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Chemistry LibreTexts

15.E: Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

15.1: The Concept of Equilibrium

Conceptual Problems

  1. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the amounts or concentrations of the reactants and the products?
  2. Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.
  3. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level?
  4. Which of these systems exists in a state of chemical equilibrium?
    1. oxygen and hemoglobin in the human circulatory system
    2. iodine crystals in an open beaker
    3. the combustion of wood
    4. the amount of C14 in a decomposing organism

Conceptual Answer

1. When a reaction is described as "having reached equilibrium" this means that the forward reaction rate is now equal to the reverse reaction rate. In regards to the amounts or concentrations of the reactants and the products, there is no change due to the forward reaction rate being equal to the reverse reaction rate.

2. It is not correct to say that the reaction has "stopped" when it has reached equilibrium because it is not necessarily a static process where it can be assumed that the reaction rates cancel each other out to equal zero or be "stopped" but rather a dynamic process in which reactants are converted to products at the same rate products are converted to reactants. For example, a soda has carbon dioxide dissolved in the liquid and carbon dioxide between the liquid and the cap that is constantly being exchanged with each other. The system is in equilibrium and the reaction taking place is: CO2(g)+2H2O(l)H2CO3(aq).

3. Chemical equilibrium is described as a dynamic process because there is a movement in which the forward and reverse reactions occur at the same rate to reach a point where the amounts or concentrations of the reactants and products are unchanging with time. Chemical equilibrium can be described in a saturated solution of NaCl as on the microscopic level Na+ and Cl ions continuously leave the surface of an NaCl crystal to enter the solution, while at the same time Na+ and Cl ions in solution precipitate on the surface of the crystal. At the macroscopic level, the salt can be seen to dissolve or not dissolve depending whether chemical equilibrium was established.

4.

a. Exists in a state of equilibrium as the chemical reaction that occurs in the body is: Hb(aq)+4H2O(l)Hb(O2)4(aq).

b. Exists in a state of equilibrium as the chemical reaction occurs is: H2(g)+I2(g)2HI(g)

c. Does not exist in a state of equilibrium as it is not a reversible process. The chemical reaction that takes place is: 6C10H15O7(s)+heatC50H10O(s)+10CH2O(g).

d. Does not exist in a state of chemical equilibrium as it is not a reversible process. The chemical reaction that takes place is: CH2O+O2H2O(l)+CO2(g)+nutrients.

15.2: The Equilibrium Constant

Conceptual Problems

  1. For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?
  2. Which of the following equilibriums are homogeneous and which are heterogeneous?
    1. 2HF(g)H2(g)+F2(g)
    2. C(s)+2H2(g)CH4(g)
    3. H2C=CH2(g)+H2(g)C2H6(g)
    4. 2Hg(l)+O2(g)2HgO(s)
  3. Classify each equilibrium system as either homogeneous or heterogeneous.
    1. NH4CO2NH2(s)2NH3(g)+CO2(g)
    2. C(s)+O2(g)CO2(g)
    3. 2Mg(s)+O2(g)2MgO(s)
    4. AgCl(s)Ag+(aq)+Cl(aq)
  4. If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?
  5. Industrial production of NO by the reaction N2(g)+O2(g)2NO(g) is carried out at elevated temperatures to drive the reaction toward the formation of the product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?
  6. How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?
  7. What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?
  8. Write the equilibrium constant expressions for K and Kp for each reaction.
    1. CO(g)+H2O(g)CO2(g)+H2(g)
    2. PCl3(g)+Cl2(g)PCl5(g)
    3. 2O3(g)3O2(g)
  9. Write the equilibrium constant expressions for K and Kp as appropriate for each reaction.
    1. 2NO(g)+O2(g)2NO2(g)

    2. 12H2(g)+12I2(g)HI(g)

    3. cisstilbene(soln)transsilbene(soln)

  10. Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?
  11. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.
    1. 2S(s)+3O2(g)2SO3(g)
    2. C(s)+CO2(g)2CO(g)
    3. 2ZnS(s)+3O2(g)2ZnO(s)+2SO2(g)
  12. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.
    1. 2HgO(s)2Hg(l)+O2(g)
    2. H2(g)+I2(s)2HI(g)
    3. NH4CO2NH2(s)2NH3(g)+CO2(g)
  13. At room temperature, the equilibrium constant for the reaction 2A(g)B(g) is 1. What does this indicate about the concentrations of A and B at equilibrium? Would you expect K and Kp to vary significantly from each other? If so, how would their difference be affected by temperature?
  14. For a certain series of reactions, if [OH][HCO3][CO23]=K1 and [OH][H2CO3][HCO3]=K2, what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.
  15. In the equation for an enzymatic reaction, ES represents the complex formed between the substrate S and the enzyme protein E. In the final step of the following oxidation reaction, the product P dissociates from the ESO2 complex, which regenerates the active enzyme:
E+SESK1
ES+O2ESO2K2
ESO2E+PK3

Give the overall reaction equation and show that K=K1×K2×K3.

Conceptual Answers

1. By reversing the reactants and products for an equilibrium reaction, the equilibrium constant becomes: K=1K.

2.

a. This equilibrium is homogenous as all substances are in the same state.

b. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is homogeneous as all substances are in the same state.

d. This equilibrium is heterogeneous as not all substances are in the same state.

3.

a. This equilibrium is heterogeneous as not all substances are in the same state.

b. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is heterogeneous as not all substances are in the same state.

d. This equilibrium is heterogeneous as not all substances are in the same state.

4. According to Le Chatelier’s principle, equilibrium will shift in the direction to counteract the effect of a constraint (such as concentration of a reactant, pressure, and temperature). Thus, in an endothermic reaction, the equilibrium shifts to the right-hand side when the temperature is increased which increases the equilibrium constant and the equilibrium shifts to the left-hand side when the temperature is decreased which decreases the equilibrium constant.

5. After sufficient industrial production of NO by the reaction of N2(g)+O2(g)2NO(g) at elevated temperatures to drive the reaction toward the formation of the product, the reaction mixture is cooled quickly because it quenches the reaction and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.

6. To differentiate between a system that has reached equilibrium and one that is reacting slowly that changes in concentrations are difficult to observe we can use Le Chatelier’s principle to observe any shifts in the reaction upon addition of a constraint (such as concentration, pressure, or temperature).

7. The relationship between the equilibrium constant, the concentration of each component of a system, and the rate constants for the forward and reverse reactions considering a reaction of a general form: aA+bBcC+dD is K=[C]c[D]d[A]a[B]b=kfkr

8.

a.

K=[CO2][H2][CO][H2O]

Kp=(PCO2)(PH2)(PCO)(PH2O)

b.

K=[PCl5][PCl3][Cl2]

Kp=(PPCl5)(PCl3)(PCl2)

c.

K=[O2]3[O3]2

Kp=(PO2)3(PO3)2

9.

a.

K=[NO2]2[NO]2[O2]

Kp=(PNO2)2(PNO)2(PO2)

b.

K=[HI][H2]12[O2]

Kp=(PHI)(PH2)12(PO2)

c.

K=transstilbenecisstilbene

Kp=(Ptransstilbene)(Pcisstilbene)

10. It is incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression because they are not reactive enough or cause a change in the concentrations of the ions or the species that exist in the gas phase.

11.

  1.  

K=[SO3]2[O2]3

Kp=(PSO3)2(PO2)3

b.

K=[CO]2[CO2]

Kp=(PCO)2(PCO2)

c.

K=[SO2]2[O2]3

Kp=(PSO2)2(PO2)3

12.

a.

K=[O2]

Kp=(PO2)

b.

K=[HI]2[H2]

Kp=(PHI)2(PH2)

c.

K=[NH3]2[CO2]

Kp=(PNH3)2(PCO2)

13.

K=[B][A]21=[B][A]2[A]2=[B][A]=B

K and Kp vary by RT, but it largely depends on T as R is constant. A raise or decrease in temperature would cause a difference.

Kp=K(RT)Δn=K(RT)1=KRT

Δn=(totalmolesofgasontheproductside)(totalofmolesonthereactantside)=12=1

14.

K=K1K2=[OH][HCO3][CO23][OH][H2CO3][HCO3]=[HCO3]2[CO23][H2CO3]

CO23(g)+H2CO3(g)2HCO3(g)

15.

K=K1×K2×K3=[ES][E][S]×[ESO2][ES][O2]×[E][P][ESO2]=[P][S][O2]

S+O2P

Numerical Problems

  1. Explain what each of the following values for K tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: K=0.892; K=3.25×108; K=5.26×1011. Are products or reactants favored at equilibrium?
  2. Write the equilibrium constant expression for each reaction. Are these equilibrium constant expressions equivalent? Explain.
    1. N2O4(g)2NO2(g)
    2. 12N2O4(g)NO2(g)
  3. Write the equilibrium constant expression for each reaction.
    1. 12N2(g)+32H2(g)NH3(g)
    2. 13N2(g)+H2(g)23NH3(g)

How are these two expressions mathematically related to the equilibrium constant expression for

N2(g)+3H2(g)2NH3(g)?

  1. Write an equilibrium constant expression for each reaction.
    1. C(s)+2H2O(g)CO2(g)+2H2(g)
    2. SbCl3(g)+Cl2(g)SbCl5(g)
    3. 2O3(g)3O2(g)
  2. Give an equilibrium constant expression for each reaction.

a. 2NO(g)+O2(g)2NO2(g)

b. 12H2(g)+12I2(g)HI(g)

c. CaCO3(s)+2HOCl(aq)Ca2+(aq)+2OCl(aq)+H2O(l)+CO2(g)

6. Calculate K and Kp for each reaction.

  1. 2NOBr(g)2NO(g)+Br2(g): at 727°C, the equilibrium concentration of NO is 1.29 M, Br2 is 10.52 M, and NOBr is 0.423 M.
  2. C(s)+CO2(g)2CO(g): at 1,200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm CO2 and 76.8 atm CO, and the vessel contains 3.55 g of carbon.

7. Calculate K and Kp for each reaction.

  1. N2O4(g)2NO2(g): at the equilibrium temperature of −40°C, a 0.150 M sample of N2O4 undergoes a decomposition of 0.456%.
  2. CO(g)+2H2(g)CH3OH(g): an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of 6.71×102 atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7,193 g of methanol.

8. Determine K and Kp (where applicable) for each reaction.

  1. 2H2S(g)2H2(g)+S2(g): at 1065°C, an equilibrium mixture consists of 1.00×103 M H2, 1.20×103 M S2, and 3.32×103 M H2S.
  2. Ba(OH)2(s)2OH(aq)+Ba2+(aq): at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.

9. Determine K and Kp for each reaction.

  1. 2NOCl(g)2NO(g)+Cl2(g): at 500 K, a 24.3 mM sample of NOCl has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of NOCl.
  2. Cl2(g)+PCl3(g)PCl5(g): at 250°C, a 500 mL reaction vessel contains 16.9 g of Cl2 gas, 0.500 g of PCl3, and 10.2 g of PCl5 at equilibrium.

10. The equilibrium constant expression for a reaction is [CO2]2[SO2]2[O2]. What is the balanced chemical equation for the overall reaction if one of the reactants is Na2CO3(s)?

11. The equilibrium constant expression for a reaction is [NO][H2O]32[NH3][O2]54. What is the balanced chemical equation for the overall reaction?

12. Given K=kfkr, what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?

13. The value of the equilibrium constant for 2H2(g)+S2(g)2H2S(g) is 1.08×107 at 700°C. What is the value of the equilibrium constant for the following related reactions

  1. H2(g)+12S2(g)H2S(g)
  2. 4H2(g)+2S2(g)4H2S(g)
  3. H2S(g)H2(g)+12S2(g)

Numerical Answers

1. In the given equilibrium reaction where K=0.8921 has a concentration of the reactants that is approximately equal to the concentration of the products so neither formation of the reactants or products is favored. In the given equilibrium reaction where K=3.25×108>1 has a concentration of the products that is relatively small compared to the concentration of the reactants so the formation of the products is favored. In the given equilibrium reaction where K=5.26×1011<1 has a concentration of products that is relatively large compared to that of the concentration of the reactants so the formation of the reactants is favored.

2.

a. K=[NO2]2[N2O4]

b. K=[NO2][N2O4]12

Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the K value for a. We would need to square it to get the K value for b.

3.

  1. K=[NH3][N2]12[H2]32

  2. K

K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}

K'=K^{\dfrac{1}{2}}

K''=K^{\dfrac{1}{3}}

4.

a. K=\dfrac{[H_2]^2[CO_2]}{[H_{2}O]^2}

b. K=\dfrac{[SbCl_5]}{[SbCl_3][Cl_2]}

c. K=\dfrac{[O_2]^3}{[O_3]^2}

5.

  1. K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}
  2. K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[I_2]^{\dfrac{1}{2}}}
  3. K=\dfrac{[Ca^{2+}][OCl^−]^2[CO_2]}{[HOCl]^2}

6.

a.

K=\dfrac{[NO]^2[Br]}{[NOBr]^2}=\frac{[1.29\,M]^2[10.52\,M]}{[0.423\,M]^2}=97.8

K_p=K(RT)^{Δn}=(97.8)((0.08206\frac{L\cdot atm}{mol\cdot K})((727+273.15)K))^{3-2}=8.03x10^{4}

b.

K_p=K(RT)^{Δn}=K(RT)^{2-1}=K(RT) \rightarrow K=\dfrac{K_p}{RT}=\frac{63.1}{(0.08206\frac{L\cdot atm}{mol\cdot K})(1,200\,K)}=6.41

K=\dfrac{(P_{CO})^2}{P_{CO_2}}=\frac{(76.8\,atm)^2}{93.5\,atm}=63.1

7.

  1.  

K=\dfrac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{[0.001368\,M]}{[0.149316\,M]}=1.25x10^{-5}

[NO_2]=(2)(0.150\,M)(0.00456)=0.001368\,M

[N_2O_4]=(0.150\,M)(1-0.00456)=0.149316\,M

K_p=K(RT)^{Δn}=(1.25x10^{-5})((0.08206\frac{L\cdot atm}{mol\cdot K})((-40+273.15)K))^{2-1}=2.39x10^{-4}

b.

K=\dfrac{[CH_{3}OH]}{[CO][H_{2}]^2}=\frac{[14.5\,M]}{[1.05\,M][1.21\,M]^2}=9.47

[CH_{3}OH]=7,193\,g\,CH_{3}OH \times \frac{1\,mol\,CH_{3}OH}{32.04\,g\,CH_{3}OH} \times \frac{1}{15.5\,L}=14.5\,M

[CO]=457.7\,g\,CO \times \frac{1\,mol\,CO}{28.01\,g\,CO} \times \frac{1}{15.5\,L}=1.05\,M

[H_{2}]=37.8\,g\,H_{2} \times \frac{1\,mol\,H_{2}}{2.02\,g\,H_{2}} \times \frac{1}{15.5\,L}=1.21\,M

K_p=K(RT)^{Δn}=(9.47)((0.08206\frac{L\cdot atm}{mol\cdot K})((227+273.15)K))^{1-3}=5.62x10^{-2}

8.

a.

K=\dfrac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}=\frac{[1.00 \times 10^{-3}\,M]^2[1.20 \times 10^{-3}\,M]}{[3.32 \times\ 10^{-3}\,M]^2}=1.09 \times 10^{-4}

K_p=K(RT)^{Δn}=(1.09 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})((1065+273.15)K))^{3-2}=1.20 \times 10^{-2}

b.

K=[OH^{-}]^2[Ba^{2+}]=[0.2136\,M]^2[0.1068\,M]=4.87 \times 10^{-3}

[OH^{-}]=0.0534\,mol\,OH^{-} \times \frac{1}{0.25\,L}=0.2136\,M

[Ba^{2+}]=0.0267\,mol\,Ba^{2+} \times \frac{1}{0.25\,L}=0.1068\,M

K_p=K(RT)^{Δn}=(4.87 \times 10^{-3})((0.08206\frac{L\cdot atm}{mol\cdot K})((25+273.15)K))^{3-1}=2.92

9.

a.

K=\dfrac{[NO]^2[Cl_{2}]}{[NOCl]^2}=4.59 \times 10^{-4}

K_p=K(RT)^{Δn}=(4.59 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})(500K))^{3-2}=1.88 \times 10^{-2}

b.

K=\dfrac{[PCl_5]}{[Cl_{2}][PCl_{3}]}=\frac{[9.80 \times 10^{-2}\,M]}{[4.77 \times 10^{-1}\,M][7.28 \times 10^{-3}\,M]}=28.2

[PCl_{5}]=10.2\,g\,PCl_{5} \times \frac{1\,mol\,PCl_{5}}{208.2388\,g\,PCl_{5}} \times \frac{1}{0.5\,L}=9.80 \times 10^{-2}\,M

[Cl_{2}]=16.9\,g\,Cl_{2} \times \frac{1\,mol\,Cl_{2}}{70.9\,g\,Cl_{2}} \times \frac{1}{0.5\,L}=4.77 \times 10^{-1}\,M

[PCl_{3}]=0.500\,g\,PCl_{3} \times \frac{1\,mol\,PCl_{3}}{137.33\,g\,PCl_{3}} \times \frac{1}{0.5\,L}=7.28 \times 10^{-3}\,M

K_p=K(RT)^{Δn}=(28.2)((0.08206\frac{L\cdot atm}{mol\cdot K})(250+273.15)K)^{1-2}=6.57 \times 10^{-1}

10. 2\,SO_{2}\,(g)+O_{2}\,(g)+2\,Na_{2}CO_{3}\,(s) \rightleftharpoons 2\,CO_{2}\,(g)+2\,Na_{2}SO_{4}\,(s)

11. NH_{3}\,(g) + \frac{5}{4}\,O_{2}\,(g)⇌NO \,(g)+\frac{3}{2}\,H_{2}O\,(g)

12. If the reaction rate of the forward reaction is doubled the magnitude of the equilibrium constant is doubled. If the reaction rate of the reverse reaction for the overall reaction is described by a factor of 3 the magnitude of the equilibrium constant is increased by a factor of 3.
13. K’= \dfrac{[H_{2}S]^2}{[H_{2}]^2[S_{2}]}= 1.08 \times 10^{7}

a. K= \dfrac{[H_{2}S]}{[H_{2}][S_{2}]^\frac{1}{2}}=K’^{\frac{1}{2}}=(1.08 \times 10^{7})^{\frac{1}{2}}=3.29 \times 10^{3} b. K= \dfrac{[H_{2}S]^4}{[H_{2}]^4[S_{2}]^2}=K’^{2}=(1.08 \times 10^{7})^{2}=1.17 \times 10^{14} c. K= \dfrac{[H_{2}][S_2]^\frac{1}{2}}{[H_{2}S]}=K’^{-\frac{1}{2}}= (1.08 \times 10^{7})^{-\frac{1}{2}}=3.04 \times 10^{-4}

15.3: Interpreting & Working with Equilibrium Constants


Conceptual Problems

  1. Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.
  2. Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A+2B \rightleftharpoons C for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

Conceptual Answers

1. The magnitude of the equilibrium constant for a reaction depends on the form in which the chemical reaction is written. For example, writing a chemical reaction in different but chemically equivalent forms causes the magnitude of the equilibrium constant to be different but can be related by comparing their respective magnitudes.

2.

a. When K is very large the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion.

K= \dfrac{[C]}{[A][B]^2}=\frac{[C]}{very\,small}=\frac{1}{0}= \infty \rightarrow [C]= \infty

b. When K is very small the reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of the reactants.

K=\dfrac{[C]}{[A][B]^2}=\frac{very\,small}{[A][B]^2}=\frac{0}{1}=0 \rightarrow [C]=0

Simplifying assumptions should not be used if the equilibrium constant is not known to be very large or very small.

Numerical Problems

Please be sure you are familiar with the quadratic formula before proceeding to the Numerical Problems.

  1. In the equilibrium reaction A+B \rightleftharpoons C, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction 2\,A \rightleftharpoons B+C?
  2. The following table shows the reported values of the equilibrium P_{O_2} at three temperatures for the reaction Ag_{2}O\,(s) \rightleftharpoons 2\,Ag\,(s)+ \frac{1}{2}\,O_{2}\,(g), for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?
T (°C) P_{O_2}\;(mmHg)
150 182
184 143
191 126
  1. Given the equilibrium system N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g), what happens to K_p if the initial pressure of N_2O_4 is doubled? If K_p is 1.7 \times 10^{−1} at 2300°C, and the system initially contains 100% N_2O_4 at a pressure of 2.6 \times 10^2 atm, what is the equilibrium pressure of each component?
  2. At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g) At equilibrium, [H_2] = 0.047\;M and [HI] = 0.345\;M What are K and K_p for this reaction?
  3. Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g) with K_p = 1.3 \times 10^{−4}. If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?
  4. Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction A\,(s) \rightleftharpoons 2\,B\,(g)+C\,(g), what is K_p?
  5. The decomposition of ammonium carbamate to NH_3 and CO_2 at 40°C is written as NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g). If the partial pressure of NH_3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO_2? What is the total gas pressure of the system? What is K_p?
  6. At 375 K, K_p for the reaction SO_{2}Cl_{2}\,(g) \rightleftharpoons SO_{2}\,(g)+Cl_{2}\,(g) is 2.4, with pressures expressed in atmospheres. At 303 K, K_p is 2.9 \times 10^{−2}.
    1. What is K for the reaction at each temperature?
    2. If a sample at 375 K has 0.100 M Cl_2 and 0.200 M SO_2 at equilibrium, what is the concentration of SO_2Cl_2?
    3. If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
  7. For the gas-phase reaction a\,A \rightleftharpoons b\,B, show that K_p = K(RT)^{Δn} assuming ideal gas behavior.
  8. For the gas-phase reaction I_2 \rightleftharpoons 2\,I, show that the total pressure is related to the equilibrium pressure by the following equation: P_T=\sqrt{K_{p}P_{I_{2}}} + P_{I_{2}}
  9. Experimental data on the system Br_{2}\,(l) \rightleftharpoons Br_{2}\,(aq) are given in the following table. Graph Br_{2}\,(aq) versus moles of Br_{2}\,(l) present; then write the equilibrium constant expression and determine K.
Grams Br_{2} in 100 mL Water Br_{2} (M)
1.0 0.0626
2.5 0.156
3.0 0.188
4.0 0.219
4.5 0.219
  1. Data accumulated for the reaction (\n-butane(g) \rightleftharpoons isobutane(g)\) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?
Moles n-butane Moles Isobutane
0.5 1.25
1.0 2.5
1.50 3.75
  1. Solid ammonium carbamate (NH_{4}CO_{2}NH_{2}) dissociates completely to ammonia and carbon dioxide when it vaporizes: NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g) At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is K_p? If the concentration of CO_2 is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the NH_{3} concentration is necessary for the system to restore equilibrium?
  2. The equilibrium constant for the reaction COCl_{2}\,(g) \rightleftharpoons CO\,(g)+Cl_{2}\,(g) is K_p = 2.2 \times 10^{−10} at 100°C. If the initial concentration of COCl_{2} is 3.05 \times 10^{−3}\; M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?
  3. Aqueous dilution of IO_{4}^{−} results in the following reaction: IO^−_{4}\,(aq)+2\,H_{2}O_(l)\, \rightleftharpoons H_4IO^−_{6}\,(aq) with K = 3.5 \times 10^{−2}. If you begin with 50 mL of a 0.896 M solution of IO_4^− that is diluted to 250 mL with water, how many moles of H_4IO_6^− are formed at equilibrium?
  4. Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation: I_{2}\,(g)+Br_{2}\,(g) \rightleftharpoons 2\,IBr\,(g) with K_p = 1.2 \times 10^2. If you begin the reaction with 7.4 g of I_2 vapor and 6.3 g of Br_2 vapor in a 1.00 L container, what is the concentration of IBr\,(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?
  5. For the reaction 2\,C\,(s) + \,N_{2}\,(g)+5\,H_{2}\, \rightleftharpoons 2\,CH_{3}NH_{2}\,(g) with K = 1.8 \times 10^{−6}. If you begin the reaction with 1.0 mol of N_2, 2.0 mol of H_2, and sufficient C\,(s) in a 2.00 L container, what are the concentrations of N_2 and CH_3NH_2 at equilibrium? What happens to K if the concentration of H_2 is doubled?

Numerical Answers

1. In both cases, the equilibrium constant will remain the same as it does not depend on the concentrations.

2. No, the data is not consistent with what I would expect to occur because enthalpy is positive indicating that the reaction is endothermic thus heat is on the left side of the reaction. As the temperature is raised P_{O_2} would be expected to increase to counteract the constraint.

3.

If the initial pressure of N_2O_4 was doubled then K_p is one half of the original value.

K_p=\dfrac{(P_{NO_{2}})^2}{(P_{N_{2}O_{4}})} \rightarrow 1.7 \times 10^{-1} = \frac{(2x)^2}{2.6 \times 10^2 -x} \rightarrow 44.2-0.17x=4x^2 \rightarrow x \approx 3.303\,atm

  P_{N_{2}O_{4}} P_{NO_{2}}
I 2.6 \times 10^{2} 0
C -x +2x
E 2.6 \times 10^2-x 2x

P_{N_{2}O_{4}}=2.6 \times 10^2-x=260-3.303=2.6 \times 10^2\,atm

P_{NO_{2}}=2x=(2)(3.303)=6.6\,atm

4.

K=\frac{[HI]^2}{[H_2][I_2]}=\frac{0.345\,M}{(0.047\,M)(0.047\,M)}=157

K_p=K(RT)^{Δn}=(157)((0.08206\frac{L\cdot atm}{mol\cdot K})(430+273.15)K)^{2-2}=157

5.

Maximum\;Percent\;Yield=\frac{Actual}{Theoretical} \times 100\%=\frac{212.593}{376.127} \times 100\%=56.52\% \approx 57\%

PV=nRT \rightarrow P=\frac{(1.999\;mol)(0.08206\frac{L\cdot atm}{mol\cdot K})(300+273.15)K}{0.250\;L}=376.127\;atm

[CO]=56.0\;g\;CO \times \frac{1\;mol\;CO}{28.01\,g\,CO}=1.999\,mol\,CO

K_p=\frac{P_{CH_{3}OH}}{(P_{CO})(P_{H_2})^{2}} \rightarrow 1.3 \times 10^{-4}= \frac{x}{(376.02-x)(100^2)} \rightarrow 1.3= \frac{x}{376.07-x} \rightarrow 488.965-1.3x=x \rightarrow 488.965=2.3x \rightarrow x=212.593\,atm

K_p=\frac{(P_{CH_{3}OH})}{(P_{CO})(P_{H_2})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(376.127-357.320)(P_{H_{2}})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(18.80635)(P_{H_{2}})^2} \rightarrow 0.002444(P_{H_{2}})^2=357.320 \rightarrow (P_{H_{2}})=382.300 \approx 3.8 \times 10^2\,atm

Minimum\,Percent\,Yield=\frac{Actual}{Theoretical} \times 100 \% \rightarrow 95\%=\frac{x}{376.127} \times 100\% \rightarrow x=357.320\,atm

  CO 2\,H_2 CH_{3}OH
I 376.127 100 0
C -x -2x +x
E 376.127-x 100 (maintained) x

6. K_p=\frac{(P_B)^2(P_C)}{P_A}=\frac{[2x]^2[x]}{[0.969-x]}=\frac{4x^3}{0.969-x}

  A 2\,B C
I 0.969 0 0
C -x +2x +x
E 0.969-x 2x x

7.

P_{CO_{2}}=P_{tot}-P_{NH_{3}}=P_{tot}-0.242\,atm

P_{tot}=P_{NH_3}+P_{CO_2}=0.242\,atm+P_{CO_2}

K_p=(P_{NH_3})^2(P_{CO_2})=(0.242\,atm)^2(P_{CO_2})

8.

a.

At\,375\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(375\,K))^{2-1}}=7.80 \times 10^{-2}

At\,303\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(303\,K))^{2-1}}=1.17 \times 10^{-3}

b. K=\frac{[SO_{2}][Cl_{2}]}{[SO_{2}Cl_{2}]} \rightarrow [SO_{2}Cl_{2}]=\frac{[0.200\,M][0.100\,M]}{7.80 \times 10^{-2}}=2.56 \times 10^{-1}\,M

c. If the sample given in part b is cooled to 303 K, the pressure inside the bulb would decrease.

9.

K_p=\frac{(P_B)^b}{(P_A)^a}=\frac{((\frac{n_B}{V})(RT))^b}{((\frac{n_A}{V})(RT))^a}=\frac{[B]^{b}(RT)^b}{[A]^a(RT)^a}=K(RT)^{b-a}=K(RT)^{Δn}

PV=nRT \rightarrow P=\frac{n}{V}RT

K=\frac{[B]^{b}}{[A]^{a}}

Δn=b-a

10.

P_T=P_I+P_{I_2}=\sqrt{K_pP_{I_2}}+P_{I_2}

K_p=\frac{(P_I)^2}{P_{I_2}} \rightarrow (P_I)^2=K_p(P_{I_2}) \rightarrow P_I=\sqrt{K_pP_{I_2}}

11.

The graph should be a positive linear correlation.

Br_2\,(l)\,(M)\,(x-axis) Br_2\,(aq)\,(M)\,(y-axis)\,(same\,value\,for\,K)
6.26 \times 10^-2 0.0626
1.56 \times 10^-1 0.156
1.88 \times 10^-2 0.188
2.50 \times 10^-2 0.219
2.82 \times 10^-2 0.219

[Br_2\,(l)]=1.0\,g\,Br_2 \times \frac{1\,mol\,Br_2}{159.808\,g\,Br_2} \times \frac{1}{0.1\,L}=6.26 \times 10^{-2}

K=\frac{[Br_2\,(aq)]}{[Br_2\,(l)]}=\frac{[Br_2\,(aq)]}{1}=[Br_2\,(aq)]

12. K=\frac{[isobutane]}{[n-butane]}=\frac{x}{1-x}

  n-butane isobutane
I 1 0
C -x +x
E 1-x x

13.

P_{NH_3}=2x=2(0.0387)=7.73 \times 10^{-2}\,atm

P_{CO_2}=x=3.87 \times 10^{-2}\,atm

K_p=(P_{NH_3})^2(P_{CO_2})=(2x)^2(x)=4x^3=4(0.0387)^3=2.32 \times 10^{-4}

P_{tot}=P_{NH_3}+P_{CO_2} \rightarrow 0.116=2x+x \rightarrow 0.116=3x \rightarrow x=0.0387

If the concentration of CO_{2} is doubled and then equilibrates to its initial equilibrium partial

pressure +x atm, the concentration of NH_{3} should also be doubled for the system to restore

equilibrium.

  NH_{3} CO_{2}
I 0 0
C 2x x
E 2x x

14.

P_{COCl_{2}}=9.34 \times 10^{-2}-x=9.34 \times 10^{-2}-9.34 \times 10^{-22}=9.34 \times 10^{-2}\,atm

P_{CO}=x=9.34 \times 10^{-22}\,atm

P_{Cl_{2}}=x=9.34 \times 10^{-22}\,atm

Assume that the equilibrium mainly lies on the reactants side because the K_p value is less than 1.

K_p=\frac{(P_{CO})(P_{Cl_{2}})}{(P_{COCl_{2}})} \rightarrow 2.2 \times 10^{-10} =\frac{x^{2}}{9.34 \times 10^{-2}-x} \rightarrow 2.0548 \times 10^{-11}-2.2 \times 10^{-10}x=x^{2} \rightarrow x^{2}+2.2 \times 10^{-10}x-2.0548 \times 10^{-11}=0 \rightarrow x=9.34 \times 10^{-22}

PV=nRT \rightarrow P=\frac{nRT}{V}=(3.05 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(100+273.15)K=9.34 \times 10^{-2}

  COCl_{2} CO Cl_{2}
I 9.34 \times 10^{-2} 0 0
C -x +x +x
E 9.34 \times 10^{-2}-x x x

15. [H_{4}IO_{6}^{-}]=x=1.6 \times 10^{-3}\,mol

K=\frac{[H_{4}IO_{6}^{-}]}{[IO_{4}^{-}]} \rightarrow 3.5 \times 10^{-2} =\frac{x}{(0.0448-x)} \rightarrow x=1.568 \times 10^{-3}

IO_{4}^{-}:50\,mL\,IO_{4}^{-} \times \frac{1\,L\,IO_{4}^{-}}{1,000\,mL\,IO_{4}^{-}} \times \frac{0.896\,mol\,IO_{4}^{-}}{1\,L\,IO_{4}^{-}}=0.0448\,mol

  IO_{4}^{-} H_{4}IO_{6}^{-}
I 0.0448 0
C -x +x
E 0.0448-x x

16. PV=nRT \rightarrow \frac{P}{RT}=\frac{n}{V} \rightarrow \frac{12.9468}{(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K}=3.5 \times 10^{-1}\,M

K_p=\frac{(P_{IBr})^2}{(P_{I_{2}})(P_{Br_{2}})} \rightarrow 1.2 \times 10^{-2} = \frac{2x}{(1.096-x)(1.479-x)}=\frac{2x}{x^2-2.575x+1.62098} \rightarrow 0.012x^2-0.0309x+0.0194518=2x \rightarrow 0.012x^2-2.0309+0.0194518=0 \rightarrow x=12.9468

  I_{2} Br_{2} IBr
I 1.096 1.479 0
C -x -x 2x
E 1.096-x 1.479-x 2x

PV=nRT \rightarrow P=\frac{nRT}{V}=(2.92 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.096\,atm

[I_{2}]=7.4\,g\,I_{2} \times \frac{1\,mol\,I_{2}}{253\,g\,I_{2}} \times \frac{1}{1.00\,L}=2.92 \times 10^{-2}\,M

PV=nRT \rightarrow P={nRT}{V}=(3.94 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.479\,atm

[Br_{2}]=6.3\,g\,Br_{2} \times \frac{1\,mol\,Br_{2}}{159.808\,g\,Br_{2}}=3.94 \times 10^{-2}\,M

17.

[N_{2}]=0.5-12x=0.5-12(0.000471330)=0.494\,M

[CH_{3}NH_{2}]=2x=2(0.0004713300=9.43 \times 10^{-4}\,M

If\,the\,concentration\,of\,H_{2}\,is\,doubled\,,then\,K=\frac{[CH_{3}NH_{2}]^{2}}{[N_2][H_2]^5}=\frac{(9.43 \times 10^{-4})^{2}}{(0.494)(1.998)^5}=5.65 \times 10^{-8}

2 \times [H_{2}]=2(1.0-2.5x)=2(1.0-2.5(0.000471330))=1.998\,M

K=\frac{[CH_{3}NH_{2}]^2}{[N_2][H_2]^5} \rightarrow 1.8 \times 10^{-6}=\frac{(2x)^2}{(0.5-x)(1.0-5x)^5} \rightarrow x=0.000471330\,M

  C N_2

H_2

CH_{3}NH_{2}
I - 0.5 1.0 0
C - -x -5x 2x
E - 0.5-x 1.0-5x 2x

[N_2]=1.00\,mol\,N_2 \times \frac{1}{2.00\,L}=0.5\,M

[H_2]=2.00\,mol\,H_2 \times \frac{1}{2.00\,L}=1.0\,M

 

15.6: Applications of Equilibrium Constants


Conceptual Problems

1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.8 and Figure 15.9 as your guides, sketch the shape of each graph using appropriate labels.

  1. H_2O\,(l) \rightleftharpoons H_2O\,(g)
  2. 2\,MgO\,(s) \rightleftharpoons 2\,Mg\,(s)+O_{2}\,(g)
  3. 2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)
  4. 2\,PbS\,(g)+3\,O_{2}\,(g) \rightleftharpoons 2\,PbO\,(s)+2\,SO_{2}\,(g)

2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  1. 2\,NaHCO_{3}\,(s) \rightleftharpoons Na_2CO_{3}\,(s) + CO_{2}\,(g)+ H_2O\,(g): [CO_2] is doubled.
  2. N_2F_{4}\,(g) \rightleftharpoons 2\,NF_{2}\,(g): [NF_{2}] is decreased by a factor of 2.
  3. H_{2}\,(g) + I_{2}\,(g) \rightleftharpoons 2\,HI\,(g): [I_2] is doubled.

3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  1. CS_{2}\,(g) + 4\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g) + 2\,H_2S\,(g): [CS_2] is doubled.
  2. PCl_{5}\,(g) \rightleftharpoons PCl_{3}\,(g) + Cl_{2}\,(g): [Cl_2] is decreased by a factor of 2.
  3. 4\,NH_{3}\,(g) + 5\,O_{2}\,(g) \rightleftharpoons 4\,NO\,(g) + 6\,H_2O\,(g): [NO] is doubled.

Conceptual Answer

1.

a. According to Figure 15.8, we could obtain a graph with the x-axis labeled [H_2O]\,(l)\,(M) and y-axis labeled [H_2O]\,(g)\,(M). The graph should have a positive linear correlation. For any equilibrium concentration of H_2O\,(g), there is only one equilibrium H_2O\;(l). Because the magnitudes of the two concentrations are directly proportional, a large [H_2O]\,(g) at equilibrium requires a large [H_2O]\,(l) and vice versa. In this case, the slope of the line is equal to K.

b. According to Figure 15.9, we could obtain a graph with the x-axis labeled [O2]\,(M) and y-axis labeled [MgO]\,(M). Because O_2\,(g) is the only one in gaseous form, the graph would depend on the concentration of O_2.

c. According to Figure 15.8, we could obtain a graph with the x-axis labeled [O_3]\,(M) and y-axis labeled [O2]\,(M). The graph should have a positive linear correlation. For every 3\,O_2\,(g) there is 2\,O_3\,(g). Because the magnitudes of the two concentrations are directly proportional, a large [O3]\,(g) at equilibrium requires a large [O_2]\,(g) and vice versa. In this case, the slope of the line is equal to K.

d. According to figure 15.8, we could obtain a graph with the x-axis labeled [O2]\,(M) and y-axis labeled [SO2]\,(M). The graph should have a positive linear correlation. For every 3\,O_{2}\,(g) there is 2\,SO_2\,(g). Because the magnitudes of the two concentrations are directly proportional, a large O_2\,(g) at equilibrium requires a large SO_2\,(g) and vice versa. In this case, the slope of the line is equal to K.

2.

a.

K=[Na_{2}CO_{3}][CO_2][H_{2}O]

If [CO_2] is doubled, [H_2O] should be halved if the system is to maintain equilibrium.

b.

K=\frac{[NF_2]^2}{[N_{2}F_{4}]}

If [NF_2] is decreased by a factor of 2, then [N_{2}F_{4}] must also be decreased by a factor of 2 if the system is to maintain equilibrium.

c.

K=\frac{[HI]^2}{[H_2][I_{2}]}

If [I_{2}] is doubled then [HI] must also be doubled if the system is to maintain equilibrium.

3.

  1.  

K=\dfrac{[CH_4][H_2S]^2}{[CS_2][H_2]^4}

If [CS_2] is doubled then [H_2] must be decreased by a factor of 2√4≅ 1.189 if the system is to maintain equilibrium.

b.

K=\dfrac{[PCl_3]}{[Cl_2][PCl_5]}

If [Cl_2] is halved then [PCl_5] must also be halved if the system is to maintain equilibrium.

c.

K=\dfrac{[NO]^4[H_2O]^6}{[NH_3][O_2]^5}

If [NO] is doubled then [H_2O] must also be multiplied by 22/3≅1.587 if the system is to maintain equilibrium.

Numerical Problems

  1. The data in the following table were collected at 450°C for the reaction N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g):
  Equilibrium Partial Pressure (atm)
P (atm) NH_3 N_2 H_2
30 (equilibrium) 1.740 6.588 21.58
100 15.20 19.17 65.13
600 321.6 56.74 220.8

The reaction equilibrates at a pressure of 30 atm. The pressure on the system is first increased to 100 atm and then to 600 atm. Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?

  1. For the reaction A \rightleftharpoons B+C, K at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?
Experiment A B C
1 2.50 M 2.50 M 2.50 M
2 1.30 atm 1.75 atm 14.15 atm
3 12.61 mol 18.72 mol 6.51 mol
  1. The following two reactions are carried out at 823 K:

CoO\,(s)+H_{2}\,(g) \rightleftharpoons Co\,(s)+H_2O\,(g) \text{ with } K=67

CoO\,(s)+CO\,(g) \rightleftharpoons Co\,(s)+CO_{2}\,(g) \text{ with } K=490

  1. Write the equilibrium expression for each reaction.
  2. Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of H_2 or CO plus 0.500 mol CoO.
  3. Using the information provided, calculate Kp for the following reaction: H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_2O\,(g)
  4. Describe the shape of the graphs of [reactants] versus [products] as the amount of CoO changes.
  1. Hydrogen iodide (HI) is synthesized via H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g), for which K_p = 54.5 at 425°C. Given a 2.0 L vessel containing 1.12 \times 10^{−2}\,mol of H_2 and 1.8 \times 10^{−3}\,mol of I_2 at equilibrium, what is the concentration of HI? Excess hydrogen is added to the vessel so that the vessel now contains 3.64 \times 10^{−1}\,mol of H_2. Calculate Q and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?

Answers

  1. The system is not at equilibrium at each of these higher pressures. To reach equilibrium, the reaction will proceed to the right to decrease the pressure because the equilibrium partial pressure is less than the total pressure.

K_p=\frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3}=\frac{[15.20]^2}{[19.17][65.13]^3}=4.4 \times 10^{-5}

K_p=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^3}=\frac{[321.6]^2}{[56.74][220.8]^3}= 1.7 \times 10^{-4}

2.

1. K=\frac{[B][C]}{[A]}=\frac{[2.50][2.50]}{[2.50]}=2.50

2. K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{19.0}{((0.08206\frac{L\cdot atm}{mol\cdot K})(200+273.15)K)^{2-1})}=0.49

K_p=\frac{(P_B)(P_C)}{(P_A)}=\frac{(1.75)(14.15)}{1.30}=19.0

3. K=\frac{[B][C]}{[A]^{2}}=\frac{(18.72)(6.51)}{12.61}=9.7

Experiment 1 is about the same as the given K value and thus considered to be about equilibrium. The second experiment has a K value that is about 1 so neither the formation of the reactants or products is favored. The third experiment has a K value that is larger than 1 so the formation of the products is favored.

3.

a.

K=\frac{[H_{2}O]}{[H_{2}]}

K=\frac{[CO_{2}]}{[Co]}

b.

[H_2]=[CO]=0.316\,mol\,H_{2} \times \frac{1}{1.00\,L}=0.316\,M

[CoO]=0.5\,mol\,CoO \times \frac{1}{1.00\,L}=0.5\,M

Reaction 1:

PV=nRT \rightarrow P=\frac{nRT}{V}=(4.65 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=0.314

[H_{2}]=0.316-x=0.316-0.311=4.65x10^{-3}

PV=nRT \rightarrow P=\frac{nRT}{V}=(0.311)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.0

[H_{2}O]=x=0.311

  H_{2} H_{2}O
I 0.316 0
C -x +x
E 0.316-x x

K=\frac{[H_{2}O]}{[H_{2}]} \rightarrow 67=\frac{x}{0.316-x} \rightarrow x=0.311

Reaction 2:

PV=nRT \rightarrow P=\frac{nRT}{V}=(0.001)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=6.75 \times 10^{-2}\,atm

[CO]=0.316-x=0.316-0.315=0.001\,M

PV=nRT \rightarrow P=\frac{nRT}{V}=(0.315)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.3\,atm

[CO_{2}]=x=0.315\,M

  CO CO_{2}
I 0.316 0
C -x +x
E 0.316-x x

K=\frac{[CO_{2}]}{[Co]} \rightarrow 490=\frac{x}{0.316-x} \rightarrow x=0.315

c.

H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_{2}O\,(g)

K_p=\frac{(P_{CO})(P_{H_{2}O})}{(P_{H_{2}})(P_{CO_{2}})}=\frac{(6.75 \times 10^{-2})(21)}{(0.314)(21.3)}=0.21

d. The shape of the graphs [reactants] versus [products] does not change as the amount of CoO changes because it is a solid.

4.

PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{n}{V}=\frac{0.101798}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.5 \times 10^{-4}\,M\,HI

[HI]=2x=2(0.050899)=0.101798\,atm

K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(3.2 \times 10^{-1}-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.050899\,atm

  H_{2} I_{2} HI
I 3.2 \times 10^{-1} 5.16 \times 10^{-2} 0
C -x -x +2x
E 3.2 \times 10^{-1}-x 5.16 \times 10^{-2}-x 2x

PV=nRT \rightarrow P=\frac{nRT}{V}=(5.6 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=3.2 \times 10^{-1}\,atm

[H_{2}]=1.12 \times 10^{-2}\,mol\,H_{2} \times \frac{1}{2.0\,L}=5.6 \times 10^{-3}\,M

PV=nRT \rightarrow P=\frac{nRT}{V}=(9.0 \times 10^{-4})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=5.16 \times 10^{-2}\,atm

[I_2]= 1.8 \times 10^{-3}\,mol\,I_{2} \times \frac{1}{2.0\,L}=9.0 \times 10^{-4}\,M

For excess hydrogen:

Q=\frac{[HI]}{[H_{2}][I_{2}]}=\frac{1.8 \times 10^{-3}}{(594.410)(1.09 \times 10^{-3})}=2.8 \times 10^{-3}

The reaction will proceed to the right to reach equilibrium.

PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{0.103162}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.8 \times 10^{-3}\,M

[HI]=2x=2(0.051581)=0.103162\,atm

PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{10.375}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=594.410\,M

[H_{2}]=10.427-x=10.427-0.051581=10.375\,atm

PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{1.9 \times 10^{-5}}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.09 \times 10^{-3}\,M

[I_{2}]=5.16 \times 10^{-2}-x=5.16 \times 10^{-2}-0.051581=1.9 \times 10^{-5}\,atm

K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(10.427-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.051581\,atm

  H_{2} I_{2} HI
I 10.427 5.16 \times 10^{-2} 0
C -x -x +2x
E 10.427-x 5.16 \times 10^{-2}-x 2x

PV=nRT \rightarrow P=\frac{nRT}{V}=(0.182\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=10.427\,atm

[H_{2}]=3.64 \times 10^{-1}\,mol\,H_{2} \times \frac{1}{2.0\,L}=0.182\,M


15.E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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