Equilibrium

Perfectly balanced. As all things should be.

“Equilibrium is a state, not of rest, but of balanced activities.”

– James Kendall, Smith’s College Chemistry, 1925 (Kendall 1925)

Key concepts:

• All closed systems result in equilibrium
• A force drives a system to equilibrium
• At equilibrium, the driving force is balanced by an opposing force and changes in the system are no longer visible

Basics

Balans

As we have seen in the previous chapter, the following reaction

\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\]

follows a two step mechanism, one of which is shown below.

\[\begin{align*} 2\mathrm{NO_2}(g) &\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{(NO_2)_2}(g) &&\mathrm{(fast)} \\[1.5ex] \mathrm{(NO_2)_2}(g) &\overset{k_2}\longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g) &&\mathrm{(slow)} \end{align*}\]

We focus our attention on the first step. Below is given this step with the products written in two different ways but are equivalent for the purpose of our discussion.

\[\begin{align*} 2\mathrm{NO_2}(g) &\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{(NO_2)_2}(g) \quad ~~\mathrm{or}\\[1.5ex] 2\mathrm{NO_2}(g) &\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{N_2O_4}(g) \end{align*}\]

Note that the NO₂ dimer can adopt different configurations and is sometimes written as N₂O₄. We will proceed with the latter for clarity/distinction from the reactants.

When the reaction initially begins (where [NO₂]₀ > 0 and [N₂O₄]₀ = 0) a force drives the reaction toward the right (towards product). There is initially no opposing force (i.e. any force to drive the reaction left towards reactants) as there is no N₂O₄ present to do so. However, upon the creation of any N₂O₄, the opposing force exists and begins to grow with increasing N₂O₄.

As the system evolves, we will be able to see a decrease in the amount of reactants and an increase in the amount of products. The changes in these concentrations become smaller and smaller until, at some point in the reaction, the forward and reverse forces will become equal. At this point, the changes in the amount of reactants and products equals zero. As stated by James Kendall, this is the point at which equilibrium is reached. The system is not at rest. The reaction continues to occur (reactants are produced/consumed and products are produced and consumed), but the visible changes in the amounts of reactants and products ceases to exist. They are balanced.

Activity: Find the nearest cinder block wall. Push as hard as you can against it. Continue to push against the wall and do not stop. You and the wall rapidly reach equilibrium. The force you applied is rapidly opposed in a very short time span and you reach equilibrium even though you continue to apply a force to the wall.

Equilibrium Constant

The equilibrium constant, K, denotes a ratio of products to reactants for a system that is at equilibrium and, for this class, will be treated as dimensionless. We construct our equilibrium constant for the reversible reaction

\[2\mathrm{NO_2}(g) \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{N_2O_4}(g)\]

from the ratio of the reverse and forward rate laws.

\[\begin{align*} \mathrm{rate}_{\mathrm{f}} &= k_1[\mathrm{NO_2}]^2 \\[1.5ex] \mathrm{rate}_{\mathrm{r}} &= k_{-1}[\mathrm{N_2O_4}] \end{align*}\]

At equilibrium, these two rates are equal.

\[\begin{align*} \mathrm{rate}_{\mathrm{f}} &= \mathrm{rate}_{\mathrm{r}} \quad \therefore \\[1.5ex] k_1[\mathrm{NO_2}]^2 &=k_{-1}[\mathrm{N_2O_4}] \end{align*}\]

Since the rates are equal, they cancel. We can rearrange the expression by placing the rate constants on one side.

\[\begin{align*} k_1[\mathrm{NO_2}]^2 &=k_{-1}[\mathrm{N_2O_4}] \\[1.5ex] \dfrac{k_1}{k_{-1}} &= \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}\\[1.5ex] K &= \dfrac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2} \end{align*}\]

The ratio of rate constants becomes a numerical value referred to as K, the equilibrium constant for the reaction. The final equation is called the equilibrium expression and, as seen here, is expressed as the ratio of products to reactants. Notice the exponent “2” in the denominator comes from the stoichiometric coefficient from the balanced chemical equation.

Equilibrium Expression

For a general reaction

\[aA + bB \rightleftharpoons cC + dD\]

the equilibrium expression is written in terms of molar concentrations as

\[K = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]

Note: Solids and pure liquids never appear in an equilibrium expression.

Gaseous Reactions

Equilibrium expressions can also be expressed as pressures of gases. For a general, gas-phase reaction

\[aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\]

the equilibrium expression can be written as

\[K_p = \dfrac{P_C^cP_D^d}{P_A^aP_B^b}\]

Pressure to concentration relation

K expressed as pressures can be related to K expressed as concentrations via

\[K_P = K (RT)^{\Delta n}\]

where R is the gas constant, T is temperature, and Δn is the change in moles of gas for a reaction given as

\[\Delta n = \sum(n_{\mathrm{gas~products}}) - (n_{\mathrm{gas~reactants}})\]

Derivation

For a reaction

\[aA(g) \rightleftharpoons bB(g)\]

the equilibrium expression is

\[K = \dfrac{[\mathrm{B}]^b}{[\mathrm{A}]^a} ~~~\mathrm{or}~~~ K_P = \dfrac{(P_\mathrm{B})^b}{(P_\mathrm{A})^a}\]

Assuming an ideal gas, we can express each gas (A and B) in terms of their pressures such that

\[\begin{align*} PV = nRT ~~\rightarrow~~ &P_\mathrm{A}V = n_\mathrm{A}RT ~~\rightarrow~~ P_\mathrm{A} = \left(\frac{n_\mathrm{A}}{V}\right)RT \\[1.5ex] &P_\mathrm{B}V = n_\mathrm{B}RT ~~\rightarrow~~ P_\mathrm{B} = \left(\frac{n_\mathrm{B}}{V}\right)RT \end{align*}\]

Note that moles over volume is equal to molarity as shown below.

\[\dfrac{n}{V} = M\] We can therefore rewrite the ideal gas expressions with our square-braket notation for molarity such as

\[\begin{align*} P_\mathrm{A} &= [\mathrm{A}]RT \\[1.5ex] P_\mathrm{B} &= [\mathrm{B}]RT \end{align*}\]

Now, substitute these terms into the K_P equation to give

\[K_P = \dfrac{(P_\mathrm{B})^b}{(P_\mathrm{A})^a} = \dfrac{([\mathrm{B}]RT)^b}{([\mathrm{A}]RT)^a}\] and recognize that

\[\dfrac{[\mathrm{B}]^b}{[\mathrm{A}]^a} = K\] Therefore

\[\begin{align*} K_P &= \dfrac{([\mathrm{B}]RT)^b}{([\mathrm{A}]RT)^a} \\[2ex] &= K \dfrac{(RT)^b}{(RT)^a} \\[2ex] &= K (RT)^{b-a} \\[2ex] K_P &= K (RT)^{\Delta n} \end{align*}\]

Reaction Quotient

The reaction quotient, Q, is equivalent to K in its construction (e.g. it is the ratio of products to reactants) and, for this class, will be treated as dimensionless.

\[Q = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\] However, it is the ratio of these particles when the reaction is not at equilibrium. Q is only equal to K when the reaction is at equilibrium. Q can be expressed with pressures just like K.

Visualizing Q and K

Imagine a “reaction line” ranging from reactants (left) to products (right). Anywhere between the limits of the line represents a mixture of reactant and products.

A reaction can “exist” anywhere along this line. It can adopt any ratio of products to reactants. Now, imagine that we have a reaction that is ready to occur (it has not started yet; we have only reactants and no products.

\[\mathrm{reactants} \longrightarrow \mathrm{products}\]

The reaction quotient, Q, would exist all the way to the left. Imagine Q as a movable slider such as one found on an audio mixing board, or a slider on your car’s climate control panel. It can move along the reaction line (left and right).

Now, this hypothetical reaction has an equilibrium constant, K, under some conditions, that is small (less than 1). We map out Q and K on our reaction line below.

Notice the location of Q (all the way to the left as we have no products yet, the reaction is yet to start). Notice the location of K. Since K < 1, the reaction is “reactant favored” meaning, at equilibrium, there are more reactants than products.

Now we begin the reaction. The reaction proceeds and it proceeds “to the right”. Q begins to slide to the right. It cannot proceed left because there is no product yet with which to turn into reactant. A driving force pushes the reaction right to form product. This force is opposed upon the creation of product. The opposing force balances the driving force when equilibrium is reached (i.e. when Q reaches K).

When Q reaches K, the reaction does not stop. It continues in both directions making reactants and products. However, the change in the amount of reactants and products are equal.

Next, notice how far Q had to move to reach equilibrium, K. It only moved a small amount relative to going all the way to products. I refer to this distance as x and x ends up being “small” since Q started near K. This distance represents the amount of reactants that were converted to products.

Let’s try this again. Consider another hypothetical reaction where K is large. We start the reaction with all reactants. Note the positions of Q and K. Here, K is large (> 1) and the reaction is product favored (i.e. at equilibrium, there are more products than reactants).

The reaction begins and the reaction proceeds to the right (Q moves right). The reaction progresses until Q equals K.

We see that Q traveled a further distance (more reactants were converted to products before equilibrium was reached). I refer to this as x being “large” since Q started far away from K.

Starting with Product

Let us visualize this hypothetical reaction, with K < 1, once again but this time we start with all product. This changes the position of Q. Now it is located all the way to the right. K is less than 1 so it sits closer to reactants.

The reaction begins and proceeds to the left. Q moves left until it reaches equilibrium, K.

Notice the magnitude of x now; it is large since Q started far away from equilibrium.

If K were large (> 1) for this reaction, we get the following starting positions

and the following result to reach equilibrium.

In this case, x is small since Q started close to equilibrium.

Direction of Reaction

We can determine which way a reaction will proceed (toward equilibrium) by identifying where we begin the reaction and where K is.

1. If starting with all reactants, the reaction proceeds right.

2. If starting with all products, the reaction proceeds left.

3. If starting with a mixture of reactants and products

   1. If Q < K, the reaction proceeds right
   2. If Q > K, the reaction proceeds left

Summary:

1. Visualize K and where it would exist on the reaction line
2. Visualize where the reaction begins (e.g. where Q is)
3. Determine which direction the reaction will proceed (left or right) by comparing Q to K
4. Visualize how far Q has to move to reach K (e.g. how big x is)

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.

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

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.

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.

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

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!

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

Solutions are found here. Don’t peek before you try it!

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.

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

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!

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

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.

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

Le Chatelier’s Principle

Begin with a reaction at equilibrium. Now do something to it.

• Add/remove some reactant
• Add/remove some product
• Change the temperature
• Change the volume/pressure (for gaseous reactions)

It is no longer at equilibrium. What happens? Which way will the reaction proceed?

This principle is named after French chemist Henry Louis Le Chatelier.

Le Chatelier and Ammonia Synthesis

Ammonia is a highly produced inorganic chemical where 80% of its total production is used in agricultural fertilizers. Le Chatelier was on the cusp of realizing a process for the synthesis of ammonia but was beaten to the punch by Fritz Haber and Carl Bosch, two German scientists, in the early 20th century. The Haber-Bosch process (or Haber process) is currently used today to produce ammonia on a wide scale and 176 million tons of it is produced every year. It is one of the most important chemical processes in modern times but is extremely energy intensive requiring pressures of 150-300 bar and 350-500 °C.

\[3\mathrm{H_2} + \mathrm{N_2} \longrightarrow 2\mathrm{NH_3}\]

The production of ammonia and its use in fertilizers allowed for a tremendous growth in agricultural output and gave rise to an explosion in the world’s population.

Figure 1: The world’s population from 1900-2015. Source

Le Chatelier lamented not having fully realized the synthesis of ammonia. Published correspondence to the editor of the Journal of Chemical Education demonstrates this (Silverman 1938).

---

LE CHATELIER AND THE SYNTHESIS OF AMMONIA

To the Editor

Dear Sir:

In the excellent article by Professor Silverman on Henry Le Chatelier in the December issue of this journal, there is one statement to which I must take exception. On page 556 we read, “It is not strange that he [Le Chatelier] should have accomplished the synthesis of ammonia from the elements in 1901, anticipating Fritz Haber, who is usually the only one mentioned in connection with the process.” Now Le Chatelier himself in his last book “De la Methode dans les Sciences Experimentales,” published in 1936, devotes three pages (pp. 73-6) to this synthesis in which he says that he tried to accomplish the direct union of hydrogen and nitrogen under a pressure of 200 atm. at a temperature of 600 ° in the presence of metallic iron. A terrific explosion occurred which nearly killed an assistant. Some time later Le Chatelier found that the explosion was due to the presence of air in the apparatus used. And thus it was left for Haber to succeed where a number of noted French chemists, including Thdnard, Sainte Claire Deville and even Berthelot had failed. At the end of his career Le Chatelier, with a disarming frankness, tells us, “I let the discovery of the ammonia synthesis slip through my hands. It was the greatest blunder of my scientific career. I should have realized this synthesis five years before Haber…..”

In calling attention to this statement I may say that those of your readers who are familiar with French will find a great many interesting observations and reflections in this scientific testament of a great Frenchman.

– H. S. van Klooster

Changing Concentrations

Consider the following reaction at equilibrium

\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]

and its equilibrium expression

\[K = \dfrac{[\mathrm{ZnCl_2}][\mathrm{H_2}]}{[\mathrm{HCl}]^2}\]

Let us add, for example, some HCl to the mixture. The reaction is no longer at equilibrium, however, the reaction will proceed to once again reach equilibrium. Which way will the reaction go?

We must consider the reaction quotient, Q.

\[Q = \dfrac{[\mathrm{ZnCl_2}][\mathrm{H_2}]}{[\mathrm{HCl}]^2}\]

Upon the addition of HCl, the denominator in the expression increases, therefore, Q < K. The reaction will proceed right (consume reactants and produce products) to reach equilibrium.

Let us now add some product to this reaction that is at equilibrium. Perhaps we add a bit of ZnCl₂. We notice that the numerator in the reaction quotient is now larger and Q > K. The reaction will therefore proceed left (consume products and produce reactants) to reach equilibrium once again.

Practice

Predict the direction of the given reaction (currently at equilibrium) for each of the following scenarios:

\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]

1. Add some HCl
2. Add some Zn
3. Remove some HCl
4. Remove some Zn
5. Add some ZnCl₂
6. Add some H₂
7. Remove some ZnCl₂
8. Remove some H₂

Answer

\[\mathrm{Zn}(s) +2\mathrm{HCl}(aq) \rightleftharpoons \mathrm{ZnCl_2}(aq) + \mathrm{H_2}(aq)\]

1. Add some HCl - right
2. Add some Zn - no shift
3. Remove some HCl - left
4. Remove some Zn - no shift
5. Add some ZnCl₂ - left
6. Add some H₂ - left
7. Remove some ZnCl₂ - right
8. Remove some H₂ - right

Changing Temperature

Exothermic Reaction

Consider the following reaction at equilibrium.

\[\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) \qquad \Delta H = -98.9~\mathrm{kJ~mol^{-1}}\]

The reaction is exothermic (ΔH < 0) meaning heat is generated (i.e. is a product of the reaction). If we treat heat (Δ) as a product of reaction, the reaction could be written as

\[\mathrm{SO_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \rightleftharpoons \mathrm{SO_3}(g) + \Delta\] Now we apply the same principles as we did with changing concentrations!

If the temperature of the reaction mixture was raised (i.e. heat is added), the reaction will shift left to consume the heat and reach equilibrium.

If the temperature of the reaction mixture was lowered (i.e. heat is removed), the reaction will shift right to produce heat and reach equilibrium.

Endothermic Reaction

Consider the following endothermic reaction at equilibrium.

\[\mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g) \qquad \Delta H = 248.1~\mathrm{kJ~mol^{-1}}\]

We can write the reaction as follows (indicating heat, Δ, as a reactant)

\[\Delta + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{H_2}(g) + \dfrac{1}{2}\mathrm{O_2}(g)\]

If the temperature was raised, the reaction will proceed to the right. If the temperature was lowered, the reaction will proceed to the left.

Changing Volume/Pressure

Recall from the ideal gas law that pressure and volume are inversely proportional.

Consider the following gaseous reaction at equilibrium

\[4\mathrm{NH}_{3}(g) + 7\mathrm{O}_{2}(g) \rightleftharpoons 4\mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\]

If the pressure of the reaction mixture was increased (perhaps by reducing the volume of the vessel that the gas was contained in), which way will the reaction proceed to reach equilibrium?

Ask yourself, what exhibits more pressure in a unit volume, a gas containing 11 moles of particles or a gas containing 10 moles of particles? Clearly it is the gas containing the larger number of particles!

Applying that concept here leads us to conclude that the reaction will proceed right since there are only 10 moles of gaseous products for the reaction vs. the 11 moles of gaseous reactants!

Decreasing the pressure (increasing the volume) of the reaction will cause the reaction to shift to the left!

Transforming K

This section is a bit of a mathematical exercise. We can change a reversible chemical reaction and rationalize the change in the equilibrium constant.

Reversing the reaction

For example, the reaction

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

has the following equilibrium expression

\[K = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = 4.46\times 10^{-3}\] indicating that the reaction is reactant, N₂O₄, is favored. If we reverse the reaction such that

\[2\mathrm{NO}_2 (g) \rightleftharpoons \mathrm{N_2O_4}(g) \] the equilibrium expression is

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

and the resulting equilibrium constant is

\[K' = \dfrac{1}{K} = \dfrac{1}{4.46\times 10^{-3}} = 224.22\]

and the product, N₂O₄, is favored.

Multiplying the reaction

Take the following reaction

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

and double everything (multiply by two) to give

\[2\mathrm{N_2O_4}(g) \rightleftharpoons 4\mathrm{NO}_2 (g)\]

Let us compare the original equilibrium expression with the new one.

\[\begin{align*} K &= \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \\[1.5ex] K' &= \dfrac{[\mathrm{NO_2}]^4}{[\mathrm{N_2O_4}]^2} \end{align*}\]

By multiplying every species by 2, the exponents in the equilibrium expression were multiplied by two. The new equilibrium constant is then

\[K' = K^2 = (4.46\times 10^{-3})^2 = 1.99\times 10^{-5}\]

We generalize this by stating that multiplying a reaction by n gives rise to a new equilibrium constant such that

\[K' = K^n\]

Adding reactions together

For a multi-step reaction

\[\begin{align*} \mathrm{A} &\rightleftharpoons \mathrm{2B} &&\qquad K_1 = \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]} \\[1.5ex] \mathrm{2B} &\rightleftharpoons \mathrm{3C} &&\qquad K_2 = \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2} \end{align*}\]

when added together gives

\[\mathrm{A} \rightleftharpoons \mathrm{3C}\]

giving a new equilibrium constant such that

\[\begin{align*} K' &= K_1 \times K_2 \\[2ex] &= \dfrac{[\mathrm{B}]^2}{[\mathrm{A}]} \times \dfrac{[\mathrm{C}]^3}{[\mathrm{B}]^2} \\[1ex] &= \dfrac{[\mathrm{C}]^3}{[\mathrm{A}]} \end{align*}\]

Therefore, multiply the equilibrium constants of both reactions together to give the final equilibrium constant.

Gibbs Energy

Gibbs free energy (or Gibbs energy; G) is a thermodynamic potential that is minimized when a chemical system reaches chemical equilibrium. The Gibbs free energy change for a reaction, ΔG, is used to define if a reaction or process is spontaneous or non-spontaneous.

Sign	Spontaneity	Favor
ΔG < 0	spontaneous	product favored
ΔG > 0	non-spontaneous	reactant favored
ΔG = 0	at equilibrium	none

The Gibbs free energy equation determines the Gibbs energy for a reaction by taking into account the enthalpy, ΔH, and entropy, ΔS, of reaction.

\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]

Note: The “degree” symbols indicate that the thermodynamic values were obtained for substances in their standard state (T = 25 °C, P = 1 bar, c = 1 M). Appendix G in Chemistry 2e reports these values for many substances. Also note that T in the Gibbs free energy equation can be any temperature, not simply 298.15 K.

Let us look up the thermodynamic values for the following reaction.

\[\mathrm{2NO_2}(g) \rightleftharpoons 2\mathrm{NO}(g) + \mathrm{O_2}(g)\]

All values in the table below are given in kJ mol^–1.

Species	ΔHf°	S°
NO2(g)	33.2	0.2401
NO(g)	90.25	0.2108
O2	0	0.2052

We can determine the enthalpy of reaction by

\[\Delta H_{\mathrm{rxn}} = \Sigma \left (\Delta H_{\mathrm{products}} \right ) - \Sigma \left (\Delta H_{\mathrm{reactants}} \right )\] Therefore,

\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \left [ (2 \times 90.25) + (0) \right ] - \left [ (2\times 33.2) \right ] \\[1.5ex] &= 114.1~\mathrm{kJ~mol^{-1}} \end{align*}\]

This reaction is endothermic! We repeat this process for the entropy of reaction.

\[\Delta S_{\mathrm{rxn}} = \Sigma \left (\Delta S_{\mathrm{products}} \right ) - \Sigma \left (\Delta S_{\mathrm{reactants}} \right )\] to give

\[\begin{align*} \Delta S_{\mathrm{rxn}} &= \left [ (2 \times 0.2108) + (0.2052) \right ] - \left [ (2\times 0.2401) \right ] \\[1.5ex] &= 0.1466~\mathrm{kJ~mol^{-1}} \end{align*}\]

The reaction experiences an increase in entropy. This is expected becuase we go from 2 moles of gas to 3 moles of gas!

Now we determine the Gibbs free energy for the reaction using the Gibbs free energy equation.

\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ} \\ &= 114.1~\mathrm{kJ~mol^{-1}} - T(0.1466~\mathrm{kJ~mol^{-1}}) \end{align*}\]

I tabulate the Gibbs free energy (in kJ mol^–1 for this reaction at three different temperatures, T, in the table below.

T (°C)	T (K)	ΔG°
-200	73.15	103.38
25	298.15	70.39
600	873.15	-13.9

We see that at low temperatures, the Gibbs free energy is negative. The reaction is non-spontaneous at low temperatures. As we increase the temperature, the reaction eventually becomes spontaneous (ΔG < 0).

According to Le Chatelier’s principle, increasing the heat for an endothermic reaction should drive the equilibrium to the right. We can determine the equilibrium constant for this reaction at the three temperatures of interest using

\[\Delta G^{\circ} = -RT\ln K\]

where R is the gas constant, T is the temperature, and K is the equilibrium constant. Rearranging for K gives

\[K = e^{\frac{-\Delta G^{\circ}}{RT}}\]

I’ve added the equilibrium constants in our table below.

T (°C)	T (K)	ΔG°	K
-200	73.15	103.38	1.51e-74
25	298.15	70.39	4.65e-13
600	873.15	-13.9	6.79

We see for this endothermic reaction that the equilibrium constant gets larger (favors product) as the temperature goes up! This fits with Le Chatelier’s principle! Also, note how ΔG is decreasing with increasing temperature for this reaction. At high temperatures, this reaction is spontaneous and products is favored. We note the correlation

\[\downarrow \Delta G \propto K \uparrow\]

How fast does the reaction go at these temperatures? According to Collision theory, as temperature increases, the rate should increase. Let us revisit the Arrhenius equation

\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\]

For this reaction, A = 2×10⁹ and E_a = 112.55 kJ mol^–1 (from (D. L. Baulch 1973)). I update our table to include the rate constant, k in M^–1 s^–1, at our three chosen temperatures.

T (°C)	T (K)	ΔG°	K	k
-200	73.15	103.38	1.51e-74	8.49e-72
25	298.15	70.39	4.65e-13	3.82e-11
600	873.15	-13.9	6.79	369.6

We see the rate constant is very small at low temperatures and therefore, the reaction is very slow. At elevated temperatures, the rate constant is much larger and the reaction is much faster!

Let us now consider the following reaction

\[\mathrm{N_2O}(g) + \mathrm{O}(g) \rightleftharpoons \mathrm{N_2}(g) + \mathrm{O_2}(g)\] This reaction is exothermic and increasing the temperature should shift the reaction to the left. The thermodynamic values are presented below.

Species	ΔHf°	S°
N2O(g)	81.6	0.22
O	249.17	0.1611
N2	0	0.1916
O2	0	0.2052

The enthalpy of reaction is

\[\Delta H_{\mathrm{rxn}} = -330.77~\mathrm{kJ~mol^{-1}}\]

and the entropy change is

\[\Delta S_{\mathrm{rxn}} = 0.3379~\mathrm{kJ~K^{-1}}\] We see that this reaction experiences an increase in entropy. I compute our numbers again (ΔG, K, and k) and present them below for our three chosen temperatures plus an additional temperature (1000 °C).

T (°C)	T (K)	ΔG°	K	k
-200	73.15	-355.49	7.15E+253	2.82E-73
25	298.15	-431.51	4.00E+75	3.17E-10
600	873.15	-625.81	2.75E+37	1.00E+04
1500	1773.15	-929.92	1.79E+25	3.60E+07

For this exothermic reaction, the rate increases with increasing temperature. The equilibrium is driven to the left towards reactants at higher temperature (K increases) and the reaction becomes more spontaneous (ΔG decreases).

Figure 2: Temperature drives the reaction left or right. Credit: Ohio State University