7.4: Equilibrium Calculations

Learning Objectives

• Determine the changes in concentration or pressure that occur as a system approaches equilibrium.
• Calculate equilibrium concentrations or pressures using initial conditions and the equilibrium expression.
• Compute equilibrium constants (\(K_c\) or \(K_p\)​) from equilibrium data.

This section explores equilibrium calculations, which are fundamental in many scientific and technological fields. One important application is in pharmaceutical science, particularly in drug development and administration. When a drug is ingested or injected, it often undergoes multiple chemical equilibria that influence its concentration in the body. Understanding the quantitative aspects of these equilibria is crucial for determining the correct dosage needed to achieve the desired therapeutic effect.

Equilibrium calculations focus on how reactant and product concentrations change as a system progresses from its initial state to equilibrium. These changes are directly related to the stoichiometry of the reaction. For example, consider the decomposition of ammonia:

$$\[\ce{2 NH3(g) \rightleftharpoons N2(g) + 3 H2(g)} \nonumber \]

Regardless of the initial amounts of reactants and products, the reaction will always follow the same stoichiometric relationships between changes in concentration, as dictated by the balanced equation.

For instance, if the nitrogen concentration increases by an amount $x$:

\[\Delta\left[ N_2\right]=+x \nonumber \]

then the corresponding changes in the other species concentrations are:

\[\begin{align*}
\Delta\left[ \ce{H2} \right] &=\Delta\left[ \ce{N2} \right]\left(\frac{3 ~\text{mol} ~\ce{H2} }{1~\text{mol}~\ce{N2} }\right)=+3 x \\[4pt]
\Delta\left[ \ce{NH3} \right] &=-\Delta\left[ \ce{N2} \right]\left(\frac{2 ~\text{mol}~\ce{NH3}}{1~\text{mol} ~\ce{N2} }\right)=-2 x
\end{align*} \]

The negative sign indicates a decrease in concentration. If nitrogen is being formed in the reaction and its concentration is increasing, ammonia must be reacting and its concentration must decrease accordingly.

Example \(\PageIndex{1}\): Determining Relative Changes in Concentration

For the following reactions, find the missing terms representing concentration changes for the reactants and products to reach equilibrium.

1. \[\underset{x}{\ce{C2H2(g)}} + \underset{?}{\ce{2 Br2(g)}} \ce{<=>} \underset{?}{\ce{C2H2Br4(g)}} \nonumber \]
2. \[\underset{?}{\ce{I2(aq)}} + \underset{?}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber \]
3. \[\underset{x}{\ce{C3H8(g)}} + \underset{?}{\ce{5O2(g)}} \ce{<=>} \underset{?}{\ce{3CO2(g)}} + \underset{?}{\ce{4H2O(g)}} \nonumber \]

Solution

1. \[\underset{x}{\ce{C2H2(g)}} + \underset{2x}{\ce{2 Br2(g)}} \ce{<=>} \underset{-x}{\ce{C2H2Br4(g)}} \nonumber \]
2. \[\underset{-x}{\ce{I2(aq)}} + \underset{-x}{\ce{I^{-}(aq)}} \ce{<=>} \underset{x}{\ce{I3^{-}(aq)}} \nonumber \]
3. \[\underset{x}{\ce{C3H8(g)}} + \underset{5x}{\ce{5O2(g)}} \ce{<=>} \underset{-3x}{\ce{3CO2(g)}} + \underset{-4x}{\ce{4H2O(g)}} \nonumber \]

Exercise \(\PageIndex{1}\)

For the following reactions, find the missing terms representing concentration changes for the reactants and products to reach equilibrium.

1. \[\underset{?}{\ce{2SO2(g)}} + \underset{x}{\ce{O2(g)}} \ce{<=>} \underset{?}{\ce{2SO3(g)}} \nonumber \]
2. \[\underset{?}{\ce{C4H8(g)}} \ce{<=>} \underset{-2x}{\ce{2C2H4(g)}} \nonumber \]
3. \[\underset{?}{\ce{4NH3(g)}} + \underset{-7x}{\ce{7O2(g)}} \ce{<=>} \underset{?}{\ce{4NO2(g)}} + \underset{?}{\ce{6H2O(g)}} \nonumber \]

Answer

(a) 2x, x, −2x; (b) x, −2x; (c) −4x, −7x, 4x, 6x

Calculation of an Equilibrium Constant

The equilibrium constant (\(K\)) for a reaction is determined using the equilibrium concentrations for \(K_c\) or partial pressures for \(K_p\) of its reactants and products. If these values are known, the calculation is straightforward: they are substituted into the equilibrium expression for \(K\).  However, in many cases, the equilibrium concentrations are not given directly and must be determined from the initial conditions. A systematic way to organize information for these types of problems is by using an ICE table.

An ICE table helps track how the system moves from its Initial concentrations (I) through the Changes (C) that occur as the reaction progresses to Equilibrium (E). It follows the stoichiometry of the reaction, making it easier to set up and solve equilibrium expressions.

Note: An ICE table is not a special formula but rather a structured way to apply the fundamental definition of equilibrium—a state in which the rates of the forward and reverse reactions are equal, and concentrations no longer change.  It is a mathematical representation of what happens in all equilibrium reactions: the system starts from some initial conditions, undergoes changes as it approaches equilibrium, and reaches a final equilibrium state where the concentrations of reactants and products no longer change.

Example \(\PageIndex{2}\): Calculation of an Equilibrium Constant

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

\[\ce{I2(aq) + I^{-}(aq) \rightleftharpoons I_3^{-}(aq)} \nonumber \]

A solution with the initial concentrations of \(\ce{I2}\) and \(\ce{I^{−}}\) of \(1.000 \times 10^{−3}~\text{M}\) results in an equilibrium concentration of \(\ce{I2}\) of \(6.61 \times 10^{−4} ~\text{M}\).  What is the equilibrium constant for the reaction?

Solution

To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:

\[K_c=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \nonumber \]

We are given the initial concentrations of the reactants and the equilibrium concentration of I₂. Let's use this information to derive terms for the equilibrium concentrations of the species, presenting all the information in an ICE table where the change in concentrations are relate the stoichiometry of the reaction.

At equilibrium the concentration of I₂ is \(6.61 \times 10^{−4}~\text{M}\) so that

\[\begin{align*}
1.000 \times 10^{-3}-x & =6.61 \times 10^{-4}\\[4pt]
x &=1.000 \times 10^{-3}-6.61 \times 10^{-4}\\[4pt]
&=3.39 \times 10^{-4} M
\end{align*} \nonumber \]

The ICE table may now be updated with numerical values for all its concentrations:

Finally, substitute the equilibrium concentrations into the K expression and solve:

\[\begin{align*}
K_c &=\frac{\left[ \ce{I3^{-}} \right]}{\left[ \ce{I2} \right]\left[ \ce{I^{-}} \right]} \\[4pt]
&=\frac{3.39 \times 10^{-4} M}{\left(6.61 \times 10^{-4} M\right)\left(6.61 \times 10^{-4} M\right)}=776
\end{align*} \nonumber \]

Exercise \(\PageIndex{2}\)

Consider the following reaction:

\[\ce{PCl3(g) +Cl2(g) \rightleftharpoons PCl5(g) } \nonumber \]

0.480 atm of PCl₃ and 0.400 atm of Cl₂ gases are added to an empty flask.  At equilibrium, the partial pressure of PCl₅ is 0.320 atm.  What is K_p for the reaction?

Answer

	PCl3(g)  +	Cl2(g)    ⇌	PCl5(g)
Initial	0.480 atm	0.400 atm	0 atm
Change	- x	-x	+x
Equilibrium	0.480 - x	0.400 - x	x = 0.320

\(P_{\ce{PCl3}}=0.480-0.320=0.160\; \text{atm}\)
\(P_{\ce{Cl2}}=0.400-0.320=0.080\; \text{atm}\)
\(P_{\ce{PCl5}}=0.320\; \text{atm}\)
\(K_p=\frac{P_{\ce{PCl5}}}{P_{\ce{PCl3}}P_{\ce{Cl2}}}=\frac{0.320}{(0.160)(0.080)}=25\)

Calculation of a Missing Equilibrium Concentration

When the equilibrium constant and all but one of the equilibrium concentrations are known, the missing equilibrium concentration can be determined algebraically using the equilibrium expression.  An example of this type of calculation is shown in the next example.

Example \(\PageIndex{3}\): Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the K_c for the reaction,

\[\ce{N2(g) + O2(g) \rightleftharpoons 2 NO(g)} \nonumber \]

is \(4.1 \times 10^{−4}\). Calculate the equilibrium concentration of NO(g). The equilibrium concentrations of N₂ and O₂ at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

Solution

Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

\[\begin{align*}
K_c &=\frac{[ NO ]^2}{\left[ N_2\right]\left[ O_2\right]} \\[4pt]
[ NO ]^2&=K_c\left[ N_2\right]\left[ O_2\right] \\[4pt]
[ NO ]&=\sqrt{K_c\left[ N_2\right]\left[ O_2\right]} \\[4pt]
&=\sqrt{\left(4.1 \times 10^{-4}\right)(0.036)(0.0089)} \\[4pt]
&=\sqrt{1.31 \times 10^{-7}} \\[4pt]
&=3.6 \times 10^{-4}
\end{align*} \nonumber \]

[NO] is \(3.6 \times 10^{−4} ~\text{mol/L}\) at equilibrium under these conditions.

Exercise \(\PageIndex{3}\)

The equilibrium constant K_c for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is \(6.00 \times 10^{−2}\).

\[\ce{2 NH3(g) <=> N2(g) + 3 H2(g)} \nonumber \]

What is the equilibrium concentration of ammonia, [NH₃], if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively?

Answer

\(K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}=6.00 \times 10^{−2}\)
\([NH_3]=\sqrt{\frac{[N_2][H_2]^3}{K_c}}=\sqrt{\frac{(4.26)(2.09)^3}{6.00 \times 10^{−2}}}=25.5 M\)

Calculation of Equilibrium Concentrations from Initial Concentrations

Let's now look at the calculation of equilibrium concentrations from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:

1. Identify the direction in which the reaction will proceed to reach equilibrium.
2. Set up an ICE table to organize initial concentrations, changes in concentration, and equilibrium concentrations.
3. Calculate the concentration changes based on stoichiometry and solve for the equilibrium concentrations.
4. Verify your results by substituting the calculated equilibrium concentrations into the equilibrium expression to check consistency with K.

The final two example exercises in this chapter demonstrate the application of this strategy in detail.

Example \(\PageIndex{4}\): Calculation of Equilibrium Concentrations

At a certain temperature, the equilibrium constant K_c for the decomposition of \(\ce{PCl5(g)}\) into \(\ce{PCl3(g)}\) and \(\ce{Cl2(g)}\) is 0.0211. What are the equilibrium concentrations of \(\ce{PCl5}\), \(\ce{PCl3}\), and \(\ce{Cl2}\) in a mixture that initially contained only \(\ce{PCl5}\) at a concentration of 1.00 M?

\(\ce{PCl5(g) \rightleftharpoons PCl3(g) + Cl2(g)}\)

Solution

Step 1.

Determine the direction the reaction proceeds.

Because only the reactant is present initially Q_c = 0 and the reaction will proceed to the right.

Step 2:

Set up an ICE table.

Step 3.

Solve for the change and the equilibrium concentrations.

Substituting the equilibrium concentrations into the equilibrium constant equation gives

\[\begin{align*}
K_c &=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2} \right]}{\left[ \ce{PCl5} \right]}=0.0211 \\[4pt]
&=\frac{(x)(x)}{(1.00-x)} \\[4pt]
0.0211&=\frac{(x)(x)}{(1.00-x)} \\[4pt]
0.0211(1.00-x) &=x^2 \\[4pt]
x^2+0.0211 x - 0.0211 &=0
\end{align*} \]

We can solve for x using the quadratic equation.  An equation of the form \(ax^2 + bx + c = 0\) can be rearranged to solve for \(x\):

\[x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \nonumber \]

In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:

\[\begin{align*}
x&=\frac{-0.0211 \pm \sqrt{(0.0211)^2-4(1)(-0.0211)}}{2(1)} \\[4pt]
&=\frac{-0.0211 \pm \sqrt{\left(4.45 \times 10^{-4}\right)+\left(8.44 \times 10^{-2}\right)}}{2} \\[4pt]
&=\frac{-0.0211 \pm 0.291}{2}
\end{align*} \]

The two roots of the quadratic are, therefore,

\[x=\frac{-0.0211+0.291}{2}=0.135 \nonumber \]

and

\[x=\frac{-0.0211-0.291}{2}=-0.156 \nonumber \]

We will only use x = 0.135 M because the other solution results in a negative concentration which is not physically possible.

The equilibrium concentrations are

\[\begin{align*}
[ \ce{PCl5}] &= 1.00-0.135 = 0.87 ~\text{M} \\[4pt]
[ \ce{PCl3} ] &= x=0.135 ~\text{M} \\[4pt]
[ \ce{Cl2} ] &= x =0.135 ~\text{M}
\end{align*} \nonumber \]

Step 4.

Verify the calculated equilibrium concentrations, by substituting into the equilibrium expression:

\[K_c=\frac{\left[ \ce{PCl3} \right]\left[ \ce{Cl2}\right]}{\left[ \ce{PCl5} \right]}=\frac{(0.135)(0.135)}{0.87}=0.021 \nonumber \]

The equilibrium constant computed using the equilibrium concentrations is close to the given K_c​ value, with minor differences due to rounding. This confirms that the calculation is accurate.

Exercise \(\PageIndex{4}\)

A 1.00-L flask is filled with 1.00 mole of H₂ and 2.00 moles of I₂. What are the equilibrium concentrations of H₂, I₂, and HI?

\[\ce{H2(g) + I2(g) <=> 2 HI(g)} \;\; K_c = 50.5\nonumber \]

Answer

	H2(g)  +	I2(g)        ⇌	2HI(g)
Initial	1.00 M	2.00 M	0 M
Change	- x	-x	+2x
Equilibrium	1.00 M- x	2.00 - x	2x

\(K_c=\frac{[HI]^2}{[I_2][H_2]}=\frac{(2x)^2}{(1.00-x)(2.00-x)}=50.5 \)
\(4x^2=50.5(2.00-3.00x+x^2)\)
\(0=46.5x^2-151.5x+101\)
Using quadratic equation we get x = 2.32 M and 0.935 M.  2.32 M would give negative concentrations, so only 0.934 M is physically reasonable.
[H₂]=1.00 M - 0.935 M = 0.065 M
[I₂] = 2.00 M - 0.935 M = 1.065 M
[HI] = 2(0.935 M) = 1.87 M
Double-checking:
\(K_c=\frac{[HI]^2}{[I_2][H_2]}=\frac{(1.87)^2}{(0.065)(1.065)}=50.5 \)

Simplifying Assumption (a.k.a. 'The Approximation')

In the previous examples, we solved for equilibrium concentrations with the quadratic formula. This method is always valid but time-consuming. Fortunately, a faster approach can often give similar answers. Our chemical intuition guides us to when an approximation might be effective. After solving with the shorter method, we can easily check our answer to ensure the concentrations we found are sufficiently accurate.

Consider the following reaction:

\[\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\;\; K_c = 7.6 \times 10^{-4} \nonumber\]

Since \(K_c\) is very small (≪1), this reaction favours the reactants (the denominator is much larger than the numerator in the \(K_c\) expression). This means that, at equilibrium, the mixture will mostly be \(\ce{N2O4}\) with only small amounts of \(\ce{NO2}\) present.

Now consider a specific example where the initial concentration of \(\ce{N2O4}\) is 0.15 M, no products are present at the start, and we must determine the concentrations of each species at equilibrium to two significant figures. The ICE table is:

	N2O4(g)  ⇌	2NO2 (g)
Initial	0.15 M	0 M
Change	- x	+2x
Equilibrium	0.15 M - x	2x

 

 

 

 

Here is where our chemical intuition helps. Because \(K_c\) for this reaction is very small, the equilibrium mixture will be mostly reactants. In addition, we are starting with only \(\ce{N2O4}\). In other words, the initial conditions in this problem are already close to the equilibrium conditions. Therefore, the change in [\(\ce{N2O4}\)] (the -x) will likely be small compared with the initial value (0.15 M). 

Why does this matter?  From the ICE table, the expression for the equilibrium concentration of \(\ce{N2O4}\) is \(0.15 – x\), which has two terms (a binomial).  If x << 0.15,  this approximation is likely to be valid:

(0.15−x) ≈ 0.15

At this point, are we certain this approximation will work? No. However, this assumption makes the calculation so much simpler that it’s time-efficient to try it and then check afterwards. 

The equilibrium expression is:

\[K_c=\frac{[\ce{NO2}]^2}{[\ce{N2O4}]}=\frac{(2x)^2}{(0.15-x)} =7.6 \times 10^{-4}\nonumber\]

With the approximation (0.15−x) ≈ 0.15, we can write:

\[K_c=\frac{(2x)^2}{(0.15)} =7.6 \times 10^{-4}\nonumber\]

This makes solving for x much simpler than using the quadratic equation:

\[\frac{4x^2}{(0.15)} =7.6 \times 10^{-4} \quad x=0.0053  \nonumber\]

Before moving on, check whether the assumption was reasonable. We assumed that x was 'small' relative to 0.15 M. Was it?

 \[\frac{(0.0053)}{0.15}\times 100\% = 3.6\% \nonumber\]

The change is fairly small, only 3.6% of the initial concentration. In general, whether this precision is 'sufficient' depends on the situation - perhaps good enough for mixing a quick solution, but not for high-accuracy analytical techniques.

In CHM 135, we use this rule for equilibrium approximations: If the neglected term is less than 5% of the other term in the binomial, the approximation is valid. Here, since  3.6% < 5%, using the simplifying assumption is justified.

To complete the problem, we'd use the value of x that we found to determine the concentrations at equilibrium:

[\ce{N2O4}\)]=0.15−(0.0053)=0.1447 M = 0.14 M (two significant figures)

[\(\ce{NO2}\)]=2x=2(0.0053)=0.011 M

In CHM 135, if an assumption is made in an equilibrium calculation, it must be explicitly stated and justified by checking against the initial quantity. The 5% rule provides a useful guideline for determining whether an approximation is reasonable. If the assumption is not valid, a more precise method (such as the quadratic formula) should be used.

Example \(\PageIndex{5}\): Calculation of Equilibrium Concentrations 

The decomposition of phosgene occurs via the following reaction:

\[\ce{COCl2(g) \rightleftharpoons CO(g) + Cl2(g)} \quad \quad K_p=3.4 \times 10^{-10} \text{ at 375 K} \nonumber \]

If a sealed vessel containing 3.5 atm of COCl₂ comes to equilibrium, what are the partial pressures of all of the species? 

Solution

	COCl2(g)  ⇌	CO (g)   +	Cl2(g)
Initial	3.5 atm	0 atm	0 atm
Change	- x	+x	+x
Equilibrium	3.5 atm - x	x	x

Substitute the equilibrium pressure terms into the K_p expression

\(K_p=\frac{P_{CO}P_{Cl_2}}{P_{COCl_2}}=\frac{(x)(x)}{3.5-x}=3.4 \times 10^{-10} \)

Since ${}_{}$$$ is very small, we can assume that the change in pressure of $COCl2$ is negligible compared to the initial pressure. Thus, we can approximate$3.5 - x \approx$, because $x$ is much smaller than the initial pressure.

\(\frac{x^2}{3.5}=3.4 \times 10^{-10} \\ x=3.4 \times 10^{-5} \text{atm} \)
Let's check the validity of our assumption:

\(\frac{3.4 \times 10^{-5}}{3.5}x100\%=0.001\%\),

which is much smaller than 5%, confirming the assumption is valid.

The equilibrium pressures are \(P_{COCl_2}=3.5 \text{ atm}, P_{CO}=P_{Cl_2}=3.4 \times 10^{-5} \text{atm}\).

Checking our calculation gives:
\(K_p=\frac{P_{CO}P_{Cl_2}}{P_{COCl_2}}=\frac{(3.4 \times 10^{-5})(3.4 \times 10^{-5})}{3.5}=3.3 \times 10^{-10} \).  This value is very close to the K_p value.

Exercise \(\PageIndex{5}\)

At 700 K, K_c = 4.25 x 10^-9 for the reaction:

\[\ce{2HBr(g) \rightleftharpoons H2(g) + Br2(g)} \nonumber \]

0.018 M of HBr gas is placed in a sealed flask.  What are the equilibrium concentrations of all of the species?

Answer

	2HBr(g) ⇌	H2(g)   +	Br2(g)
Initial	0.018 M	0 M	0 M
Change	- 2x	+x	+x
Equilibrium	0.018M- 2x	x	x

\(K_c=\frac{[H_2][Br_2]}{[HBr]^2}=\frac{(x)(x)}{(0.018-2x)^2}=4.25\times10^{-9}\)

Since K_c is very small, we can assume that the change in concentration of HBr is negligible compared to the initial concentration. Thus, we can approximate$3.5 - x \approx$, because $x$ is much smaller than the initial concentration:

\(\frac{(x)(x)}{(0.018)^2}=4.25\times 10^{-9} \\ x^2=1.38\times 10^{-12}\\x=1.2\times 10^{-6} M\)

Let's check the validity of our assumption, the percentage change in [HBr] is

\(\frac{1.2 \times 10^{-6}M}{0.018M} \times 100\%=0.0065\%\),

which is much smaller than 5%, confirming the assumption is valid.

The equilibrium concentrations are [HBr] = 0.018 M, [H₂] = [Br₂] = 1.2 x 10^-6 M.

Double-checking our calculation gives:

\(\frac{(1.2\times10^{-6})(1.2\times10^{-6})}{(0.018)^2}=4.4\times10^{-9} \) This value is very close to the K_c value.

Summary

In equilibrium problems, we use concentrations or pressures to calculate the equilibrium constant, ${}_{}$$$ or ${}_{}$$$, or use equilibrium constants to determine equilibrium concentrations or pressures. The changes in concentration (or pressure) as the system moves toward equilibrium are proportional to the stoichiometric coefficients in the balanced chemical equation. The sign of the change is positive when the concentration (or pressure) increases and negative when it decreases.

ICE (Initial, Change, Equilibrium) tables provide a structured way to organize information for equilibrium problems. The stoichiometric coefficients of the balanced equation determine the relative changes in the reactant and product concentrations.

To simplify calculations, when $K$ is very small, we can assume that the change in concentration or pressure is negligible compared to the initial value. This assumption is valid when the change is less than 5% of the initial concentration (or pressure). If the assumption cannot be made, a more detailed calculation often involving the quadratic equation may be necessary.