4.5 Solving Equilibrium Problems

Find x


An “equilibrium problem” in general chemistry typically involves one to solve for the equilibrium concentrations of the reaction given some starting information. This section introduces a variety of equilibrium type problems, each one illustrating the method in solving for the equilibrium concentrations.

4.5.1 ICE Table

One tool we use to solve equilibrium problems is the ICE table. ICE is an acronym that stands for Initial pressure/concentration, Change in pressure/concentration, and Equilibrium pressure/concentration. The table is then populated with values that have units of either Molarity or pressures. The purpose of the ICE table is to simply keep our data organized. That’s it. By adhering to the table format, we can conveniently keep track of what we are doing. It is a very useful tool! Here’s how it works.

Take the following reversible reaction.

\[\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO}_2 (g)\]

We would write our ICE table as follows

  N2O4 ⇌ 2NO2

I


C


E


4.5.2 Example 1

The following reaction begins with [N2O4] = 1.00 M.

\[\mathrm{N_2O_4}(g)\rightleftharpoons 2\mathrm{NO}_2 (g) \qquad K = 4.46\times 10^{-3}\] Determine the equilibrium concentrations of the reaction.


Write out ICE Table

Here I populate the initial concentration row using the information given to me by the problem.

  N2O4 ⇌ 2NO2

I

1

0

C


E



Determine the direction of reaction

Since the reaction is beginning with all reactant, the reaction must move right. The change in concentration of the reactants is therefore going to be a negative quantity (e.g. –x) because reactants will be consumed. We don’t know the magnitude of x but we know it will be negative. The “1” in front of the x is due to the stoichiometric coefficient from the balanced chemical equation.

The change in concentration of the products must therefore be +2x. Since the reaction will proceed right, products will be produced and its change in concentration is positive.

Stoichiometrically, for every 1 mole of reactant that is consumed, 2 moles of product is produced. Our ICE table now looks like the following

  N2O4 ⇌ 2NO2

I

1

0

C

-x

+2x

E



Fill in the Equilibrium Row

The equilibrium row of the table is simply the first two rows (the I and C rows) added together. What this tells us is this. The equilibrium concentration of NO2 is the initial concentration minus how far the reaction proceeds to reach equilibrium (–x). Since the reaction is moving right, the final concentration of N2O4 must be less than the initial concentration.

On the other hand, initially there is zero NO2 product. The reaction is going to proceed right and product will be produced. Therefore, the initial concentration of the product is 0. However, its concentration will be the magnitude of 2x at equilibrium.

  N2O4 ⇌ 2NO2

I

1

0

C

-x

+2x

E

1.000-x

2x


Write out the Equilibrium Expression

Next, write out the equilibrium expression.

\[\dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = K\]


Substitute in Equilibrium Concentration Terms

Substitute in the equilibrium concentrations from the table as well as the equilibrium constant into the equilibrium expression.

\[\dfrac{(2x)^2}{(1.00 - x)} = 4.46\times 10^{-3}\] The next step is to solve for x. There can be multiple ways to do this, some of which are more straightforward than others. I will introduce these methods below.


4.5.2.1 Quadratic Method

Sometimes we can solve for x by using the quadratic formula if our equilibrium expression gives us a quadratic equation. A quadratic equation takes on the form

\[\color{red}{a}x^2 + \color{blue}{b}x + \color{green}{c} = 0\] Notice how there is an x2 and x term in the same equation. Here, x cannot be isolated and solved for. We must use the quadratic equation (below) to solve for x.

\[x = \dfrac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2-4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}\]

Let us rearrange our equilibrium expression into the form of a quadratic.

\[\begin{align*} \dfrac{(2x)^2}{(1.000 - x)} &= 4.46\times 10^{-3} \\[1.5ex] 4x^2 &= 4.46\times 10^{-3}(1.000 -x) \\[1ex] &= 4.46\times 10^{-3} - 4.46\times 10^{-3}x \\[1ex] \color{red}{4}x^2 &+ \color{blue}{4.46\times 10^{-3}}x \color{green}{-4.46\times 10^{-3}} = 0 \end{align*}\]

Next, solve the equation using the quadratic equation.

\[\begin{align*} x &= \dfrac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2-4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}\\[2ex] &= \dfrac{-\color{blue}{4.46\times 10^{-3}} \pm\sqrt{(\color{blue}{4.46\times 10^{-3}})^2 - 4(\color{red}{4})(\color{green}{-4.46\times 10^{-3}})}}{2(\color{red}{4})}\\[2ex] &\approx 0.03284~~\mathrm{or}~~-0.03395 \end{align*}\]

Here, we see two solutions (two roots) from the ± operation. We will always disregard the negative root. Therefore,

\[x = 0.03284\]

Determine equilibrium concentrations

Now that we know what x is, plug it back into the equilibrium row of the ICE table.

\[\begin{align*} [\mathrm{N_2O_4}] &= 1.000 - x \\[1.5ex] &= 1.000- 0.03284 \\[1.5ex] &= 0.9672~M \\[1.5ex] [\mathrm{NO_2}] &= 2x \\[1.5ex] &= 2(0.03284~M) \\[1.5ex] &= 0.06568 \end{align*}\]


4.5.2.2 Small x Approximation Method

Another method we can use to solve for x is the small x approximation. Remember our “reaction line” exercises presented earlier? Our reaction is starting near equilibrium (Q is located at all reactants and K is small therefore Q is starting near K). The magnitude of x, in this case, is small. It might be useful to invoke the small x approximation because of this.

For this method, we revisit our equilibrium expression.

\[\dfrac{(2x)^2}{(1.00 \color{red}{- x})} = 4.46\times 10^{-3}\]

Notice the x in the denominator of the fraction. If x were a very small number, the denominator will essentially not change (it will essentially be equal to 0.100 if x is indeed “small enough”). Therefore, we can throw out the “–x term in the denominator to give

\[\dfrac{(2x)^2}{(1.00)} = 4.46\times 10^{-3}\] This allows us to directly solve for x as we will not have a quadratic equation to solve for.

\[\begin{align*} \dfrac{(2x)^2}{(1.000)} &= 4.46\times 10^{-3} \\[1.5ex] (2x)^2 &= 4.46\times 10^{-3} \\[1.5ex] 2x &= \sqrt{4.46\times 10^{-3}} \\[1.5ex] x &= \dfrac{\sqrt{4.46\times 10^{-3}}}{2}\\[1.5ex] &= 0.0334 \end{align*}\]


Test x!

We assumed that x was small but was it? We must test x to ensure that the small x approximation was a good approximation.

If x is within 5% of the number it was being subtracted from (or added to), the small x approximation is good enough.

Note: The “5%” threshold is somewhat arbitrary. However, I use the “5%” rule for the purpose of this class (as many other classes/textbooks do). However, in practice, this threshold may not be small enough!

To test x, take the value of x and divide by the number it was being subtracted from (or added to). Multiply by 100% and see what the result is!

\[\begin{align*} \dfrac{0.0334}{1.000} \times 100\% = 3.34\% \end{align*}\]

Since x was smaller than 5% of “1.000”, we move accept that the approximation was reasonable and use this value for x to determine the equilibrium concentrations.

Note: If x was 5% or larger, we would have to STOP, backtrack, and solve the equilibrium expression using the quadratic method.


Determine equilibrium concentrations

\[\begin{align*} [\mathrm{N_2O_4}] &= 1.000~M - x \\[1.5ex] &= 1.000~M - 0.0334~M \\[1.5ex] &= 0.9666~M \\[1.5ex] [\mathrm{NO_2}] &= 2x \\[1.5ex] &= 2(0.0334~M) \\[1.5ex] &= 0.0668~M \end{align*}\]

Compare the equilibrium concentrations from the small x approximation to those determined with the quadratic. They are very similar!


4.5.2.3 Practice

Try to solve the equilibrium concentrations for the same reaction with different starting amounts!

Experiment [N2O4]i [NO2]i [N2O4]eq [NO2]eq K

1

0.67

0

4.65 × 10–3

2

0.446

0.05

4.66 × 10–3

3

0.5

0.03

4.60 × 10–3

4

0.6

0.04

4.60 × 10–3

5

0

0.2

4.63 × 10–3
Solution


4.5.3 Example 2

Sometimes you do not end up with a quadratic and/or have to use the small x approximation!

Take this problem for example. For the following reaction

\[\begin{align*} \mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons \mathrm{2HI}(g) \qquad K = 54.3 \end{align*}\]

the initial concentrations are

\[\begin{align*} [\mathrm{H_2}]_{\mathrm{i}} = 0.500~M \quad [\mathrm{I_2}]_{\mathrm{i}} = 0.500~M \end{align*}\]

Solve for the equilibrium concentrations.


Fill out the ICE table.

  H2 + I2 ⇌ 2HI

I

0.5

0.5

0

C

-x

-x

+2x

E

0.50-x

0.50-x

2x

Write out the equilibrium expression and solve for x.

\[\begin{align*} \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} &= K \\[1.5ex] \dfrac{(2x)^2}{(0.500-x)(0.500-x)} &= 54.3 \\[1.5ex] \dfrac{(2x)^2}{(0.500-x)^2} &= 54.3 \\[1.5ex] \sqrt{\dfrac{(2x)^2}{(0.500-x)^2}} &= \sqrt{54.3} \\[1.5ex] \dfrac{2x}{0.500-x} &= 7.369 \\[1.5ex] 2x &= 7.369 (0.500 - x) \\ 2x &= 3.6845 - 7.369x \\ 9.369x &= 3.6845 \\ x &= 0.3933 \end{align*}\]

Plug in x into the equilibrium terms and solve for the equilibrium expressions.

\[\begin{align*} [\mathrm{H_2}] &= 0.500 - x = 0.500 - 0.3933 = 0.1067~M \\ [\mathrm{I_2}] &= 0.500 - x = 0.500 - 0.3933 = 0.1067~M \\ [\mathrm{HI}] &= 2x = 2(0.3933) = 0.7866~M \end{align*}\]


4.5.4 Example 3

We can solve equilibrium problems by starting with some equilibrium concentrations instead of only initial concentrations!

For the following reaction

\[\begin{align*} \mathrm{N_2}(g) + \mathrm{O_2}(g) \rightleftharpoons \mathrm{2NO}(g) \qquad K = 4.1\times 10^{-4} \end{align*}\]

determine the equilibrium concentration of NO given

\[\begin{align*} [\mathrm{N_2}]_{\mathrm{eq}} &= 0.05~M \\[1.5ex] [\mathrm{O_2}]_{\mathrm{eq}} &= 0.002~M \end{align*}\]


Fill out an ICE table.

  N2 + O2 ⇌ 2NO

I




C




E

0.05

0.002

?

We have two equilibrium concentrations to fill in. We don’t know any of the other information.

Write out the equilibrium expression and solve for the NO concentration!

\[\begin{align*} \dfrac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} &= K \\[1.5ex] \dfrac{[\mathrm{NO}]^2}{(0.05)(0.002)} &= 4.1\times 10^{-4} \\[1.5ex] [\mathrm{NO}]^2 &= 4.1\times 10^{-4} \times (0.05)(0.002) \\[1.5ex] \sqrt{[\mathrm{NO}]^2} &= \sqrt{4.1\times 10^{-4} \times (0.05)(0.002)} \\[1.5ex] [\mathrm{NO}] &= 2.02\times 10^{-4}~M \end{align*}\]


4.5.5 Example 4

Sometimes we can start an equilibrium problem with both initial and equilibrium concentrations and solve for the equilibrium constant!

For the reaction

\[\begin{align*} \mathrm{I_2}(aq) + \mathrm{I^-}(aq) \rightleftharpoons \mathrm{I_3^-}(aq) \end{align*}\]

determine the equilibrium constant, K, given

\[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} &= 0.1~M \\ [\mathrm{I^-}]_{\mathrm{i}} &= 0.08~M \\ [\mathrm{I_2}]_{\mathrm{eq}} &= 0.05~M \end{align*}\]


Fill out the ICE table.

  I2 + I– ⇌ I3–

I

0.1

0.08

0

C

-x

-x

+x

E

0.05

0.080-x

x


Recognize that we have an initial concentration and an equilibrium concentration for I2. We can directly solve for x.

\[\begin{align*} [\mathrm{I_2}]_{\mathrm{i}} - x &= [\mathrm{I_2}]_{\mathrm{eq}} \\[1.5ex] x &= [\mathrm{I_2}]_{\mathrm{i}} - [\mathrm{I_2}]_{\mathrm{eq}} \\ &= 0.1~M - 0.05~M \\ &= 0.05~M \end{align*}\]

Write out the equilibrium expression, substitute in the equilibrium terms, and then plug in x. Solve for K.

\[\begin{align*} K &= \dfrac{[\mathrm{I_3^-}]}{[\mathrm{I_2}][\mathrm{I^-}]} \\[1.5ex] &= \dfrac{x}{(0.05)(0.08-x)} \\[1.5ex] &= \dfrac{0.05}{(0.05)(0.08-0.05)} \\[1.5ex] &= 33.33 \end{align*}\]




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