17.3 Equilibrium Constants
Learning Objectives
 Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
 Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
 Relate the magnitude of an equilibrium constant to properties of the chemical system
Now that we have a symbol (⇌) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:
We can write the reaction quotient (Q) for this equation. When evaluated using concentrations, it is called Q_{c}. We use brackets to indicate molar concentrations of reactants and products.
The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction [latex]2\text{ NO}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{O}_4(g)[/latex] is given by this expression:
Example 17.3a
Writing Reaction Quotient Expressions
Write the expression for the reaction quotient for each of the following reactions:
 [latex]3\text{ O}_2(g)\;{\rightleftharpoons}\;2\text{ O}_3(g)[/latex]
 [latex]\text{N}_2(g)\;+\;3\text{ H}_2(g)\;{\rightleftharpoons}\;2\text{ NH}_3(g)[/latex]
 [latex]4\text{ NH}_3(g)\;+\;7\text{ O}_2(g)\;{\rightleftharpoons}\;4\text{ NO}_2(g)\;+\;6\text{ H}_2\text{O}(g)[/latex]
Solution
 [latex]Q_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}[/latex]
 [latex]Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}[/latex]
 [latex]Q_c = \frac{[\text{NO}_2]^4[\text{H}_2\text{O}]^6}{[\text{NH}_3]^4[\text{O}_2]^7}[/latex]
Exercise 17.3a
Write the expression for the reaction quotient for each of the following reactions:
 [latex]2\text{ SO}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{ SO}_3(g)[/latex]
 [latex]\text{C}_4\text{H}_8(g)\;{\rightleftharpoons}\;2\text{ C}_2\text{H}_4(g)[/latex]
 [latex]2\text{ C}_4\text{H}_{10}(g)\;+\;13\text{ O}_2(g)\;{\rightleftharpoons}\;8\text{ CO}_2(g)\;+\;10\text{ H}_2\text{O}(g)[/latex]
Check Your Answer^{[1]}
The numeric value of Q_{c} for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Q_{c} is determined. When pure reactants are mixed, Q_{c} is initially zero because there are no products present at that point. As the reaction proceeds, the value of Q_{c} increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (Figure 17.3a). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.
When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as K_{c}.
That a reaction quotient always assumes the same value at equilibrium can be expressed as:
This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.
Example 17.3b
Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
When 0.10 mol NO_{2} is added to a 1.0L flask at 25 °C, the concentration changes so that at equilibrium, [NO_{2}] = 0.016 M and [N_{2}O_{4}] = 0.042 M.
 What is the value of the reaction quotient before any reaction occurs?
 What is the value of the equilibrium constant for the reaction?
Solution
 Before any product is formed, [latex][\text{NO}_2] = \frac{0.10\;\text{mol}}{1.0\;\text{L}} = 0.10\;M[/latex], and [N_{2}O_{4}] = 0 M. Thus,
[latex]Q_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} = \frac{0}{0.10^2} = 0[/latex]  At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium,
[latex]K_c = Q_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^{2}} = \frac{0.042}{0.016^2} = 1.6\;\times\;10^{2}[/latex].
The equilibrium constant is 1.6 × 10^{2}.
Note that dimensional analysis would suggest the unit for this K_{c} value should be M^{−1}. However, it is common practice to omit units for K_{c} values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so K_{c} values are truly unitless.
Exercise 17.3b
Check Your Learning Exercise (Text Version)
For the reaction, 2 SO_{2(g) }+ O_{2(g) }↔ 2 SO_{3(g)} the concentrations at equilibrium are [SO_{2}] = 0.90 M, [O_{2}] = 0.35 M, and [SO_{3}] = 1.1 M. What is the value of the equilibrium constant, K_{c}?

 1.1
 4.3
 8.6
 0.90
Check Your Answer^{[2]}
Source: “Exercise 17.3b” is adapted from “Example 13.22” in General Chemistry 1 & 2, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson, licensed under CC BY 4.0.
The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for K_{c} indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of K_{c }– much less than 1 – indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.
Once a value of K_{c} is known for a reaction, it can be used to predict directional shifts when compared to the value of Q_{c}. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 17.3b illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H_{2}O, CO_{2}, and H_{2} react to reach compositions adhering to the same equilibrium (the value of Q_{c} changes until it equals the value of K_{c}). This value is 0.640, the equilibrium constant for the reaction under these conditions.
It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 17.3b when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.
Example 17.3c
Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
Reactants/Products  Experiment 1  Experiment 2  Experiment 3 

[CO]_{i}  0.0203 M  0.011 M  0.0094 M 
[H_{2}O]_{i}  0.0203 M  0.0011 M  0.0025 M 
[CO_{2}]_{i}  0.0040 M  0.037 M  0.0015 M 
[H_{2}]_{i}  0.0040 M  0.046 M  0.0076 M 
Solution
Experiment 1:
Q_{c} < K_{c} (0.039 < 0.64)
The reaction will shift to the right.
Experiment 2:
Q_{c} > K_{c} (140 > 0.64)
The reaction will shift to the left.
Experiment 3:
Q_{c} < K_{c} (0.48 < 0.64)
The reaction will shift to the right.
Exercise 17.3c
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
 A 1.00L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:
[latex]2\text{ NO}(g)\;+\;\text{Cl}_2(g)\;{\rightleftharpoons}\;2\text{ NOCl}(g)\;\;\;\;\;\;\;K_c = 4.6\;\times\;10^4[/latex]  A 5.0L flask containing 17 g of NH_{3}, 14 g of N_{2}, and 12 g of H_{2}:
[latex]\text{N}_2(g)\;+\;3\text{ H}_2(g)\;{\rightleftharpoons}\;2\text{ NH}_3(g)\;\;\;\;\;\;\;K_c = 0.060[/latex]  A 2.00L flask containing 230 g of SO_{3}(g):
[latex]2\text{ SO}_3(g)\;{\rightleftharpoons}\;2\text{ SO}_2(g)\;+\;\text{O}_2(g)\;\;\;\;\;\;\;K_c = 0.230[/latex]
Check Your Answer^{[3]}
In Example 17.3c, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing K_{c} values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:
 Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
 For relatively dilute solutions, a substance’s activity and its molar concentration are roughly equal.
 Activities for pure condensed phases (solids and liquids) are equal to 1.
As a consequence of this last consideration, Q_{c} and K_{c} expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression’s value). Several examples of equilibria yielding such expressions will be encountered in this section.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquidphase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.
[latex]\begin{array}{rll @{{}={}} ll} \text{C}_2\text{H}_2(aq)\;+\;2\text{ Br}_2(aq)\;{\rightleftharpoons}\;\text{C}_2\text{H}_2\text{Br}_4(aq) & K_c & = \frac{[\text{C}_2\text{H}_2\text{Br}_4]}{[\text{C}_2\text{H}_2][\text{Br}_2]^2} \\[0.5em] \text{I}_2(aq)\;+\;\text{I}^{}(aq)\;{\rightleftharpoons}\;\text{I}_3^{\;\;}(aq) & K_c & = \frac{[\text{I}_3^{\;\;}]}{[\text{I}_2][\text{I}^{}]} \\[0.5em] \text{Hg}_2^{\;\;2+}(aq)\;+\;\text{NO}_3^{\;\;}(aq)\;+\;3\text{ H}_3\text{O}^{+}(aq) \\[0.5em] {\rightleftharpoons}\;2\text{ Hg}^{2+}(aq)\;+\;\text{HNO}_2(aq)\;+\;4\text{ H}_2\text{O}(l) & K_c & = \frac{[\text{Hg}^{2+}]^2[\text{HNO}_2]}{[\text{Hg}_2^{\;\;2+}][\text{NO}_3^{\;\;}][\text{H}_3\text{O}^{+}]^3} \\[0.5em] \text{HF}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{F}^{}(aq) & K_c & = \frac{[\text{H}_3\text{O}^{+}][\text{F}^{}]}{[\text{HF}]} \\[0.5em] \text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{}(aq) & K_c & = \frac{[\text{NH}_4^{\;\;+}][\text{OH}^{}]}{[\text{NH}_3]} \end{array}[/latex]
In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H_{2}O(l) is the solvent for these solutions, its concentration does not appear as a term in the K_{c} expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.
Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.
[latex]\begin{array}{rlrl @{{}\rightleftharpoons{}} rl} \text{C}_2\text{H}_6(g) & \rightleftharpoons\text{C}_2\text{H}_4(g)\;+\;\text{H}_2(g) & K_c & = \frac{[\text{C}_2\text{H}_4][\text{H}_2]}{[\text{C}_2\text{H}_6]} \\[0.5em] 3\text{ O}_2(g) & \rightleftharpoons 2\text{ O}_3(g) & K_c & = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \\[0.5em] \text{N}_2(g)\;+\;3\text{ H}_2(g) & \rightleftharpoons 2\text{ NH}_3(g) & K_c & = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \\[0.5em] \text{C}_3\text{H}_8(g)\;+\;5\text{ O}_2(g) & \rightleftharpoons 3\text{ CO}_2(g)\;+\;4\text{ H}_2\text{O}(g) & K_c & = \frac{[\text{CO}_2]^3[\text{H}_2\text{O}]^4}{[\text{C}_3\text{H}_8][\text{O}_2]^5} \end{array}[/latex]
Note that the concentration of H_{2}O(g) has been included in the last example because water is not the solvent in this gasphase reaction and its concentration (and activity) changes.
Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, [latex]\frac{n}{V}[/latex].
Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.
Using the partial pressures of the gases, we can write the reaction quotient for the system
[latex]\text{C}_2\text{H}_6(g)\;{\rightleftharpoons}\;\text{C}_2\text{H}_4(g)\;+\;\text{H}_2(g)[/latex]
by following the same guidelines for deriving concentrationbased expressions:
In this equation we use Q_{P} to indicate a reaction quotient written with partial pressures: [latex]P_{\text{C}_2\text{H}_6}[/latex] is the partial pressure of C_{2}H_{6}; [latex]P_{\text{H}_2}[/latex], the partial pressure of H_{2}; and [latex]P_{\text{C}_2\text{H}_6}[/latex], the partial pressure of C_{2}H_{4}. At equilibrium:
The subscript P in the symbol K_{P} designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, K_{P}, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.
Conversion between a value for K_{c}, an equilibrium constant expressed in terms of concentrations, and a value for K_{P}, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).
The equation relating K_{c} and K_{P} is derived as follows. For the gasphase reaction [latex]m\text{A}\;+\;n\text{B}\;{\rightleftharpoons}\;x\text{C}\;+\;y\text{D}[/latex]:
The relationship between K_{c} and K_{P} is
In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gasphase reaction [latex]m\text{A}\;+\;n\text{B}\;{\leftrightharpoons}\;x\text{C}\;+\;y\text{D}[/latex], we have
Example 17.3d
Calculation of K_{P}
Write the equations for the conversion of K_{c} to K_{P} for each of the following reactions:
 [latex]\text{C}_2\text{H}_6(g)\;{\rightleftharpoons}\;\text{C}_2\text{H}_4(g)\;+\;\text{H}_2(g)[/latex]
 [latex]\text{CO}(g)\;+\;\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;\text{CO}_2(g)\;+\;\text{H}_2(g)[/latex]
 [latex]\text{N}_2(g)\;+\;3\text{ H}_2(g)\;{\rightleftharpoons}\;2\text{ NH}_3(g)[/latex]
 K_{c} is equal to 0.28 for the following reaction at 900 °C:
[latex]\text{CS}_2(g)\;+\;4\text{ H}_2(g)\;{\rightleftharpoons}\;\text{CH}_4(g)\;+\;2\text{ H}_2\text{S}(g)[/latex]
What is K_{P} at this temperature?
Solution
 Δn = (2) − (1) = 1
K_{P} = K_{c} (RT)^{Δn} = K_{c} (RT)^{1} = K_{c} (RT) 
Δn = (2) − (2) = 0
K_{P} = K_{c} (RT)^{Δn} = K_{c} (RT)^{0} = K_{c} 
Δn = (2) − (1 + 3) = −2
K_{P} = K_{c} (RT)^{Δn} = K_{c} (RT)^{−2} = [latex]\frac{K_c}{(RT)^2}[/latex] 
K_{P} = K_{c} (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 × 10^{−5}
Exercise 17.3d
Write the equations for the conversion of K_{c} to K_{P} for each of the following reactions, which occur in the gas phase:
 [latex]2\text{ SO}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{ SO}_3(g)[/latex]
 [latex]\text{N}_2\text{O}_4(g)\;{\rightleftharpoons}\;2\text{ NO}_2(g)[/latex]
 [latex]\text{C}_3\text{H}_8(g)\;+\;5\text{ O}_2(g)\;{\rightleftharpoons}\;3\text{ CO}_2(g)\;+\;4\text{ H}_2\text{O}(g)[/latex]
 At 227 °C, the following reaction has K_{c} = 0.0952:
[latex]\text{CH}_3\text{OH}(g)\;{\rightleftharpoons}\;\text{CO}(g)\;+\;2\text{ H}_2(g)[/latex]
What would be the value of K_{P} at this temperature
Check Your Answer^{[4]}
Heterogeneous Equilibria
A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).
Some heterogeneous equilibria involve chemical changes; for example:
[latex]\begin{array}{rl @{{}\rightleftharpoons{}} ll} \text{PbCl}_2(s) & \rightleftharpoons\text{Pb}^{2+}(aq)\;+\;2\text{ Cl}^{}(aq) & K_c & = [\text{Pb}^{2+}][\text{Cl}^{}]^2 \\[0.5em] \text{CaO}(s)\;+\;\text{CO}_2(g) & \rightleftharpoons\text{CaCO}_3(s) & K_c & = \frac{1}{[\text{CO}_2]} \\[0.5em] \text{C}(s)\;+\;2\text{ S}(g) & \rightleftharpoons\text{CS}_2(g) & K_c & = \frac{[\text{CS}_2]}{[\text{S}]^2} \end{array}[/latex]
Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:
We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:
Links to Interactive Learning Tools
Explore Equilibrium Constant Expression from the Physics Classroom.
Key Equations
 [latex]Q = \frac{[\text{C}]^x[\text{D}]^y}{[\text{A}]^m[\text{B}]^n}\;\;\;\;\;\;\;\text{where}\;m\text{A}\;+\;n\text{B}\;{\rightleftharpoons}\;x\text{C}\;+\;y\text{D}[/latex]
 [latex]Q_P = \frac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n}\;\;\;\;\;\;\;\text{where}\;m\text{A}\;+\;n\text{B}\;{\rightleftharpoons}\;x\text{C}\;+\;y\text{D}[/latex]
 P = MRT
 K_{P} = K_{c} (RT)^{Δn}
Attribution & References
 (a) [latex]Q_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}[/latex]; (b) [latex]Q_c = \frac{[\text{C}_2\text{H}_4]^2}{[\text{C}_4\text{H}_8]}[/latex]; (c) [latex]Q_c = \frac{[\text{CO}_2]^8[\text{H}_2\text{O}]^{10}}{[\text{C}_4\text{H}_{10}]^2[\text{O}_2]^{13}}[/latex] ↵
 b) 4.3 ↵
 (a) Q_{c} = 6.45 × 10^{3}, shifts right. (b) Q_{c} = 0.12, shifts left. (c) Q_{c} = 0, shifts right ↵


 K_{P} = K_{c} (RT)^{−1};
 K_{P} = K_{c} (RT);
 K_{P} = K_{c} (RT);
 160 or 1.6 × 10^{2}

ratio of the product of molar concentrations (or pressures) of the products to that of the reactants, each concentration (or pressure) being raised to the power equal to the coefficient in the equation
value of the reaction quotient for a system at equilibrium
equilibrium constant for reactions based on concentrations of reactants and products
when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant
equilibria within a single phase
written as: KP
equilibrium constant for gasphase reactions based on partial pressures of reactants and products
equilibria between reactants and products in different phases