1. a. The rates of the forward and reverse reactions are equal at equilibrium.

Transcription

1 CHATER THIRTEEN CHEMICAL EQUILIBRIUM For Review 1. a. The rates of the forward and reverse reactions are equal at equilibrium. b. There is no net change in the composition (as long as temperature is constant). See Figure 1.5 for an illustration of the concentration vs. time plot for this reaction. In Fig. 1.5, A = H, B = N, and C = NH. Notice how the reactant concentrations decrease with time until they reach equilibrium where the concentrations remain constant. The product concentration increases with time until equilibrium is reached. Also note that H decreases faster than N ; this is due to the :1 mole ratio in the balanced equation. H should be used up three times faster than N. Similarly, NH should increase at rate twice the rate of decrease of N. This is shown in the plot. Reference Figure 1. for the reaction rate vs. time plot. As concentrations of reactants decrease, the rate of the forward reaction decreases. Also, as product concentration increases, the rate of the reverse reaction increases. Eventually they reach the point where the rate that reactants are converted into products exactly equals the rate that products are converted into reactants. This is equilibrium and the concentrations of reactants and products do not change.. The law of mass action is a general description of the equilibrium condition; it defines the equilibrium constant expression. The law of mass action is based on experimental observation. K is a constant; the value (at constant temperature) does not depend on the initial conditions. Table 1.1 illustrates this nicely. Three experiments were run with each experiment having a different initial condition (only reactants present initially, only products present initially, and some of reactants and products present initially). In all three experiments, the value of K calculated is the same in each experiment (as it should be). Equilibrium and rates of reaction (kinetics) are independent of each other. A reaction with a large equilibrium constant value may be a fast reaction or a slow reaction. The same is true for a reaction with a small equilibrium constant value. Kinetics is discussed in detail in Chapter 1 of the text. The equilibrium constant is a number that tells us the relative concentrations (pressures) of reactants and products at equilibrium. An equilibrium position is a set of concentrations that satisfy the equilibrium constant expression. More than one equilibrium position can satisfy the same equilibrium constant expression. 6

2 CHATER 1 CHEMICAL EQUILIBRIUM 7 From Table 1.1, each of the three experiments have different equilibrium positions; that is, each experiment has different equilibrium concentrations. However, when these equilibrium concentrations are inserted into the equilibrium constant expression, each experiment gives the same value for K. The equilibrium position depends on the initial concentrations one starts with. Since there are an infinite number of initial conditions, there are an infinite number of equilibrium positions. However, each of these infinite equilibrium positions will always give the same value for the equilibrium constant (assuming temperature is constant).. NOCl(g) NO(g) + Cl (g) K = The expression for K is the product concentrations divided by the reactant concentrations. When K has a value much less than one, the product concentrations are relatively small and the reactant concentrations are relatively large. NO(g) N (g) + O (g) K = When K has a value much greater than one, the product concentrations are relatively large and the reactant concentrations are relatively small. In both cases, however, the rate of the forward reaction equals the rate of the reverse reaction at equilibrium (this is a definition of equilibrium).. The difference between K and K p are the units used to express the amounts of reactants and products present. K is calculated using units of molarity. K p is calculated using partial pressures in units of atm (usually). Both have the same form; the difference is the units used to determine the values. K p is only used when the equilibria involves gases; K can be used for gas phase equilibria and for solution equilibria. K p = K(RT) n where n = moles gaseous products in the balanced equation moles gaseous reactants in the balanced equation. K = K p when n = 0 (when moles of gaseous products = moles of gaseous reactants. K K p when n 0. When a balanced equation is multiplied by a factor n, K new = (K original ) n. So if a reaction is tripled, K new = K original. If a reaction is reversed, K new = 1/K original. Here K p, new = 1/K p, original. 5. When reactants and products are all in the same phase, these are homogeneous equilibria. Heterogeneous equilibria involve more than one phase. In general, for a homogeneous gas phase equilibria, all reactants and products are included in the K expression. In heterogeneous equilibria, equilibrium does not depend on the amounts of pure solids or liquids present. The amount of solids and liquids present are not included in K expressions; they just have to be present. On the other hand, gases and solutes are always included in K expressions. Solutes have (aq) written after them. 6. For the gas phase reaction: aa + bb cc + dd [C] [D] the equilibrium constant expression is: K = a [A] [B] c d b

3 8 CHATER 1 CHEMICAL EQUILIBRIUM and the reaction quotient has the same form: Q = [C] [A] c a [D] [B] d b The difference is that in the expression for K we use equilibrium concentrations, i.e., [A], [B], [C], and [D] are all in equilibrium with each other. Any set of concentrations can be plugged into the reaction quotient expression. Typically, we plug initial concentrations into the Q expression and then compare the value of Q to K to see how far we are from equilibrium. If Q = K, the reaction is at equilibrium with these concentrations. If Q K, then the reaction will have to shift either to products or to reactants to reach equilibrium. For Q > K, the net change in the reaction to get to equilibrium must be a conversion of products into reactants. We say the reaction shifts left to reach elquilibrium. When Q < K, the net change in the reaction to get to equilibrium must be a conversion of reactants into products; the reaction shifts right to reach equilibrium. 7. The steps to solve equilibrium problems are outlined at the beginning of section 1.6. The ICE table is a convenient way to summarize an equilibrium problem. We make three rows under the balanced reaction. The initials in ICE stand for initial, change, and equilibrium. The first step is to fill in the initial row, then deduce the net change that must occur to reach equilibrium. You then define x or x or x as the change (molarity or partial pressure) that must occur to reach equilibrium. After defining your x, fill in the change column in terms of x. Finally, sum the initial and change columns together to get the last row of the ICE table; the equilibrium concentrations (or partial pressures). The ICE table again summarizes what must occur for a reaction to reach equilibrium. This is vital in solving equilibrium problems. 8. The assumption comes from the value of K being much less than 1. For these reactions, the equilibrium mixture will not have a lot of products present; mostly reactants are present at equilibrium. If we define the change that must occur in terms of x as the amount (molarity or partial pressure) of a reactant that must react to reach equilibrium, then x must be a small number because K is a very small number. We want to know the value of x in order to solve the problem, so we don t assume x = 0. Instead, we concentrate on the equilibrium row in the ICE table. Those reactants (or products) have equilibrium concentrations in the form of 0.10 x or x or.5 x, etc., is where an important assumption can be made. The assumption is that because K << 1, x will be small (x << 1) and when we add x or subtract x from some initial concentration, it will make little or no difference. That is, we assume that 0.10 x 0.10 or x 0.5 or.5 x.5; we assume that the initial concentration of a substance is equal to the equilibrium concentration. This assumption makes the math much easier, and usually gives a value of x that is well within 5% of the true value of x (we get about the same answer with a lot less work). We check the assumptions for validity using the 5% rule. From doing a lot of these calculations, it is found that when a assumption like 0.0 x 0.0 is made, if x is less than 5% of the number the assumption was made against, then our final answer is within acceptable error limits of the true value of x (as determined when the equation is solved exactly). For our example above (0.0 x 0.0), if (x/0.0) 100 5%, then our assumption is valid by the 5% rule. If the error is greater than 5%, then we must solve the equation exactly or use a math trick called the method of successive approximations. See Appendix A1. for details regarding the method of successive approximations as well as for a review in solving quadratic equations exactly.

4 CHATER 1 CHEMICAL EQUILIBRIUM 9 9. LeChatlier s rinciple: if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in the direction that tends to reduce that change. a. When a gaseous (or solute) reactant is added to a reaction at equilibrium, the reaction shifts right to use up some of the added reactant. Adding NOCl causes the equilibrium to shift to the right. b. If a gaseous (or solute) product is added to a system at equilibrium, the reaction shifts left to use up some of the added product. Adding NO(g) causes the equilibrium to shift left. c. Here a gaseous reactant is removed (NOCl), so the reaction shifts left to produce more of the NOCl. d. Here a gaseous product is removed (Cl ), so the reaction shifts right to produce more of the Cl. e. In this reaction, moles of gaseous reactants are converted into moles of gaseous products. If the volume of the container is decreased, the reaction shifts to the side that occupies a smaller volume. Here, the reactions shift left to side with the fewer moles of reactant gases present. For all of these changes, the value of K does not change. As long as temperature is constant, the value of K is constant. H (g) + O (g) H O(l); In this reaction, the amount of water present has no effect on the equilibrium; H O(l) just has to be present whether it s grams or grams. The same is true for solids. When solids or liquids are in a reaction, addition or removal of these solids or liquids has no effect on the equilibrium (the reaction remains at equilibrium). Note that for this example, if the temperature is such that H O(g) is the product, then the amount of H O(g) present does affect the equilibrium. A change in volume will change the partial pressure of all reactants and products by the same factor. The shift in equilibrium depends on the number of gaseous particles on each side. An increase in volume will shift the equilibrium to the side with the greater number of particles in the gas phase. A decrease in volume will favor the side with lesser gas phase particles. If there are the same number of gas phase particles on each side of the reaction, a change in volume will not shift the equilibrium. When we change the pressure by adding an unreactive gas, we do not change the partial pressures (or concentrations) of any of the substances in equilibrium with each other. This is because the volume of the container did not change. If the partial pressures (and concentrations) are unchanged, the reaction is still at equilibrium. 10. In an exothermic reaction, heat is a product. When the temperature increases, heat (a product) is added and the reaction shifts left to use up the added heat. For an exothermic reaction, the value of K decreases as temperature increases. In an endothermic reaction, heat is a reactant. Heat (a reactant) is added when the temperature increases and the reaction shifts right to use up the added heat in order to reestablish equilibrium. The value of K increases for an endothermic reaction as temperature increases.

5 50 CHATER 1 CHEMICAL EQUILIBRIUM Questions A decrease in temperature corresponds to the removal of heat. Here, the value of K increases as T decreases. This indicates that as heat is removed, more products are produced. Heat must be a product, so this is an exothermic reaction. 9. No, equilibrium is a dynamic process. Both reactions: H O + CO H + CO and H + CO H O + CO are occurring, but at equal rates. Thus, 1 C atoms will be distributed between CO and CO. 10. No, it doesn't matter from which direction the equilibrium position is reached. Both experiments will give the same equilibrium position since both experiments started with stoichiometric amounts of reactants or products. [H ][CO ] 11. H O(g) + CO(g) H (g) + CO (g) K = [H O][CO] =.0 K is a unitless number since there is an equal number of moles of product gases as compared to moles of reactant gases in the balanced equation. Therefore, we can use units of molecules per liter instead of moles per liter to determine K. We need to start somewhere, so let s assume molecules of CO react. If molecules of CO react, then molecules of H O must react, and molecules each of H and CO are formed. We would have 6 = molecules CO, 8 = 5 molecules H O, 0 + = molecules H, and 0 + = molecules CO present. This will be an equilibrium mixture if K =.0: K = molecules H moleculesco L L 5 molecules H O moleculesco 5 L L Because this mixture does not give a value of K =.0, this is not an equilibrium mixture. Let s try molecules of CO reacting to reach equilibrium. molecules CO remaining = 6 = molecules CO; molecules H O remaining = 8 = molecules H O; molecules H present = 0 + = molecules H ; molecules CO present = 0 + = molecules CO molecules H moleculesco L L K =. 0 molecules HO moleculesco L L Since K =.0 for this reaction mixture, we are at equilibrium.

6 CHATER 1 CHEMICAL EQUILIBRIUM When equilibrium is reached, there is no net change in the amount of reactants and products present since the rates of the forward and reverse reactions are equal to each other. The first diagram has A B molecules, A molecules and 1 B molecule present. The second diagram has A B molecules, A molecules, and B molecules. Therefore, the first diagram cannot represent equilibrium since there was a net change in reactants and products. Is the second diagram the equilibrium mixture? That depends on whether there is a net change between reactants and products when going from the second diagram to the third diagram. The third diagram contains the same number and type of molecules as the second diagram, so the second diagram is the first illustration that represents equilibrium. The reaction container initially contained only A B. From the first diagram, A molecules and 1 B molecule are present (along with A B molecules). From the balanced reaction, these A molecules and 1 B molecule were formed when A B molecules decomposed. Therefore, the initial number of A B molecules present equals + = 6 molecules A B. 1. K and K p are equilibrium constants as determined by the law of mass action. For K, concentration units of mol/l are used, and for K p, partial pressures in units of atm are used (generally). Q is called the reaction quotient. Q has the exact same form as K or K p, but instead of equilibrium concentrations, initial concentrations are used to calculate the Q value. The use of Q is when it is compared to the K value. When Q = K (or when Q p = K p ), the reaction is at equilibrium. When Q K, the reaction is not at equilibrium and one can deduce the net change that must occur for the system to get to equilibrium. 1. H (g) + I (g) HI(g) K = [HI] [H ][I ] H (g) + I (s) HI(g) K = [HI] [H ] (Solids are not included in K expressions.) Some property differences are: a. the reactions have different K expressions. b. for the first reaction, K = K p (since n = 0), and for the second reaction, K K p (since n 0). c. a change in the container volume will have no effect on the equilibrium for reaction 1, whereas a volume change will effect the equilibrium for reaction (shifts the reaction left or right depending on whether the volume is decreased or increased). 15. We always try to make good assumptions that simplify the math. In some problems, we can set-up the problem so that the net change, x, that must occur to reach equilibrium is a small number. This comes in handy when you have expressions like 0.1 x or x, etc. When x is small, we can assume that it makes little difference when subtracted from or added to some relatively big number. When this is the case, 0.1 x 0.1 and x 0.77, etc. If the assumption holds by the 5% rule, the assumption is assumed valid. The 5% rule refers to x (or x or x, etc.) that is assumed small compared to some number. If x (or x or x, etc.) is less than 5% of the number the assumption was made against, then the assumption

7 5 CHATER 1 CHEMICAL EQUILIBRIUM will be assumed valid. If the 5% rule fails to work, one can use a math procedure called the method of successive approximations to solve the quadratic or cubic equation. Of course, one could always solve the quadratic or cubic equation exactly. This is generally a last resort (and is usually not necessary). 16. Only statement e is correct. Addition of a catalyst has no effect on the equilibrium position; the reaction just reaches equilibrium more quickly. Statement a is false for reactants that are either solids or liquids (adding more of these has no effect on the equilibrium). Statement b is false always. If temperature remains constant, then the value of K is constant. Statement c is false for exothermic reactions where an increase in temperature decreases the value of K. For statement d, only reactions which have more reactant gases than product gases will shift left with an increase in container volume. If the moles of gas are equal or if there are more moles of product gases than reactant gases, the reaction will not shift left with an increase in volume. The Equilibrium Constant 17. a. K = [NO] [N ][O ] b. K = [NO] [N O ] [SiCl ][H] c. K = [SiH ][Cl ] [Cl ] [Br ] d. K = [Br ] [Cl ] 18. a. K p = NO N O b. K p = NO N O SiCl H c. K p = SiH Cl Cl d. K p = Br Br Cl 19. K = [NH] = for N (g) + H (g) NH (g) [N ][H ] When a reaction is reversed, then K new = 1/K original. When a reaction is multiplied through by a value of n, then K new = (K original ) n. [NH] a. 1/ N (g) + / H (g) NH (g) K= = K 1/ = (1. 10 ) 1/ = / / [N ] [H ] [N][H] 1 1 b. NH (g) N (g) + H (g) K = = 77 [NH ] K 1.10 c. NH (g) 1/ N (g) + / H (g) K = 1/ [N] [H] [NH ] / 1 K 1/ / [NH] d. N (g) + 6 H (g) NH (g) K = 6 [N ] [H ] = (K) = (1. 10 ) = 1.7 =

8 CHATER 1 CHEMICAL EQUILIBRIUM 5 0. H (g) + Br (g) HBr(g) K p = a. HBr 1/ H + 1/ Br b. HBr H + Br ' K p = HBr =.5 10 ( )( ) H Br ( '' (H )( K p = HBr H Br ) 1/ ) ( HBr Br 1/ ) 1 K p 1/ 1 1 =.9 Kp / = c. 1/ H + 1/ Br HBr ''' HBr K p = 1/ 1/ ( ) ( ) H Br = (K p ) 1/ = 190 [N][HO] 1. NO(g) + H (g) N (g) + H O(g) K = [NO] [H ] (5. 10 )(.9 10 ) K = 5 (8.110 ) (.110 ) = [NCl ]. K = [N ][Cl ] = (1. 10 (0.19) ) (.10 ) = [NO] =.5 10 mol.0 L = M; [Cl ] =. mol = 0.80 M.0 L [NOCl] = 1.0 mol.0 L [NO] [Cl = 0. M; K = ] ( ) (0.80) = 1.7 [NOCl] (0.) [N O] =.0010 mol ; [N ] =.00L.8010 mol ; [O ] =.00L mol.00l K =.0010 [N O].00 ( ) = [N ] [O ] (1.010 ) (1.510 ) If the given concentrations represent equilibrium concentrations, then they should give a value of K = (0.00) = (.0010 ) (0.005)

9 5 CHATER 1 CHEMICAL EQUILIBRIUM Because the given concentrations when plugged into the equilibrium constant expression give a value equal to K ( ), this set of concentrations is a system at equilibrium. NO 5. K p = NO NH 6. K p = N O H ( = (0.55) ( ) (.5 10 = ) (0.85)(.110 ) ) = 6. = (0.167) (0.55)( ) = When the given partial pressures are plugged into the K p expression, the value does not equal the K p value of Therefore, one can conclude that the given set of partial pressures does not represent a system at equilibrium. 7. K p = K(RT) Δn where Δn = sum of gaseous product coefficients sum of gaseous reactant coefficients. For this reaction, Δn = 1 =. K = [CO][H] (0.)(1.1) = 1.9 [CH OH] 0.15 K = K(RT) = 1.9 ( L/atm/K mol 600. K) = K = K(RT) Δn, K = K p (RT) Δn Δn = = 1; K = 0.5 ( ) = 9. Solids and liquids do not appear in equilibrium expressions. Only gases and dissolved solutes appear in equilibrium expressions. a. K = [H O] ; K p = [NH ] [CO ] HO NH CO b. K = [N ][Br ] ; K p = Br N c. K = [O ] ; K p = [H O] O d. K= ; K p = [H ] HO H 0. K p = K(RT) Δn where Δn equals the difference in the sum of the coefficients between gaseous products and gaseous reactants (Δn = mol gaseous products mol gaseous reactants). When Δn = 0, then K p = K. In Exercise 1.9, only reaction d has Δn = 0 so only reaction d has K p = K. 1. Because solids do not appear in the equilibrium constant expression, K = 1/[O ].

10 CHATER 1 CHEMICAL EQUILIBRIUM mol 1 [O ] = ; K =.0 L [O ] H. K p = H O H O H 1 = ) ( ; tot =, 6. torr = 15.0 torr +, 1 H H = = 1. torr Because l atm = 760 torr: K = (1./ 760) (15.0 / 760) =.07 Equilibrium Calculations [N][O ]. NO(g) N (g) + O (g) K = =. 10 [NO] Use the reaction quotient Q to determine which way the reaction shifts to reach equilibrium. For the reaction quotient, initial concentrations given in a problem are used to calculate the value for Q. If Q < K, then the reaction shifts right to reach equilibrium. If Q > K, then the reaction shifts left to reach equilibrium. If Q = K, then the reaction does not shift in either direction because the reaction is at equilibrium. a. [N ]=.0 mol =.0 M; [O ] = 1.0 L.6 mol 1.0 L [N] o[o] o (.0)(.6) Q = = [NO] (0.0) o =.6 M; [NO] = 0.0mol = 0.0 M 1.0 L Q > K so the reaction shifts left to produce more reactants in order to reach equilibrium. b. [N ]= 0.6mol = 0.1 M; [O ] =.0 L.0 mol.0 L =.0 M; [NO] = 0.0mol = M.0 L Q = (0.1)(.0) =. 10 = K; at equilibrium (0.016) c. [N ] =. mol = 0.80 M; [O ] =.0 L 1.7 mol.0 L = 0.57 M; [NO] = 0.060mol = 0.00 M.0 L (0.80)(0.57) Q = = < K; Reaction shifts right to reach equilibrium. (0.00). As in Exercise 1., determine Q for each reaction, and compare this value to K p (. 10 ) to determine which direction the reaction shifts to reach equilibrium. Note that, for this reaction, K = K p since Δn = 0.

11 56 CHATER 1 CHEMICAL EQUILIBRIUM N O (0.11)(.0) a. Q = (0.010) NO =. 10 Q < K p so the reaction shifts right to reach equilibrium. (0.6)(0.67) b. Q = =.0 10 > K p (0.078) Reaction shifts left to reach equilibrium. (0.51)(0.18) c. Q = (0.006) =. 10 = K p ; at equilbrium 5. CaCO (s) CaO(s) + CO (g) K p = CO = 1.0 a. Q = CO ; We only need the partial pressure of CO to determine Q since solids do not appear in equilibrium expressions (or Q expressions). At this temperature all CO will be in the gas phase. Q =.55 so Q > K p ; Reaction will shift to the left to reach equilibrium; the mass of CaO will decrease. b. Q = 1.0 = K p so the reaction is at equilibrium; mass of CaO will not change. c. Q = 1.0 = K p so the reaction is at equilibrium; mass of CaO will not change. d. Q = 0.11 < K p ; The reaction will shift to the right to reach equilibrium; the mass of CaO will increase. 6. CH CO H + C H 5 OH CH CO C H 5 + H O K = [CH CO CH5][HO] =. [CH CO H][C H OH] 5 a. Q = b. Q = c. Q = d. Q = (0.)(0.10) (0.010)(0.010) = 0 > K; Reaction will shift left to reach equilibrium so the concentration of water will decrease. (0.)(0.000) =. = K; Reaction is at equilibrium, so the concentration of (0.000)(0.10) water will remain the same. (0.88)(0.1) = 0.0 < K; Because Q < K, the concentration of water will in- (0.0)(6.0) crease because the reaction shifts right to reach equilibrium. (.)(.) (0.88)(10.0) =. = K; At equilibrium, so the water concentration is unchanged. (.0)[H O] e. K =. =, [H O] = 0.55 M (0.10)(5.0)

12 CHATER 1 CHEMICAL EQUILIBRIUM 57 f. Water is a product of the reaction, but it is not the solvent. Thus, the concentration of water must be included in the equilibrium expression since it is a solute in the reaction. Only when water is the solvent do we not include it in the equilibrium expression. [H] [O] ( ) [O] 7. K =,. 10, [O ] = M [H O] (0.11) 8. K = NOBr NO Br, (0.768) 109, NO = atm NO 9. SO (g) + NO (g) SO (g) + NO(g) K = [SO ][NO] [SO ][NO ] To determine K, we must calculate the equilibrium concentrations. The initial concentrations are:.00mol [SO ] o = [NO] o = 0; [SO ] o = [NO ] o = =.00 M 1.00L Next, we determine the change required to reach equilibrium. At equilibrium, [NO] = 1.0 mol/1.00 L = 1.0 M. Since there was zero NO present initially, 1.0 M of SO and 1.0 M NO must have reacted to produce 1.0 M NO as well as 1.0 M SO, all required by the balanced reaction. The equilibrium concentration for each substance is the sum of the initial concentration plus the change in concentration necessary to reach equilibrium. The equilibrium concentrations are: [SO ] = [NO] = M = 1.0 M; [SO ] = [NO ] =.00 M M = 0.70 M We now use these equilibrium concentrations to calculate K: K = [SO ][NO] (1.0)(1.0) =. [SO ][NO ] (0.70)(0.70) 0. S 8 (g) S (g) K = S S 8 Initially: S = 1.00 atm and 8 S = 0 atm Change: Since 0.5 atm of S 8 remain at equilibrium, then 1.00 atm 0.5 atm = 0.75 atm of S 8 must have reacted in order to reach equilibrium. Since there is a :1 mol ratio between S and S 8 (from the balanced reaction), then (0.75 atm) =.0 atm of S must have been produced when the reaction went to equilibrium (moles and pressure are directly related at constant T and V). Equilibrium: S 8 = 0.5 atm, S = atm =.0 atm; Solving for K p : K = (.0) 0.5 =. 10

13 58 CHATER 1 CHEMICAL EQUILIBRIUM 1. When solving equilibrium problems, a common method to summarize all the information in the problem is to set up a table. We call this table the ICE table since it summarizes initial concentrations, changes that must occur to reach equilibrium and equilibrium concentrations (the sum of the initial and change columns). For the change column, we will generally use the variable x, which will be defined as the amount of reactant (or product) that must react to reach equilibrium. In this problem, the reaction must shift right to reach equilibrium because there are no products present initially. Therefore, x is defined as the amount (in units of mol/l) of reactant SO that reacts to reach equilibrium; we use the coefficients in the balanced equation to relate the net change in SO to the net change in SO and O. The general ICE table for this problem is: SO (g) SO (g) + O (g) [SO ] [O] K = [SO ] Initial 1.0 mol/.0 L 0 0 Let x mol/l of SO react to reach equilibrium Change x +x +x/ Equil..0 x x x/ From the problem, we are told that the equilibrium SO concentration is.0 mol/.0 L = 1.0 M ([SO ] e = 1.0 M). From the ICE table, [SO ] e = x so x = 1.0. Solving for the other equilibrium concentrations: [SO ] e =.0 - x = =.0 M; [O ] = x/ = 1.0/ = 0.50 M. [SO ] [O] (1.0) (0.50) K = = [SO ] (.0) Alternate Method: Fractions in the change column can be avoided (if you want) by defining x differently. If we were to let x mol/l of SO react to reach equilibrium, then the ICE table is: [SO ] [O] SO (g) SO (g) + O (g) K = [SO ] Initial.0 M 0 0 Let x mol/l of SO react to reach equilibrium Change x +x +x Equil..0 x x x Solving: x = [SO ] e = 1.0 M, x = 0.50 M; [SO ] e =.0 (0.50) =.0 M; [O ] e = x = 0.50 M These are exactly the same equilibrium concentrations as solved for previously, thus K will be the same (as it must be). The moral of the story is to define x in a manner that is most comfortable for you. Your final answer is independent of how you define x initially.. NH (g) N (g) + H (g) [N] [H] K = [NH] Initial.0 mol/.0 L 0 0 Let x mol/l of NH react to reach equilibrium Change x +x +x Equil..0 x x x

14 CHATER 1 CHEMICAL EQUILIBRIUM 59 From the problem: [NH ] e =.0 mol/.0 L = 1.0 M =.0 x, x = 0.50 M [N ] = x = 0.50 M; [H ] = x = (0.50 M) = 1.5 M [N] [H] (0.50)(1.5) K = = 1.7 [NH ] (1.0). H (g) + N (g) NH (g) Initial [H ] o [N ] o 0 x mol/l of N reacts to reach equilibrium Change x x +x Equil [H ] o x [N ] o x x From the problem: [NH ] e =.0 M = x, x =.0 M; [H ] e = 5.0 M = [H ] o x; [N ] e = 8.0 M = [N ] o x 5.0 M = [H ] o (.0 M), [H ] o = 11.0 M; 8.0 M = [N ] o.0 M, [N ] o = 10.0 M. N (g) + H (g) NH (g) Initial 1.00 atm.00 atm 0 x atm of N reacts to reach equilibrium Change x x +x equil x.00 x x From set-up: TOT =.00 atm = N H NH.00 atm = (1.00 x) + (.00 x) + x =.00 x x = atm; H =.00 x =.00 (0.500) = 0.50 atm 5. Q = 1.00, which is less than K. Reaction shifts to the right to reach equilibrium. Summarizing the equilibrium problem in a table: SO (g) + NO (g) SO (g) + NO(g) K =.75 Initial M M M M x mol/l of SO reacts to reach equilibrium Change x x +x +x Equil x x x x lug the equilibrium concentrations into the equilibrium constant expression: [SO ][NO] (0.800 x) K =.75 [SO ][NO ] (0.800 x) ; Take the square root of both sides and solve for x:

15 60 CHATER 1 CHEMICAL EQUILIBRIUM x = 1.9, x = x,.9 x = 0.75, x = 0.6 M x The equilibrium concentrations are: [SO ] = [NO] = x = = 1.06 M; [SO ] = [NO ] = x = 0.5 M 6. Q = 1.00, which is less than K. Reaction shifts right to reach equilibrium. [HI] H (g) + I (g) HI(g) K [H ][I ] = 100. Initial 1.00 M 1.00 M 1.00 M x mol/l of H reacts to reach equilibrium Change x x +x Equil x 1.00 x x (1.00 x) K = 100. = (1.00 x) ; Taking the square root of both sides: 10.0 = 1.00 x, x = x, 1.0 x = 9.0, x = 0.75 M 1.00 x [H ] = [I ] = = 0.5 M; [HI] = (0.75) =.50 M 7. Because only reactants are present initially, the reaction must proceed to the right to reach equilibrium. Summarizing the problem in a table: N (g) + O (g) NO(g) K p = Initial 0.80 atm 0.0 atm 0 x atm of N reacts to reach equilibrium Change x x +x Equil x 0.0 x x K p = = N NO O (x), 0.050( x + x ) = x (0.80 x)(0.0 x) x = x x,.95 x x = 0 Solving using the quadratic formula (see Appendix 1. of the text): b(b ac) x a 1/ [(0.050) (.95)( (.95) x =.9 10 atm or x = atm; Only x =.9 10 atm makes sense (x cannot be negative), so the equilibrium NO partial pressure is: )] 1/

16 CHATER 1 CHEMICAL EQUILIBRIUM 61 NO = x = (.9 10 atm) = atm 8. H O(g) + Cl O(g) HOCl(g) K = = a. The initial concentrations of H O and Cl O are: [HOCl] [H O][Cl O] 1.0 g HO 1mol = mol/l; 1.0 L 18.0g.0 g Cl O 1mol =. 1.0 L 86.90g H O(g) + Cl O(g) HOCl(g) Initial M. 10 M 0 x mol/l of H O reacts to reach equilibrium Change x x +x Equil x. 10 x x 10 mol/l K = = (x), 1.1 ( x)(.10 x) x x = x.91 x x = 0 (We carried extra significant figures.) Solving using the quadratic formula: / ( ) =.6 10 or A negative answer makes no physical sense; we can't have less than nothing. So x =.6 10 M. [HOCl] = x = 9. [H O] = M; [Cl O] =. 10 x = = M b. H O(g) + Cl O(g) HOCl(g) 10 x = = 1.8 Initial mol/.0 L = 0.50 M x mol/l of HOCl reacts to reach equilibrium Change +x +x x Equil. x x 0.50 x [HOCl] (0.50 x) K = = [H O][Cl O] x The expression is a perfect square, so we can take the square root of each side: 10 M

17 6 CHATER 1 CHEMICAL EQUILIBRIUM 0.50 x 0.0 =, 0.0 x = 0.50 x,.0 x = 0.50 x x = 0.17 (We carried extra significant figures.) x = [H O] = [Cl O] = 0.17 = 0. M; [HOCl] = 0.50 x = = 0.07 M 9. SO (g) + O (g) SO (g) K p = 0.5 Initial 0.50 atm 0.50 atm 0 x atm of SO reacts to reach equilibrium Change x x +x Equil x 0.50 x x K p = 0.5 = SO SO O = (x) (0.50 x) (0.50 x) This will give a cubic equation. Graphing calculators can be used to solve this expression. If you don t have a graphing calculator, an alternative method for solving a cubic equation is to use the method of successive approximations (see Appendix 1. of the text). The first step is to guess a value for x. Because the value of K is small (K < 1), then not much of the forward reaction will occur to reach equilibrium. This tells us that x is small. Lets guess that x = atm. Now we take this estimated value for x and substitute it into the equation everywhere that x appears except for one. For equilibrium problems, we will substitute the estimated value for x into the denominator, then solve for the numerator value of x. We continue this process until the estimated value of x and the calculated value of x converge on the same number. This is the same answer we would get if we were to solve the cubic equation exactly. Applying the method of successive approximations and carrying extra significant figures: x [0.50 (0.050)] [0.50 (0.050)] x (0.0) (0.5) = 0.5, x = x [0.50 (0.067)] [0.50 (0.067)] x (0.66) (0.) = 0.5, x = x, 0.5, x = 0.06; (0.8) (0.) (0.7) x (0.7) = 0.5, x = 0.06 The next trial gives the same value for x = 0.06 atm. We are done except for determining the equilibrium concentrations. They are: SO = 0.50 x = 0.50 (0.06) = 0.76 = 0.8 atm O = 0.50 x = 0.8 = 0. atm; SO = x = 0.1 = 0.1 atm 50. a. The reaction must proceed to products to reach equilibrium. Summarizing the problem in

18 CHATER 1 CHEMICAL EQUILIBRIUM 6 a table where x atm of N O reacts to reach equilibrium: N O (g) NO (g) K p = 0.5 Initial.5 atm 0 Change x +x Equil..5 x x K p = NO NO ( x), x = x, x x 1.15 = 0. 5 x We carried extra significant figures in this expression (as will be typical when we solve an expression using the quadratic formula). Solving using the quadratic formula (Appendix 1. of text): x = 0.5 [(0.5) ()( 1.15)] () 1/ , x = 0.50 (Other value is negative.) NO = x = 1.0 atm; N O =.5 x =.0 atm b. The reaction must shift to reactants (shifts left) to reach equilibrium. N O (g) NO (g) Initial atm Change +x x Equil. x 9.0 x K p = ( 9.0 x) x = 0.5, x 6.5 x + 81 = 0 (carrying extra sig. figs.) Solving using quadratic formula: x = ( 6.5) [( 6.5) () ()( 81)] 1/, x =.0 atm The other value, 5.1, is impossible. N O = x =.0 atm; NO = 9.0 x = 1.0 atm c. No, we get the same equilibrium position starting with either pure N O or pure NO in stoichiometric amounts. 51. a. The reaction must proceed to products to reach equilibrium since only reactants are present initially. Summarizing the problem in a table:

19 6 CHATER 1 CHEMICAL EQUILIBRIUM NOCl(g) NO(g) + Cl (g) K = 1.6 Initial.0 mol.0 L 0 0 x mol/l of NOCl reacts to reach equilibrium Change x +x +x Equil. 1.0 x x x 5 K = = [NO] [Cl ] (x) ( x) [NOCl] (1.0 x) 10 5 If we assume that 1.0 x 1.0 (from the small size of K, we know that not a lot of products are present at equilibrium so x is small), then: x = 1.0, x = ; Now we must check the assumption. 1.0 x = 1.0 (0.016) = 0.97 = 1.0 (to proper significant figures) Our error is about %, i.e., x is.% of 1.0 M. Generally, if the error we introduce by making simplifying assumptions is less than 5%, we go no further and the assumption is said to be valid. We call this the 5% rule. Solving for the equilibrium concentrations: [NO] = x = 0.0 M; [Cl ] = x = M; [NOCl] = 1.0 x = 0.97 M 1.0 M Note: If we were to solve this cubic equation exactly (a long and tedious process), we would get x = This is the exact same answer we determined by making a simplifying assumption. We saved time and energy. Whenever K is a very small value, always make the assumption that x is small. If the assumption introduces an error of less than 5%, then the answer you calculated making the assumption will be considered the correct answer. b. NOCl(g) NO(g) + Cl (g) Initial 1.0 M 1.0 M 0 x mol/l of NOCl reacts to reach equilibrium Change x +x +x Equil. 1.0 x x x = ( 1.0 x) ( x) (1.0) ( x) (1.0 x) (1.0) (assuming x << 1.0) 5 x = ; Assumptions are great (x is. 10 % of 1.0). [Cl ] = M and [NOCl] = [NO] = 1.0 M

20 CHATER 1 CHEMICAL EQUILIBRIUM 65 c. NOCl(g) NO(g) + Cl (g) Initial.0 M M x mol/l of NOCl reacts to reach equilibrium Change x +x +x Equil..0 x x x = (x) (1.0 x) x (.0 x).0 (assuming x << 1.0) Solving: x =.0 10 ; Assumptions good (x is 0.% of 1.0 and x is 0.% of.0). [Cl ] = x = 1.0 M; [NO] = (.0 10 ) = M; [NOCl] =.0 M 5. N O (g) NO (g) K = [NO] =.0 [N O ] 10 7 Initial 1.0 mol/10.0 L 0 x mol/l of N O reacts to reach equilibrium Change x +x Equil x x K = [NO ] [N O ] = (x) = x 7 10 ; Because K has a small value, assume that x is small compared to 0.10 so that 0.10 x Solving: x 0.10, x =.0 10, x = M x Checking the assumption by the 5% rule: = 0.10% Because this number is less than 5%, we will say that the assumption is valid. [N O ] = = 0.10 M; [NO ] = x = ( ) =.0 10 M 5. CO (g) CO(g) + O (g) K = [CO] [CO [O ] ] = Initial.0 mol/5.0 L 0 0 x mol/l of CO reacts to reach equilibrium Change x +x +x Equil. 0.0 x x x K = = [CO] [CO [O ] ] (x) ( x) = (0.0 x) ; Assuming x << 0.0:

21 66 CHATER 1 CHEMICAL EQUILIBRIUM x (0.0), = x, x =. 10 M 0.16 Checking assumption: ( ) 100 =.%; Assumption valid by the 5% rule. [CO ] = 0.0 x = 0.0 (. [CO] = x = (. 10 ) = ) = 0.9 M 10 M; [O ] = x =. 10 M 5. COCl (g) CO(g) + Cl (g) K p = Initial 1.0 atm 0 0 x atm of COCl reacts to reach equilibrium Change x +x +x Equil. 1.0 x x x CO COCl Cl = = CO CoCl Cl = x 1.0 x x 1.0 (Assuming 1.0 x 1.0.) x = atm; Assumption good (x is % of 1.0). COCl = 1.0 x = = 1.0 atm; CO = Cl = x = atm 55. This is a typical equilibrium problem except that the reaction contains a solid. Whenever solids and liquids are present, we basically ignore them in the equilibrium problem. NH OCONH (s) NH (g) + CO (g) K p =.9 Initial 0 0 Some NH OCONH decomposes to produce x atm of NH and x atm of CO. Change +x +x Equil. x x K p =.9 10 = NH = (x) (x) = x CO 10 x = / = atm; NH = x = 0.18 atm; CO = x = atm total = NH = 0.18 atm atm = 0.7 atm CO 56. NH Cl(s) NH (g) + HCl(g) K = NH HCl

22 CHATER 1 CHEMICAL EQUILIBRIUM 67 For this system to reach equilibrium, some of the NH Cl(s) decomposes to form equal moles of NH (g) and HCl(g) at equilibrium. Because mol HCl = mol NH, the partial pressures of each gas must be equal to each other. At equilibrium: total = NH + HCl and total =. atm = NH,. atm = Le Chatelier's rinciple NH = HCl NH = HCl ; K p = (.)(.) = a. No effect; Adding more of a pure solid or pure liquid has no effect on the equilibrium position. b. Shifts left; HF(g) will be removed by reaction with the glass. As HF(g) is removed, the reaction will shift left to produce more HF(g). c. Shifts right; As H O(g) is removed, the reaction will shift right to produce more H O(g). 58. When the volume of a reaction container is increased, the reaction itself will want to increase its own volume by shifting to the side of the reaction which contains the most molecules of gas. When the molecules of gas are equal on both sides of the reaction, then the reaction will remain at equilibrium no matter what happens to the volume of the container. a. Reaction shifts left (to reactants) since the reactants contain molecules of gas compared to molecules of gas on the product side. b. Reaction shifts right (to products) since there are more product molecules of gas () than reactant molecules (1). c. No change since there are equal reactant and product molecules of gas. d. Reaction shifts right. e. Reaction shifts right to produce more CO (g). One can ignore the solids and only concentrate on the gases because gases occupy a relatively large volume compared to solids. We make the same assumption when liquids are present (only worry about the gas molecules). 59. a. right b. right c. no effect; He(g) is neither a reactant nor a product. d. left; The reaction is exothermic; heat is a product: CO(g) + H O(g) H (g) + CO (g) + Heat Increasing T will add heat. The equilibrium shifts to the left to use up the added heat. e. no effect; There are equal numbers of gas molecules on both sides of the reaction, so a change in volume has no effect on the equilibrium position.

23 68 CHATER 1 CHEMICAL EQUILIBRIUM 60. a. The moles of SO will increase because the reaction will shift left to use up the added O (g). b. Increase; Because there are fewer reactant gas molecules than product gas molecules, the reaction shifts left with a decrease in volume. c. No effect; The partial pressures of sulfur trioxide, sulfur dioxide, and oxygen are unchanged. d. Increase; Heat + SO SO + O ; Decreasing T will remove heat, shifting this endothermic reaction to the left. e. Decrease 61. a. left b. right c. left d. no effect (reactant and product concentrations are unchanged) e. no effect; Since there are equal numbers of product and reactant gas molecules, a change in volume has no effect on this equilibrium position. f. right; A decrease in temperature will shift the equilibrium to the right since heat is a product in this reaction (as is true in all exothermic reactions). 6. a. shift to left b. shift to right; The reaction is endothermic (heat is a reactant), thus an increase in temperature will shift the equilibrium to the right. c. no effect d. shift to right e. shift to right; Because there are more gaseous product molecules than gaseous reactant molecules, the equilibrium will shift right with an increase in volume. 6. An endothermic reaction, where heat is a reactant, will shift right to products with an increase in temperature. The amount of NH (g) will increase as the reaction shifts right so the smell of ammonia will increase. 6. As temperature increases, the value of K decreases. This is consistent with an exothermic reaction. In an exothermic reaction, heat is a product and an increase in temperature shifts the equilibrium to the reactant side (as well as lowering the value of K). Additional Exercises 65. O(g) + NO(g) NO (g) K = 1/ = NO (g) + O (g) NO(g) + O (g) K = 1/ = O (g) + O(g) O (g) K = ( )( ) =

24 CHATER 1 CHEMICAL EQUILIBRIUM a. N (g) + O (g) NO(g) K p = 1 NO = atm; n NO = V RT 1 10 = 16 N NO O NO (0.8)(0.) (1 10 atm)( L) = L atm (98K) molk 10 mol NO molecules 10 cm mol NO molecules NO cm 1 10 mol NO b. There is more NO in the atmosphere than we would expect from the value of K. The answer must lie in the rates of the reaction. At 5 C the rates of both reactions: N + O NO and NO N + O are so slow that they are essentially zero. Very strong bonds must be broken; the activation energy is very high. Nitric oxide is produced in high energy or high temperature environments. In nature, some NO is produced by lightning, and the primary manmade source is from automobiles. The production of NO is endothermic (ΔH = +90 kj/mol). At high temperatures, K will increase, and the rates of the reaction will also increase, resulting in a higher production of NO. Once the NO gets into a more normal temperature environment, it doesn t go back to N and O because of the slow rate. 67. a. AsH (g) As(s) + H (g) Initial 9.0 torr 0 Equil. 9.0 x x Using Dalton s Law of artial ressure: total = 88.0 torr = AsH H = 9.0 x + x, x = 96.0 torr H = x = (96.0) = 88 torr b. AsH = 9.0 (96.0) = 00.0 torr K p = H (0.79) (0.6) AsH = atm = 0.79 atm 760torr 1atm = 0.6 atm 760torr 68. FeSCN + (aq) Fe + (aq) + SCN (aq) K = 9.1 Initial.0 M 0 0 x mol/l of FeSCN + reacts to reach equilibrium Change x +x +x Equil..0 x x x 10

25 70 CHATER 1 CHEMICAL EQUILIBRIUM = [Fe ][SCN ] x x [FeSCN ].0 x.0 (Assuming.0 x.0.) 69. a. x =. 10 M; Assumption good by the 5% rule (x is.% of.0). [FeSCN + ] =.0 x =.0. Cl 5 = nrt V 10 =.0 M; [Fe + ] = [SCN ] = x =. 10 M.50g Cl L atm 600. K 08.g / mol molk = 1.16 atm 0.500L b. Cl 5 (g) Cl (g) + Cl (g) K p = Cl Cl5 Cl = 11.5 Initial 1.16 atm 0 0 x atm of Cl 5 reacts to reach equilibrium Change x +x +x Equil x x x x K p = = 11.5, x x 1. = 0; Using the quadratic formula: x = 1.06 atm 1.16 x Cl 5 = = 0.10 atm c. Cl Cl = 1.06 atm; Cl 5 = 0.10 atm tot = Cl 5 Cl Cl = =. atm d. ercent dissociation = x = = 91.% 70. SO Cl (g) Cl (g) + SO (g) Initial = initial pressure of SO Cl Change x +x +x Equil. 0 x x x total = atm = 0 x + x + x = 0 + x x 100 = 1.5, 0 = 8.00 x 0 Solving: = 0 + x = 9.00 x, x = atm x = atm = Cl SO ; 0 x = = atm = SO Cl K p = Cl SO (0.100) = SOCl 10 atm

26 CHATER 1 CHEMICAL EQUILIBRIUM K = [HF] (0.00) = 0.; 0.00 mol F /5.00 L = M F added [H ][F ] (0.0500)(0.0100) From LeChatelier s principle, added F causes the reaction to shift right to reestablish equilibrium. H (g) + F (g) HF(g) Initial M M 0.00 M x mol/l of F reacts to reach equilibrium Change x x +x Equil x x x (0.00 x) K = 0. = (0.500 x) ; Taking the square root of the equation: 17.9 = 0.00 x, x = x, 19.9 x = 0.95, x = 0.09 mol/l x [HF] = (0.09) = 0.50 M; [H ] = [F ] = = M 7. a. Doubling the volume will decrease all concentrations by a factor of one-half. 1 [FeSCN ] eq Q = = K, Q > K 1 1 [Fe ] eq [SCN ] eq The reaction will shift to the left to reestablish equilibrium. b. Adding Ag + will remove SCN through the formation of AgSCN(s). The reaction will shift to the left to produce more SCN -. c. Removing Fe + as Fe(OH) (s) will shift the reaction to the left to produce more Fe +. d. Reaction will shift to the right as Fe + is added. 7. H + + OH H O; Sodium hydroxide (NaOH) will react with the H + on the product side of the reaction. This effectively removes H + from the equilibrium, which will shift the reaction to the right to produce more H + and CrO. Because more CrO is produced, the solution turns yellow. 7. N (g) + H (g) NH (g) + heat a. This reaction is exothermic, so an increase in temperature will decrease the value of K (see Table 1. of text.) This has the effect of lowering the amount of NH (g) produced at equilibrium. The temperature increase, therefore, must be for kinetics reasons. As temperature increases, the reaction reaches equilibrium much faster. At low temperatures, this reaction is very slow, too slow to be of any use.

27 7 CHATER 1 CHEMICAL EQUILIBRIUM b. As NH (g) is removed, the reaction shifts right to produce more NH (g). c. A catalyst has no effect on the equilibrium position. The purpose of a catalyst is to speed up a reaction so it reaches equilibrium more quickly. d. When the pressure of reactants and products is high, the reaction shifts to the side that has fewer gas molecules. Since the product side contains molecules of gas as compared to molecules of gas on the reactant side, then the reaction shifts right to products at high pressures of reactants and products. 75. Cl 5 (g) Cl (g) + Cl (g) K = At equilibrium, [Cl 5 ] = [Cl ]. [Cl [Cl ][Cl 5 ] ] = = [Cl ][Cl ], [Cl ] = (.5 [Cl ] 10 ) = M 76. CaCO (s) CaO(s) + CO (g) K p = 1.16 = Some of the 0.0 g of CaCO will react to reach equilibrium. The amount that reacts is the quantity of CaCO required to produce a CO pressure of 1.16 atm (from the K p expression). CO CO V n CO = 1.16atm10.0 L = = 0.1 mol CO RT L atm 107K K mol 1molCaCO g mass CaCO reacted = 0.1 mol CO = 1. g CaCO molco molcaco mass % CaCO reacted = 1. g 0.0 g 100 = 66.0% 77. Cl 5 (g) Cl (g) + Cl (g) K p = Cl Cl / Cl5 Initial = initial Cl 5 pressure Change x +x +x Equil. 0 x x x total = 0 x + x + x = 0 + x = 58.7 torr 0 = n Cl 5 V RT =.156g 08.g / mol L atm 5. K K mol.000l = 0.90 atm = 189. torr x = total 0 = = torr

28 CHATER 1 CHEMICAL EQUILIBRIUM 7 = torr = 0.0 atm Cl Cl Cl 5 K p = = = 19.7 torr = atm (0.0) = g C 5 H 6 O 1molC5H 6O 11.10g total mol gas at equilibrium = n TOT = = 0.09 mol C 5 H 6 O initially TOT V 1.6atm.50L = RT L atm 7K K mol = mol C 5 H 6 O (g) C H 6 (g) + CO(g) Initial 0.09 mol 0 0 Let x mol C 5 H 6 O react to reach equilibrium Change x + x +x Equil x x x mol total = 0.09 x + x + x = x, x = mol K = = [C H6][CO] [C H O ] molCH6 (0.0186) molco.50l.50l ( ) molc5h6o.50l = Challenge roblems 79. There is a little trick we can use to solve this problem in order to avoid solving a cubic equation. Because K for this reaction is very small, the dominant reaction is the reverse reaction. We will let the products react to completion by the reverse reaction, then we will solve the forward equilibrium problem to determine the equilibrium concentrations. Summarizing these steps in a table: NOCl(g) NO(g) + Cl (g) K = 1.6 Before 0.0 M 1.0 M Let 1.0 mol/l Cl react completely. (K is small, reactants dominate.) Change React completely After New initial conditions x mol/l of NOCl reacts to reach equilibrium Change x +x +x Equil..0 x x x K = (x) ( x) = (.0 x) x.0 (assuming.0 x.0)

29 7 CHATER 1 CHEMICAL EQUILIBRIUM x 5 = , x =.5 10 ; Assumption good by the 5% rule (x is.5% of.0). [NOCl] = = 1.95 M =.0 M; [NO] = M; [Cl ] = 0.05 M Note: If we do not break this problem into two parts (a stoichiometric part and an equilibrium part), we are faced with solving a cubic equation. The set-up would be: NOCl(g) NO(g) + Cl (g) Initial 0.0 M 1.0 M Change +.0 y y Equil. y.0 y 1.0 y (.0 y) (1.0 y) = (y) If we say that y is small to simplify the problem, then: = ; We get y = 50. This is impossible! y To solve this equation, we cannot make any simplifying assumptions; we have to solve a cubic equation. If you don t have a graphing calculator, this is difficult. Alternatively, we can use some chemical common sense and solve the problem as illustrated above. 80. a. NO(g) + Br (g) NOBr(g) Initial 98. torr 1. torr 0 x torr of NO reacts to reach equilibrium Change x x +x Equil. 98. x 1. x x total = NO = (98. x) + (1. x) + x = 19.7 x Br NOBr total = = 19.7 x, x = 9. torr; NO = 98. (9.) = 0.0 torr = atm Br = = 1.1 torr = atm; NOBr = (9.) = 58. torr = atm K p = NOBr NO Br (0.0768) = 1 (0.056) (0.0159) b. NO(g) + Br (g) NOBr(g) Initial 0.0 atm 0.0 atm 0 x atm of NO reacts to reach equilibrium Change x x +x Equil. 0.0 x 0.0 x x