19.2: Equilibrium Constant
Lesson Objectives
The student will:
 write equilibrium constant expressions.
 use equilibrium constant expressions to solve for unknown concentrations.
 use known concentrations to solve for the equilibrium constants.
 explain what the value of \begin{align*}K\end{align*}
K means in terms of relative concentrations of reactants and products.
Vocabulary
 equilibrium constant (K)
Introduction
You were introduced to two types of equations in the chapter on “Chemical Kinetics.” One consisted of reactions where the overall equation and the reaction mechanism were exactly the same. In those reactions, the reaction occurred with a single collision. The other type of reactions involved a reaction mechanism consisting two or more steps, and the overall equation was the sum of the equations for the reaction mechanism.
For those reactions that consist of a single collision, the reaction rate can be expressed by inserting the concentrations of the reactants into the expression:



\begin{align*}\mathrm{Reaction \ Rate} = k_F[A][B]\end{align*}
Reaction Rate=kF[A][B] , where \begin{align*}A\end{align*}A and \begin{align*}B\end{align*}B are the reactants

\begin{align*}\mathrm{Reaction \ Rate} = k_F[A][B]\end{align*}

For the reactions that have reaction mechanisms of two or more steps, the relationship between the concentration of a particular reactant and the reaction rate is more complex. The rate of these reactions can still be expressed in a similar manner, but which reactants go into the expression and how they are involved must be determined experimentally.
In the previous section, you were introduced to the possibility of a reverse reaction. For the reaction between \begin{align*}A\end{align*}



\begin{align*}A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}\end{align*}
A(g)+B(g)→C(g)+D(g)

\begin{align*}A_{(g)} + B_{(g)} \rightarrow C_{(g)} + D_{(g)}\end{align*}

but we also could have a reverse reaction:



\begin{align*}A_{(g)} + B_{(g)} \leftarrow C_{(g)} + D_{(g)}\end{align*}
A(g)+B(g)←C(g)+D(g)

\begin{align*}A_{(g)} + B_{(g)} \leftarrow C_{(g)} + D_{(g)}\end{align*}

If this reverse reaction occurs, it would be the same simple collision that was involved in the forward reaction. Therefore, the reaction rate of the reverse reaction could be expressed as:



\begin{align*}\mathrm{Reaction \ Rate} = k_R[C][D]\end{align*}
Reaction Rate=kR[C][D] , where the products \begin{align*}C\end{align*}C and \begin{align*}D\end{align*}D have become reactants

\begin{align*}\mathrm{Reaction \ Rate} = k_R[C][D]\end{align*}

When this reaction reaches equilibrium and the two rates become constant and equal,



\begin{align*}Rate_{forward} = Rate_{reverse}\end{align*}
Rateforward=Ratereverse and \begin{align*}k_F[A][B] = k_R[C][D]\end{align*}kF[A][B]=kR[C][D]

\begin{align*}Rate_{forward} = Rate_{reverse}\end{align*}




\begin{align*}k_F[A][B] = k_R[C][D]\end{align*}
kF[A][B]=kR[C][D]

\begin{align*}k_F[A][B] = k_R[C][D]\end{align*}

Algebraic manipulation of the expression yields



\begin{align*}\frac {k_F} {k_R} = \frac {[C][D]} {[A][B]}\end{align*}
kFkR=[C][D][A][B]

\begin{align*}\frac {k_F} {k_R} = \frac {[C][D]} {[A][B]}\end{align*}

Since both \begin{align*}k_F\end{align*}



\begin{align*}K = \frac {[C][D]} {[A][B]}\end{align*}
K=[C][D][A][B]

\begin{align*}K = \frac {[C][D]} {[A][B]}\end{align*}

The Equilibrium Constant
The equilibrium constant (\begin{align*}K\end{align*}
The equilibrium constant for any reaction can be written by following a few simple rules. We have already learned how to write the expression for a simple generic reaction:


 \begin{align*}A_{(aq)} + B_{(aq)} \rightleftharpoons C_{(aq)} + D_{(aq)} \ \text{and} \ K = \frac{[C] [D]} {[A] [B]}\end{align*}

But suppose the reaction was \begin{align*}2 \ A_{(aq)} + B_{(aq)} \rightleftharpoons C_{(aq)} + D_{(aq)}\end{align*}. Writing the equation with a coefficient of 2 in front of the \begin{align*}A\end{align*} is exactly the same as writing the equation this way:


 \begin{align*}A_{(aq)} + A_{(aq)} + B_{(aq)} \rightleftharpoons C_{(aq)} + D_{(aq)}\end{align*}

In this second form of the equation, we have just written \begin{align*}A\end{align*} twice instead of using a coefficient to show that two molecules are involved in the reaction. If we write the equilibrium constant expression from this last equation, it would look like:


 \begin{align*}K = \frac{[C] [D]} {[A] [A] [B]}\end{align*}

With our knowledge of algebra, we know that writing \begin{align*}[A] \times [A]\end{align*} is the same as writing \begin{align*}[A]^2\end{align*}. Therefore, the equilibrium constant would become


 \begin{align*}K = \frac{[C] [D]} {[A]^2 [B]}\end{align*}

As you can see, the coefficient in the equation has become an exponent in the equilibrium constant. The general rule for writing an equilibrium constant expression is to write the coefficient of each species in the reaction as an exponent in the equilibrium constant expression. So, for the reaction:


 \begin{align*}a \ A_{(g)} + b \ B_{(g)} \rightleftharpoons c \ C_{(g)} + d \ D_{(g)}\end{align*}

where the lower case \begin{align*}a, \ b, \ c,\end{align*} and \begin{align*}d\end{align*} are coefficients, the equilibrium constant expression would be:


 \begin{align*}K = \frac{[C]^c [D]^d} {[A]^a [B]^b}\end{align*}

Here is an example equation and its equilibrium constant expression.


 \begin{align*}2 \ NO_{2(g)} \rightleftharpoons 2 \ NO_{(g)} + O_{2(g)}\end{align*}



 \begin{align*}K = \frac{[NO]^2 [O_2]} {[NO_2]^2}\end{align*}

There are cases where the concentration of a particular reactant or product does not change when it is used or produced. The concentration of a substance is the moles of the substance divided by the volume (in liters) occupied. In the case of a substance dissolved in a beaker of water, its volume is the same as the volume of the solution. If half of the substance is used up in a reaction, the remaining half is still dissolved in the same volume of water. Therefore, its concentration will be cut in half. In the case of a solid reactant, however, it is not dissolved in the water, so its volume is the volume of the solid itself. If half of the solid is used in a reaction, the amount of moles remaining has been cut in half, but the volume of the solid has also been cut in half. Its concentration, then, stays exactly the same. This is also true of any solid or liquid that is not dissolved in a solution.
If the state of a reactant or product is indicated by \begin{align*}(aq)\end{align*} or \begin{align*}(g)\end{align*}, its concentration can change, but if its state is indicated by \begin{align*}(s)\end{align*} or \begin{align*}(l)\end{align*}, its concentration cannot change. Since the concentrations of these substances cannot change, their numerical value in the equilibrium constant expression will be a constant. To simplify the equilibrium expression, these constants can be algebraically combined with \begin{align*}K\end{align*} to produce a new constant, \begin{align*}K'\end{align*}. To avoid confusion, chemists have decided to always omit solids and liquids from the equilibrium expression, and it is assumed that the constant concentration values for those reaction components are already combined into the equilibrium constant. Here are some examples.
\begin{align*}\mathrm{CO}_{(g)} + 3 \ \mathrm{H}_{2(g)} \rightleftharpoons \mathrm{CH}_4(g) + \mathrm{H}_2\mathrm{O}_{(g)}\end{align*}


 \begin{align*}K = \frac {[\mathrm{CH}_4][\mathrm{H}_2\mathrm{O}]} {[\mathrm{CO}][\mathrm{H}_2]^3}\end{align*}
 Note the coefficient of 3 becomes an exponent.

\begin{align*}2 \ \mathrm{TiCl}_{3(s)} + 2 \ \mathrm{HCl}_{(g)} \rightleftharpoons 2 \ \mathrm{TiCl}_{4(s)} + \mathrm{H}_{2(g)}\end{align*}


 \begin{align*}K = \frac {[\mathrm{H}_2]} {[\mathrm{HCl}]^2}\end{align*}
 Note the two solids are omitted and the coefficients become exponents.

\begin{align*}\mathrm{P}_{4(s)} + 6 \ \mathrm{Cl}_{2(g)} \rightleftharpoons 4 \ \mathrm{PCl}_{3(s)}\end{align*}


 \begin{align*}K = \frac {1} {[\mathrm{Cl}_2]^6}\end{align*}
 Note the solids are omitted and the coefficient becomes an exponent.

\begin{align*}\mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{OH}^_{(aq)}\end{align*}


 \begin{align*}K = [\mathrm{H}^+][\mathrm{OH}^]\end{align*}
 Note the liquid water is omitted and the denominator is omitted when it is 1.

This video shows and example of how to plug values into an equilibrium constant expression (9c): http://www.youtube.com/watch?v=lZbrwtY5K3k (2:48).
Mathematics with Equilibrium Expressions
The mathematics involved with equilibrium expressions can range from the straightforward to the complex. For example, look at the sample question below.
Example:


 \begin{align*}\mathrm{SO}_{2(g)}+ \mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)} + \mathrm{NO}_{(g)}\end{align*}

Determine the value of \begin{align*}K\end{align*} when the equilibrium concentrations are: \begin{align*}[\mathrm{SO}_2] = 1.20 \ \mathrm{M}; \ [\mathrm{NO}_2] = 0.60 \ \mathrm{M}; \ [\mathrm{NO}] = 1.6 \ \mathrm{M}; \ [\mathrm{SO}_3] = 2.2 \ \mathrm{M}\end{align*}. Are the reactants or products favored? Explain your answer.
Solution:
Step 1: Write the equilibrium constant expression:


 \begin{align*}K = \frac {[\mathrm{SO}_3][\mathrm{NO}]} {[\mathrm{SO}_2][\mathrm{NO}_2]}\end{align*}

Step 2: Substitute in given values and solve:


 \begin{align*}K = \frac {(2.2)(1.6)} {(1.20)(0.60)} = 4.9\end{align*}

The equilibrium constant value is the ratio of the concentrations of the products over the reactants. Therefore, a value of 4.9 for the equilibrium constant indicates that there are more products (numerator) than there are reactants (denominator). If the equilibrium constant is 1 or nearly 1, it indicates that the molarities of the reactants and products are about the same. If the equilibrium constant value is a large number, it indicates that the great majority of the material is in the form of products at equilibrium, and and we say “the products are strongly favored.” If the equilibrium constant is small, it indicates that the reactants are strongly favored.
Example:
At a given temperature, the reaction \begin{align*}\mathrm{CO}_{(g)} + \mathrm{H}_2\mathrm{O}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)} + \mathrm{CO}_{2(g)}\end{align*} produces the following concentrations at equilibrium: \begin{align*}[\mathrm{CO}] = 0.200 \ \mathrm{M}; \ [\mathrm{H}_2\mathrm{O}] = 0.500 \ \mathrm{M}; \ [\mathrm{H}_2] = 0.320 \ \mathrm{M}; \ [\mathrm{CO}_2] = 0.420 \ \mathrm{M}\end{align*}. Find \begin{align*}K\end{align*}.
Solution:
Step 1: Write the equilibrium expression:


 \begin{align*}K = \frac {[\mathrm{H}_2][\mathrm{CO}_2]} {[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}\end{align*}

Step 2: Substitute the given values and solve:


 \begin{align*}K = \frac {(0.320)(0.420)} {(0.200)(0.500)} = 1.34\end{align*}

Example:
For the same reaction and the same temperature as the previous example, the concentrations of the substances at equilibrium become: \begin{align*}[\mathrm{CO}] = ?; \ [\mathrm{H}_2\mathrm{O}] = 0.100 \ \mathrm{M}; \ [\mathrm{H}_2] = 0.100 \ \mathrm{M}; \ [\mathrm{CO}_2] = 0.100 \ \mathrm{M}\end{align*}. What is the concentration of \begin{align*}\mathrm{CO}\end{align*}?
Solution:
Since it is the same reaction at the same temperature, the \begin{align*}K\end{align*} value will be the same, 1.34.
Step 1: Rearrange the equilibrium constant expression to solve for the unknown:


 \begin{align*}K = \frac {[\mathrm{H}_2][\mathrm{CO}_2]} {[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]} \ \text{so} \ [\mathrm{CO}] = \frac {[\mathrm{H}_2][\mathrm{CO}_2]} {K \ [\mathrm{H}_2\mathrm{O}]}\end{align*}

Step 2: Substitute and solve:


 \begin{align*}[\mathrm{CO}] = \frac {(0.100)(0.100)} {(1.34)(0.100)} = 0.0746 \ \text{M}\end{align*}

For an introduction to chemical equilibrium (9b) see: http://www.youtube.com/watch?v=vCNENmgNJYg (10:29), http://www.youtube.com/watch?v=KlNl12Q_aJY (9:03).
Lesson Summary
 The equilibrium constant, \begin{align*}K\end{align*}, is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.
 If the value of \begin{align*}K\end{align*} is greater than \begin{align*}1\end{align*}, the products in the reaction are favored; if the value of \begin{align*}K\end{align*} is less than \begin{align*}1\end{align*}, the reactants in the reaction are favored; if \begin{align*}K\end{align*} is equal to \begin{align*}1\end{align*}, neither reactants nor products are favored.
Review Questions
 Why are solids and liquids not included in the equilibrium constant expression?
 What does the value of \begin{align*}K\end{align*} mean in terms of the amount of reactants and products?
 What is the correct equilibrium constant expression for the following reaction: \begin{align*}2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{SO}_{3(g)}\end{align*}?
 \begin{align*}K = \frac {[\mathrm{SO}_2]^2[\mathrm{O}_2]} {[\mathrm{SO}_3]^2}\end{align*}
 \begin{align*}K = \frac {[\mathrm{SO}_3]^2} {[\mathrm{SO}_2]^2[\mathrm{O}_2]}\end{align*}
 \begin{align*}K = \frac {[2 \ \mathrm{SO}_2][\mathrm{O}_2]} {[2 \ \mathrm{SO}_3]}\end{align*}
 \begin{align*}K = \frac {[2 \ \mathrm{SO}_3]} {[2 \ \mathrm{SO}_2][\mathrm{O}_2]}\end{align*}
 What is the correct equilibrium constant expression for the following reaction: \begin{align*}\mathrm{Cu(OH)}_{2(s)} \rightleftharpoons \mathrm{Cu}^{2+}_{(aq)} + 2 \ \mathrm{OH}^_{(aq)}\end{align*}?
 \begin{align*}K = \frac {[\mathrm{Cu}_{2+}][\mathrm{OH}_]^2} {[\mathrm{Cu(OH)}^2]}\end{align*}
 \begin{align*}K = \frac {[\mathrm{Cu(OH)}^2]} {[\mathrm{Cu}^{2+}][\mathrm{OH}^]^2}\end{align*}
 \begin{align*}K = \frac {1} {[\mathrm{Cu}^{2+}][\mathrm{OH}^]^2}\end{align*}
 \begin{align*}K = [\mathrm{Cu}^2+][\mathrm{OH}^]^2\end{align*}
 Consider the following equilibrium system: \begin{align*}2 \ \mathrm{NO}_{(g)} + \mathrm{Cl}_{2(g)} \rightleftharpoons 2 \ \mathrm{NOCl}_{(g)}\end{align*}. At a certain temperature, the equilibrium concentrations are as follows: \begin{align*}[\mathrm{NO}] = 0.184 \ \mathrm{mol/L}, \ [\mathrm{Cl}_2] = 0.165 \ \mathrm{mol/L}\end{align*}, and \begin{align*}[\mathrm{NOCl}] = 0.060 \ \mathrm{mol/L}\end{align*}. What is the equilibrium constant for this reaction?
 \begin{align*}0.506\end{align*}
 \begin{align*} 0.648\end{align*}
 \begin{align*}1.55\end{align*}
 \begin{align*}1.97\end{align*}
 For the reaction \begin{align*}2 \ \mathrm{MgCl}_{2(s)} + \ \mathrm{O}_{2(g)} \rightleftharpoons 2 \ \mathrm{MgO}_{(s)} + 2 \ \mathrm{Cl}_{2(g)}\end{align*}, the equilibrium constant was found to be 3.86 at a certain temperature. If \begin{align*}0.560 \ \mathrm{mol} \ \mathrm{O}_{2(g)}\end{align*} is placed in a \begin{align*}1.00 \ \mathrm{L}\end{align*} container, what is the concentration of \begin{align*}\mathrm{Cl}_{2(g)}\end{align*} at equilibrium?
 \begin{align*}1.47 \ \mathrm{mol/L}\end{align*}
 \begin{align*}2.16 \ \mathrm{mol/L}\end{align*}
 \begin{align*}2.88 \ \mathrm{mol/L}\end{align*}
 not enough information is available
 Write the equilibrium constant expressions for each of the following equations:
 \begin{align*}\mathrm{CH}_3\mathrm{NH}_{2(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_{3(aq)}^+ + \mathrm{OH}^_{(aq)}\end{align*}
 \begin{align*}2 \ \mathrm{CaSO}_{4(s)} \rightleftharpoons 2 \ \mathrm{CaO}_{(s)} + 2 \ \mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)}\end{align*}
 \begin{align*}2 \ \mathrm{Fe}^{3+}_{(aq)} + 3 \ \mathrm{S}^{2}_{(aq)} \rightleftharpoons \mathrm{Fe}_2\mathrm{S}_{3(s)}\end{align*}
 \begin{align*}\mathrm{Hg}_{(l)} + \mathrm{H}_2\mathrm{S}_{(g)} \rightleftharpoons \mathrm{HgS}_{(s)} + \mathrm{H}_{2(g)} \end{align*}
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