Chapter 14: Fundamental Equilibrium Concepts

# 14.2 Equilibrium Constants

### Learning Outcomes

- Express reaction quotients for chemical equations representing homogeneous and heterogeneous reactions
- Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products at equilibrium
- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
- Manipulate equilibrium constants to reflect changes in the chemical equation or to find the equilibrium constant for coupled equilibria

The status of a reversible reaction is conveniently assessed by evaluating its **reaction quotient ( Q)**. For a reversible reaction described by

[latex]m\text{A}+n\text{B}+\rightleftharpoons x\text{C}+y\text{D}[/latex]

the reaction quotient is derived directly from the stoichiometry of the balanced equation as

[latex]\displaystyle{Q}_{c}=\frac{{\left[\text{C}\right]}^{x}{\left[\text{D}\right]}^{y}}{{\left[\text{A}\right]}^{m}{\left[\text{B}\right]}^{n}}[/latex]

where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:

[latex]Q_p=\dfrac{\text{P}^x_C\text{P}^y_D}{\text{P}^m_A\text{P}^n_B}[/latex]

Note that the reaction quotient equations above are a simplification of more rigorous expressions that use *relative* values for concentrations and pressures rather than *absolute* values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing *Q*. In most cases, this will introduce only modest errors in calculations involving reaction quotients.

### Example 14.2.1: Writing Reaction Quotient Expressions

Write the expression for the reaction quotient for each of the following reactions:

- [latex]\ce{3O2}\left(g\right)\rightleftharpoons2\ce{O3}\left(g\right)[/latex]
- [latex]\ce{N2}\left(g\right)+\ce{3H2}\left(g\right)\rightleftharpoons\ce{2NH3}\left(g\right)[/latex]
- [latex]\ce{4NH3}\left(g\right)+\ce{7O2}\left(g\right)\rightleftharpoons\ce{4NO2}\left(g\right)+\ce{6H2O}\left(g\right)[/latex]

## Show Solution

- [latex]\displaystyle{Q}_{c}=\frac{{\left[{O}_{3}\right]}^{2}}{{\left[{O}_{2}\right]}^{3}}[/latex]
- [latex]\displaystyle{Q}_{c}=\frac{{\left[\ce{NH3}\right]}^{2}}{\left[\ce{N2}\right]{\left[\ce{H2}\right]}^{3}}[/latex]
- [latex]\displaystyle{Q}_{c}=\frac{{\left[\ce{NO2}\right]}^{4}{\left[\ce{H2O}\right]}^{6}}{{\left[\ce{NH3}\right]}^{4}{\left[\ce{O2}\right]}^{7}}[/latex]

#### Check Your Learning

The numerical value of *Q* varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:

[latex]\ce{2SO2}\text{(g)} + \ce{O2}\text{(g)} \rightleftharpoons \ce{2SO3}\text{(g)}[/latex]

Two different experimental scenarios are depicted in Figure 14.2.1, one in which this reaction is initiated with a mixture of reactants only, [latex]\ce{SO2}[/latex] and [latex]\ce{O2}[/latex], and another that begins with only product, [latex]\ce{SO3}[/latex]. For the reaction that begins with a mixture of reactants only, *Q* is initially equal to zero:

[latex]Q_c = \dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]} = \dfrac{0^2}{[\ce{SO2}]^2[\ce{O2}]} = 0[/latex]

As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of *Q _{c}*), product concentration increases (as does the numerator of

*Q*), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of

_{c}*Q*.

_{c}If the reaction begins with only product present, the value of *Q _{c}* is initially undefined (immeasurably large, or infinite):

[latex]Q_c = \dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]} = \dfrac{[\ce{SO3}]^2}{0} \rightarrow \infty[/latex]

In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of *Q _{c}* decrease with time, the reactant concentrations and the denominator of

*Q*increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.

_{c}The constant value of *Q* exhibited by a system at equilibrium is called the **equilibrium constant, K**:

[latex]K=Q \qquad\text{at equilibrium}[/latex]

Comparison of the data plots in Figure 14.2.1 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the **law of mass action**

When a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant

At a given temperature, the reaction quotient for a system at equilibrium is constant.

### Example 14.2.2: Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

[latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)[/latex]

When 0.10 mol [latex]\ce{NO2}[/latex] is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO_{2}] = 0.016 *M* and [latex]\ce{N2O4}[/latex] = 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?

## Show Solution

- Before any product is formed, [latex]\left[{\ce{NO}}_{2}\right]=\dfrac{0.10\text{mol}}{1.0\text{L}}=0.10M[/latex], and [N
_{2}O_{4}] = 0*M*. Thus,

[latex]{Q}_{c}=\dfrac{\left[{\ce{N}}_{2}{\ce{O}}_{4}\right]}{{\left[{\ce{NO}}_{2}\right]}^{2}}=\dfrac{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}=\dfrac{\left[{\ce{N}}_{2}{\ce{O}}_{4}\right]}{{\left[{\ce{NO}}_{2}\right]}^{2}}=\dfrac{0.042}{{0.016}^{2}}=1.6\times {10}^{2}\text{.}[/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*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 chapter, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so

_{c}*K*values are truly unitless.

_{c}#### Check Your Learning

By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large *K* will reach equilibrium when most of the reactant has been converted to product, whereas a small *K* indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of *K* does *not* indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.

The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing *Q* to *K* for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.

To further illustrate this important point, consider the reversible reaction shown below:

[latex]\ce{CO}\left(g\right)+\ce{H2O}\left(g\right)\rightleftharpoons\ce{CO2}\left(g\right)+\ce{H2}\left(g\right)\qquad{K}_{c}=0.640\qquad\text{T}=800^{\circ}\text{C}[/latex]

The bar charts in Figure 14.2.2 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.

### Example 14.2.3: Predicting the Direction of Reaction

Given below are the starting concentrations of reactants and products for three experiments involving this reaction:

[latex]\ce{CO}(g)+\ce{H2O}(g)\rightleftharpoons\ce{CO2}(g)+\ce{H2}(g)[/latex]

[latex]{K}_{c}=0.64[/latex]

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 |
---|---|---|---|

[latex]\ce{CO}[/latex]_{i} |
0.0203 M |
0.011 M |
0.0094 M |

[latex]\ce{H2O}[/latex]_{i} |
0.0203 M |
0.0011 M |
0.0025 M |

[latex]\ce{CO2}[/latex]_{i} |
0.0040 M |
0.037 M |
0.0015 M |

[latex]\ce{H2}[/latex]_{i} |
0.0040 M |
0.046 M |
0.0076 M |

## Show Solution

#### Experiment 1

[latex]\displaystyle{Q}_{c}=\frac{\left[\ce{CO2}\right]\left[\ce{H2}\right]}{\left[\ce{CO}\right]\left[\ce{H2O}\right]}=\frac{\left(0.0040\right)\left(0.0040\right)}{\left(0.0203\right)\left(0.0203\right)}=0.039\text{.}[/latex]

*Q _{c}* <

*K*(0.039 < 0.64)

_{c}The reaction will shift to the right in the forward direction.

#### Experiment 2

[latex]\displaystyle{Q}_{c}=\frac{\left[\ce{CO2}\right]\left[\ce{H2}\right]}{\left[\ce{CO}\right]\left[\ce{H2O}\right]}=\frac{\left(0.037\right)\left(0.046\right)}{\left(0.011\right)\left(0.0011\right)}=1.4\times {10}^{2}[/latex]

*Q _{c}* >

*K*(140 > 0.64)

_{c}The reaction will shift to the left in the reverse direction.

#### Experiment 3

[latex]\displaystyle{Q}_{c}=\frac{\left[\ce{CO2}\right]\left[\ce{H2}\right]}{\left[\ce{CO}\right]\left[\ce{H2O}\right]}=\frac{\left(0.0015\right)\left(0.0076\right)}{\left(0.0094\right)\left(0.0025\right)}=0.48[/latex]

*Q _{c}* <

*K*(0.48 < 0.64)

_{c}The reaction will shift to the right.

#### Check Your Learning

## 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). By this definition, homogenous equilibria take place in *solutions*. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

- [latex]\ce{C2H2}\left(aq\right)+\ce{2Br2}\left(aq\right)\rightleftharpoons\ce{C2H2Br4}\left(aq\right)\qquad{K}_{c}=\frac{\left[\ce{C2H2Br4}\right]}{\left[\ce{C2H2}\right]{\left[\ce{Br2}\right]}^{2}}[/latex]
- [latex]\ce{I2}\left(aq\right)+\ce{I-}\left(aq\right)\rightleftharpoons\ce{I3-}\left(aq\right)\qquad{K}_{c}=\frac{\left[\ce{I3-}\right]}{\left[\ce{I2}\right]\left[\ce{I-}\right]}[/latex]
- [latex]\ce{HF}\left(aq\right)+\ce{H2O}\left(l\right)\rightleftharpoons\ce{H3O+}\left(aq\right)+\ce{F-}\left(aq\right)\qquad{K}_{c}=\frac{\left[\ce{H3O+}\right]\left[\ce{F-}\right]}{\left[\ce{HF}\right]}[/latex]
- [latex]\ce{NH3}\left(aq\right)+\ce{H2O}\left(l\right)\rightleftharpoons\ce{NH4+}\left(aq\right)+\ce{OH-}\left(aq\right)\qquad{K}_{c}=\frac{\left[\ce{NH4+}\right]\left[\ce{OH-}\right]}{\left[\ce{NH3}\right]}[/latex]

These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is *not* included in the reaction quotient. The reason for this omission is related to the more rigorous form of the *Q* (or *K*) expression mentioned previously in this chapter, in which *relative concentrations for liquids and solids are equal to 1 and needn’t be included*. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

The equilibria below all involve gas-phase solutions:

- [latex]\ce{C2H6}\left(g\right)\rightleftharpoons\ce{C2H4}\left(g\right)+\ce{H2}\left(g\right)\qquad{K}_{c}=\frac{\left[\ce{C2}\ce{H4}\right]\left[\ce{H2}\right]}{\left[\ce{C2H6}\right]}[/latex]
- [latex]\ce{3O2}\left(g\right)\rightleftharpoons2\ce{O3}\left(g\right)\qquad{K}_{c}=\frac{{\left[\ce{O3}\right]}^{2}}{{\left[\ce{O2}\right]}^{3}}[/latex]
- [latex]\ce{N2}\left(g\right)+\ce{3H2}\left(g\right)\rightleftharpoons2\ce{NH3}\left(g\right)\qquad{K}_{c}=\frac{{\left[\ce{NH3}\right]}^{2}}{\left[\ce{N2}\right]{\left[\ce{H2}\right]}^{3}}[/latex]
- [latex]\ce{C3H8}\left(g\right)+\ce{5O2}\left(g\right)\rightleftharpoons\ce{3CO2}\left(g\right)+\ce{4H2O}\left(g\right)\qquad{K}_{c}=\frac{{\left[\ce{CO2}\right]}^{3}{\left[\ce{H2O}\right]}^{4}}{\left[\ce{C3H8}\right]{\left[\ce{O2}\right]}^{5}}[/latex]

Note that the concentration of [latex]\ce{H2O}[/latex](*g*) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (K* _{c}*) or partial pressures (K

_{p}) of the reactants and products. A relation between these two

*K*values may be simply derived from the ideal gas equation and the definition of molarity:

[latex]\begin{array} { }PV&=&nRT \\ P&=&\left(\dfrac{n}{V}\right)RT\\ &=&MRT\end{array}[/latex]

where P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.

For the gas-phase reaction [latex]m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]:

[latex]\begin{array}{ }{K}_{P}&=&\dfrac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}\\ &=&\dfrac{{\left(\left[\text{C}\right]\times RT\right)}^{x}{\left(\left[\text{D}\right]\times RT\right)}^{y}}{{\left(\left[\text{A}\right]\times RT\right)}^{m}{\left(\left[\text{B}\right]\times RT\right)}^{n}}\\ &=&\dfrac{{\left[\text{C}\right]}^{x}{\left[\text{D}\right]}^{y}}{{\left[\text{A}\right]}^{m}{\left[\text{B}\right]}^{n}}\times \dfrac{{\left(RT\right)}^{x+y}}{{\left(RT\right)}^{m+n}}\\ &=&{K}_{c}{\left(RT\right)}^{\left(x+y\right)-\left(m+n\right)}\\ &=&{K}_{c}{\left(RT\right)}^{\Delta n}\end{array}[/latex]

And so, the relationship between *K _{c}* and

*K*is

_{P}[latex]{K}_{P}={K}_{c}{\left(RT\right)}^{\Delta n}[/latex]

where [latex]\Delta[/latex]*n* is the difference between the sum of the molar amounts of product and reactant fases, in this case:

[latex]\Delta n=\left(x+y\right)-\left(m+n\right)[/latex]

### Example 14.2.4: Calculation of *K*_{P}

_{P}

Write the equations for the conversion of *K _{c}* to

*K*for each of the following reactions:

_{P}- [latex]\ce{C2H6}\left(g\right)\rightleftharpoons\ce{C2H4}\left(g\right)+\ce{H2}\left(g\right)[/latex]
- [latex]\ce{CO}\left(g\right)+\ce{H2O}\left(g\right)\rightleftharpoons\ce{CO2}\left(g\right)+\ce{H2}\left(g\right)[/latex]
- [latex]\ce{N2}\left(g\right)+\ce{3H2}\left(g\right)\rightleftharpoons\ce{2NH3}\left(g\right)[/latex]
*K*is equal to 0.28 for the following reaction at 900 °C: [latex]\ce{CS2}\left(g\right)+\ce{4H2}\left(g\right)\rightleftharpoons\ce{CH4}\left(g\right)+\ce{2H2S}\left(g\right)[/latex]. What is_{c}*K*at this temperature?_{P}

## Show Solution

- [latex]\Delta[/latex]
*n*= (2) − (1) = 1*K*=_{P}*K*(_{c}*RT*)^{[latex]\Delta[/latex]n}=*K*(_{c}*RT*)^{1}=*K*(_{c}*RT*) - [latex]\Delta[/latex]
*n*= (2) − (2) = 0*K*=_{P}*K*(_{c}*RT*)^{[latex]\Delta[/latex]n}=*K*(_{c}*RT*)^{0}=*K*_{c} - [latex]\Delta[/latex]
*n*= (2) − (1 + 3) = -2*K*=_{P}*K*(_{c}*RT*)^{[latex]\Delta[/latex]n}=*K*(_{c}*RT*)^{−2}= [latex]\frac{{K}_{c}}{{\left(RT\right)}^{2}}[/latex] *K*=_{P}*K*(RT)_{c}^{[latex]\Delta[/latex]n}= (0.28)[(0.0821)(1173)]^{−2}= 3.0 × 10^{−5}

#### Check Your Learning

## Heterogeneous Equilibria

A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:

- [latex]\ce{PbCl2}\left(s\right)\rightleftharpoons\ce{Pb^2+}\left(aq\right)+\ce{2Cl-}\left(aq\right)\qquad{K}_{c}=\left[\ce{Pb^2+}\right]{\left[\ce{Cl^-}\right]}^{2}[/latex]
- [latex]\ce{CaO}\left(s\right)+\ce{CO2}\left(g\right)\rightleftharpoons\ce{CaCO3}\left(s\right)\qquad{K}_{c}=\frac{1}{\left[\ce{CO2}\right]}[/latex]
- [latex]\ce{C}\left(s\right)+\ce{2S}\left(g\right)\rightleftharpoons\ce{CS2}\left(g\right)\qquad{K}_{c}=\frac{\left[\ce{CS2}\right]}{{\left[\ce{S}\right]}^{2}}[/latex]
- [latex]\ce{Br2}\left(l\right)\rightleftharpoons\ce{Br2}\left(g\right)\qquad{K}_{c}=\left[\ce{Br2}\right][/latex]

Two of the above examples include terms for gaseous species only in their equilibrium constants, and so *K _{p}* expressions may also be written:

- [latex]\ce{CaO}\left(s\right)+\ce{CO2}\left(g\right)\rightleftharpoons\ce{CaCO3}\left(s\right)\qquad{K}_{P}=\frac{1}{{P}_{\ce{CO2}}}[/latex]
- [latex]\ce{C}\left(s\right)+\ce{2S}\left(g\right)\rightleftharpoons\ce{CS2}\left(g\right)\qquad{K}_{P}=\frac{{P}_{C{S}_{2}}}{{\left({P}_{S}\right)}^{2}}[/latex]

### Coupled Equilibria

The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more *coupled* equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.

- Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.[latex]\text{A} \rightleftharpoons \text{B}[/latex] [latex]\text{K}_c = \dfrac{[B]}{[A]}[/latex][latex]\text{B} \rightleftharpoons \text{A}[/latex] [latex]\text{K}_c' = \dfrac{[A]}{[B]}[/latex][latex]\text{K}_{c'} = \dfrac{1}{\text{K}_c}[/latex]
- Changing the stoichiometric coefficients in an equation by some factor
*x*results in an exponential change in the equilibrium constant by that same factor:[latex]\text{A} \rightleftharpoons \text{B}[/latex] [latex]\text{K}_{c} = \dfrac{[B]}{[A]}[/latex][latex]\text{xA} \rightleftharpoons \text{xB}[/latex] [latex]\text{K}_c' = \dfrac{[B]^x}{[A]^x}[/latex][latex]\text{K}_{c'} = \text{K}_c^x[/latex] - Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:[latex]\text{A} \rightleftharpoons \text{B}[/latex] [latex]\text{K}_{c1} = \dfrac{[B]}{[A]}[/latex][latex]\text{B} \rightleftharpoons \text{C}[/latex] [latex]\text{K}_{c2} = \dfrac{[C]}{[B]}[/latex]
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:

[latex]\text{A} + \text{B} \rightleftharpoons \text{B} + \text{C}[/latex]

[latex]\text{A} + \cancel{\text{B}} \rightleftharpoons \cancel{\text{B}} + \text{C}[/latex]

[latex]\text{A} \rightleftharpoons \text{C}[/latex] [latex]\text{K}_c' = \dfrac{[C]}{[A]}[/latex]

Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:

[latex]\text{K}_{c1}\text{K}_{c2} = \dfrac{[B]}{[A]} \text{x} \dfrac{[C]}{[B]} = \dfrac{[\cancel{\text{B}}][C]}{[A][\cancel{\text{B}}]} = \dfrac{[C]}{[A]} = \text{K}_c'[/latex]

[latex]\text{K}_c' = \text{K}_{c1}\text{K}_{c2}[/latex]

Example 14.2.5 demonstrates the use of this strategy in describing coupled equilibrium processes.

### EXAMPLE 14.2.5: Equilibrium Constants for Coupled Reactions

A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:

[latex]\ce{2NH3}\text{(g)} + \ce{3I2}\text{(g)} \rightleftharpoons \ce{N2}\text{(g)} + \ce{6HI}\text{(g)}[/latex]

Use the information below to calculate Kc for this reaction.

[latex]\ce{N2}\text{(g)} + \ce{3H2}\text{(g)} \rightleftharpoons \ce{2NH3}\text{(g)}[/latex] [latex]\text{K}_{c1} = 0.50 \text{at } 400^{\circ} \text{C}[/latex]

[latex]\ce{H2}\text{(g)} + \ce{I2}\text{(g)} \rightleftharpoons \ce{2HI}(g)[/latex] [latex]\text{K}_{c2} = 50 \text{at } 400^{\circ} \text{C}[/latex]

## Show Solution

The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.

Reverse the first coupled reaction equation:

[latex]\ce{2NH3}\text{(g)} \rightleftharpoons \ce{N2}\text{(g)} + \ce{3H}_2(g)[/latex] [latex]\text{K}_{c1'} = \frac{1}{\text{K}_{c1}} = \frac{1}{0.50} = 2.0[/latex]

Multiply the second coupled reaction by 3:

[latex]\ce{3H2}\text{(g)} + \ce{3I2}\text{(g)} \rightleftharpoons \ce{6HI}(g)[/latex] [latex]\text{K}_{c2'} = \text{K}_{c2}^3 = 50^3 = 1.2 \text{x} 10^5[/latex]

Finally, add the two revised equations:

[latex]\ce{2NH3}\text{(g)} + \cancel{\ce{3H2}\text{(g)}} + \ce{3I2}\text{(g)} \rightleftharpoons \ce{N2}\text{(g)} + \cancel{\ce{3H2}\text{(g)}} + \ce{6HI}\text{(g)}[/latex]

[latex]\ce{2NH3}\text{(g)} + \ce{3I2}\text{(g)} \rightleftharpoons \ce{N2}\text{(g)} + \ce{6HI}\text{(g)}[/latex]

[latex]\text{K}_c = \text{K}_{c1'}\text{K}_{c2'} = (2.0)(1.2 \text{x} 10^5) = 2.5\text{x} 10^5[/latex]

#### Check Your Learning

### 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

- [latex]Q=\dfrac{{\left[\text{C}\right]}^{\text{x}}{\left[\text{D}\right]}^{\text{y}}}{{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}}\text{ where }m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]
- [latex]{Q}_{P}=\dfrac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}\text{ where }m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}[/latex]
- [latex]P=MRT[/latex]
- [latex]{K}_{P}={K}_{c}{\left(RT\right)}^{\Delta n}[/latex]

## Glossary

**equilibrium constant ( K): **value of the reaction quotient for a system at equilibrium

**heterogeneous equilibria: **equilibria between reactants and products in different phases

**homogeneous equilibria: **equilibria within a single phase

** K_{c}: **equilibrium constant for reactions based on concentrations of reactants and products

** K_{P}:** equilibrium constant for gas-phase reactions based on partial pressures of reactants and products

**law of mass action: **when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant

**reaction quotient ( Q): **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

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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

when a reversible reaction has attained equilibrium at a given temperature, the reaction quotient remains constant

equilibria within a single phase

equilibrium constant for reactions based on concentrations of reactants and products

equilibrium constant for gas-phase reactions based on partial pressures of reactants and products

equilibria between reactants and products in different phases