Key Concepts and Summary
The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.
A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.
Key Equations
| $Q_c=\frac{[C]^x[D]^y}{[A]^m[B]^n} \qquad for\;the\;reaction\; mA+nB⇌xC+yD$ |
| $Q_P=\frac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n} \qquad for\;the\; reaction\;mA+nB⇌xC+yD$ |
| Kc = Qc at equilibrium (also Kp = Qp) |
| KP = Kc (RT)Δn |
Practice Problems: Equilibrium Constants
Defining Equilibrium – K and Q
Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
Solution
The equilibrium constant (K) is defined only at the point where the reaction is at equilibrium. Since this state is independent of the starting conditions of the reaction, there is only one value for K at a given temperature (K can vary with T). The reaction quotient (Q) can be measured at any time during a reaction, and so can have any value from 0 to (approaching) infinity.
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in the Equilibrium topic page.
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.
If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4? $$2\,NO_2\;(g) \rightleftharpoons N_2O_4\;(g)$$
Solution
The concentrations of reactants and products at equilibrium are not directly tied to the starting conditions of the reaction – the reaction proceeds until the ratio of products and reactants matches the equilibrium constant. From only the final amounts of product and reactant, we cannot tell what the starting conditions of the reaction were.
Once you have studied the section on Equilibrium Calculations, you can confirm this: A final mixture of [NO2]=0.76 M and [N2O4]=0.13 M can be achieved by either starting with 1 M N2O4 and 0 NO2, or 0.34 M each of N2O4 and NO2 (given that the KC for the reaction is 0.21 at 100°C). Simply by observing the final state of the reaction mixture, we cannot tell what the starting conditions were, as there will always be multiple possibilities.
Writing and calculating K and Q
Among the solubility rules commonly discussed in high school chemistry 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 $$CuCl\;(s) \rightleftharpoons Cu^+\;(aq)+ Cl^−\;(aq)$$ Is Kc > 1, < 1, or ≈ 1 for this reaction? Explain.
Solution
$$K_c = [Cu^+][Cl^-]$$ Since we know CuCl is insoluble (from the rules above), we know that the reaction does not proceed in reverse (as written) to an appreciable extent: very little will dissolve into its ions when placed in water (so $[Cu^+]$ and $[Cl^-]$ are both very small. (Realistically this will be >0, but very close to 0 for an insoluble salt). Since the equilibrium for this reaction lies “to the left” as written, we should have a very small K value. K < 1 for this reaction.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $$Pb^{2+}\;(aq)+2\, Cl^−\;(aq) \rightleftharpoons PbCl_2(s)$$ Is Kc > 1, < 1, or ≈ 1? Explain.
Solution
$$K_c = \frac{1}{[Pb^{2+}][Cl^-]^2}$$ This reaction is written so that the solid is on the right-hand side of the equation. Since PbCl2 is insoluble, in this case the equilibrium must lie to the right, again producing very few ions and preferring the solid. Kc > 1 for this reaction.
Another statement from the “solubility rules”: “Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble”.
(a) Write the expression for the equilibrium constant for the reaction represented by the equation $$Ca^{2+}\;(aq)+CO_3^{2−}\,(aq)\rightleftharpoons CaCO_3\,(s)$$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Solution
$$K_c = \frac{1}{[Ca^{2+}][CO_3 ^{2-}]^2}$$ Since CaCO3 is insoluble, the equilibrium must lie to the right, producing very few ions and preferring the solid. Kc > 1 for this reaction.
(b) Write the expression for the equilibrium constant for the reaction represented by the equation $$Ba_3(PO_4)_2 \rightleftharpoons 3\, Ba^{2+}\, (aq)+2\, PO_4^{3−}\, (aq)$$ Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Solution
$$K_c = [Ba^{2+}]^3 [PO_4 ^{3-}]^2$$ Since $Ba_3(PO_4)_2$ is insoluble, the equilibrium must lie to the left. Therefore, we should have a very small K value. K < 1 for this reaction.
Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $KI\;(aq)+I_2\;(aq) \rightleftharpoons KI_3\;(aq)$ give the same expression for the reaction quotient. $KI_3$ is composed of the ions $K^+$ and $I_3^-$.
Solution
| Overall reaction: |
$KI\;(aq)+I_2\;(aq) \rightleftharpoons KI_3\;(aq)$ | $K_{C1} = \frac{[KI_3]}{[KI][I_2]}$ |
| Total ionic equation: |
$\require{cancel} K^+\, (aq) + I^- \, (aq) + I_2 \, (aq) \rightleftharpoons K^+ \, (aq) + I_3 ^- \, (aq) $ | $\require{cancel} K_{C2} = \frac{\cancel{[K^+]}[I_3 ^- ]}{\cancel{[K^+]}[I^-][I_2]} = \frac{[I_3 ^- ]}{[I^-][I_2]}$ |
| Net ionic equation: |
$I^- \, (aq) + I_2 \, (aq) \rightleftharpoons I_3 ^- \, (aq) $ | $K_{C3} = \frac{[I_3 ^- ]}{[I^-][I_2]}$ |
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
(a) $CH_4\;(g)+Cl_2\;(g) \rightleftharpoons CH_3Cl\;(g)+HCl\;(g)$
(b) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2\, NO\; (g)$
(c) $2\, SO_2\;(g)+O_2(g) \rightleftharpoons 2\, SO_3\;(g)$
(d) $BaSO_3\;(s) \rightleftharpoons BaO(s)+SO_2\;(g)$
(e) $P_4\;(g)+5\, O_2\;(g) \rightleftharpoons P_4O_{10}\;(s)$
(f) $Br_2\;(g) \rightleftharpoons 2\, Br\;(g)$
(g) $CH_4\;(g)+2\, O_2\;(g) \rightleftharpoons CO_2\;(g)+H_2O\;(l)$
(h)$CuSO_4\cdot 5\, H_2O\;(s) \rightleftharpoons CuSO_4\;(s)+5\, H_2O\;(g)$
Solution
(a)$\quad Q_c=\frac{[CH_3Cl][HCl]}{[CH_4][Cl_2]}$
(b)$\quad Q_c=\frac{[NO]^2}{[N_2][O_2]}$
(c)$\quad Q_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$
(d)$\quad Q_c=[SO_2]$
(e)$\quad Q_c=\frac{1}{[O_2]^5 [P_4]}$
(f) $\quad Q_c=\frac{[Br]^2}{[Br_2]}$
(g)$\quad Q_c=\frac{[CO_2]}{[CH_4][O_2]^2}$
(h)$\quad Q_c=[H_2O]^5$
Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions:
(a) $N_2\;(g)+3\, H_2\;(g) \rightleftharpoons 2NH_3\;(g)$
(b) $4\, NH_3\;(g)+5\, O_2\;(g) \rightleftharpoons 4\, NO\;(g)+6\, H_2O\;(g)$
(c) $N_2O_4\;(g) \rightleftharpoons 2\, NO_2\;(g)$
(d) $CO_2\;(g)+H_2\;(g) \rightleftharpoons CO\;(g)+H_2O\;(g)$
(e) $NH_4Cl\;(s) \rightleftharpoons NH_3\;(g)+HCl\;(g)$
(f) $2\, Pb(NO_3)_2\;(s) \rightleftharpoons 2\, PbO\;(s)+4\, NO_2\;(g)+O_2\;(g)$
(g) $2\, H_2\;(g)+O_2\;(g) \rightleftharpoons 2\, H_2O\;(l)$
(h) $S_8\;(g) \rightleftharpoons 8\, S\;(g)$
Solution
(a)$\quad Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
(b)$\quad Q_c = \frac{[NO]^4 [H_2 O]^6}{[NH_3]^4 [O_2]^5}$
(c)$\quad Q_c = \frac{[NO_2]^2}{[N_2 O_4]}$
(d)$\quad Q_c = \frac{[H_2 O][CO]}{[CO_2][H_2]}$
(e)$\quad Q_c = [NH_3][HCl]$
(f)$\quad Q_c = [NO_2]^4 [O_2]$
(g)$\quad Q_c = \frac{1}{[H_2]^2[O_2]}$
(d)$\quad Q_c = \frac{[S]^8}{[S_8]}$
Write the expression of the reaction quotient for the ionization of HOCN in water.
Solution
First, write the reaction equation for the ionization:
$$ HOCN (aq) \rightleftharpoons H^+ \, (aq) + OCN^- \, (aq) $$
If you do not recognize the ions of a compound right away, try looking for it (or its derivatives) in the tables of acids and bases (Appendix H for acids and Appendix I for bases) or even see if you recognize the ions in the reactions provided in the Appendix of reduction potentials. With practice, you will learn to recognize most common complex ions on sight.
Here, we can find HOCN as a weak acid, which also provides reinforcement that our reactant should be molecular (aqueous) HOCN, not a solid or another form. You can also write this equation with $H_3 O^+$ instead of $H^+$ if you include $H_2 O\, (l)$ as a reactant to keep the reaction balanced.
Knowing the reaction, we can write the reaction quotient:
$$\mathbf{Q_C = \frac{[H^+][OCN^-]}{[HOCN]}}$$
Write the reaction quotient expression for the ionization of NH3 in water.
Solution
As with the HOCN above, we can find ammonia in Appendix I to help build the reaction: $$NH_3\, (aq) + H_2O\, (l) \rightleftharpoons NH_4 ^+ \, (aq) + OH^- \, (aq)$$ and the reaction quotient: $$ \mathbf{Q_C = \frac{[NH_4 ^+][OH^- ]}{[NH_3]}}$$
What is the approximate value of the equilibrium constant $K_P$ for the change $C_2H_5OC_2H_5\;(l) \rightleftharpoons C_2H_5OC_2H_5\;(g)$ at 25 °C, if the equilibrium vapor pressure for this substance is 0.760 bar at 25 °C.
Solution
From the reaction given, the expression for $K_P$ is: $$ K_P = P_{C_2H_5OC_2H_5} \\ \mathbf{K_P = 0.760}$$
Predicting the Direction of Reaction Progress
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene:$$3C_2H_2(g) \rightleftharpoons C_6H_6\;(g)$$. Which value of Kc would make this reaction most useful commercially? Kc ≈ 0.01, Kc ≈ 1, or Kc ≈ 10. Explain your answer.
Solution
Since $K_c=\frac{[C_6H_6]}{[C_2H_2]^3}$, a value of $K_c$ ≈ 10 means that C6H6 predominates over C2H2 (i.e. more benzene would exist at equilibrium than acetylene). In such a case, the reaction could be useful commercially, as long as the rate of reaction is also reasonable. Smaller values of K would indicate less benzene being produced, which makes the reaction less useful for producing benzene.
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
If the yield should be near 100%, the reaction must be very product-favoured, and Kc > 1.
For a precipitation reaction (A reaction producing a solid from its aqueous ions, such as $Ag^+ \, (aq) + Cl^- \, (aq) \rightleftharpoons AgCl\, (s)$) to be useful in a gravimetric analysis, the product of the reaction must not redissolve to any appreciable extent as it is produced. Is Kc > 1, < 1, or ≈ 1 for a useful precipitation reaction?
Solution
If the solid does not redissolve, the reaction must be very product-favoured (as written), and Kc > 1. When you reach the study on solubility reactions, you will find that most are written as dissolutions (i.e. $AgCl\, (s) \rightleftharpoons Ag^+ \, (aq) + Cl^- \, (aq) $, with a named K as “Ksp“). In this form, a successful gravimetric reaction would require a reaction with a very small Ksp.
The initial concentrations or pressures of reactants and products are given for each of the following systems. Determine the direction in which each system will proceed to reach equilibrium.
(a) $2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_c=17$; [NH3] = 0.20 M, [N2] = 1.00 M, [H2] = 1.00 M
Solution
$$Q_C = \frac{[N_2][H_2]^3}{[NH_3]^2} \\ Q_C = \frac{[1.00\, \text{M}][1.00\, \text{M}]^3}{[0.20\, \text{M}]^2} \\ Q_C = 25$$ Q > K. Reaction proceeds to the left (←).
(b) $2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_p=6.8×10^4$; NH3 = 3.0 bar, N2 = 2.0 bar, H2 = 1.0 bar
Solution
$$Q_P = \frac{P_{(N_2)}P_{(H_2)} ^3}{P_{(NH_3)} ^2} \\ Q_P = \frac{[2.0\, \text{bar}][1.0\, \text{bar}]^3}{[3.00\, \text{bar}]^2} \\ Q_P = 0.22$$ Q < K. Reaction proceeds to the right (→).
(c) $2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_c=0.230$; [SO3] = 0.00 M, [SO2] = 1.00 M, [O2] = 1.00 M
Solution
$$Q_C = \frac{[SO_2]^2[O_2]}{[SO_3]^2} \\ Q_C = \frac{[1.00\, \text{M}]^2 [1.00\, \text{M}]}{[0.00\, \text{M}]^2} \\ Q_C \rightarrow \infty \text{(undefined large number)}$$ Q > K. Reaction proceeds to the left (←).
(d)$2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_p=16.5$; SO3 = 1.00 bar, SO2 = 1.00 bar, O2 = 1.00 bar
Solution
$$Q_P = \frac{P_{(O_2)}P_{(SO_2)} ^2}{P_{(SO_3)} ^2} \\ Q_P = \frac{[1.00\, \text{bar}][1.00\, \text{bar}]^2}{[1.00\, \text{bar}]^2} \\ Q_P = 1.00$$ Q < K. Reaction proceeds to the right (→).
(e) $2NO\;(g)+Cl_2(g) \rightleftharpoons 2NOCl\;(g)\qquad K_c=4.6×10^4$; [NO] = 1.00 M, [Cl2] = 1.00 M, [NOCl] = 0 M
Solution
$$Q_C = \frac{[NOCl]^2}{[NO]^2 [Cl_2]} \\ Q_C = \frac{[0\, \text{M}]^2 }{[1.00\, \text{M}]^2 [1.00\, \text{M}]} \\ Q_C = 0$$ Q < K. Reaction proceeds to the right (→).
(f) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO\;(g)\qquad K_p=0.050$; NO = 10.0 bar, N2 = O2 = 5 bar
Solution
$$Q_P = \frac{P_{(NO)} ^2}{P_{(N_2)}P_{(O_2)}} \\ Q_P = \frac{[10.0\, \text{bar}]^2}{[5\, \text{bar}][5\, \text{bar}]} \\ Q_P = 4$$ Q > K. Reaction proceeds to the left (←).
The initial concentrations or pressures of reactants and products are given for each of the following systems. Determine the direction in which each system will proceed to reach equilibrium.
(a)$2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_c=17$; [NH3] = 0.50 M, [N2] = 0.15 M, [H2] = 0.12 M
Solution
$$Q_C = \frac{[N_2][H_2]^3}{[NH_3]^2} \\ Q_C = \frac{[0.15\, \text{M}][0.12\, \text{M}]^3}{[0.50\, \text{M}]^2} \\ Q_C = 1.0 \times 10^{-3}$$ Q < K. Reaction proceeds to the right (→).
(b) $2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_p=6.8×10^4$; NH3 = 2.00 bar, N2 = 10.00 bar, H2 = 10.00 bar
Solution
$$Q_P = \frac{P_{(N_2)}P_{(H_2)} ^3}{P_{(NH_3)} ^2} \\ Q_P = \frac{[10.00\, \text{bar}][10.00\, \text{bar}]^3}{[2.00\, \text{bar}]^2} \\ Q_P = 2.50 \times 10^{3}$$ Q < K. Reaction proceeds to the right (→).
(c) $2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_c=0.230$; [SO3] = 2.00 M, [SO2] = 2.00 M, [O2] = 2.00 M
Solution
$$Q_C = \frac{[SO_2]^2[O_2]}{[SO_3]^2} \\ Q_C = \frac{[2.00\, \text{M}]^2 [2.00\, \text{M}]}{[2.00\, \text{M}]^2} \\ Q_C = 2.00$$ Q > K. Reaction proceeds to the left (←).
(d) $2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_p=16.5$; SO2 = 1.00 bar, O2 = 1.130 bar, SO3 = 0 bar
Solution
$$Q_P = \frac{P_{(O_2)}P_{(SO_2)} ^2}{P_{(SO_3)} ^2} \\ Q_P = \frac{[1.130\, \text{bar}][1.00\, \text{bar}]^2}{[0\, \text{bar}]^2} \\ Q_P \rightarrow \infty \text{(undefined large number)}$$ Q > K. Reaction proceeds to the left (←).
(e) $2NO\;(g)+Cl_2(g) \rightleftharpoons 2NOCl\;(g)\qquad K_p=2.5×10^3$; NO = 1.00 bar, Cl2 = 1.00 bar, NOCl = 0 bar
Solution
$$Q_P = \frac{P_{(NOCl)} ^2}{P_{(NO)} ^2 P_{(Cl_2)}} \\ Q_P = \frac{[0\, \text{bar}]^2 }{[1.00\, \text{bar}]^2 [1.00\, \text{bar}]} \\ Q_P = 0$$ Q < K. Reaction proceeds to the right (→).
(f)$N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO\;(g)\qquad K_c=0.050$ [N2] = 0.100 M, [O2] = 0.200 M, [NO] = 1.00 M
Solution
$$Q_C = \frac{[NO]^2}{[N_2][O_2]} \\ Q_C = \frac{[1.00\, \text{M}]^2}{[0.100\, \text{M}][0.200\, \text{M}]} \\ Q_C = 50$$ Q > K. Reaction proceeds to the left (←).
The following reaction has $K_P= 4.50×10^{−5}$ at 720 K. $$N_2\;(g)+3H_2\;(g) \rightleftharpoons 2NH_3\;(g)$$
If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? $P_{(NH_3)}= 93\;bar$, $P_{(N_2)}$ = 48 bar, and $P_{(H_2)}$ = 52 bar
Solution
$$Q_P = \frac{P_{NH_3} ^2}{{P_{H_2 }^3} {P_{N_2}}} \\ Q_P = \frac{(93\, \text{bar}) ^2}{(52 \, \text{bar}) ^3 (48 \, \text{bar})} \\ Q_P = 1.2 \times 10^{-3} \\ Q > K $$ Reaction will proceed to the left (←) to reach equilbirium.
A reaction vessel is prepared as described below, for this reaction:
$$SO_2Cl_2\;(g) \rightleftharpoons SO_2\;(g)+Cl_2\;(g)$$
[SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078.
When this reaction reaches equilibrium, will there be more or less SO2Cl2 present than at the start of the reaction?
Solution
$$Q_C = \frac{[Cl_2][SO_2]}{[SO_2 Cl_2]} \\ Q_C = \frac{[0.16\, \text{M}][0.050\, \text{M}]}{[0.12\, \text{M}]} \\ Q_C = 0.067 \\ Q < K $$ Reaction will proceed to the right (→) to reach equilbirium. SO2Cl2 will be consumed; the amount present will decrease during the reaction.
Relating Kc and KP
Which of the systems below are homogeneous equilibria? Which are heterogeneous equilibria?
(a) $CH_4\;(g)+Cl_2\;(g) \rightleftharpoons CH_3Cl\;(g)+HCl\;(g)$
(b) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2\,NO(g)$
(c) $2\, SO_2\;(g)+O_2(g) \rightleftharpoons 2\, SO_3\;(g)$
(d) $BaSO_3\;(s) \rightleftharpoons BaO(s)+SO_2(g)$
(e) $P_4\;(g)+5\, O_2\;(g) \rightleftharpoons P_4O_{10}\;(s)$
(f) $Br_2\;(g) \rightleftharpoons 2\, Br\;(g)$
(g) $CH_4\;(g)+2\, O_2(g) \rightleftharpoons CO_2\;(g)+H_2O\;(l)$
(h)$CuSO_4\cdot 5\, H_2O\;(s) \rightleftharpoons CuSO_4\;(s)+5\, H_2O\;(g)$
Solution
(a) homogenous
(b) homogenous
(c) homogenous
(d) heterogeneous
(e) heterogeneous
(f) homogenous
(g) heterogeneous
(h) heterogeneous
Which of the systems described below are homogeneous equilibria? Which are heterogeneous equilibria?
(a) $2\, NH_3\;(g) \rightleftharpoons N_2\;(g)+3\, H_2\;(g)$
(b) $2\, NH_3\;(g) \rightleftharpoons N_2\;(g)+3\, H_2\;(g)$
(c) $2\, SO_3\;(g) \rightleftharpoons 2\, SO_2\;(g)+O_2\;(g)$;
(d)$2\, SO_3\;(g) \rightleftharpoons 2\, SO_2\;(g)+O_2\;(g)$;
(e) $2\, NO\;(g)+Cl_2(g) \rightleftharpoons 2\, NO\;(g)$;
(f) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2\, NO\;(g)$;
Solution
All of these reactions are homogeneous (all gas-phase).
For which of the reactions below does Kc (calculated using concentrations) equal KP (calculated using pressures)?
(a) $CH_4\;(g)+Cl_2\;(g) \rightleftharpoons CH_3Cl\;(g)+HCl\;(g)$
(b) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO(g)$
(c) $2SO_2\;(g)+O_2(g) \rightleftharpoons 2SO_3\;(g)$
(d) $BaSO_3\;(s) \rightleftharpoons BaO(s)+SO_2(g)$
(e) $P_4\;(g)+5O_2\;(g) \rightleftharpoons P_4O_{10}\;(s)$
(f) $Br_2\;(g) \rightleftharpoons 2Br\;(g)$
(g) $CH_4\;(g)+2O_2(g) \rightleftharpoons CO_2\;(g)+H_2O\;(l)$
(h)$CuSO_4·5H_2O\;(s) \rightleftharpoons CuSO_4\;(s)+5H_2O\;(g)$
Conversion from $K_C$ to $K_P$ is done via:
$$K_P = K_C (RT)^{\Delta n}$$
In order for $K_P = K_C$ then, we will need $\Delta n$ to be 0, so that $(RT)^{ \Delta n} = 1 $. Reactions with an equal number of moles of gas in the products and in the reactants will meet this requirement. This situation occurs in reactions (a) and (b).Solution
For which of the reactions below does Kc (calculated using concentrations) equal KP (calculated using pressures)?
(a) $2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_c=17$
(b) $2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_c=0.230$
(c) $2NO\;(g)+Cl_2(g) \rightleftharpoons 2NO\;(g)\qquad K_c=4.6×10^4$
(d) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO\;(g)\qquad K_p=0.050$
Solution
Conversion from $K_C$ to $K_P$ is done via: $$K_P = K_C (RT)^{\Delta n}$$ In order for $K_P = K_C$ then, we will need $\Delta n$ to be 0, so that $(RT)^{ \Delta n} = 1 $. Reactions with an equal number of moles of gas in the products and in the reactants will meet this requirement. This situation occurs in reaction (d).
Convert the values of Kc to values of KP or the values of KP to values of Kc.
(a) $N_2\;(g)+3H_2\;(g) \rightleftharpoons 2NH_3\;(g)\qquad K_c=0.50\;at\;400.°C$
Solution
$$K_P = K_C (RT)^{\Delta n} \\ K_P = 0.50 (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 673\,\text{K})^{(2 – 4)} \\ \mathbf{K_P= 1.6×10^{-4}}$$
(b) $H_2\;(g)+I_2\;(g) \rightleftharpoons 2HI\;(g)\qquad K_c=50.2\;at\;448°C$
Solution
$$K_P = K_C (RT)^{\Delta n} \\ K_P = 50.2 (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 716\,\text{K})^{(2-2)} \\ \mathbf{K_P= 50.2}$$
(c) $Na_2SO_4·10H_2O\;(s) \rightleftharpoons Na_2SO_4\;(s)+10H_2O\;(g)\qquad K_P=4.08×10^{-25}\;at\;25°C$
Solution
$$K_P = K_C (RT)^{\Delta n} \\ 4.08×10^{-25} = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 298\,\text{K})^{(10-0)} \\ \mathbf{K_C=4.66×10^{-39}}$$
(d) $H_2O\;(l) \rightleftharpoons H_2O\;(g)\qquad K_P=0.122\;at\;50.°C$
Solution
$$K_P = K_C (RT)^{\Delta n} \\ 0.122 = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 323\,\text{K})^{(1-0)} \\ \mathbf{K_C=4.54×10^{-3}}$$
Convert the values of Kc to values of KP or the values of KP to values of Kc.
(a) $Cl_2\;(g)+Br_2\;(g) \rightleftharpoons 2BrCl\;(g)\qquad K_c=4.7×10^{-2}\;at\;25°C$
$$K_P = K_C (RT)^{\Delta n} \\
K_P = 4.7×10^{-2} (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 298\,\text{K})^{(2-2)} \\
\mathbf{K_P= 4.7×10^{-2}}$$Solution
(b) $2SO_2\;(g)+O_2(g) \rightleftharpoons 2SO_3\;(g)\qquad K_P=48.2\;at\;500°C$
$$K_P = K_C (RT)^{\Delta n} \\
48.2 = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 773\,\text{K})^{(2-3)} \\
\mathbf{K_C= 3.10×10^{3}}$$Solution
(c) $CaCl_2·6H_2O\;(s) \rightleftharpoons CaCl_2\;(s)+6H_2O\;(g)\qquad K_P=5.09×10^{-44}\;at\;25°C$
$$K_P = K_C (RT)^{\Delta n} \\
5.09×10^{-44} = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 298\,\text{K})^{(6-0)} \\
\mathbf{K_C= 2.19×10^{-52}}$$Solution
(d) $2H_2\;(l)+O_2\;(g) \rightleftharpoons 2H_2O\;(g)\qquad K_P=0.196\;at\;60°C$
$$K_P = K_C (RT)^{\Delta n} \\
0.196 = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 333\,\text{K})^{(2-1)} \\
\mathbf{K_C= 7.08×10^{-3}}$$Solution