CHAPTER 14 CHEMICAL EQUILIBRIUM

Transcription

1 CHATER 14 CHEMICAL EQUILIBRIUM roblem Categories Biological: Conceptual: 14.1, 14., 14.9, 14.5, 14.54, 14.55, 14.56, 14.57, 14.58, 14.59, 14.60, 14.61, 14.6, 14.66, 14.67, 14.68, 14.69, 14.81, 14.91, , , Descriptive: 14.85, 14.90, Environmental: 14.65, Industrial: 14.79, 14.95, 14.97, Difficulty Level Easy: 14.14, 14.15, 14.16, 14.17, 14.18, 14.19, 14.0, 14.1, 14., 14.9, 14.0, 14.6, 14.5, 14.54, 14.55, 14.56, 14.58, 14.59, 14.60, 14.6, 14.67, 14.77, 14.9, 14.94, 14.98, 14.99, Medium: 14.1, 14., 14.5, 14.6, 14.8, 14.1, 14., 14.5, 14.9, 14.40, 14.41, 14.4, 14.4, 14.44, 14.45, 14.48, 14.57, 14.61, 14.6, 14.64, 14.65, 14.66, 14.68, 14.69, 14.70, 14.74, 14.75, 14.78, 14.81, 14.8, 14.84, 14.85, 14.87, 14.90, 14.9, 14.9, 14.95, 14.97, , , , 14.11, Difficult: 14.4, 14.7, 14.46, 14.47, 14.71, 14.7, 14.7, 14.76, 14.79, 14.80, 14.8, 14.86, 14.88, 14.89, 14.91, 14.96, , 14.10, , , , , , , 14.11, , , , c [B] [A] (1) With c 10, products are favored at equilibrium. Because the coefficients for both A and B are one, we epect the concentration of B to be 10 times that of A at equilibrium. Choice (a) is the best choice with 10 B molecules and 1 A molecule. () With c 0.10, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we epect the concentration of A to be 10 times that of B at equilibrium. Choice (d) is the best choice with 10 A molecules and 1 B molecule. You can calculate c in each case without knowing the volume of the container because the mole ratio between A and B is the same. Volume will cancel from the c epression. Only moles of each component are needed to calculate c Note that we are comparing similar reactions at equilibrium two reactants producing one product, all with coefficients of one in the balanced equation. (a) The reaction, A + C AC has the largest equilibrium constant. Of the three diagrams, there is the most product present at equilibrium. The reaction, A + D AD has the smallest equilibrium constant. Of the three diagrams, there is the least amount of product present at equilibrium When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 1 1 '

2 CHATER 14: CHEMICAL EQUILIBRIUM The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations into the equilibrium constant epression to calculate c. Step 1: Calculate the concentrations of the components in units of mol/l. The molarities can be calculated by simply dividing the number of moles by the volume of the flask..50 mol [H ] 0.08 M 1.0 L mol 6 [S ] M 1.0 L 8.70 mol [HS] 0.75 M 1.0 L Step : Once the molarities are known, c can be found by substituting the molarities into the equilibrium constant epression. [HS] (0.75) 7 c [H ] [S ] (0.08) ( ) If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same? Using Equation (14.5) of the tet: c (0.081 T) Δn where, Δn 1 and T (17 + 7) 1546 (.4 10 )( ) Strategy: The relationship between c and is given by Equation (14.5) of the tet. What is the change in the number of moles of gases from reactant to product? Recall that Δn moles of gaseous products moles of gaseous reactants What unit of temperature should we use? Solution: The relationship between c and is given by Equation (14.5) of the tet. c (0.081 T) Δn Rearrange the equation relating and c, solving for c. c (0.081 ) Δ n T Because T 6 and Δn 1, we have: c.5 10 Δn (0.081 ) (0.081)(6 ) T

3 88 CHATER 14: CHEMICAL EQUILIBRIUM We can write the equilibrium constant epression from the balanced equation and substitute in the pressures. Do we need to know the temperature? (0.050) N O (0.15)(0.) The equilibrium constant epressions are: (a) [NH ] c [N ][H ] [NH ] c 1 [N ] [H ] Substituting the given equilibrium concentration gives: (a) (0.5) (0.11)(1.91) c 0.08 (0.5) c (0.11) (1.91) Is there a relationship between the c values from parts (a) and? 14.1 The equilibrium constant epression for the two forms of the equation are: [I] ' [I ] c and c [I ] [I] The relationship between the two equilibrium constants is 1 1 ' 4 c c.8 10 can be found as shown below. c '(0.081 T) Δn ( )( ) Because pure solids do not enter into an equilibrium constant epression, we can calculate directly from the pressure that is due solely to CO (g). CO Now, we can convert to c using the following equation. c (0.081 T) Δn c (0.081 T ) Δ n c (1 0) ( )

4 CHATER 14: CHEMICAL EQUILIBRIUM We substitute the given pressures into the reaction quotient epression. Q Cl Cl Cl 5 (0.)(0.111) (0.177) The calculated value of Q is less than for this system. The system will change in a way to increase Q until it is equal to. To achieve this, the pressures of Cl and Cl must increase, and the pressure of Cl 5 must decrease. Could you actually determine the final pressure of each gas? 14.4 Strategy: Because they are constant quantities, the concentrations of solids and liquids do not appear in the equilibrium constant epressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH and CO. Note that for every 1 atm of CO produced, atm of NH will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calculate the partial pressures of NH and CO at equilibrium. Solution: The equilibrium constant epression for the reaction is NH CO The total pressure in the flask (0.6 atm) is a sum of the partial pressures of NH and CO. T NH + CO 0.6 atm Let the partial pressure of CO. From the stoichiometry of the balanced equation, you should find that NH CO. Therefore, the partial pressure of NH. Substituting into the equation for total pressure gives: T NH + CO atm 0.11 atm CO NH 0.4 atm Substitute the equilibrium pressures into the equilibrium constant epression to solve for. NH CO (0.4) (0.11) Of the original 1.05 moles of Br, 1.0% has dissociated. The amount of Br dissociated in molar concentration is: 1.05 mol [Br ] M L Setting up a table: Br (g) Br(g) Initial (M): 1.05 mol L Change (M): (0.019) Equilibrium (M):

5 90 CHATER 14: CHEMICAL EQUILIBRIUM [Br] (0.058) c [Br ] If the CO pressure at equilibrium is atm, the balanced equation requires the chlorine pressure to have the same value. The initial pressure of phosgene gas can be found from the ideal gas equation. nrt ( mol)(0.081 L atm/mol )(800 ) V (1.50 L) 1.1 atm Since there is a 1:1 mole ratio between phosgene and CO, the partial pressure of CO formed (0.497 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is: CO(g) + Cl (g) COCl (g) Initial (atm): Change (atm): Equilibrium (atm): The value of is then found by substitution. COCl 0.81 (0.497). CO Cl 14.7 Let be the initial pressure of Br. Using the balanced equation, we can write epressions for the partial pressures at equilibrium. Br (1 0.4) Br 0.17 The sum of these is the total pressure atm 0.1 atm The equilibrium pressures are then Br 0.66(0.1) 0.14 atm 0.4(0.1) atm Br 0.17(0.1) 0.06 atm We find by substitution. Br Br ( ) (0.071) (0.06) ( ) (0.14) The relationship between and c is given by c (RT) Δn

6 CHATER 14: CHEMICAL EQUILIBRIUM 91 We find c (for this system Δn +1) c.8 10 Δn 1 ( RT ) RT ( ) In this problem, you are asked to calculate c. Step 1: Calculate the initial concentration of Cl. We carry an etra significant figure throughout this calculation to minimize rounding errors..50 mol [Cl] M 1.50 L Step : Let's represent the change in concentration of Cl as. Setting up a table: Cl(g) (g) + Cl (g) Initial (M): Change (M): + + Equilibrium (M): If 8.0 percent of the Cl has dissociated at equilibrium, the amount reacted is: (0.80)(1.667 M) M In the table above, we have represented the amount of Cl that reacts as. Therefore, M 0.4 M The equilibrium concentrations of Cl,, and Cl are: [Cl] (1.67 )M ( )M 1.00 M [] M [Cl ] 0.4 M Step : The equilibrium constant c can be calculated by substituting the above concentrations into the equilibrium constant epression. [] [Cl ] (0.4668) (0.4) c 0.05 [Cl] (1.00) 14.9 The target equation is the sum of the first two. H S H + + HS HS H + + S H S H + + S Since this is the case, the equilibrium constant for the combined reaction is the product of the constants for the component reactions (Section 14. of the tet). The equilibrium constant is therefore: c c ' c "

7 9 CHATER 14: CHEMICAL EQUILIBRIUM What happens in the special case when the two component reactions are the same? Can you generalize this relationship to adding more than two reactions? What happens if one takes the difference between two reactions? 14.0 ' " 14.1 Given: ( )( ) ' CO CO '' COCl COCl For the overall reaction: COCl ' " 14 ( ) (1. 10 )( ) CO Cl To obtain SO as a reactant in the final equation, we must reverse the first equation and multiply by two. For the equilibrium, SO (g) S(s) + O (g) ''' c ' 5 c Now we can add the above equation to the second equation to obtain the final equation. Since we add the two equations, the equilibrium constant is the product of the equilibrium constants for the two reactions. SO (g) S(s) + O (g) S(s) + O (g) SO (g) SO (g) + O (g) SO (g) ''' 106 c '' 18 c ''' '' c c c (a) Assuming the self-ionization of water occurs by a single elementary step mechanism, the equilibrium constant is just the ratio of the forward and reverse rate constants. r 1 5 kf k k k The product can be written as: [H + ][OH ] [H O] What is [H O]? It is the concentration of pure water. One liter of water has a mass of 1000 g (density 1.00 g/ml). The number of moles of H O is: 1mol 1000 g 55.5 mol 18.0 g

8 CHATER 14: CHEMICAL EQUILIBRIUM 9 The concentration of water is 55.5 mol/1.00 L or 55.5 M. The product is: [H + ][OH ] ( )(55.5) We assume the concentration of hydrogen ion and hydroide ion are equal. [H + ] [OH ] ( ) 1/ M 14.6 At equilibrium, the value of c is equal to the ratio of the forward rate constant to the rate constant for the reverse reaction. kf kf c k k f (1.6)( ) 0.64 The forward reaction is third order, so the units of k f must be: rate k f [A] [B] k r rate M/s 1/ M s (concentration) M f k f 0.64 /M s 14.9 Given: SO SO O Initially, the total pressure is ( ) atm or 1.11 atm. As the reaction progresses from left to right toward equilibrium there will be a decrease in the number of moles of molecules present. (Note that moles of SO react with 1 mole of O to produce moles of SO, or, at constant pressure, three atmospheres of reactants forms two atmospheres of products.) Since pressure is directly proportional to the number of molecules present, at equilibrium the total pressure will be less than 1.11 atm Strategy: We are given the initial concentrations of the gases, so we can calculate the reaction quotient (Q c ). How does a comparison of Q c with c enable us to determine if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium? Solution: Recall that for a system to be at equilibrium, Q c c. Substitute the given concentrations into the equation for the reaction quotient to calculate Q c. [NH 0 ] [0.48] Q c [N ] 0[H ] 0 [0.60][0.76] 0.87 Comparing Q c to c, we find that Q c < c (0.87 < 1.). The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. Therefore, [NH ] will increase and [N ] and [H ] will decrease at equilibrium.

9 94 CHATER 14: CHEMICAL EQUILIBRIUM The balanced equation shows that one mole of carbon monoide will combine with one mole of water to form hydrogen and carbon dioide. Let be the depletion in the concentration of either CO or H O at equilibrium (why can serve to represent either quantity?). The equilibrium concentration of hydrogen must then also be equal to. The changes are summarized as shown in the table. H + CO H O + CO Initial (M): Change (M): + + Equilibrium (M): (0.000 ) (0.000 ) [H The equilibrium constant is: O][CO] c [H ][CO ] 0.54 Substituting, (0.000 ) 0.54 Taking the square root of both sides, we obtain: (0.000 ) M The number of moles of H formed is: mol/l 10.0 L 0.17 mol H 14.4 Strategy: The equilibrium constant is given, and we start with pure. The partial pressure of O at equilibrium is 0.5 atm. From the stoichiometry of the reaction, we can determine the partial pressure of at equilibrium. nowing and the partial pressures of both O and, we can solve for the partial pressure of. Solution: Since the reaction started with only pure, the equilibrium concentration of must be twice the equilibrium concentration of O, due to the :1 mole ratio of the balanced equation. Therefore, the equilibrium partial pressure of is ( 0.5 atm) 0.50 atm. We can find the equilibrium pressure by rearranging the equilibrium constant epression, then substituting in the known values. O O (0.50) (0.5) atm 14.4 Notice that the balanced equation requires that for every two moles of HBr consumed, one mole of H and one mole of Br must be formed. Let be the depletion in the concentration of HBr at equilibrium. The equilibrium concentrations of H and Br must therefore each be. The changes are shown in the table. H + Br HBr Initial (M): Change (M): + + Equilibrium (M): (0.67 )

10 CHATER 14: CHEMICAL EQUILIBRIUM 95 The equilibrium constant relationship is given by: c [HBr] [H ][Br ] Substitution of the equilibrium concentration epressions gives Taking the square root of both sides we obtain: (0.67 ) 6 c The equilibrium concentrations are: [H ] [Br ] M [HBr] 0.67 ( ) 0.67 M If the depletion in the concentration of HBr at equilibrium were defined as, rather than, what would be the appropriate epressions for the equilibrium concentrations of H and Br? Should the final answers be different in this case? Strategy: We are given the initial amount of I (in moles) in a vessel of known volume (in liters), so we can calculate its molar concentration. Because initially no I atoms are present, the system could not be at equilibrium. Therefore, some I will dissociate to form I atoms until equilibrium is established. Solution: We follow the procedure outlined in Section 14.4 of the tet to calculate the equilibrium concentrations. Step 1: The initial concentration of I is mol/.0 L M. The stoichiometry of the problem shows 1 mole of I dissociating to moles of I atoms. Let be the amount (in mol/l) of I dissociated. It follows that the equilibrium concentration of I atoms must be. We summarize the changes in concentrations as follows: I (g) I(g) Initial (M): Change (M): + Equilibrium (M): ( ) Step : Write the equilibrium constant epression in terms of the equilibrium concentrations. nowing the value of the equilibrium constant, solve for. c [I] ( ) [I ] ( ) 4 + ( ) ( ) 0

11 96 CHATER 14: CHEMICAL EQUILIBRIUM The above equation is a quadratic equation of the form a + b + c 0. The solution for a quadratic equation is b ± b 4ac a Here, we have a 4, b , and c Substituting into the above equation, ( ) ± ( ) 4(4)( ) (4) 5 ( ) ± ( ) M or M The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer. Step : Having solved for, calculate the equilibrium concentrations of all species. [I] ()( M) M [I ] ( ) [ ( )] M M Tip: We could have simplified this problem by assuming that was small compared to We could then assume that By making this assumption, we could have avoided solving a quadratic equation Since equilibrium pressures are desired, we calculate. c (0.081 T) Δn ( )( ) COCl (g) CO(g) + Cl (g) Initial (atm): Change (atm): + + Equilibrium (atm): (0.760 ) (0.760 ) atm At equilibrium: COCl ( )atm atm CO 0.5 atm Cl 0.5 atm

12 CHATER 14: CHEMICAL EQUILIBRIUM (a) The equilibrium constant, c, can be found by simple substitution. [HO][CO] (0.040)(0.050) c 0.5 [CO ][H ] (0.086)(0.045) The magnitude of the reaction quotient Q c for the system after the concentration of CO becomes 0.50 mol/l, but before equilibrium is reestablished, is: (0.040)(0.050) Qc (0.50)(0.045) The value of Q c is smaller than c ; therefore, the system will shift to the right, increasing the concentrations of CO and H O and decreasing the concentrations of CO and H. Let be the depletion in the concentration of CO at equilibrium. The stoichiometry of the balanced equation then requires that the decrease in the concentration of H must also be, and that the concentration increases of CO and H O be equal to as well. The changes in the original concentrations are shown in the table. CO + H CO + H O Initial (M): Change (M): + + Equilibrium (M): (0.50 ) (0.045 ) ( ) ( ) The equilibrium constant epression is: c [HO][CO] ( )( ) [CO ][H ] (0.50 )(0.045 ) ( ) ( ) 0 The positive root of the equation is The equilibrium concentrations are: [CO ] ( ) M 0.48 M [H ] ( ) M 0.00 M [CO] ( ) M M [H O] ( ) M M The equilibrium constant epression for the system is: ( CO) CO The total pressure can be epressed as: + total CO CO If we let the partial pressure of CO be, then the partial pressure of CO is: CO total (4.50 )atm

13 98 CHATER 14: CHEMICAL EQUILIBRIUM Substitution gives the equation: ( CO) CO (4.50 ) 1.5 This can be rearranged to the quadratic: The solutions are 1.96 and.48; only the positive result has physical significance (why?). The equilibrium pressures are CO 1.96 atm CO ( ).54 atm The initial concentrations are [H ] 0.80 mol/5.0 L 0.16 M and [CO ] 0.80 mol/5.0 L 0.16 M. H (g) + CO (g) H O(g) + CO(g) Initial (M): Change (M): + + Equilibrium (M): [HO][CO] c 4. [H ][CO ] (0.16 ) Taking the square root of both sides, we obtain: M The equilibrium concentrations are: [H ] [CO ] ( ) M 0.05 M [H O] [CO] 0.11 M 14.5 (a) Addition of more Cl (g) (a reactant) would shift the position of equilibrium to the right. (c) Removal of SO Cl (g) (a product) would shift the position of equilibrium to the right. Removal of SO (g) (a reactant) would shift the position of equilibrium to the left (a) Removal of CO (g) from the system would shift the position of equilibrium to the right. (c) Addition of more solid Na CO would have no effect. [Na CO ] does not appear in the equilibrium constant epression. Removal of some of the solid NaHCO would have no effect. Same reason as.

14 CHATER 14: CHEMICAL EQUILIBRIUM (a) This reaction is endothermic. (Why?) According to Section 14.5, an increase in temperature favors an endothermic reaction, so the equilibrium constant should become larger. (c) This reaction is eothermic. Such reactions are favored by decreases in temperature. The magnitude of c should decrease. In this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the equilibrium constant Strategy: A change in pressure can affect only the volume of a gas, but not that of a solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier's principle, the system will adjust to partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: V nrt so n. Solution: (a) Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases because liquids and solids are virtually incompressible. ressure change should have no effect on this system. (c) (d) (e) Same situation as (a). Only the product is in the gas phase. An increase in pressure should favor the reaction that decreases the total number of moles of gas. The equilibrium should shift to the left, that is, the amount of B should decrease and that of A should increase. In this equation there are equal moles of gaseous reactants and products. A shift in either direction will have no effect on the total number of moles of gas present. There will be no change when the pressure is increased. A shift in the direction of the reverse reaction (left) will have the result of decreasing the total number of moles of gas present (a) A pressure increase will favor the reaction (forward or reverse?) that decreases the total number of moles of gas. The equilibrium should shift to the right, i.e., more I will be produced at the epense of I. (c) If the concentration of I is suddenly altered, the system is no longer at equilibrium. Evaluating the magnitude of the reaction quotient Q c allows us to predict the direction of the resulting equilibrium shift. The reaction quotient for this system is: [I ] 0 Q c [I] 0 Increasing the concentration of I will increase Q c. The equilibrium will be reestablished in such a way that Q c is again equal to the equilibrium constant. More I will form. The system shifts to the left to establish equilibrium. The forward reaction is eothermic. A decrease in temperature at constant volume will shift the system to the right to reestablish equilibrium Strategy: (a) What does the sign of ΔH indicate about the heat change (endothermic or eothermic) for the forward reaction? The stress is the addition of Cl gas. How will the system adjust to partially offset the stress? (c) The stress is the removal of Cl gas. How will the system adjust to partially offset the stress? (d) The stress is an increase in pressure. The system will adjust to decrease the pressure. Remember, pressure is directly proportional to moles of gas. (e) What is the function of a catalyst? How does it affect a reacting system not at equilibrium? at equilibrium?

15 400 CHATER 14: CHEMICAL EQUILIBRIUM Solution: (a) The stress applied is the heat added to the system. Note that the reaction is endothermic (ΔH > 0). Endothermic reactions absorb heat from the surroundings; therefore, we can think of heat as a reactant. heat + Cl 5 (g) Cl (g) + Cl (g) The system will adjust to remove some of the added heat by undergoing a decomposition reaction (from left to right) (c) (d) (e) The stress is the addition of Cl gas. The system will shift in the direction to remove some of the added Cl. The system shifts from right to left until equilibrium is reestablished. The stress is the removal of Cl gas. The system will shift to replace some of the Cl that was removed. The system shifts from left to right until equilibrium is reestablished. The stress applied is an increase in pressure. The system will adjust to remove the stress by decreasing the pressure. Recall that pressure is directly proportional to the number of moles of gas. In the balanced equation we see 1 mole of gas on the reactants side and moles of gas on the products side. The pressure can be decreased by shifting to the side with the fewer moles of gas. The system will shift from right to left to reestablish equilibrium. The function of a catalyst is to increase the rate of a reaction. If a catalyst is added to the reacting system not at equilibrium, the system will reach equilibrium faster than if left undisturbed. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of reactant and product, or the equilibrium constant (a) Increasing the temperature favors the endothermic reaction so that the concentrations of SO and O will increase while that of SO will decrease. Increasing the pressure favors the reaction that decreases the number of moles of gas. The concentration of SO will increase. (c) (d) (e) Increasing the concentration of SO will lead to an increase in the concentration of SO and a decrease in the concentration of O. A catalyst has no effect on the position of equilibrium. Adding an inert gas at constant volume has no effect on the position of equilibrium There will be no change in the pressures. A catalyst has no effect on the position of the equilibrium (a) If helium gas is added to the system without changing the pressure or the temperature, the volume of the container must necessarily be increased. This will decrease the partial pressures of all the reactants and products. A pressure decrease will favor the reaction that increases the number of moles of gas. The position of equilibrium will shift to the left. If the volume remains unchanged, the partial pressures of all the reactants and products will remain the same. The reaction quotient Q c will still equal the equilibrium constant, and there will be no change in the position of equilibrium For this system, [CO ]. This means that to remain at equilibrium, the pressure of carbon dioide must stay at a fied value as long as the temperature remains the same. (a) If the volume is increased, the pressure of CO will drop (Boyle's law, pressure and volume are inversely proportional). Some CaCO will break down to form more CO and CaO. (Shift right)

16 CHATER 14: CHEMICAL EQUILIBRIUM 401 (c) (d) (e) Assuming that the amount of added solid CaO is not so large that the volume of the system is altered significantly, there should be no change at all. If a huge amount of CaO were added, this would have the effect of reducing the volume of the container. What would happen then? Assuming that the amount of CaCO removed doesn't alter the container volume significantly, there should be no change. Removing a huge amount of CaCO will have the effect of increasing the container volume. The result in that case will be the same as in part (a). The pressure of CO will be greater and will eceed the value of. Some CO will combine with CaO to form more CaCO. (Shift left) Carbon dioide combines with aqueous NaOH according to the equation CO (g) + NaOH(aq) NaHCO (aq) This will have the effect of reducing the CO pressure and causing more CaCO to break down to CO and CaO. (Shift right) (f) Carbon dioide does not react with hydrochloric acid, but CaCO does. CaCO (s) + HCl(aq) CaCl (aq) + CO (g) + H O(l) (g) The CO produced by the action of the acid will combine with CaO as discussed in (d) above. (Shift left) This is a decomposition reaction. Decomposition reactions are endothermic. Increasing the temperature will favor this reaction and produce more CO and CaO. (Shift right) 14.6 (i) The temperature of the system is not given. (ii) It is not stated whether the equilibrium constant is or c (would they be different for this reaction?). (iii) A balanced equation is not given. (iv) The phases of the reactants and products are not given (a) Since the total pressure is 1.00 atm, the sum of the partial pressures of and Cl is 1.00 atm partial pressure of Cl 1.00 atm 0.64 atm 0.6 atm The stoichiometry of the reaction requires that the partial pressure of be twice that of Cl. Hence, the partial pressure of is 0.4 atm and the partial pressure of Cl is 0.1 atm. The equilibrium constant is found by substituting the partial pressures calculated in part (a) into the equilibrium constant epression. Cl Cl (0.64) (0.4) (0.1) (a) N O (.0)(0.01) atm (0.78)(0.1) atm

17 40 CHATER 14: CHEMICAL EQUILIBRIUM (c) (d) Since increases with temperature, it is endothermic. Lightening. The electrical energy promotes the endothermic reaction The equilibrium epression for this system is given by: CO H O (a) In a closed vessel the decomposition will stop when the product of the partial pressures of CO and H O equals. Adding more sodium bicarbonate will have no effect. In an open vessel, CO (g) and H O(g) will escape from the vessel, and the partial pressures of CO and H O will never become large enough for their product to equal. Therefore, equilibrium will never be established. Adding more sodium bicarbonate will result in the production of more CO and H O The relevant relationships are: c [B] [A] and B A c (0.081 T) Δn c (0.081 T) Δn +1 We set up a table for the calculated values of c and. T ( C) c 00 (0.84) 56.9 (0.015) 56.9( ) (0.764) (0.171) (0.74) (0.50).41.41( ) ( ) 116 Since c (and ) decrease with temperature, the reaction is eothermic (a) The equation that relates and c is: c (0.081 T) Δn For this reaction, Δn c 8 10 (0.081 T ) ( ) Because of a very large activation energy, the reaction of hydrogen with oygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a miture of these gases results in the eplosive formation of water.

18 CHATER 14: CHEMICAL EQUILIBRIUM Using data from Appendi we calculate the enthalpy change for the reaction. Δ H ΔHf(Cl) ΔHf() Δ Hf(Cl ) (51.7 kj/mol) (90.4 kj/mol) (0) 77.4 kj/mol The enthalpy change is negative, so the reaction is eothermic. The formation of Cl will be favored by low temperature. A pressure increase favors the reaction forming fewer moles of gas. The formation of Cl will be favored by high pressure (a) Calculate the value of by substituting the equilibrium partial pressures into the equilibrium constant epression. B (0.60) (0.60) 1.7 A The total pressure is the sum of the partial pressures for the two gaseous components, A and B. We can write: A + B 1.5 atm and B 1.5 A Substituting into the epression for gives: (1.5 A ) 1.7 A 1.7A + A Solving the quadratic equation, we obtain: and by difference, A 0.69 atm B 0.81 atm Check that substituting these equilibrium concentrations into the equilibrium constant epression gives the equilibrium constant calculated in part (a). B A (0.69) (a) The balanced equation shows that equal amounts of ammonia and hydrogen sulfide are formed in this decomposition. The partial pressures of these gases must just be half the total pressure, i.e., 0.55 atm. The value of is NH H S (0.55) 0.16 We find the number of moles of ammonia (or hydrogen sulfide) and ammonium hydrogen sulfide. nnh V (0.55 atm)(4.000 L) mol RT (0.081 L atm/ mol)(97 ) n NH4HS 1mol g mol (before decomposition) 51.1 g

19 404 CHATER 14: CHEMICAL EQUILIBRIUM From the balanced equation the percent decomposed is mol 100% mol 48.% (c) If the temperature does not change, has the same value. The total pressure will still be atm at equilibrium. In other words the amounts of ammonia and hydrogen sulfide will be twice as great, and the amount of solid ammonium hydrogen sulfide will be: [0.105 (0.058)]mol mol NH 4 HS 14.7 Total number of moles of gas is: mol of gas You can calculate the partial pressure of each gaseous component from the mole fraction and the total pressure Χ T 0.0 atm atm Χ 0.0 atm atm 1.0 O O T 0.96 Χ 0.0 atm 0.19 atm 1.0 T Calculate by substituting the partial pressures into the equilibrium constant epression. O (0.19) (0.0078) (0.009) Since the reactant is a solid, we can write: NH ( ) CO The total pressure is the sum of the ammonia and carbon dioide pressures. + total NH CO From the stoichiometry, Therefore: NH CO atm total CO CO CO CO NH atm 0.1 atm Substituting into the equilibrium epression: (0.1) (0.106)

20 CHATER 14: CHEMICAL EQUILIBRIUM Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1 L. H + Cl HCl Initial (M): Change (M): + + Equilibrium (M): ( ) (.59 ) Substitute the equilibrium concentrations into the equilibrium constant epression, then solve for. Since Δn 0, c. Solving the quadratic equation, [HCl] (.59 ) c 19 [H ][Cl ] ( ) 0.10 Having solved for, calculate the equilibrium concentrations of all species. [H ] 0.57 M [Cl ] 0.10 M [HCl].9 M Since we assumed that the vessel had a volume of 1 L, the above molarities also correspond to the number of moles of each component. From the mole fraction of each component and the total pressure, we can calculate the partial pressure of each component. Total number of moles mol 0.57 H atm 0.10 Cl atm HCl atm Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentrations. The initial concentration of I (g) is mol/0.48 L M. The amount of I that dissociates is (0.05)(0.115 M) M. We carry etra significant figures throughout this calculation to minimize rounding errors. I I Initial (M): Change (M): ()(0.0085) Equilibrium (M): Substitute the equilibrium concentrations into the equilibrium constant epression to solve for c. [I] ( ) 4 c [I ] c (0.081 T) Δn ( )( )

21 406 CHATER 14: CHEMICAL EQUILIBRIUM This is a difficult problem. Epress the equilibrium number of moles in terms of the initial moles and the change in number of moles (). Net, calculate the mole fraction of each component. Using the mole fraction, you should come up with a relationship between partial pressure and total pressure for each component. Substitute the partial pressures into the equilibrium constant epression to solve for the total pressure, T. The reaction is: N + H NH Initial (mol): 1 0 Change (mol): Equilibrium (mol): (1 ) ( ) Mole fraction of NH mol of NH total number of moles Χ NH (1 ) + ( ) mol Substituting into the following mole fraction equations, the mole fractions of N and H can be calculated. Χ Χ N H (0.5) (0.5) 4 4 (0.5) The partial pressures of each component are equal to the mole fraction multiplied by the total pressure. NH 0.1 T N 0.0 T H 0.59 Substitute the partial pressures above (in terms of T ) into the equilibrium constant epression, and solve for T. NH H N 4 (0.1) T (0.59 T) (0.0 T) T T atm T For the balanced equation: [H ] [S ] c [HS] [HS] [S ] c (.5 10 ).4 10 M [H ]

22 CHATER 14: CHEMICAL EQUILIBRIUM We carry an additional significant figure throughout this calculation to minimize rounding errors. The initial molarity of SO Cl is: 1molSOCl 6.75 g SO Cl 15.0 g SOCl [SOCl ] M.00 L The concentration of SO at equilibrium is: mol [SO ] M.00 L Since there is a 1:1 mole ratio between SO and SO Cl, the concentration of SO at equilibrium ( M) equals the concentration of SO Cl reacted. The concentrations of SO Cl and Cl at equilibrium are: SO Cl (g) SO (g) + Cl (g) Initial (M): Change (M): Equilibrium (M): Substitute the equilibrium concentrations into the equilibrium constant epression to calculate c. [SO ][Cl ] (0.0175)(0.0175) c [SOCl ] ( ) For a 100% yield,.00 moles of SO would be formed (why?). An 80% yield means.00 moles (0.80) 1.60 moles SO is formed. The amount of SO remaining at equilibrium ( )mol 0.40 mol The amount of O reacted 1 (amount of SO reacted) ( )mol 0.80 mol The amount of O remaining at equilibrium ( )mol 1.0 mol Total moles at equilibrium moles SO + moles O + moles SO ( )mol.0 moles SO total total O total total SO total total SO SO O 0.1 (0.500 total) (0.15 total) (0.75 total) total 8 atm

23 408 CHATER 14: CHEMICAL EQUILIBRIUM I (g) I(g) Assuming 1 mole of I is present originally and α moles reacts, at equilibrium: [I ] 1 α, [I] α. The total number of moles present in the system (1 α) + α 1 + α. From roblem (a) in the tet, we know that is equal to: 4α (1) 1 α If there were no dissociation, then the pressure would be: 1mol L atm 1.00 g (147 ) 5.8 g mol nrt V L 0.95 atm observed pressure 1.51 atm 1 + α calculated pressure 0.95 atm 1 α Substituting in equation (1) above: 4 α (4)(0.584) α 1 (0.584) anting decreases the concentration of CO because CO is ehaled during respiration. This decreases the concentration of carbonate ions, shifting the equilibrium to the left. Less CaCO is produced. Two possible solutions would be either to cool the chickens' environment or to feed them carbonated water According to the ideal gas law, pressure is directly proportional to the concentration of a gas in mol/l if the reaction is at constant volume and temperature. Therefore, pressure may be used as a concentration unit. The reaction is: N + H NH Initial (atm): Change (atm): + Equilibrium (atm): (0.86 ) (0.7 ) NH H N ( ) (0.7 ) (0.86 ) At this point, we need to make two assumptions that is very small compared to 0.7 and that is very small compared to Hence, and ( ) (0.7) (0.86)

24 CHATER 14: CHEMICAL EQUILIBRIUM 409 Solving for atm The equilibrium pressures are: N H [0.86 (.0 10 )]atm atm [0.7 ()(.0 10 )]atm 0.66 atm NH ()(.0 10 atm) atm Was the assumption valid that we made above? Typically, the assumption is considered valid if is less than 5 percent of the number that we said it was very small compared to. Is this the case? 14.8 (a) The sum of the mole fractions must equal one. ΧCO + Χ CO 1 and Χ 1 Χ CO According to the hint, the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass. (Χ CO 8.01 g) + [(1 Χ CO ) g] 5 g Solving, X CO 0.56 and Χ CO 0.44 CO Solving for the pressures total CO + CO 11 atm CO Χ CO total (0.56)(11 atm) 6. atm CO Χ CO total (0.44)(11 atm) 4.8 atm CO (6.) CO (a) The equation is: fructose glucose Initial (M): Change (M): Equilibrium (M): Calculating the equilibrium constant, [glucose] 0.11 c 1.16 [fructose] 0.11 ercent converted amount of fructose converted 100% original amount of fructose % 5.7% If you started with radioactive iodine in the solid phase, then you should fine radioactive iodine in the vapor phase at equilibrium. Conversely, if you started with radioactive iodine in the vapor phase, you should find radioactive iodine in the solid phase. Both of these observations indicate a dynamic equilibrium between solid and vapor phase.

25 410 CHATER 14: CHEMICAL EQUILIBRIUM (a) There is only one gas phase component, O. The equilibrium constant is simply O 0.49 atm From the ideal gas equation, we can calculate the moles of O produced by the decomposition of CuO. n V RT (0.49 atm)(.0 L) (0.081 L atm/ mol)(197 ) O mol O From the balanced equation, 4molCuO (9. 10 mol O ).7 10 mol CuO decomposed 1molO Fraction of CuO decomposed amount of CuO lost original amount of CuO.7 10 mol mol (c) (d) If a 1.0 mol sample were used, the pressure of oygen would still be the same (0.49 atm) and it would be due to the same quantity of O. Remember, a pure solid does not affect the equilibrium position. The moles of CuO lost would still be.7 10 mol. Thus the fraction decomposed would be: If the number of moles of CuO were less than.7 10 mol, the equilibrium could not be established because the pressure of O would be less than 0.49 atm. Therefore, the smallest number of moles of CuO needed to establish equilibrium must be slightly greater than.7 10 mol If there were 0.88 mole of CO initially and at equilibrium there were 0.11 moles, then ( ) moles 0.77 moles reacted. + CO + CO Initial (mol): Change (mol): Equilibrium (mol): ( ) Solving for the equilibrium constant: (0.77)(0.77) c 1.7 ( )(0.11) In the balanced equation there are equal number of moles of products and reactants; therefore, the volume of the container will not affect the calculation of c. We can solve for the equilibrium constant in terms of moles We first must find the initial concentrations of all the species in the system mol [H ] 0.40 L 0.98 M mol [I ] 0.40 L M mol [HI] 0.40 L 0.69 M

26 CHATER 14: CHEMICAL EQUILIBRIUM 411 Calculate the reaction quotient by substituting the initial concentrations into the appropriate equation. Q c 0 [HI] (0.69) [H ] [I ] (0.98)(0.410) We find that Q c is less than c. The equilibrium will shift to the right, decreasing the concentrations of H and I and increasing the concentration of HI. We set up the usual table. Let be the decrease in concentration of H and I. H + I HI Initial (M): Change (M): + Equilibrium (M): (0.98 ) (0.410 ) ( ) The equilibrium constant epression is: c [HI] ( ) [H ][I ] (0.98 )(0.410 ) 54. This becomes the quadratic equation The smaller root is 0.8 M. (The larger root is physically impossible.) Having solved for, calculate the equilibrium concentrations. [H ] ( ) M M [I ] ( ) M 0.18 M [HI] [ (0.8)] M 0.85 M Since we started with pure A, then any A that is lost forms equal amounts of B and C. Since the total pressure is, the pressure of B + C The pressure of B C 0.4. B C (0.4 )(0.4 ) A The gas cannot be (a) because the color became lighter with heating. Heating (a) to 150 C would produce some HBr, which is colorless and would lighten rather than darken the gas. The gas cannot be because Br doesn't dissociate into Br atoms at 150 C, so the color shouldn't change. The gas must be (c). From 5 C to 150 C, heating causes N O 4 to dissociate into, thus darkening the color ( is a brown gas). N O 4 (g) (g) Above 150 C, the breaks up into colorless and O. (g) (g) + O (g) An increase in pressure shifts the equilibrium back to the left, forming, thus darkening the gas again. (g) + O (g) (g)

27 41 CHATER 14: CHEMICAL EQUILIBRIUM Since the catalyst is eposed to the reacting system, it would catalyze the A B reaction. This shift would result in a decrease in the number of gas molecules, so the gas pressure decreases. The piston would be pushed down by the atmospheric pressure. When the cover is over the bo, the catalyst is no longer able to favor the forward reaction. To reestablish equilibrium, the B A step would dominate. This would increase the gas pressure so the piston rises and so on. Conclusion: Such a catalyst would result in a perpetual motion machine (the piston would move up and down forever) which can be used to do work without input of energy or net consumption of chemicals. Such a machine cannot eist Given the following: [NH ] c [N ][H ] 1. (a) Temperature must have units of elvin. c (0.081 T) Δn (1.)( ) ( 4) (c) Recalling that, Therefore, Since the equation is equivalent to forward 1 reverse ' 1 c N (g) + H (g) NH (g) 1 [N (g) + H (g) NH (g)] then, ' c for the reaction: 1 N (g) + H (g) NH (g) equals Thus, ( c) 1 for the reaction: N (g) + H (g) NH (g) 1 ' c ( c) (d) For in part : (0.8)( ) and for in part (c): (1.1)( ) (a) Color deepens increases (c) decreases (d) increases (e) unchanged

28 CHATER 14: CHEMICAL EQUILIBRIUM The vapor pressure of water is equivalent to saying the partial pressure of H O(g) HO p c Δn 1 (0.081 T ) ( ) otassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right We can calculate the average molar mass of the gaseous miture from the density. M drt Let M be the average molar mass of and N O 4. The above equation becomes: M drt (. g/l)(0.081 L atm/ mol)(47 ) 1. atm M 50.4 g/mol The average molar mass is equal to the sum of the molar masses of each component times the respective mole fractions. Setting this up, we can calculate the mole fraction of each component. M ΧM + Χ M 50.4 g/mol 4 4 Χ (46.01 g/mol) + (1 Χ )(9.01 g/mol) 50.4 g/mol Χ We can now calculate the partial pressure of from the mole fraction and the total pressure. Χ T (0.905)(1. atm) 1.18 atm We can calculate the partial pressure of N O 4 by difference. T 4 4 ( ) atm 0.1 atm Finally, we can calculate for the dissociation of N O atm 1 4 (1.) (a) Since both reactions are endothermic (ΔH is positive), according to Le Châtelier s principle the products would be favored at high temperatures. Indeed, the steam-reforming process is carried out at very high temperatures (between 800 C and 1000 C). It is interesting to note that in a plant that uses natural gas (methane) for both hydrogen generation and heating, about one-third of the gas is burned to maintain the high temperatures.

29 414 CHATER 14: CHEMICAL EQUILIBRIUM In each reaction there are more moles of products than reactants; therefore, we epect products to be favored at low pressures. In reality, the reactions are carried out at high pressures. The reason is that when the hydrogen gas produced is used captively (usually in the synthesis of ammonia), high pressure leads to higher yields of ammonia. (i) The relation between c and is given by Equation (14.5) of the tet: c (0.081 T) Δn Since Δn 4, we write: (18)( ) (ii) Let be the amount of CH 4 and H O (in atm) reacted. We write: CH 4 + H O CO + H Initial (atm): Change (atm): + + Equilibrium (atm): The equilibrium constant is given by: CO H CH H O ( )( ) (15 )(15 ) (15 ) Taking the square root of both sides, we obtain: which can be epressed as 5. + (.7 10 ) ( ) 0 Solving the quadratic equation, we obtain 1 atm (The other solution for is negative and is physically impossible.) At equilibrium, the pressures are: CH 4 HO (15 1) (15 1) CO 1 atm H (1 atm) atm atm 9 atm (a) shifts to right shifts to right (c) no change (d) no change (e) no change (f) shifts to left

30 CHATER 14: CHEMICAL EQUILIBRIUM NH HCl NH HCl. (1.1)(1.1) atm The equilibrium is: N O 4 (g) (g) ( ) Volume is doubled so pressure is halved. Let s calculate Q and compare it to Q < 0.0 Equilibrium will shift to the right. Some N O 4 will react, and some will be formed. Let amount of N O 4 reacted. N O 4 (g) (g) Initial (atm): Change (atm): + Equilibrium (atm): Substitute into the epression to solve for ( ) At equilibrium: Check: (0.01) atm atm (0.100) close enough to (a) React Ni with CO above 50 C. ump away the Ni(CO) 4 vapor (shift equilibrium to right), leaving the solid impurities behind. Consider the reverse reaction: Ni(CO) 4 (g) Ni(s) + 4CO(g) H Hf Hf 4 Δ 4 Δ (CO) Δ [Ni(CO) ] ΔH (4)( kj/mol) (1)( 60.9 kj/mol) kj/mol

31 416 CHATER 14: CHEMICAL EQUILIBRIUM The decomposition is endothermic, which is favored at high temperatures. Heat Ni(CO) 4 above 00 C to convert it back to Ni (a) Molar mass of Cl g/mol 1mol L atm.50 g ( 5 ) 08. g mol nrt 1.0 atm V L Cl 5 Cl + Cl Initial (atm) Change (atm) + + Equilibrium (atm) At equilibrium: Cl atm (c) (d) T (1.0 ) atm 0.69 atm 1.0 atm (a) conc. B A time conc. B A time

32 CHATER 14: CHEMICAL EQUILIBRIUM 417 (c) conc. B A time (a) Hg mmhg atm (equil. constants are epressed without units) c Δn 1 T ( ) (0.081 ) Volume of lab (6.1 m)(5. m)(.1 m) 100 m [Hg] c mol 00.6 g 1 L 1 cm Total mass of Hg vapor 100 m 1 L 1 mol 1000 cm 0.01 m. g The concentration of mercury vapor in the room is:. g 0.0 g/m 100 m mg/m Yes! This concentration eceeds the safety limit of 0.05 mg/m. Better clean up the spill! Initially, at equilibrium: [ ] M and [N O 4 ] M. At the instant the volume is halved, the concentrations double. [ ] ( M) M and [N O 4 ] (0.487 M) M. The system is no longer at equilibrium. The system will shift to the left to offset the increase in pressure when the volume is halved. When a new equilibrium position is established, we write: N O M M c 4 + [ ] ( ) [N O ] (0.974 ) Solving M (impossible) and M At the new equilibrium, [N O 4 ] M [ ] ( 0.017) M As we can see, the new equilibrium concentration of is greater than the initial equilibrium concentration ( M). Therefore, the gases should look darker!

33 418 CHATER 14: CHEMICAL EQUILIBRIUM There is a temporary dynamic equilibrium between the melting ice cubes and the freezing of water between the ice cubes (a) A catalyst speeds up the rates of the forward and reverse reactions to the same etent. (c) (d) A catalyst would not change the energies of the reactant and product. The first reaction is eothermic. Raising the temperature would favor the reverse reaction, increasing the amount of reactant and decreasing the amount of product at equilibrium. The equilibrium constant,, would decrease. The second reaction is endothermic. Raising the temperature would favor the forward reaction, increasing the amount of product and decreasing the amount of reactant at equilibrium. The equilibrium constant,, would increase. A catalyst lowers the activation energy for the forward and reverse reactions to the same etent. Adding a catalyst to a reaction miture will simply cause the miture to reach equilibrium sooner. The same equilibrium miture could be obtained without the catalyst, but we might have to wait longer for equilibrium to be reached. If the same equilibrium position is reached, with or without a catalyst, then the equilibrium constant is the same First, let's calculate the initial concentration of ammonia. 1molNH 14.6 g 17.0 g NH [NH ] 0.14 M 4.00 L Let's set up a table to represent the equilibrium concentrations. We represent the amount of NH that reacts as. NH (g) N (g) + H (g) Initial (M): Change (M): + + Equilibrium (M): 0.14 Substitute into the equilibrium constant epression to solve for. [N ][H ] c [NH ] ( )( ) 7 (0.14 ) (0.14 ) Taking the square root of both sides of the equation gives: Rearranging, Solving the quadratic equation gives the solutions: M and 0.44 M The positive root is the correct answer. The equilibrium concentrations are: [NH ] 0.14 (0.086) 0.04 M [N ] M [H ] (0.086) 0.6 M