Equilibrium Constants: Summary - UCalgary Chemistry Textbook

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

$Q_c=\frac{[C]^x[D]^y}{[A]^m[B]^n} \qquad for\;the\;reaction\; mA+nB⇌xC+yD$

$Q_P=\frac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n} \qquad for\;the\; reaction\;mA+nB⇌xC+yD$

K_c = Q_c at equilibrium (also K_p = Q_p)

K_P = K_c (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 K_C 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 K_c > 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 K_c > 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. K_c > 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 K_c > 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. K_c > 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 K_c > 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:	$I^- \, (aq) + I_2 \, (aq) \rightleftharpoons I_3 ^- \, (aq)$	$K_{C3} = \frac{[I_3 ^- ]}{[I^-][I_2]}$

The K values for the total and net ionic equations simplify to the same functions directly. To see why these are equivalent to K_C1 for the overall equation, consider the values for the concentrations. Since the only source of

$I^-$ in the reaction is the

$KI$ (and

$KI$ is soluble: it fully dissociates into its component ions),

$[I^-] = [KI]$ .

$[I_3 ^-] = [KI_3]$ for the same reason. All three expressions are therefore equivalent, and will give the same value of

$K_C$ when calculated.

Write the mathematical expression for the reaction quotient, Q_c, for each of the following reactions:

(a) $CH_4\;(g)+Cl_2\;(g) \rightleftharpoons CH_3Cl\;(g)+HCl\;(g)$

(b) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2\, NO\; (g)$

(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) $\quad Q_c=\frac{[NO]^2}{[N_2][O_2]}$

(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, Q_c, for each of the following reactions:

(a) $N_2\;(g)+3\, H_2\;(g) \rightleftharpoons 2NH_3\;(g)$

(b) $4\, NH_3\;(g)+5\, O_2\;(g) \rightleftharpoons 4\, NO\;(g)+6\, H_2O\;(g)$

(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) $\quad Q_c = \frac{[NO]^4 [H_2 O]^6}{[NH_3]^4 [O_2]^5}$

(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 K_c would make this reaction most useful commercially? K_c ≈ 0.01, K_c ≈ 1, or K_c ≈ 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 K_c > 1, < 1, or ≈ 1 for a titration reaction?

Solution

If the yield should be near 100%, the reaction must be very product-favoured, and K_c > 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 K_c > 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 K_c > 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 “K_sp“). In this form, a successful gravimetric reaction would require a reaction with a very small K_sp.

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 (→).

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 (→).

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. K_c 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 K_c and K_P

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) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2\,NO(g)$

(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) $2\, NH_3\;(g) \rightleftharpoons N_2\;(g)+3\, H_2\;(g)$

(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 K_c (calculated using concentrations) equal K_P (calculated using pressures)?

(a) $CH_4\;(g)+Cl_2\;(g) \rightleftharpoons CH_3Cl\;(g)+HCl\;(g)$

(b) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO(g)$

(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) $CuSO_4·5H_2O\;(s) \rightleftharpoons CuSO_4\;(s)+5H_2O\;(g)$

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 reactions (a) and (b).

For which of the reactions below does K_c (calculated using concentrations) equal K_P (calculated using pressures)?

(a) $2NH_3\;(g) \rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad K_c=17$

(b) $2SO_3\;(g) \rightleftharpoons 2SO_2\;(g)+O_2\;(g)\qquad K_c=0.230$

(d) $N_2\;(g)+O_2\;(g) \rightleftharpoons 2NO\;(g)\qquad K_p=0.050$

Solution

Convert the values of K_c to values of K_P or the values of K_P to values of K_c.

(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

$K_P = K_C (RT)^{\Delta n} \\ K_P = 50.2 (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 716\,\text{K})^{(2-2)} \\ \mathbf{K_P= 50.2}$

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 K_c to values of K_P or the values of K_P to values of K_c.

(a) $Cl_2\;(g)+Br_2\;(g) \rightleftharpoons 2BrCl\;(g)\qquad K_c=4.7×10^{-2}\;at\;25°C$

Solution

$K_P = K_C (RT)^{\Delta n} \\ K_P = 4.7×10^{-2} (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 298\,\text{K})^{(2-2)} \\ \mathbf{K_P= 4.7×10^{-2}}$

(b) $2SO_2\;(g)+O_2(g) \rightleftharpoons 2SO_3\;(g)\qquad K_P=48.2\;at\;500°C$

Solution

$K_P = K_C (RT)^{\Delta n} \\ 48.2 = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 773\,\text{K})^{(2-3)} \\ \mathbf{K_C= 3.10×10^{3}}$

Solution

$K_P = K_C (RT)^{\Delta n} \\ 5.09×10^{-44} = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 298\,\text{K})^{(6-0)} \\ \mathbf{K_C= 2.19×10^{-52}}$

(d) $2H_2\;(l)+O_2\;(g) \rightleftharpoons 2H_2O\;(g)\qquad K_P=0.196\;at\;60°C$

Solution

$K_P = K_C (RT)^{\Delta n} \\ 0.196 = K_C (0.0831451\, \text{L bar K}^{-1}\text{mol}^{-1} \times 333\,\text{K})^{(2-1)} \\ \mathbf{K_C= 7.08×10^{-3}}$

Key Concepts and Summary

Key Equations

Practice Problems: Equilibrium Constants

Defining Equilibrium – K and Q

Writing and calculating K and Q

Predicting the Direction of Reaction Progress

Relating Kc and KP

Relating K_c and K_P