl(g). 10. If the reaction quotient (Q) for a reaction is greater than Kc then a. the reaction must move to the right to reach equilibrium. b. the reaction must move to the left to reach equilibrium. c. the reaction is at equilibrium. d. the temperature must rise to reach equilibrium. 11. For the reaction: 3H(g) + N(g) <===> NH3(g) the relationship between K and Kp at a given temperature T is: a. K = 1/Kp b. K = Kp(RT)- c. Kp = K(RT)- d. K = Kp e. none of these 1. Consider the equilibrium 3H(g) + N(g) <===> NH3(g) at a certain temperature. An equilibrium mixture in a 4.00-L vessel contains 1.60 mol NH3(g), mol N(g), and 1.0 mol H(g). What is the value of Kc? a b. 9.6 c d e Consider the reaction SCl(l) + CCl4(l) <===> CS(g) + 3Cl(g); H = kj If the above reactants and products are contained in a closed vessel and the reaction system is at equilibrium, the number of moles of CS(g) can be increased by a. removing some SCl(l) from the system. b. removing some CCl4(l) from the system. c. increasing the pressure of the reaction vessel. d. increasing the temperature of the reaction system. e. adding some Cl(g) to the system. 14. Which of the following equilibria would not be affected by pressure changes at constant temperature? a. FeO(s) + CO(g) <===> Fe(s) + CO(g) b. CaCO3(s) <===> CaO(s) + CO(g) c. HO(g) <===> H(g) + O(g) d. NO(g) + O(g) <===> NO(g) e. PCl5(l) <===> PCl3(g) + Cl(g) 15. According to La Chatelier's Principle, for the following reaction at equilibrium: NO(g) + Cl(g) <===> ClNO(g) If the pressure of the reaction container is decreased: a. the number of ClNO molecules increases

23 b. the number of Cl molecules increases c. the number of NO molecules decreases d. the number of NO molecules stays constant e. the number of Cl molecules decreases 16. Consider following system at equilibrium: PCl5(g) <==> PCl3(g) + Cl(g); H = +500 kj Which of the following changes will shift the equilibrium to the LEFT? a. increasing temperature b. increasing volume c. increasing pressure d. removing Cl(g) e. adding PCl5(g) 17. A 1.00-mol sample of HI is placed in a 1-L vessel at 460oC, and the reaction system is allowed to come to equilibrium. The HI partially decomposes, forming 0.11 mol H and 0.11 mol I. What is the equilibrium constant for the reaction H(g) + I(g) <===> HI(g) at 460oC? a b. 7.1 c. 8.1 d e For the reaction N(g) + 3 H(g) <+==> NH3(g) an equilibrium mixture at 98 K in a 4.00 L container has 1.60 mol of NH3, mol of N, and 1.0 of mol H. The value of Kc is a b c d For the reaction Ni(s) + 4 CO(g) <=== Ni(CO) 4 (g), the equilibrium constant expression is 0. A weak acid is 5% ionized. Therefore we can say that this reaction is -favored, because. a. product; the amount of products >> the amount of reactants b. product; the amount of products << the amount of reactants c. reactant; the amount of products >> the amount of reactants d. reactant; the amount of products << the amount of reactants e. neither; not enough information is available to reach a conclusion 1. Two reactants are combined and the system eventually reaches equilibrium. Which statement about the rates of the forward and reverse reactions is correct? a. From the beginning of the reaction until equilibrium, the rate of the forward reaction decreased and the rate of reverse reaction increased. b. From the beginning of the reaction until equilibrium, the rate of the forward reaction and the rate of the reverse reaction increased. c. From the beginning of the reaction until equilibrium, the rate of the forward reaction and the rate of the reverse reaction decreased.

24 d. From the beginning of the reaction until equilibrium, the rate of the forward reaction increased and the rate of the reverse reaction decreased. e. From the beginning of the reaction until equilibrium, the rate of the forward reaction remained the same as the rate of the reverse reaction.. Consider the reaction 4NH 3 (g) + 5O (g) <===> 4NO(g) + 6 H O(g) The equilibrium constant expression for this reaction is a. 4[ NH3 ] 5[ O ] K = c 4 NO 6 H O [ ] [ ] b. 4[ NO] 6[ H O] K = c 4[ NH3 ] 5[ O ] 4 5 c. [ NH3 ] [ O ] K = c 4 [ NO] [ H ] 6 O 4 6 d. [ NO] [ H O] K = c 4 [ NH ] [ ] 5 3 O e. [ NO][ H O] K = c [ NH ][ O ] 3 3. Consider the reaction 4NH 3 (g) + 5O (g) <===> 4NO(g) + 6H O(g) The equilibrium constant expression for reverse of this reaction is a. 4[NH 3] 5[O ] Kc = 4[NO]6[H O] b. 4[NO]6[H O] Kc = 4 [NH 3]5[O ] c. 4 5 [NH 3] [O ] K c = 4 6 [NO] [H O] d. 4 6 [NO] [H O] K c = 4 5 [NH 3] [O ] e. [NO][H O] Kc = [NH 3][O ] 4. Consider the reaction NOCl(g) <===> NO(g) + Cl (g) If this reaction is multiplied through by what would be the equilibrium constant expression? a. b. c. K = c [NO] [C1 ] [NOC1] [NO] [C1 ] K = c [NOC1] K = c 4 [NO] [C1 ] [NOC1] 4 d. 4 [NOC1] K = c 4 [NO] [C1 ] e. [NOC1] K = c [NO] [C1 ] 5. Consider the reaction Br (g) + Cl (g) <===> BrCl(g) Calculate the value of K p at 400 C if the partial pressures of BrCl, Br, and Cl are 1.87 atm, 1.00 atm, and 0.50 atm, respectively. a. 7.0

25 b. 3.7 c..3 d. 0.7 e Consider the reaction Br (g) + Cl (g) <===> BrCl(g) Calculate the value of K p at 400 C if the partial pressures of BrCl, Br, and Cl are 3.74 atm,.00 atm, and 1.00 atm, respectively. a b c. 1.9 d. 4.7 e Consider the reaction COCl (g) <===> CO(g) + Cl (g) At equilibrium, [CO] = M; [Cl ] = M; and [COCl ] = M. Calculate the value of the equilibrium constant. a b c d e Consider the reaction A <===> B, where the value of K c is Which statement about the system at equilibrium is correct? a. The amount of A is very close to the amount of B. b. The amount of A is slightly less than the amount of B. c. The amount of A is much larger than the amount of B. d. The amount of A is much less than the amount of B. e. More information is needed to make any statement about relative amounts of the two chemicals. 9. Consider the reaction A <===> B, where the value of K c is Which statement about the system at equilibrium is correct? a. The amount of A is very close to the amount of B. b. The amount of A is slightly less than the amount of B. c. The amount of A is much larger than the amount of B. d. The amount of A is much less than the amount of B. e. More information is needed to make any statement about the amounts of the two chemicals. 30. Consider the reaction system ethanol + acetic acid <====> ethyl acetate, where [ethyl acetate] Kc = = 095. [ethanol][acetic acid] The concentrations of ethanol and acetic acid are 0.45 M and the concentration of ethyl acetate is 1.1 M. Use the reaction quotient to determine whether the system is at equilibrium. a. The value of Q is 5.43, and the system is at equilibrium. b. The value of Q is.4, and the system is at equilibrium. c. The value of Q is 0.95, and the system is at equilibrium. d. The value of Q is 5.43, and the system is not at equilibrium. e. The value of Q is.4, and the system is not at equilibrium. 31. Consider the reaction A <===> B, where the value of K is At equilibrium, the concentration of A is 0.45 M. What is the concentration of B? a b c

26 d e Consider the reaction A <===> B, where the value of K is At equilibrium, the concentration of B is 0.45 M. What is the concentration of A? a b c d e Consider the reaction 3A <===> B, where the value of K is.1. If the concentration of A is at equilibrium, what is the concentration of B? a b. 1.0 c d e Consider the equilibrium system C(s) + CO (g) <====> CO(g) If more carbon is added, the equilibrium will, and if CO is removed the equilibrium will. a. shift forward; shift reverse b. shift forward; shift forward c. shift reverse; shift reverse d. Be unchanged; shift forward e. neither can be predicted 35. Consider the equilibrium system C(s) + CO (g) <====> CO(g) If CO is added, the equilibrium will, and if CO is added the equilibrium will. a. shift forward; shift reverse b. shift forward; shift forward c. shift reverse; shift reverse d. Be unchanged; shift forward e. neither can be predicted 36. Consider the equilibrium system C(s) + CO (g) <====> CO(g) If C is removed, the equilibrium will, and if CO is added, the equilibrium will. a. shift forward; shift reverse b. shift forward; shift forward c. shift reverse; shift reverse d. Be unchanged; shift reverse e. neither can be predicted 37. Consider the equilibrium system C(s) + CO (g) <====> CO(g) If the pressure on the system is increased the equilibrium will, because. a. shift forward; higher pressure favors fewer moles of gas b. shift forward; higher pressure favors more moles of gas c. shift reverse; higher pressure favors fewer moles of gas d. shift reverse; higher pressure favors more moles of gas e. Be unchanged; of the presence of the solid C 38. Consider the equilibrium system C(s) + CO (g) <====> CO(g) If the pressure on the system is decreased the equilibrium will, because. a. shift forward; lower pressure favors fewer moles of gas b. shift forward; lower pressure favors more moles of gas c. shift reverse; lower pressure favors fewer moles of gas d. shift reverse; lower pressure favors more moles of gas

27 e. Be unchanged; of the presence of the solid C 39. Consider the endothermic reaction at equilibrium: C(s) + CO (g) <====> CO(g) If the system is heated, the equilibrium will, because. a. shift forward; increased temperature favors an endothermic reaction b. shift forward; increased temperature favors an exothermic reaction c. shift reverse; increased temperature favors an endothermic reaction d. shift reverse; increased temperature favors an exothermic reaction e. be unchanged; temperature has no effect on equilibrium 40. Consider the endothermic reaction at equilibrium: C(s) + CO (g) <====> CO(g) If the system is cooled, the equilibrium will, because. a. shift forward; decreased temperature favors an endothermic reaction b. shift forward; decreased temperature favors an exothermic reaction c. shift reverse; decreased temperature favors an endothermic reaction d. shift reverse; decreased temperature favors an exothermic reaction e. be unchanged; temperature has no effect on equilibrium 41. At room temperature, if the value of K c for a given reaction is a large number, we can conclude that the energy of the products is likely to be the energy of the reactants because. a. less than; the value of K c favors the reactants b. less than; the value of K c favors the products c. the same as; the value of K c favors neither the products nor the reactants d. greater than; the value of K c favors the reactants e. greater than; the value of K c favors the products 4. At room temperature, if the value of K c for a given reaction is a very small number, we can conclude that the energy of the products is likely to be the energy of the reactants because. a. less than; the value of K c favors the reactants b. less than; the value of K c favors the products c. the same as; the value of K c favors neither the products nor the reactants d. greater than; the value of K c favors the reactants e. greater than; the value of K c favors the products 43. In predicting which state of an equilibrium is more favored, the factor concerned with probability is called a. enthalpy b. entropy c. molarity d. temperature e. work 44. Considering only the probability factor in the gaseous reaction A + B <===> C, the side is favored because. a. product; there are more possible arrangements of molecules on the reactant side b. product; there are more possible arrangements of molecules on the product side c. reactant; there are more possible arrangements of molecules on the reactant side d. reactant; there are more possible arrangements of molecules on the product side e. More information is needed to determine which side is favored. 45. Which factor concerning a reaction system most influences the relative importance of the probability factor versus the energy factor? a. total number of moles b. volume c. density of any liquids d. temperature e. pressure 46. Consider the exothermic reaction for the manufacture of ammonia: N (g) + 3H (g) NH 3 (g) This reaction is -favored on the basis of enthalpy and -favored on the basis of entropy. a. product; product b. reactant; reactant

28 c. product; reactant d. reactant; product e. More information is needed to make this determination. 47. Consider the exothermic reaction for the manufacture of ammonia: N (g) + 3H (g) NH 3 (g). If the pressure on this system is increased, the equilibrium will shift to the ; if the temperature is increased the equilibrium will shift to the. a. right; right b. left; left c. neither; left d. right; left e. left; right Answer Key 1. Ans: d 1/[CO]. Ans: c [CO] 3 [HO] 4 Kc= [C3H8][O] 5 3. Ans: d is described by each of these statements. 4. Ans: d the rate at which reactants form products is the same as the rate at which products form reactants. 5. Ans: b. only one phase. 6. Ans: c concentrations at the start of the experiment. 7. Ans: c [HO] Kc = [H] [O] 8. Ans: b [I][CO] 5 [CO] 5 9. Ans: d PCl5(g) <==> PCl3(g) + Cl(g). 10. Ans: b the reaction must move to the left to reach equilibrium. 11. Ans: c Kp = K(RT) - 1. Ans: b Ans: d increasing the temperature of the reaction system. 14. Ans: a FeO(s) + CO(g) <===> Fe(s) + CO(g) 15. Ans: b the number of Cl molecules increases 16. Ans: c increasing pressure 17. Ans: d Ans: c Ans: a 0. Ans: D. reactant; the amount of products << the amount of reactants 1. Ans:A. From the beginning of the reaction until equilibrium, the rate of the forward reaction decreased and the rate of the reverse reaction increased.. Ans:: D 3. Ans: C 4. Ans: C 5. Ans: A Ans: E Ans: C Ans: D. The amount of A is much less than the amount of B. 9.Ans: C. The amount of A is much larger than the amount of B. 30. Ans: D. The value of Q is 5.43, and the system is not at equilibrium. 31. Ans: C Ans: C Ans: D Ans: D. be unchanged; shift forward 35. Ans: A. shift forward; shift reverse 36. Ans: D. be unchanged; shift reverse 37. Ans: C. shift reverse; higher pressure favors fewer moles of gas 38. Ans: B shift forward; lower pressure favors more moles of gas

29 39. Ans: A. shift forward; increased temperature favors an endothermic reaction 40. Ans: D. shift reverse; decreased temperature favors an exothermic reaction 41. Ans: B. less than; the value of K c favors the products 4. Ans: D. greater than; the value of K c favors the reactants 43. Ans: B. entropy 44. Ans: C. reactant; there are more possible arrangements of molecules on the reactant side 45. Ans: D. temperature 46. Ans: C. product; reactant 47. Ans: D. right; left