4.7: Equilibrium Constant Manipulations
- Page ID
- 364666
Variations in the Form of the Equilibrium Constant Expression
Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, earlier this chapter we considered the general, reversible reaction:
\[m\ce{A}+n\ce{B}+ \rightleftharpoons x\ce{C}+y\ce{D} \label{13.3.1}\]
We defined the equilibrium constant (\(K\)) for this equation to be:
\[ K=\dfrac{[\ce{C}]^x[\ce{D}]^y}{[\ce{A}]^m[\ce{B}]^n} \label{13.3.2}\]
where each of the concentrations is the equilibrium concentration. If we write the reaction in reverse, we obtain the following:
\[xC+yD \rightleftharpoons mA+nB \label{Eq10}\]
The corresponding equilibrium constant \(K′\) is as follows:
\[K'=\dfrac{[A]^m[B]^n}{[C]^x[D]^y} \label{Eq11}\]
This expression is the inverse of the expression for the original equilibrium constant, so \(K′ = 1/K\). That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction \(\ce{N2O4 <=> 2NO2}\) is as follows:
\[K=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq12}\]
but for the opposite reaction, \(2 NO_2 \rightleftharpoons N_2O_4\), the equilibrium constant K′ is given by the inverse expression:
\[K'=\dfrac{[\ce{N2O4}]}{[\ce{NO2}]^2} \label{Eq13}\]
Consider another example, the formation of water:
\[\ce{2H2(g) + O2(g) <=> 2H2O(g)}. \nonumber\]
Because \(\ce{H2}\) is a good reductant and \(\ce{O2}\) is a good oxidant, this reaction has a very large equilibrium constant (\(K = 2.4 \times 10^{47}\) at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form \(\ce{O2}\) and \(\ce{H2}\), is very small: \(K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{-48}\). As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into \(\ce{H2}\) and \(\ce{O2}\).
The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.
Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction
\[\ce{2NO2 <=> N2O4} \nonumber\]
as
\[\ce{NO2 <=> 1/2 N2O4} \nonumber\]
with the equilibrium constant K″ is as follows:
\[ K′′=\dfrac{[\ce{N2O4}]^{1/2}}{[\ce{NO2}]} \label{Eq14}\]
The values for K′ (Equation \(\ref{Eq13}\)) and K″ are related as follows:
\[ K′′=(K')^{1/2}=\sqrt{K'} \label{Eq15}\]
In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by \(n\), then the new equilibrium constant is the original equilibrium constant raised to the \(n^{th}\) power.
A Video Discussing Relationships Involving Equilibrium Constants: https://youtu.be/2vZDpXX1zr0
At 745 K, K is 0.118 for the following reaction:
\[\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber\]
What is the equilibrium constant for each related reaction at 745 K?
- \(\ce{2NH3(g) <=> N2(g) + 3H2(g)}\)
- \(\ce{1/2 N2(g) + 3/2 H2(g) <=> NH3(g)}\)
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate \(K\) for each reaction.
Solution:
The equilibrium constant expression for the given reaction of \(N_{2(g)}\) with \(H_{2(g)}\) to produce \(NH_{3(g)}\) at 745 K is as follows:
\[K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118 \nonumber\]
This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:
\[K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47 \nonumber\]
In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:
\[K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344 \nonumber\]
At 527°C, the equilibrium constant for the reaction
\[\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \nonumber \]
is \(7.9 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature:
\[\ce{SO3(g) <=> SO2(g) + 1/2 O2(g)} \nonumber \]
- Answer
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\(3.6 \times 10^{-3}\)
Equilibrium Constant Expressions for the Sums of Reactions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of \(\ce{N2}\) with \(\ce{O2}\) to give \(\ce{NO2}\). This reaction is an important source of the \(\ce{NO2}\) that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (step 1), \(\ce{N2}\) reacts with \(\ce{O2}\) at the high temperatures inside an internal combustion engine to give \(\ce{NO}\). The released \(\ce{NO}\) then reacts with additional \(\ce{O2}\) to give \(\ce{NO2}\) (step 2). The equilibrium constant for each reaction at 100°C is also given.
\(\ce{N2(g) + O2(g) <=> 2NO(g)}\;\; K_1=2.0 \times 10^{-25} \label{step 1}\)
\(\ce{2NO(g) + O2(g) <=> 2NO2(g)}\;\;\;K_2=6.4 \times 10^9 \label{step 2}\)
Summing reactions (step 1) and (step 2) gives the overall reaction of \(N_2\) with \(O_2\):
\(\ce{N2(g) + 2O2(g) <=> 2NO2(g)} \;\;\;K_3=? \label{overall reaction 3}\)
The equilibrium constant expressions for the reactions are as follows:
\[K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2}\]
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,
\[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3\]
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.
To determine \(K\) for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
The following reactions occur at 1200°C:
- \(CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{-2}\)
- \(CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4\)
Calculate the equilibrium constant for the following reaction at the same temperature.
- \(CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?\)
Strategy:
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.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
\[CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)} \nonumber\]
\[\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)} \nonumber\]
\[ CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)} \nonumber\]
The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\):
\[K_3 = K_1K_2 = (9.17 \times 10^{-2})(3.3 \times 10^4) = 3.03 \times 10^3 \nonumber\]
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.
- \(\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}\)
- \(SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}\)
- \(\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?\)
- Answer
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\(K_3 = 1.1 \times 10^{66}\)
Summary
When a reaction is written in the reverse direction, \(K\) and the equilibrium constant expression are inverted. When the coefficients of an equilibrium reaction are changed, the equilibrium constant expression and magnitude will change accordingly. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions