## Equilibrium Constants

Contents

Now that we have a symbol \(\text{(⇌)}\) 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:

\(m\text{A}+n\text{B}+⇌x\text{C}+y\text{D}\)

We can write the **reaction quotient ( Q)** for this equation. When evaluated using concentrations, it is called

*Q*. We use brackets to indicate molar concentrations of reactants and products.

_{c}\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[\text{C}\right]}^{x}\left[\text{D}\right]^{y}}{{\left[\text{A}\right]}^{m}\left[\text{B}\right]^{n}}\)

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 \(2{\text{NO}}_{2}\left(g\right)⇌{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\) is given by this expression:

\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}\)

## Example

### Writing Reaction Quotient Expressions

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

(a) \(3{\text{O}}_{2}\left(g\right)⇌2{\text{O}}_{3}\left(g\right)\)

(b) \({\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)\)

(c) \(4{\text{NH}}_{3}\left(g\right)+7{\text{O}}_{2}\left(g\right)⇌4{\text{NO}}_{2}\left(g\right)+6{\text{H}}_{2}\text{O}\left(g\right)\)

### Solution

(a) \({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{O}}_{3}\right]}^{2}}{{\left[{\text{O}}_{2}\right]}^{3}}\)

(b) \({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}\)

(c) \({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{NO}}_{2}\right]}^{4}\left[{\text{H}}_{2}\text{O}\right]^{6}}{{\left[{\text{NH}}_{3}\right]}^{4}\left[{\text{O}}_{2}\right]^{7}}\)

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*is determined. When pure reactants are mixed,

_{c}*Q*is initially zero because there are no products present at that point. As the reaction proceeds, the value of

_{c}*Q*increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (see the figure below). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.

_{c}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:

\({Q}_{c}\phantom{\rule{0.2em}{0ex}}\text{at equilibrium}={K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[\text{C}\right]^{x}\left[\text{D}\right]^{y}\text{…}}{\left[\text{A}\right]^{m}\left[\text{B}\right]^{n}\text{…}}\)

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

### Evaluating a Reaction Quotient

Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:

\(2{\text{NO}}_{2}\left(g\right)⇌{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\)

When 0.10 mol NO_{2} is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO_{2}] = 0.016 *M* and [N_{2}O_{4}] = 0.042 *M*.

(a) What is the value of the reaction quotient before any reaction occurs?

(b) What is the value of the equilibrium constant for the reaction?

### Solution

(a) Before any product is formed, \(\left[{\text{NO}}_{2}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{0.10\phantom{\rule{0.2em}{0ex}}\text{mol}}{1.0\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=0.10\phantom{\rule{0.2em}{0ex}}M,\) and [N_{2}O_{4}] = 0 *M*. Thus,

\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0}{{0.10}^{2}}\phantom{\rule{0.2em}{0ex}}=0\)

(b) At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, \({K}_{c}={Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.042}{{0.016}^{2}}\phantom{\rule{0.2em}{0ex}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}.\) 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 module, 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}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*—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.

_{c}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*. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in the figure below illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H

_{c}_{2}O, CO

_{2}, and H

_{2}react to reach compositions adhering to the same equilibrium constant (the value of

*Q*changes until it equals the value of

_{c}*K*). This value is 0.640, the equilibrium constant for the reaction under these conditions.

_{c}\(\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=0.640\phantom{\rule{5em}{0ex}}\text{T}=800\phantom{\rule{0.2em}{0ex}}\text{°C}\)

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 the figure below 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

### Predicting the Direction of Reaction

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

\(\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\)

\({K}_{c}=0.64\)

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:

\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(0.0040\right)\left(0.0040\right)}{\left(0.0203\right)\left(0.0203\right)}\phantom{\rule{0.2em}{0ex}}=0.039.\)

*Q _{c}* <

*K*(0.039 < 0.64)

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

Experiment 2:

\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(0.037\right)\left(0.046\right)}{\left(0.011\right)\left(0.0011\right)}\phantom{\rule{0.2em}{0ex}}=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\)

*Q _{c}* >

*K*(140 > 0.64)

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

Experiment 3:

\({Q}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CO}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\text{O}\right]}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(0.0015\right)\left(0.0076\right)}{\left(0.0094\right)\left(0.0025\right)}\phantom{\rule{0.2em}{0ex}}=0.48\)

*Q _{c}* <

*K*(0.48 < 0.64)

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

In the example above, 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*expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression’s value)

_{c}*.*Several examples of equilibria yielding such expressions will be encountered in this section.