Chemistry » Fundamental Equilibrium Concepts » Equilibrium Constants

Equilibrium Constants

Equilibrium Constants

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

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

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 Qc for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Qc is determined. When pure reactants are mixed, Qc is initially zero because there are no products present at that point. As the reaction proceeds, the value of Qc 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.

Three graphs are shown and labeled, “a,” “b,” and “c.” All three graphs have a vertical dotted line running through the middle labeled, “Equilibrium is reached.” The y-axis on graph a is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph a. The first is labeled, “[ S O subscript 2 ];” this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph b is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph b. The first is labeled, “[ S O subscript 2 ];” this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph c is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, “k.”

(a) The change in the concentrations of reactants and products is depicted as the \(2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{SO}}_{3}\left(g\right)\) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction \(2{\text{SO}}_{3}\left(g\right)⇌2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.

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 Kc.

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 NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 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 [N2O4] = 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 \(×\) 102.

Note that dimensional analysis would suggest the unit for this Kc value should be M−1. However, it is common practice to omit units for Kc 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 Kc values are truly unitless.

The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for Kc indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of Kc—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 Kc is known for a reaction, it can be used to predict directional shifts when compared to the value of Qc. 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, H2O, CO2, and H2 react to reach compositions adhering to the same equilibrium constant (the value of Qc changes until it equals the value of Kc). This value is 0.640, the equilibrium constant for the reaction under these conditions.

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

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.

Two sets of bar graphs are shown. The left is labeled, “Before reaction,” and the right is labeled, “At equilibrium.” Both graphs have y-axes labeled, “Concentration ( M ),” and three bars on the x-axes labeled, “Mixture 1,” “Mixture 2,” and “Mixture 3.” The y-axis has a scale beginning at 0.00 and ending at 0.10, with measurement increments of 0.02. The bars on the graphs are color coded and a key is provided with a legend. Red is labeled, “C O;” blue is labeled, “H subscript 2 O;” green is labeled, “C O subscript 2,” and yellow is labeled, “H subscript 2.” The graph on the left shows the red bar for mixture one just above 0.02 and the blue bar near 0.05. For mixture two, the green bar is near 0.05, and the yellow bar is near 0.09. For mixture 3, the red bar is near 0.01, the blue bar is slightly above that with green and yellow topping it off at 0.02. On the right graph, the bar for mixture one shows the red bar slightly above 0.01, the blue bar stacked on it rising slightly above 0.02, the green rising near 0.04, and the yellow bar reaching near 0.05. A label above this bar reads, “Q equals 0.640.” The bar for mixture two shows the red bar slightly above 0.02, the blue bar stacked on it rising near 0.05, the green rising near 0.07, and the yellow bar reaching near 0.10. A label above this bar reads “Q equals 0.640.” The bar for mixture three shows the red bar near 0.01, the blue bar stacked on it rising slightly above 0.01, the green rising near 0.02, and the yellow bar reaching 0.02. A label above this bar reads “Q equals 0.640”.

Concentrations of three mixtures are shown before and after reaching equilibrium at 800 °C for the so-called water gas shift reaction: \(\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right).\)

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/ProductsExperiment 1Experiment 2Experiment 3
[CO]i0.0203 M0.011 M0.0094 M
[H2O]i0.0203 M0.0011 M0.0025 M
[CO2]i0.0040 M0.037 M0.0015 M
[H2]i0.0040 M0.046 M0.0076 M

Solution

Experiment 1:

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

Qc < Kc (0.039 < 0.64)

The reaction will shift to the right.

Experiment 2:

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

Qc > Kc (140 > 0.64)

The reaction will shift to the left.

Experiment 3:

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

Qc < Kc (0.48 < 0.64)

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 Kc 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, Qc and Kc 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.

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