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15.5: Equilibrium Constants and Coupled Reactions

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    Coupled Equations

    Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In section 5.7: Hess's Law, we learned that for coupled reactions enthalpies are additive. In this section we shall see that for coupled reactions equilibrium constants are multiplicative. Thus just as we could calculate the enthalpy of reaction for a reaction we had not studied, we can also calculate the equilibrium constant, if we know the equilibrium constants of the coupled reactions that add up to the desired reaction.

    Multi-step Equilibrium Rules

    1. The equilibrium constant for a reaction written backwards is the reciprocal of the constant for the reaction written in the forward direction
      \[\begin{align} aA \rightleftharpoons bB \;\; & K_{forward} =\frac{[B]^b}{[A]^a} \nonumber\\ \; \nonumber \\ bB \rightleftharpoons aA \; \; & K_{backwards} = \frac{[A]^a}{[B]^b}=\frac{1}{K_{forward}} \end{align}  \]
    2. The equilibrium constant of two or more coupled reactions is the product of the equilibrium constant for each coupled step.

      \[\begin{align} aA \rightleftharpoons bB \;\;  K_{1} & =\frac{[B]^b}{[A]^a} \nonumber \\ \; \nonumber \\ bB \rightleftharpoons cC \; \; K_{2} & = \frac{[C]^c}{[B]^b} \\ \; \\ \text{the sum of reaction 1 and 2 is}  \nonumber\\ \; \nonumber\\  aA \rightleftharpoons cC \; \; K_3 & = \frac{[C]^c}{[A]^a} \end{align}  \nonumber\\ \; \nonumber \\ \text{and } K_3=K_1K_2 \; \text{ where the intermediate B cancels out.} \nonumber \\ \; \nonumber \\ K_1K_2=\frac{\cancel{\textcolor{red}{[B]^b}}}{[A]^a}\frac{[C]^c}{\cancel{\textcolor{red}{[B]^b}}} =\frac{[C]^c}{[A]^a}=K_3 \]

    To illustrate this procedure, let’s consider the reaction of \(N_2\) with \(O_2\) to give \(NO_2\). This reaction is an important source of the \(NO_2\) that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps with NO (in red to identify it) being the intermediate that allows the two reactions to be added to a third reaction. In the first reaction (step 1), \(N_2\) reacts with \(O_2\) at the high temperatures inside an internal combustion engine to give \(\textcolor{red}{NO}\). The released \(\textcolor{red}{NO}\) then reacts with additional \(O_2\) to give \(NO_2\) (step 2). The equilibrium constant for each reaction at 100°C is also given.

    \(N_{2(g)}+O_{2(g)} \rightleftharpoons \textcolor{red}{2NO_{(g)}}\;\; K_1=2.0 \times 10^{−25} \tag{step 1}\)

    \(\textcolor{red}{2NO_{(g)}}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9 \tag{step 2}\)

    Summing reactions (step 1) and (step 2) gives the overall reaction of \(N_2\) with \(O_2\):

    \(N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=? \tag{overall reaction 3}\)

    The equilibrium constant expressions for the reactions are as follows:

    \[K_1=\dfrac{\textcolor{red}{[NO]^2}}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{\textcolor{red}{[NO]^2}[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2}\]

    What is the relationship between \(K_1\), \(K_2\), and \(K_3\), all at 100°C? The expression for \(K_1\) has \([NO]^2\) in the numerator, the expression for \(K_2\) has \([NO]^2\) in the denominator, and \([NO]^2\) does not appear in the expression for \(K_3\). Multiplying \(K_1\) by \(K_2\) and canceling the \([NO]^2\) terms,

    \[ K_1K_2=\dfrac{\cancel{\textcolor{red}{[NO]^2}}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{\textcolor{red}{[NO]^2}}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3\]

    Thus the product of the equilibrium constant expressions for \(K_1\) and \(K_2\) is the same as the equilibrium constant expression for \(K_3\):

    \[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15}\]

    The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, \(ΔH\) for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.

    YouTube Demonstration

    Video\PageIndex{1}\: 4.47 min YouTube showing how to calculate K for an unknown equation if it is known for coupled equations that can be added to form the unknown equation.


    Example \(\PageIndex{1}\)

    The following reactions occur at 1200°C:

    1. \(CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{−2}\)
    2. \(CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4\)

    Calculate the equilibrium constant for the following reaction at the same temperature.

    1. \(CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?\)

    Given: two balanced equilibrium equations, values of \(K\), and an equilibrium equation for the overall reaction

    Asked for: equilibrium constant for the overall reaction


    Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of \(K\) for that equation. Calculate \(K\) for the overall equation by multiplying the equilibrium constants for the individual equations.


    The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:

    \[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)}\]

    \[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)}\]

    \[ CO_{(g)} + 2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\]

    The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\):

    \[K_3 = K_1K_2 = (9.17 \times 10^{−2})(3.3 \times 10^4) = 3.03 \times 10^3\]

    Exercise \(\PageIndex{1}\)

    In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.

    1. \(\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}\)
    2. \(SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}\)
    3. \(\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?\)

    \(K_3 = 1.1 \times 10^{66}\)


    Deep Dive

    We are now in a position to understand why the equilibrium constant can be determined can be determined from the stoichiometric coefficients, while the order of reaction had to be determined experimentally. This is because a real reaction occurs through a mechanism of one or more coupled elementary steps where the order of reaction is the stoichiometric coefficient of the actual mechanistic (elementary) step, and the values of the intermediates cancel out as the successive equilibrium constants cancel out.

    Lets review section 14.2.2, Multistep Reactions, for the following reaction:

    \[H_{2}(g) + 2ICl(g) \rightarrow I_{2}(g) + 2HCl{g}\]

    we postulated two elementary steps, where HI is the intermediate.

    \[\begin{align} H_{2} + ICl & \rightarrow HCl + \textcolor{red}{HI} \\ \textcolor{red}{HI} + ICl & \rightarrow HCl + I_{2} \end{align}\]

    Now from section 15.1.4: Kinetics and Equilibrium, we saw that at equilibria the rate forward = the rate backward,

    \[ aA\underset{k_{-1}} {\overset{k_1} {\rightleftharpoons}} bB\]

    The Rates are

    \[\underbrace{R_{F}=k_{1}[A]^{m}}_{\text{Forward Rate for mechanistic step}} \; \; \text{and} \; \; \underbrace{R_{B}=k_{-1}[B]^{b}}_{\text{Backward Rate for mechanistic step}}\]

    but if these are mechanistic steps, the orders of reaction are the stoichiometric coefficients

    \[\underbrace{R_{F}=k_{1}[A]^{a}}_{\text{Forward Rate for mechanistic step}} \; \; \text{and} \; \; \underbrace{R_{B}=k_{-1}[B]^{b}}_{\text{Backward Rate for mechanistic step}}\]

    If we now look at eq. 15.2.4 from the perspective of a mechanistic step we see that

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


    \(H_2+ICl \rightleftharpoons HCl+ \textcolor{red}{HI}\;\;\;K_1=\dfrac{[HCl]\textcolor{red}{[HI]}}{[H_2][ICl]} \tag{eq.15.5.9(above)}\)

    \(\textcolor{red}{HI}+ICl \rightleftharpoons HCl+ I_2\;\;\;K_2=\dfrac{[HCl][I_2]}{\textcolor{red}{[HI]}[ICl]} \tag{eq.15.5.10(above)}\)

    Summing the above two equations gives equation 15.5.8

    \(H_2+2ICl \rightleftharpoons I_2+2HCl \;\;\;K_3=\dfrac{[I_2][HCl]^2}{[H_2][ICl]^2} \tag{eq.15.5.8(above)}\)

    \[K_3=K_1K_2 = \left(\dfrac{[HCl]\textcolor{red}{[HI]}}{[H_2][ICl]}\right)\left(\dfrac{[HCl][I_2]}{\textcolor{red}{[HI]}[ICl]}\right)= \dfrac{[I_2][HCl]^2}{[H_2][ICl]^2}\]

    In summary: You can use the stoichiometric coefficients in calculating the powers in equilibrium constants because real reactions have a mechanism involving one or more coupled elementary steps, and the order of reaction is based on the stoichiometric coefficient of the elementary step. Intermediates cancel out because they are formed in an early step (where they are on the numerator of the rate constant) and are then consumed in a subsequent step (where they appear in the denominator), and thus the equilibrium constant is the product concentrations to the power of their coefficients, divided by the reactant concentrations to the power of their coefficients.

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:


    This page titled 15.5: Equilibrium Constants and Coupled Reactions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford.

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