CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS

Transcription

1 CHATER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS 17.1 If the rate of the forward reaction exceeds the rate of reverse reaction, products are formed faster than they are consumed. The change in reaction conditions results in more products and less reactants. A change in reaction conditions can result from a change in concentration or a change in temperature. If concentration changes, product concentration increases while reactant concentration decreases, but the K c remains unchanged because the ratio of products and reactants remains the same. If the increase in the forward rate is due to a change in temperature, the rate of the reverse reaction also increases. The equilibrium ratio of product concentration to reactant concentration is no longer the same. Since the rate of the forward reaction increases more than the rate of the reverse reaction, K c increases (numerator, [products], is larger and denominator, [reactants], is smaller). [ products ] K c reactants 17. The faster the rate and greater the yield, the more useful the reaction will be to the manufacturing process. 17. A system at equilibrium continues to be very dynamic at the molecular level. Reactant molecules continue to form products, but at the same rate that the products decompose to re-form the reactants. 17. If K is very large, the reaction goes nearly to completion. A large value of K means that the numerator is much larger than the denominator in the K expression. A large numerator, relative to the denominator, indicates that most of the reactants have reacted to become products. One cannot say with certainty whether the value of K for the phosphorus plus oxygen reaction is large or small (although it likely is large). However, it is certain that the reaction proceeds very fast The reaction is exothermic because the energy of the reactants is greater than the energy of products. The products would be favored and be present at equilibrium at higher concentrations than the reactants because a lower energy state is favored. The change in energy looks large on the graph so the product concentrations are much larger than the reactant concentrations and K is large No, the value of Q is determined by the mass action expression with arbitrary concentrations for products and reactants. Thus, its value is not constant The equilibrium constant expression is K [O ]. If the temperature remains constant, K remains constant. If the initial amount of Li O present was sufficient to reach equilibrium, the amount of O obtained will be constant, regardless of how much Li O (s) is present On the graph, the concentration of HI increases at twice the rate that H decreases because the stoichiometric ratio in the balanced equation is 1 H : HI. Q for a reaction is the ratio of concentrations of products to concentrations of reactants. As the reaction progresses the concentration of reactants decrease and the concentration of products increase, which means that Q increases as a function of time. H (g) + I (g) HI(g) Q [ HI] [ H ][ I ] 17-1

2 The value of Q increases as a function of time until it reaches the value of K. b) No, Q would still increase with time because the [I ] would decrease in exactly the same way as [H ] decreases A homogeneous equilibrium reaction exists when all the components of the reaction are in the same phase (i.e., gas, liquid, solid, aqueous). NO(g) + O (g) NO (g) A heterogeneous equilibrium reaction exists when the components of the reaction are in different phases. Ca(HCO ) (aq) CaCO (s) + H O(l) + CO (g) / N (g) + 1/ O (g) NO(g) [ NO] Q c(form) 1 1 [ N] [ O] NO(g) 1/ N (g) + 1/ O (g) Q c(decomp) 1 1 [ N] [ O] [ NO] Q c(decomp) 1/Q c(form), so the constants do differ (they are the reciprocal of each other) Yes, the Q s for the two reactions do differ. The balanced equation for the first reaction is / H (g) + 1/ N (g) NH (g) (1) The coefficient in front of NH is fixed at 1 mole according to the description. In the second reaction, the coefficient in front of N is fixed at 1 mole. H (g) + N (g) NH (g) () The reaction quotients for the two equations and their relationship are: NH Q 1 NH H N H [ N ] Q 1 Q Q Check that correct coefficients from balanced equation are included as exponents in the mass action expression. a) NO(g) + O (g) N O (g) Q c NO NO [ O ] b) SF 6 (g) + SO (g) SO F (g) [ SO ] F Q c SF6 SO 17-

3 c) SC1F 5 (g) + H (g) S F 10 (g) + HCl(g) Q c SF 10 HCl SClF5 [ H] 17.1 a) C H 6 (g) + 7 O (g) CO (g) + 6 H O(g) 6 CO HO Q c 7 CH 6 [ O] b) CH (g) + F (g) CF (g) + HF(g) CF HF Q c [ CH][ F] c) SO (g) SO (g) + O (g) SO Q c O SO 17.1 Check that correct coefficients from the balanced equation are included as exponents in the mass action expression. a) NO Cl(g) NO (g) + Cl (g) NO Q c Cl [ NOCl] b) OCl (g) Cl (g) + O (g) Q c Cl O OCl c) NH (g) + O (g) N (g) + 6 H O(g) 6 N HO Q c NH [ O] a) O (g) O (g) Q c O O b) NO(g) + O (g) NO (g) + O (g) [ NO ] Q c [ O ] NO O c) N O(g) + H (g) NH (g) + H O(g) NH [ HO] Q c NO H 17-

4 H The Q for the original reaction is Q ref S [ HS] a) The given reaction 1/ S (g) + H (g) H S(g) is the reverse reaction of the original reaction multiplied by a factor of 1/. The equilibrium constant for the reverse reaction is the inverse of the original constant. When a reaction is multiplied by a factor, K, the new equation is equal to the K of the original equilibrium raised to a power equal to the factor. For the reaction given in part a), take (1/K) 1/. [ HS] Q a (1 / Q ref ) 1/ 1 [ S ] [ H] K (1 / 1.6 x 10 - ) 1/ b) The given reaction 5 H S(g) 5 H (g) + 5/ S (g) is the original reaction multiplied by 5/. Take the original K to the 5/ power to find K of given reaction. [ N ] HO Q c(ref) [ NO] [ H ] 5 H Q a (Q ref ) 5/ S 5 [ HS] 5 K (1.6 x 10 - ) 5/.817 x x [ N ] a) Q c [Q c(ref) ] 1/ HO [ NO][ H ] Thus, K c [K c(ref) ] 1/ (6.5 x 10 ) 1/ NO H b) Q c [Q c(ref) ] - [ N ] [ HO] K c [K c(ref) ] - (6.5 x 10 ) x x The concentration of solids and pure liquids do not change, so their concentration terms are not written in the reaction quotient expression. a) Na O (s) + CO (g) Na CO (s) + O (g) [ O ] Q c [ CO ] b) H O(l) H O(g) Q c [H O(g)] Only the gaseous water is used. The (g) is for emphasis. c) NH Cl(s) NH (g) + HCl(g) Q c [NH ][HCl] a) H O(l) + SO (g) H SO (aq) [ HSO ] Q c SO b) KNO (s) KNO (s) + O (g) Q c [O ] c) S 8 (s) + F (g) 8 SF 6 (g) 8 Q c SF6 F 17-

5 17.0 Make sure not to include solids and liquids in reaction quotient. a) NaHCO (s) Na CO (s) + CO (g) + H O(g) Q c [CO ][H O] b) SnO (s) + H (g) Sn(s) + H O(g) HO Q c [ H ] c) H SO (l) + SO (g) H S O 7 (l) 1 Q c SO 17.1 a) Al(s) + NaOH(aq) + 6 H O(l) Na[Al(OH) ](aq) + H (g) Na Al OH Q c ( ) [ H ] [ NaOH] b) CO (s) CO (g) Q c [CO (g)] c) N O 5 (s) NO (g) + O (g) Q c [NO ] [O ] Only the gaseous carbon dioxide is used. The (g) is for emphasis. 17. Write balanced chemical equations for each reaction, and then write the appropriate equilibrium expression. a) HCl(g) + O (g) Cl (g) + H O(g) Cl HO Q c [ HCl] [ O ] b) As O (s) + 10 F (g) AsF 5 (l) + O (g) O Q c 10 [ F ] c) SF (g) + H O(l) SO (g) + HF(g) SO HF Q c [ SF ] d) MoO (s) + 6 XeF (g) MoF 6 (l) + Xe(g) + O (g) Xe O Q c 6 XeF 17. a) The balanced equations and corresponding reaction quotients are given below. Note the second equation must occur twice to get the appropriate overall equation. (1) Cl (g) + F (g) ClF(g) Q 1 [ ClF] [ Cl ][ F ] ClF () ClF(g) + F (g) ClF (g) Q [ ClF ][ F ] ( ClF(g) + F (g) ClF (g) Q [ ClF ][ F ] ClF ClF Overall: Cl (g) + F (g) ClF (g) Q overall [ Cl ][ F ] 17-5

6 b) The second equation occurs twice, thus it could simply be multiplied by two and its reaction quotient squared. The reaction quotient for the overall reaction, Q overall, determined from the reaction is: ClF Cl F Q overall Q overall Q 1 Q Q 1 Q Q [ ClF] [ Cl ][ F ] ClF ClF ClF x x [ ClF ][ F ] [ ClF ][ F ] [ Cl ][ F ] 17. Concentration and pressure of gas are directly proportional as long as the temperature is constant: C n/v /RT K c and K p are related by the equation K c K p (RT) - n, where n represents the change in the number of moles of gas in the reaction (moles gaseous products - moles gaseous reactants). When n is zero (no change in number of moles of gas), the term (RT) n equals 1 and K c K p. When n is not zero, meaning that there is a change in the number of moles of gas in the reaction, then K c K p a) Since n number of moles gaseous products - number of moles gaseous reactants, n is a positive integer. If n is a positive integer, then (RT) - n is less than 1. Thus, K p must be larger since it is multiplied by a fraction that is less than 1 gives K c. K c is smaller than K p. b) Assuming that RT > 1 (which occurs when T > 1. K, because x 1. 1), K p > K c if the number of moles of gaseous products exceeds the number of moles of gaseous reactants. K p < K c when the number of moles of gaseous reactants exceeds the number of moles of gaseous product a) Number of moles of gaseous reactants 0; Number of moles of gaseous products ; n - 0 b) Number of moles of gaseous reactants 1; Number of moles of gaseous products 0; n c) Number of moles of gaseous reactants 0; Number of moles of gaseous products ; n a) n gas 1 b) n gas - c) n gas First, determine n for the reaction and then calculate K c using K p K c (RT) n. a) n Number of product gas moles - Number of reactant gas moles K c K p / (RT) n (.9 x 10 - ) / [(0.081) (1000.)] This assumes that 1000 has at least significant figures. b) n Number of product gas moles - Number of reactant gas moles K c K p / (RT) n (8.5) / [(0.081) (500.)] First, determine n for the reaction and then calculate K c using K p K c (RT) n. a) n Number of product gas moles - Number of reactant gas moles - 0 K c K p / (RT) n (9) / [(0.081) (70.)] 0 9 b) n Number of product gas moles - Number of reactant gas moles - -1 K c K p / (RT) n (.5 x ) / [(0.081) (500.)] x x First, determine n for the reaction and then calculate K p using K p K c (RT) n. a) n Number of product gas moles - Number of reactant gas moles K p K c (RT) n (6.1 x 10 - ) [(0.081) (98)] b) n Number of product gas moles - Number of reactant gas moles - - K p K c (RT) n (. x 10 - ) [(0.081) (1000.)] x x First, determine n for the reaction and then calculate K p using K p K c (RT) n. a) n Number of product gas moles - Number of reactant gas moles - 0 K p K c (RT) n (0.77)[(0.081) (100.)] b) n Number of product gas moles - Number of reactant gas moles - -1 K p K c (RT) n (1.8 x )[(0.081) (570.)] x x

7 17. When Q < K, the reaction proceeds to the right to form more products. The reaction quotient and equilibrium constant are determined by [products] / [reactants]. For Q to increase and reach the value of K, the concentration of products (numerator) must increase in relation to the concentration of reactants (denominator). 17. Reactant and product concentrations are constant over time. The forward reaction rate equals the reverse reaction rate. The reaction quotient equals the equilibrium constant: Q K To decide if the reaction is at equilibrium, calculate Q p and compare it to K p. If Q p K p, then the reaction is at equilibrium. If Q p > K p, then the reaction proceeds to the left to produce more reactants. If Q p < K p, then the reaction proceeds to the right to produce more products. H Br Q p (0.010) (0.010) / (0.0).5 x 10 - > K p.18 x 10-9 HBr Q p > K p, thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants) Q p NOBr NOBr (0.10) (0.10) / (0.10) 0.10 < K p 60.6 Q p < K p Thus, the reaction is not at equilibrium and will proceed to the right (towards the products) There is insufficient information to calculate the partial pressures of each gas (T is not given). There is sufficient information to determine the concentrations and hence Q c. Convert the K p given to K c using K p K c (RT) n. Compare the Q c to the K c just calculated and make a prediction. n Number of product gas moles - Number of reactant gas moles - 0 Since n 0, K p K c.7 (note if n had any other value, we could not finish the calculation without the temperature.) [ CO ] H Q c [ CO][ H O ] 0.6 /.0 0. / / /.0.66 (unrounded) > K c.7 Q c > K c Thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants) At equilibrium, equal concentrations of CFCl and HCl exist, regardless of starting reactant concentrations. The equilibrium concentrations of CFCl and HCl would still be equal if unequal concentrations of CCl and HF were used. This occurs only when the two products have the same coefficients in the balanced equation. Otherwise, more of the product with the larger coefficient will be produced When x mol of CH reacts, x mol of H O also reacts to form x mol of CO and x mol of H. The final (equilibrium) concentration of each reactant is the initial concentration minus the amount that reacts. The final (equilibrium) concentration of each product is the initial concentration plus the amount that forms a) The approximation applies when the change in concentration from initial to equilibrium is so small that it is insignificant. This occurs when K is small and initial concentration is large. b) This approximation will not work when the change in concentration is greater than 5%. This can occur when [reactant] initial is very small, or when [reactant] change is relatively large due to a large K Since all equilibrium concentrations are given in molarities and the reaction is balanced, construct an equilibrium expression and substitute the equilibrium concentrations to find K c. [ HI] 1.87 x10 K c [ H][ I] x x

8 17. K c [ N ][ H ] NH [ 0.11 ][ 0. ] [ 0.05] The reaction table requires that the initial [Cl 5 ] be calculated: [Cl 5 ] 0.15 mol/.0 L M x [Cl 5 ] reacting (-x), and the amount of Cl and of Cl forming (+x). Concentration (M) Cl 5 (g) Cl (g) + Cl (g) Initial Change -x +x +x Equilibrium x x x 17. The reaction table requires that the initial [H ] and [F ] be calculated: [H ] 0.10 mol / 0.50 L 0.0 M; [F ] mol / 0.50 L 0.10 M x [H ] reacting Concentration (M) H (g) + F (g) HF(g) Initial Change -x -x + x Equilibrium 0.0 -x x x 17.5 Construct an equilibrium expression and solve for NOCl. NOCl K p 6.5 x 10 NOCl NO Cl K ( )( ) ( ) p NO Cl 6.5 x atm A high pressure for NOCl is expected because the K p indicates that the reaction proceeds largely to the right, i.e., to the formation of products C(s) + H (g) CH (g) CH Q p 0.6 H CH Q p H (0.6) (1.) atm 17.7 The ammonium hydrogen sulfide will decompose to produce hydrogen sulfide and ammonia gas until K p 0.11: NH HS(s) H S(g) + NH (g) Q p HS NH 0.11 HS NH NH Q p atm 17.8 H S(g) H (g) + S (g) [ H S] 0.5 mol /.0 L 0.15 M Concentration (M) H S(g) H (g) S (g) Initial Change -x + x + x Equilibrium x x x H Q c S [ x] [ x] HS 0.15 x 9.0 x

9 Assuming 0.15 M - x 0.15 M [ x] [ x ] x 9.0 x 10-8 [ 0.15] 0.15 x x 10 - M (unrounded) [ H ] x ( x 10 - M) x x 10 - M (Since (1.6 x 10 - ) / (0.15) < 0.05, the assumption is OK.) 17.9 The initial concentrations of N and O are (0.0 mol/1.0 L) 0.0 M and (0.15 mol/1.0 L) 0.15 M, respectively. N (g) + O (g) NO(g) Concentration (M) N (g) O (g) NO(g) Initial Change -x -x + x Equilibrium x x x [ NO] Q c [ N][ O ] x.10 x 10 [ 0.0 x][ 0.15 x] Assume 0.0 M - x 0.0 M and 0.15 M - x 0.15 M x.10 x 10- [ 0.0][ 0.15 ] x x 10 - M (unrounded) [NO] x ( x 10 - M) x x 10 - M (Since (1.8 x 10 - ) / (0.15) < 0.05, the assumption is OK.) NO (g) NO(g) + O (g) ressure (atm) NO (g) NO(g) O (g) Initial Change - x + x + x Equilibrium x x x NOO ( x) ( x) Q p NO ( 0.75 x ).8 x 10-1 Assume 0.75 atm - x 0.75 atm ( x )( x) ( x ).8 x 10-1 ( 0.75) ( 0.75) x.979 x 10-5 atm.0 x 10-5 atm O NO x (.979 x 10-5 atm) x x 10-5 atm NO First, determine the initial concentration of H and the change, x, to reach equilibrium. Initial (unrounded) concentrations: [HI] (0.0 mol) / (1.50 L) [H ] (0.006 mol) / (1.50 L) [I ] (0.001 mol) / (1.50 L) M Concentration (M) HI(g) H (g) + I (g) Initial Change - x +x +x Equilibrium x x x [H ] eq x x M (unrounded) [I ] eq x M I [HI] eq x M HI 17-9

10 17.5 Initial concentrations: [A] (1.75 x 10 - mol) / (1.00 L) 1.75 x 10 - M [B] (1.5 x 10 - mol) / (1.00 L) 1.5 x 10 - M [C] (6.50 x 10 - mol) / (1.00 L) 6.50 x 10 - M Concentration (M) A(g) B(g) + C(g) Initial 1.75 x x x 10 - Change - x + x +x Equilibrium 1.75 x x 1.5 x x 6.50 x x [A] eq.15 x x x x [B] eq 1.5 x x.5 x 10 - M B [C] eq 6.50 x x.5 x 10 - M C 17.5 Construct a reaction table, using [ICl] init (0.500 mol / 5.00 L) M, and substitute the equilibrium concentrations into the equilibrium expression. Concentration (M) ICl(g) I (g) + Cl (g) Initial Change - x +x +x Equilibrium x x x [ I ] Cl [ x][ x] K c [ ICl] [ x] Take the square root of each side: [ x] [ x] (unrounded) x x 1.66 x x [I ] eq [Cl ] eq M [ICl] eq x M ICl 17.5 Concentration (M) SCl (g) + C H (g) S(CH CH Cl) (g) Initial Change - x - x + x Equilibrium x x x [S(CH CH Cl) ] eq x 0.50 M [SCl ] eq x 0.5 M [C H ] eq x 0.7 M [ S(CH ] CHCl) [ 0.50] K c SCl C H (unrounded) K p ( ) ( )( + )

11 17.55 NH (g) + O (g) N (g) + 6 H O(g) To find the equilibrium constant, determine the equilibrium concentrations of each reactant and product and insert into the equilibrium expression. Since [N ] increases from 0 to 1.96 x 10 - M, the concentration of H O gas will increase by times as much (stoichiometric ratio is 6 H O : N ) and the concentration of products will decrease by a factor equivalent to the stoichiometric ratio ( for NH and / for O ). Since the volume is 1.00 L, the concentrations are equal to the number of moles present. Concentration (M) NH (g) + O (g) N (g) + 6 H O(g) Initial Change - x - x + x + 6 x Equilibrium x x + x + 6 x All intermediate concentration values are unrounded. [H O] eq (6 mol H O/ mol N ) (1.96 x 10 - ) x 10 - M [NH ] eq ( mol NH /1.00 L) - ( mol NH / mol N ) (1.96 x 10 - ) x 10 - M [O ] eq ( mol O /1.00 L) - ( mol O / mol N ) (1.96 x 10 - ) x 10 - M [ N ] HO K c NH [ O ] x x x x x x 10 If values for concentrations were rounded to calculate K c, the answer is 5.90 x ressure (atm) FeO(s) + CO(g) Fe(s) + CO (g) Initial Change - x + x Equilibrium x x CO K p CO 0.0 x 1.00 x x atm CO x atm CO A change in equilibrium conditions such as a change in concentration of a component, a change in pressure (volume), or a change in temperature Equilibrium position refers to the specific concentrations or pressures of reactants and products that exist at equilibrium, whereas equilibrium constant refers to the overall ratio of equilibrium concentrations and not to specific concentrations. Reactant concentration changes cause changes in the specific equilibrium concentrations of reactants and products (equilibrium position), but not in the equilibrium constant A positive H rxn indicates that the reaction is endothermic, and that heat is consumed in the reaction: NH Cl(s) + heat NH (g) + HCl(g) a) The addition of heat causes the reaction to proceed to the right to counterbalance the effect of the added heat. Therefore, more products form at a higher temperature and container (B) best represents the mixture. b) When heat is removed, the reaction shifts to the left to offset that disturbance. Therefore, NH and HCl molecules combine to form more reactant and container (A) best represents the mixture Equilibrium component concentration values may change but the mass action expression of these concentrations is a constant as long as temperature remains constant. Changes in component amounts, pressures (volumes), or addition of a catalyst will not change the value of the equilibrium constant a) Rate f k f [Reactants] x. An increase in reactant concentration shifts the equilibrium to the right by increasing the initial forward rate. Since K eq k f /k r and k f and k r are not changed by changes in concentration, K eq remains constant. b) A decrease in volume causes an increase in concentrations of gases. The reaction rate for the formation of fewer moles of gases is increased to a greater extent. Again, the k f and k r values are unchanged. c) An increase in temperature increases k r to a greater extent for an exothermic reaction and thus lowers the K eq value

12 17.6 An endothermic reaction can be written as: reactants + heat products. A rise in temperature favors the forward direction of the reaction, i.e., the formation of products and consumption of reactants. Since K [products]/[reactants], the addition of heat increases the numerator and decreases the denominator, making K larger than K a) Equilibrium position shifts toward products. Adding a reactant (CO) causes production of more products. b) Equilibrium position shifts toward products. Removing a product (CO ) causes production of more products. c) Equilibrium position does not shift. The amount of a solid reactant or product does not impact the equilibrium as long as there is some solid present. d) Equilibrium position shifts toward reactants. roduct is added; dry ice is solid carbon dioxide that sublimes to carbon dioxide gas. At very low temperatures, CO solid will not sublime, but since the reaction lists carbon dioxide as a gas, the assumption that sublimation takes place is reasonable a) no change b) no change c) shifts toward the products d) shifts toward the reactants An increase in container volume results in a decrease in pressure. Le Châtelier s principle states that the equilibrium will shift in the direction that forms more moles of gas to offset the decrease in pressure. a) More F forms ( moles of gas) and less F (1 mole of gas) is present. b) More C H and H form ( moles of gas) and less CH ( moles of gas) is present a) less CH OH(l); more CH OH(g) b) less CH and NH ; more HCN and H Decreasing container volume increases the concentration of all gaseous reactants and products. a) With a decrease in container volume, the [H ], [Cl ], and [HCl] will increase by the same factor, x. The increase in concentrations of reactants is offset by the increase in concentration of the product, so there is no effect on the amounts of reactants or products in this reaction. b) In this equilibrium, a decrease in container volume will increase the concentrations of the reactant gases, H and O. In this case, Q c 1/([H ] [O ]) so Q c decreases with the increased reactant concentrations. To return to equilibrium, the concentrations and amounts of H and O will decrease from their initial values before the volume was changed. More H O will form because of the shift in equilibrium position. In general, decreasing the volume of a gaseous equilibrium shifts the reactions to the side with fewer moles of gas a) more CO and H O; less C H 8 and O b) more NH and O ; less N and H O The purpose of adjusting the volume is to cause a shift in equilibrium. a) Because the number of reactant gaseous moles equals the product gaseous moles, a change in volume will have no effect on the yield. b) The moles of gaseous product ( CO) exceed the moles of gaseous reactant (1 O ). A decrease in pressure favors the reaction direction that forms more moles of gas, so increase the reaction vessel volume a) increase volume b) decrease volume An increase in temperature causes a shift in the endothermic direction of the equilibrium. a) Reverse reaction is endothermic, so amount of product decreases. b) Forward reaction is endothermic, so amounts of products increase. c) Forward reaction is endothermic, so amounts of products increase. d) Reverse reaction is endothermic, so amount of product decreases a) decrease b) decrease c) decrease d) increase 17-1

13 17.7 The van t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for K. K 98 K T 1 98 K K 500 K? T 500. K R 8.1 J/mol K o H rxn [(0. kj/mol DH) x ( mol DH)](10 J/1 kj) 6. x 10 J ln K K Hrxn R T T1 ln 6. x 10 J J/mol K 500. K 98 K K ln K / K (1.80) ( ) The van t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and solve for K. K 98 K 1.5 x 10 o T 1 98 K H rxn -18 kj/mol K 0 K? T (7 + 0.) 7 K R 8.1 J/mol K o H rxn (-18 kj/mol) (10 J / 1 kj) -1.8 x 10 5 J ln K K Hrxn R T T1 5 K 1.8 x 10 J 1 1 ln.5 x J/mol K 7 K 98 K K ln x 10 K / (.5 x 10 ) x 10 K (.5 x 10 ) ( x 10 ) x x Fe O (s) + O (g) 6 Fe O (s) K p.5 x at 98 K 1 a) K p.5 x O O.0 x atm 1 b) Q p 1 / (0.1) (unrounded) O K p > Q p thus, the reaction will proceed to the right. c) K p K c (RT) n K c K p / ((RT) n ) K c (.5 x ) / [(0.081) (98)] x x a) The forward reaction is exothermic, so it is favored by lower temperatures. There are fewer moles of gas as products than as reactants, so products are favored by higher pressure. b) Addition of O would decrease Q and have no impact on K. c) To enhance yield of SO, a low temperature is used. Reaction rates are slower at lower temperatures, so a catalyst is used to speed up the reaction. 17-1

14 17.77 a) Q c [ XY] [ X ][ Y ] b) Scene A: Q c Scene B: Q c Scenes C-E: Q c [ 0] [ 0.] [ 0.6] [ 0.1][ 0.1 ] 6 x 101 c) left-to-right. Frame A must be the earliest time. d) K x 10 1 e) Scene B, at higher temperatures, the reaction shifts to the left (forming more X and Y ). f) None, volume (pressure) has no effect on the position of the equilibrium a) (1) H S(g) + O (g) SO (g) + H O(g) () SO (g) + Cl (g) SO Cl (g) Overall: H S(g) + O (g) + Cl (g) SO Cl (g) + H O(g) SO HO [ SOCl] b) Q 1 Q HS O SO Cl SOCl HO Q overall [ HS] [ O] [ Cl] SO HO x [ HS ] [ O] Q 1 x Q [ SOCl] [ SO ] [ Cl ] SOCl HO [ HS] [ O] [ Cl] a) Initially, a total of 6.00 volumes of gas (5.00 N O ) exist at 5.00 atm and 900. K. The volumes are proportional to moles of gas, so volume fraction equals mole fraction. The initial partial pressures of the two gases are: N + H 5.00 atm N X N tot (5.00 / 6.00) (5.00 atm) atm (unrounded) H X H tot (1.00 / 6.00) (5.00 atm) 0.8 atm (unrounded) The reaction table describing equilibrium is written as follows: (atm) N (g) + O (g) NO(g) Initial Change - x - x + x Equilibrium ( x) (0.8 - x) x Since K p is very small, assume that [N ] eq x and [H ] eq x 0.8. NO ( x) K p N O (.16667)( 0.8 ) 6.70 x x.116 x 10-5 (unrounded) Check assumption: percent error (.116 x 10-5 /.16667) 100% %. The assumption is good. Therefore, [NO] eq x (.116 x 10-5 ).8 x x 10-5 atm b) Convert atm () to concentration (mol/l) using the ideal gas law, assuming that it is applicable at 900. K. n / V / RT 17-1

15 5 ( ).8 x 10 atm 0.01 g NO 1 µ g L atm 6 1molNO g ( 900. K) mol K µg/l a) More CaCO, because the forward reaction is exothermic, decreasing the temperature will cause an increase in the amount of CaCO formed. b) Less CaCO, the only gas in the equation is a reactant. Increasing the volume will cause the equilibrium to shift toward the reactant side and the amount of CaCO formed decrease. c) More CaCO, increasing the partial pressure of CO will cause more CaCO to be formed. d) No change, removing half of the initial CaCO will have no effect on the amount of CaCO formed, because CaCO is solid Set up reaction table: Concentration (M) NH COONH (s) NH + CO Initial 7.80 g 0 0 Change - + x + x Equilibrium - x x The solid is irrelevant (as long as some is present). K c 1.58 x 10-8 (x) (x); x x 10 - M (unrounded) Total concentration of gases ( x 10 - M) x 10 - M.777 x 10 - M (unrounded) To find total pressure use the ideal gas equation: nrt / V (n/v) RT MRT (.777 x 10 - M) (0.081 L atm/mol K)[( )K] atm 17.8 Calculate K c [ CO ] H K c [ CO][ H O ] Calculate new concentrations New H 0.10 M + (0.60 mol /.0 L) 0.0 Concentration (M) CO(g) + H O(g) CO (g) + H (g) Initial Change x x - x - x Equilibrium x x x x [ CO ] H K c [ CO][ HO ] 0.0 x 0.0 x 0.0 x [ x][ x] x.0 [ 0.0 x] [ x].0 x (unrounded) [CO] [H O] x M [CO ] [H ] x M 17.8 a) (1) Ni S (s) + 7 O (g) 6 NiO(s) + SO (g) () 6 NiO(s) + 6 H (g) 6 Ni(s) + 6 H O(g) () 6 Ni(s) + CO(g) 6 Ni(CO) (g) Overall: Ni S (s) + 7 O (g) + 6 H (g) + CO(g) SO (g) + 6 H O(g) + 6 Ni(CO) (g) b) As always, the solid is not included in the Q expression. 6 ( ) 6 SO HO Ni CO Q c(overall) 7 6 O H CO 17-15

16 [ SO ] Q 1 x Q x Q [ H O] ( ) 6 Ni CO x x 7 6 O H CO 6 6 SO H O ( ) Ni CO 7 6 [ O ] [ H ] [ CO] a) Convert K c to K p, substitute the given values into the equilibrium expression and solve for C. H K p K c (RT) n K c (RT) -1 (9 x 10 )[(0.081 L atm/mol K) (600. K)] x 10 (unrounded) CH5OH 00. K p x C H CH HO CH ( ).767 x 10 - x 10 - atm b) The forward direction, towards the production of ethanol, produces the least number of moles of gas and is favored by high pressure. A low temperature favors an exothermic reaction. c) Use van t Hoff s equation to find K c at a different temperature. K 1 9 x 10 o T K H rxn (-7.8 kj/mol) (10 J/1 kj) -.78 x 10 J/mol K? T 50. K R 8.1 J/mol K ln K K Hrxn R T T1 K.78 x 10 J 1 1 ln 9x J/mol K 50. K 600. K K ln.19 9x10 K / 9 x K (9 x 10 ) (.7765).19 x 10 5 x 10 5 d) No, condensing the C H 5 OH would not increase the yield. Ethanol has a lower boiling point (78.5 C) than water (100 C). Decreasing the temperature to condense the ethanol would also condense the water, so moles of gas from each side of the reaction are removed. The direction of equilibrium (yield) is unaffected when there is no net change in the number of moles of gas a) Birds flying north and south is not a good example of an equilibrium. At any one time, the rates of movement in either direction are not equal. Generally, the birds are only moving in one direction or the other. b) This situation is an example of equilibrium. Carts move in and out of the store, but the number of carts inside and outside remains about constant. c) A tug-of-war is not a good example of an equilibrium because eventually the ribbon goes all the way to one side. d) Vaporization of stew and condensation on the lid is an example of equilibrium. Water exists both as a gas and liquid and the quantities of each phase remain about constant n/v M V / RT (.0 atm) ( + ) L atm ( K) mol K H (g) + CO (g) H O(g) + CO(g) x x x x [ HO ] CO K c [ H ][ CO ] (0.5) 1/ [ x][ x] [ x][ x] [ x] [ x] M each gas (unrounded) [ x] [ x] x 0.09 (unrounded) Mass H [( ) mol/l](1.00 L) (.016 g H /mol) g H 17-16

17 17.87 To get the two equations to sum to the desired equation, the first equation must be reversed and doubled. This will result in squaring the reciprocal of its K c value. The other equation does not need to be changed. Adding the two equations means the new K c value will be the product of the individual K c values. NO (g) NO(g) + O (g) K 1 K c 1.1 x 10-5 NO(g) N (g) + O (g) K (K c ) -.0 x K (unrounded) Overall: NO (g) N (g) + O (g) K c (overall) K 1 K.77 x x 10 1 M a) BrCl(g) Br (g) + Cl (g) [ Br ] Cl b) Q A [ BrCl] c) Q A increases and Q B decreases, until Q A 1/Q B at equilibrium. d) In container A, BrCl molecules decompose to form Br and Cl with an equilibrium constant of K A, while in container B, Br and Cl molecules will form BrCl molecules with an equilibrium constant of K B. At constant temperature, K A 1/K B. Since the total Br mass and Cl mass is initially identical, identical amounts of BrCl, Cl, and Br will be formed at equilibrium in each container a) Convert K c to K p, substitute given values into the equilibrium expression and solve for SO ( n - -1). K p K c (RT) n K c (RT) -1 (1.7 x 10 8 )[(0.081 L atm / mol K) (600. K)] x 10 6 (unrounded) SO ( 00. ) K p.51 x SO SO O SO ( ) atm b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the moles equals the molarity. Concentration (M) SO (g) + O (g) SO (g) Initial Change - x -x + x Equilibrium x x x x , therefore, [SO ] M, [O ] M, and [SO ] M The change in concentration for O is half the amount of SO because 1 mol of O reacts to form moles of SO. Substitute equilibrium concentrations into the equilibrium expression and solve for K c. K c SO [ 0.000] x 10 SO [ O ] [ 0.000] [ ] The pressure of SO is estimated using the concentration of SO and the ideal gas law (although the ideal gas law is not well behaved at high pressures and temperatures). L atm ( mol) ( K) mol K SO nrt / V atm 1.0 L ( ) 17-17

18 17.90 The original concentrations are: (0.50 mol / L) M for CO and Cl. Concentration (M) CO(g) + Cl (g) COCl (g) Initial Change - x - x + x Equilibrium x x x [ COCl ] [ x] [ x] Q c [ CO][ Cl ] [ x][ x] x + x x x (unrounded) a.95 b c.55 b ± b ac x a x ( ) ( )( ) (.95) ( 7.9) ± x or (unrounded) (The value is not possible because x would be negative.) [CO] [Cl ] M [COCl ] x 0.1 M The equilibrium constant for the reaction is K p 0.0 atm CO The amount of calcium carbonate solid in the container at the first equilibrium equals the original amount, mol, minus the amount reacted to form 0.0 atm of carbon dioxide. The moles of CaCO reacted is equal to the number of moles of carbon dioxide produced. Moles CaCO moles CO n V / RT ( 0.0 atm)( 10.0 L) n mol CaCO lost (unrounded) L atm ( 85 K) mol K mol CaCO mol CaCO As more carbon dioxide gas is added, the system returns to equilibrium by converting the added carbon dioxide to calcium carbonate to maintain the partial pressure of carbon dioxide at 0.0 atm. Convert the added 0.00 atm of CO to moles. The moles of CO reacted equals the moles of CaCO formed. Add to the moles of CaCO at the first equilibrium position and convert to grams. Moles CaCO moles CO n V / RT ( 0.00 atm)( 10.0 L) n mol CaCO formed (unrounded) L atm ( 85 K) mol K Mass CaCO [( ) mol CaCO ]( g CaCO / mol CaCO ) g CaCO 17.9 a) C H (g) + O (g) CO (g) + H O(g) b) NO (g) + 6 H O(g) NH (g) + 7 O (g) 17-18

19 17.9 C H (g) + H (g) C H (g) H rxn 1 mol (5.7 kj/mol) - (1 mol (7 kj/mol) + 1 mol (0 kj/mol)) kj (unrounded) 00 ln K H 1 1 K 000 R T1 T K00 ( 17.5 kj/ mol ) J ln.9 x 10 8 J 00. K 000. K 1 kj 8.1 mol K K00 ln x 10 8 K x x 10 K 00 (.9 x 10 8 ) ( x 10 5 ) x 10.0 x The first equation is in the correct form, but the second equation must be reversed. Reversing the equation leads to the reciprocal of the K value. 1/ H (g) + 1/ O (g) HO(g) K c H(g) 1/ H (g) K' c (K c ) x 10 (unrounded) Overall: H(g) + 1/ O (g) HO(g) K c(overall) K c1 x (K c ) x S F 10 (g) SF (g) + SF 6 (g) The reaction is described by the following equilibrium expression: [ SF ] 6 K c SF SF 10 At the first equilibrium, [S F 10 ] 0.50 M and [SF ] [SF 6 ] x. Therefore, the concentrations of SF and SF 6 can be expressed mathematically as [SF ] [SF 6 ] 0.50K c At the first equilibrium, [S F 10 ].5 M and [SF ] [SF 6 ] x. Therefore, the concentrations of SF and SF 6 can be expressed mathematically as [SF ] [SF 6 ].5K c Thus, the concentrations of SF and SF 6 increase by a factor of.5kc 0.50K c K c [F ] (The liquids are not included in K c.) [F ] 10 x10.71 x 10-5 x 10-5 M a) Calculate the partial pressures of oxygen and carbon dioxide because volumes are proportional to moles of gas, so volume fraction equals mole fraction. Assume that the amount of carbon monoxide gas is small relative to the other gases, so the total volume of gases equals V + V + V CO O N CO (10.0 mol CO / 61.0 mol gas) (.0 atm) atm (unrounded) O (1.00 mol O / 61.0 mol gas) (.0 atm) atm (unrounded) a) Use the partial pressures and given K p to find CO. CO (g) CO(g) + O (g) COO CO ( ) K p 1. x 10-8 CO ( ) CO.099 x x 10-1 atm 17-19

20 b) Convert partial pressure to moles per liter using the ideal gas equation, and then convert moles of CO to grams. ( ) n CO / V / RT.099 x 10 1 atm.611 x mol/l (unrounded) L atm ( 800 K) mol K (.611 x mol/l) (8.01 g CO/mol CO) (1 pg / 10-1 g) pg CO/L a) O.N. (Fe) + O.N. (O) 0 O.N. (Fe) + (-) 0 O.N. (Fe) 8 O.N. (Fe) 8/ b) Fe + + Fe + c) (0.050 mol H O ) / (1.0 L) M H O K c [ H ] [ x] 5.1 HO x [ ] [ x] [ x] (unrounded) x ( x 10 - mol/l) (1.0L) x 10 - mol Fe O reacting (unrounded) Mass Fe O ( x 10 - ) (1.55 g Fe O / 1 mol Fe O ) g Fe O Although the yield is favored by low temperature, the rate of formation is not. In fact, ammonia forms so slowly at low temperatures that the process becomes uneconomical. In practice, a compromise is achieved that optimizes yield and rate (high pressure, continual removal of NH, increases the temperature) a) Write a reaction table given that CH (init) CO (init) 10.0 atm, substitute equilibrium values into the equilibrium expression, and solve for. H ressure (atm) CH (g) + CO (g) CO(g) + H (g) Initial Change - x - x + x + x Equilibrium x x x x COH ( x) ( x) ( x) K p CH CO ( 10.0 x)( 10.0 x ) 10.0 x.58 x 10 6 ( x) ( 10.0 x) x 10 ( ) A quadratic is necessary: x x 10 x x 10 0 (unrounded) a b x 10 c x 10 b ± b ac x a ( ) ( )( ) x 10 ± x x 10 x ( ) x (unrounded) (hydrogen) x ( ) atm (unrounded) If the reaction proceeded entirely to completion, the partial pressure of H would be 0.0 atm (pressure is proportional to moles, and twice as many moles of H form for each mole of CH or CO that reacts). The percent yield is ( atm/0.0 atm) x 100% %. 17-0

21 b) Repeat the calculations for part (a) with the new K p value. The reaction table is the same. COH ( x) ( x) ( x) K p.66 x 10 CH CO ( 10.0 x)( 10.0 x ) x ( x) ( 10.0 x) x 10 ( ) A quadratic is needed: x x 10 x x 10 0 (unrounded) a b x 10 c x 10 ( ) ( )( ) x 10 ± x x 10 x ( ) x (unrounded) (hydrogen) x (9.918) atm (unrounded) If the reaction proceeded entirely to completion, the partial pressure of H would be 0.0 atm (pressure is proportional to moles, and twice as many moles of H form for each mole of CH or CO that reacts). The percent yield is ( atm/0.0 atm) x 100% %. c) van t Hoff equation: 100 ln K H 1 1 K 100 R T1 T 6.58 x 10 H 1 1 ln.66 x 10 7 J 100. K 100. K 8.1 mol K H( x 10-6 ) H / ( x 10-6 ) x x 10 5 J/mol (The subtraction of the 1/T terms limits the answer to three significant figures.) a) C H 8 (g) + H O(g) CO(g) + 7 H (g) K' K p 1 p x CO(g) + H O(g) CO (g) + H (g) K' (K p p ) (0.69) (unrounded) (Overall): C H 8 (g) + 6 H O(g) CO (g) + 10 H (g) b) K p(overall) K' p1 x K' p x x x x ( ) ( ) CO H c) K p ( CH )( HO) 8 6 The partial pressures of each reactant are proportional to the moles, and the limiting reactant may be determined from the partial pressures. C H 8 (initial) (1.00 / 5.00) x 5.0 atm 1.0 atm 1.0 atm / H O(initial) (.00 / 5.00) x 5.0 atm.0 atm.0 atm / limiting reactant. CO (formed).0 atm x ( mol CO / 6 mol H O).0 atm H (formed).0 atm x (10 mol H / 6 mol H O) atm (unrounded) C H 8 (remaining) 1.0 atm - [.0 atm x (1 mol C H 8 / 6 mol H O)] 0. atm (unrounded) H O(remaining) 0.00 atm (limiting reactant) Total CO + H + C H 8 + H O.0 atm atm + 0. atm atm 9.0 atm d) ercent C H 8(unreacted) [0. atm / 1.0 atm] x 100%.% 0% 17-1

22 17.10 a) Multiply the second equation by to cancel the moles of CO produced in the first reaction. CH (g) + O (g) CO(g) + H (g) CO(g) + H O(g) CO (g) + H (g) CH (g) + O (g) + H O(g) CO (g) + 6 H (g) b) K p (9.x10 8 ) (1.7) x x 10 9 c) n 8-5 K c ( x 10 9 ) / [(0.081 atm L/mol K) (1000 K)].186 x x 10 d) The initial total pressure is given as 0. atm. To find the final pressure use relationship between pressure and number of moles of gas: n initial / initial n final / final Total mol of gas initial.0 mol CH mol O +.0 mol H O 5.0 mol Total mol of gas final.0 mol CO mol H 8.0 mol final (0. atm) (8.0 mol / 5.0 mol) 8 atm H (g) + N (g) NH (g) NH (1.9% / 100%) (110. atm) 5.69 atm (unrounded) % - 1.9% 58.51% N + H H + N (58.51% / 100%) (110. atm) 6.61 atm (unrounded) H (/) (6.61 atm) atm (unrounded) N (1/) (6.61 atm) atm (unrounded) K p ( NH ) ( H ) ( N ) ( 5.69) ( ) ( ) x x a) Careful reading of the problem indicates that the given K p occurs at K and that the initial pressure of N is 00. atm. Solve the equilibrium expression for N, assuming that N (eq) x is about 00. ( ) N K p 10 ( N ) - 10 The exponent, two digits past the decimal point, limits the significant figures..10 N ( 00. ) x x 10-1 atm ( ) H b) K p ( H ) The exponent, two digits past the decimal point, limits the significant figures H ( 600. ) x x 10-8 atm c) Convert pressures to moles using the ideal gas law. Convert moles to atoms using Avogadro s number. 1 ( x 10 atm) 6.0 x 10 atoms Moles N / L / RT L atm mol ( 1000.K) mol K N atoms/l 8 ( 5.87 x 10 atm) 6.0 x 10 atoms Moles H / L / RT L atm mol ( K) mol K.0 x x 10 1 H atoms/l d) The more reasonable step is N (g) + H(g) NH(g) + N(g). With only 9 N atoms in 1.0 L, the first reaction would produce virtually no NH(g) molecules. There are orders of magnitude, more N molecules than N atoms, so the second reaction is the more reasonable step. 17-

23 a) Since the volume is 1.00 L, the molarity equal the number of moles present. NH (g) N (g) + H (g) Initial Change + x -x - x Equilibrium x M x x x mol [N ] eq ( ) M 1.5 M N [H ] eq [ (0.0500)] M 1.50 M H N H K c NH [ 0.100] N [ b) K c ] H [ 1.5 ] NH 8. x c) K c in a) is the square of K c in b). The balanced equations are different; therefore, the values of K c are different a) H (g) + N (g) NH (g) K p K p ( NH ) ( N )( H ) ( 50. ) ( x)( x) 1.00 x x 10 - x atm N x (1.0016) atm H total nitrogen + hydrogen + ammonia ( atm) + ( atm) + (50. atm) atm total ( 50. ) b) K p 1.00 x 10 - ( x)( 6x) x atm N 6 x 6 (18.) atm H total nitrogen + hydrogen + ammonia (18. atm) + (110.6 atm) + (50. atm) atm total This is not a valid argument. The total pressure in (b) is greater than in (a) to produce the same amount of NH The equilibrium equation is: CO(g) + O (g) CO (g) a) CO(g) + O (g) CO (g) Initial 0 atm 0 atm 1.00 atm Change + x +x - x Equilibrium + x x x K p ( CO ) ( CO ) ( O ) ( 1.00 x) ( x) ( x) ( 1.00) ( x) ( x) 1.00 x x 10 Assume x is small compared to x 10 x.90 x 10-8 (unrounded) The assumption is justified. CO x (.90 x 10-8 ) 5.88 x x 10-8 atm 17-

24 b) Since x is so small, total is the carbon dioxide pressure (1.00 atm). c) Set up a new table: CO(g) + O (g) CO (g) Initial 0 atm 0 atm 1.00 atm Change + x + x - x Equilibrium + x x x x (5% / 100%) 0.5 x (unrounded) K p ( CO ) ( CO ) ( O ) ( 1.00 ( 0.175) ) x ( ( 0.175) ) ( 0.175) ( 1.00 x) ( x) ( ) x 10 1 d) ( x) + ( x) + (x) x atm e) K decreases as the temperature increases. The process is exothermic a) Equilibrium partial pressures for the reactants, nitrogen and oxygen, can be assumed to equal their initial partial pressures because the equilibrium constant is so small that very little nitrogen and oxygen will react to form nitrogen monoxide. After calculating the equilibrium partial pressure of nitrogen monoxide, test this assumption by comparing the partial pressure of nitrogen monoxide with that of nitrogen and oxygen. nitrogen (equilibrium) ( x) atm oxygen (equilibrium) ( x) atm NO (equilibrium) x atm ( NO ) K p.5 x 10-1 ( N )( O ) ( x) ( x)( 0.10 x) ( x) ( 0.780)( 0.10 ).5 x x 10-1 Assume x is small because K is small. x 1.66 x (unrounded) Based on the small amount of nitrogen monoxide formed, the assumption that the partial pressures of nitrogen and oxygen change to an insignificant degree holds. nitrogen (equilibrium) ( x ) atm atm N oxygen (equilibrium) ( x ) atm 0.10 atm O NO (equilibrium) (1.66 x ) atm.669 x x atm NO b) The total pressure is the sum of the three partial pressures: atm atm +.67 x atm atm c) K p K c (RT) n K c K p.5 x 10-1 because there is no net increase or decrease in the number of moles of gas in the course of the reaction a) K p K c (RT) n n - ( + 1) -1 K c K p / (RT) n ( 1. x 10 ) K c x x ( 0.081)( 57) 17-

25 b) H rxn [ H f (products)] - [ H f (reactants)] H rxn H f [NO (g)] - H f [NO(g)] - H f [O (g)] mol (. kj/mol) - mol (90.9 kj/mol) - (0.0) kj c) ln K K Hrxn R T T x kj J ln x J/mol K T 57 K 1 kj T x 10 K a) Use the equation K p K c (RT) n n 0 K p K c (RT) K c 1.6 x 10 - b) The equilibrium constant for the reverse reaction is simply the inverse of the equilibrium constant for the forward reaction: K c ' 1 / K c 1 / (1.6 x 10 - ) c) H rxn [ H f (products)] - [ H f (reactants)] H rxn H f [H (g)] + H f [I (g)] - H f [HI(g)] 1 mol (0 kj/mol) - 1 mol (0 kj/mol) - mol (5.9 kj/mol) kj d) ln K K Hrxn R T T1.0 x 10 Hrxn 1 1 ln 1.6 x J/mol K 79 K 98 K H rxn x x 10 J/mol C 5 H 11 OH + CH COOH CH COOC 5 H 11 + H O Removing water should help to increase the yield of banana oil. Both isopentyl alcohol and acetic acid are more soluble in water than isopentyl acetate. Thus, removing water will increase the concentration of both reactants and cause a shift in equilibrium toward the products a) The enzyme that inhibits F is the one that catalyzes the reaction that produces F. The enzyme is number 5 that catalyzed the reaction from E to F. b) Enzyme 8 is inhibited by I. c) If F inhibited enzyme 1, then neither branch of the reaction would take place once enough F was produced. d) If F inhibited enzyme 6, then the second branch would not take place when enough F was made The K is very small, thus the reaction will shift to the right to reach equilibrium. To simplify the calculations, assume the equilibrium shifts entirely to the left, and then a little material reacts to reach equilibrium. Shifting entirely to the left gives [H S] 0.600, and [H ] [S ] 0 H S(g) H (g) + S (g) Initial M 0 M 0 M Change - x + x +x Equilibrium x + x +x 17-5

26 H K c S [ HS] 9.0 x 10-8 [ x] [ x] [ x] [ x] [ x] [ 0.600] 9.0 x x 10-8 Assume x is small compared to M. x.008 x 10 - (unrounded) assumption justified [H S] x (.008 x 10 - ) M H S [H ] x (.008 x 10 - ).016 x x 10 - M H [S ] x.008 x x 10 - M S a) K p ( ) 10 H.08 x 10-5 O H x x x 10 - atm O b) (1) Adding more Na SO (s) will decrease the ratio of hydrated form/anhydrous form merely because you are increasing the value of the denominator, not because the equilibrium shifts. () Reducing the container size will increase the concentration of the water vapor, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. () Adding more water vapor will increase the concentration of the water vapor, which will shift the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase. () Adding N gas will not change the partial pressure of the water vapor, so the ratio of hydrated form/anhydrous form will not change The reaction is: CO(g) + H O(g) CO (g) + H (g) a) Set up a table with the initial CO and initial H mol / 0.00 L M. CO H O CO H Initial M M 0 0 Change -x -x +x +x Equilibrium x x x x [CO] equilibrium x. x 10 - M [H O] x M [CO ] [H ] [ CO ] H K c [ CO][ HO ] [ 0.00][ 0.00] b) M total [CO] + [H O] + [CO ] + [H ] (0.00 M) + (0.00 M) + ( M) + ( M) M n total M total V ( mol/l) (0.00 L) mol total L atm ( mol) (( ) K) mol K + total n total RT / V atm ( 0.00 L) c) Initially, an equal number of moles must be added mol CO d) Set up a table with the initial concentrations equal to the final concentrations from part (a), and then add mol CO / 0.00 L M to compensate for the added CO. CO H O CO H Initial 0.00 M 0.00 M M M Added CO M Change -x -x +x +x Equilibrium x x x x 17-6