CNX Chemistry SSM Ch13 Mod02 PDF

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Textbook answers chapter 13 part 2...

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OpenStax College Chemistry 13.2: Equilibrium Constants Chemistry 13: Fundamental Equilibrium Concepts 13.2: Equilibrium Constants 7. Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 13.5. Solution Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase. 9. Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg2Cl2, AgCl, PbCl2, and CuCl. (a) Write the expression for the equilibrium constant for the reaction represented by the equation –  AgCl(s)   Ag ( aq)  Cl ( aq) . Is K > 1, < 1, or ≈ 1? Explain your answer. c

(b) Write the expression for the equilibrium constant for the reaction represented by the equation 2 – Pb  ( aq)  2Cl ( aq)   PbCl2 ( s) . Is K > 1, < 1, or ≈ 1? Explain your answer. c

Solution (a) Kc = [Ag+][Cl–] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 1 Kc  2  Pb2    Cl  > 1 because PbCl2 is insoluble and formation of the solid will M; (b) reduce the concentration of ions to a low level (< 1 M). 11. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is  C6 H6 ( g) . manufactured by the catalytic conversion of acetylene to benzene: 3C2 H2 ( g )   Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer. Solution [C6 H 6 ] Kc = [C2 H2 ]3 Since , a value of K ≈ 10 means that C H predominates over C H . In such a c

6

6

2

2

case, the reaction would be commercially feasible if the rate to equilibrium is suitable. 13. For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction? Solution Kc> 1; the product must be formed in overwhelmingly large proportions. 15. Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions: CH4 ( g )  Cl2 ( g )   CH3Cl( g )  HCl( g) (a)   (b) N2 ( g )  O2 ( g )   2NO( g ) (c)

2SO 2 (g )  O 2 ( g )   2SO 3 ( g )

  (d) BaSO 3 ( s)    BaO(s)  SO2 ( g )   (e) P4 ( g)  5O2 ( g)    P4 O10 ( s)

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OpenStax College Chemistry 13.2: Equilibrium Constants   (f) Br2 ( g)    2Br( g )   (g) CH4 ( g)  2O2 ( g)    CO2 ( g)  2H2 O( l)   (h) CuSO 4 5H 2O( s)    CuSO4 ( s)  5H2 O( g ) Solution

Qc = (a)

Qc =

2 2 SO 3  NO   Qc = = 2  SO 2   O 2  ; (d) Qc= [SO2]; (e)  N2   O2  ; (c) 2  CO2  Br  Qc = =  CH4   O2  2 ; (h) Qc= [H2O]5  Br2 ; (g)

CH 3Cl  HCl  Q  CH 4   Cl2  ; (b) c 1

 P4   O 2 

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Qc

; (f) 17. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.   Kc  17 (a) 2NH3 ( g )    N 2 (g )  3H2 (g ) ; [NH ] = 0.20 M, [N ] = 1.00 M, [H ] = 3

1.00 M   (b) 2NH 3 ( g )    N 2 ( g )  3H 2 (g ) N2 = 2.0 atm, H2 = 1.0 atm

KP  6.8  104

   3 2 2 (c) [O2] = 1.00 M   (d) 2SO 3 ( g )    2SO2 (g )  O2 (g ) = 1.00 atm, O2 = 1.00 atm   (e) 2NO(g )  Cl2 ( g )    2NOCl(g ) [NOCl] = 0 M   (f) N2 ( g)  O2 ( g)    2NO(g ) 2SO ( g )    2SO ( g )  O

K c  0.230

g

KP = 16.5

2

; initial pressures: NH3 = 3.0 atm,

; [SO3] = 0.00 M, [SO2] = 1.00 M,

; initial pressures: SO3 = 1.00 atm, SO2

Kc  4.6  104

; [NO] = 1.00 M, [Cl2] = 1.00 M,

KP  0.050 ; initial pressures: NO = 10.0 atm, N2 =

O2 = 5 atm Solution Qc 

2

[N2 ][H2 ]3 (1.00)(1.00)3   25 [NH 3 ]2 (0.20)2

(a) Qc < Kc, proceeds left; PN 2 ( PH 2 ) 3 (2.0)(1.0)3 QP    0.22 ( PNH3 )2 (3.0)2 (b) QP < KP, proceeds right; [SO2 ]2 [O2 ] (1.00)2 (1.00) Qc    undefined [SO3 ]2 (0) (c) Qc > Kc, proceeds left; (PSO2 )2 PO2 (1.00)2 (1.00) QP    1.00 ( PSO3 ) 2 (1.00)2 (d) Page 2 of 4

OpenStax College Chemistry 13.2: Equilibrium Constants QP > KP, proceeds right; 2 2 ( PNOCl ) (0)  0 QP  ( PNO )2 PCl 2 (1.00)2 (1.00) (e) QP < KP, proceeds right; [NO]2 (10.0)2 Qc   4 [N ][O ] (5.00)(5.00) 2 2 Qc>Kc, proceeds left (f)

19. The following reaction has KP = 4.50  10–5 at 720 K. N2 (g ) + 3H 2 (g )   2NH3 (g ) If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? P(NH3) = 93 atm, P(N2) = 48 atm, and P(H2) = 52 Solution ( PNH3 ) 2 Qp = ( PN 2 )( PH 2 )3 . Plugging in the given The reaction quotient expression for this problem is 2 (93)  (48)  (52)3  , so Qp = 1.3 10–3. Since this value is larger values of partial pressures gives than KP (4.50  10–5), the system will shift toward the reactants to reach equilibrium. 21. Which of the systems described in Exercise 15 give homogeneous equilibria? Which give heterogeneous equilibria? Solution (a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous 23. For which of the reactions in Exercise 15 does Kc (calculated using concentrations) equal KP (calculated using pressures)? Solution When the number of gaseous components are the same on both sides of the equilibrium expression, Kc will equal KP. This situation occurs in (a) and (b). 25. Convert the values of Kc to values of KP or the values of KP to values of Kc.   Kc  0.50 at 400  C (a) N2 ( g)  3H2 ( g)    2NH3 ( g ) (b)

H2  I2   2HI

K c  50.2 at 448  C

  (c) Na 2SO 4 •10H 2O( s)    Na 2SO 4 ( s )  10H 2O(g )   K P  0.122 at 50  C (d) H2 O(l )    H2 O(g )

Solution K P  K c ( RT )  n

K P  4.08  10–25 at 25 C

, where Δn is the sum of gaseous products minus the sum of gaseous reactants. (a) Δn = (2) – (1 + 3) = –2, KP = 0.50[0.08206  673.15]–2 = 1.6  10–4; (b) Δn = (2) – (1 + 1) = 0, KP = Kc(RT)0 = Kc= 50.2; (c) Δn = (10) – (0) = 10, Kc = KP(RT)–Δn, Kc = 4.08  10–25[0.08206  298.15]–10 = 5.31  10–39; (d) Δn = (1) – (0) = 1, Kc = 0.122(0.08206  323.15)–1 = 4.60  10–3 27. What is the value of the equilibrium constant expression for the change    H 2 O( l)   H2 O(g ) at 30 C? Page 3 of 4

OpenStax College Chemistry 13.2: Equilibrium Constants Solution K P  PH2 O

. The vapor pressure of H2O at 1 atm  0.042 atm 3.18 torr  760 torr 30 ºC is 31.8 torr. Converting to atmospheres gives . K P  PH 2O  0.042 . Therefore, 29. Write the reaction quotient expression for the ionization of NH3 in water. Solution + NH3 ( aq)  H2 O( l)   NH4 ( aq)  OH ( aq) . Because the concentration of water is a The equilibrium expression for this transformation is

constant, the term [H2O] is normally incorporated into the reaction quotient as well as the final [NH4  ][OH ] Qc  [HN 3 ] equilibrium constant. This resource file is copyright 2015, Rice University. All Rights Reserved.

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