Calculating Equilibrium Constant from Pressure — Examples

Have you ever thought about how much produce is at the grocery store? We have piles of apples, bananas, lettuce, and carrots. It's crazy! But, over the past few decades, agricultural land has diminished. How do we produce mountains of food with less land? Fertilizer!

Today, we'll study gas-phase equilibrium reactions. Specifically, in the example problems, we'll investigate the Haber process. It's the most famous gas-phase reaction! The Haber process generates about two-thirds of the world's ammonia. Its the essential component in fertilizers. Roughly half the world's food supply is grown using fertilizer from the Haber process. Without the Haber process, our mountains of produce would be handfuls of produce.

Previously, we learned about calculating the equilibrium constant using concentrations. We can apply the same procedure to gases! But, we won't use molar concentration. We'll use gas pressure!

A Quick Refresher on Partial Pressures

First, let's remind ourselves about partial pressures. Start by imagining a mixture of gases. The gas mixture has some total pressure. But, we can also consider each component's pressure. But remember, in chemistry, we're unbelievably creative! So, we call each component's pressure its partial pressure. Do you remember Dalton's Law? We used it for partial pressure calculations. It says that the sum of the partial pressure equals the total. It has two forms.

• $P_{total}=\Sigma P_{1}+P_{2}+P_{3}+...$

• $P_{total}=\Sigma X_{1}P_{total}+X_{2}P_{total}+X_{3}P_{total}+...$

The first form of the law is less helpful for equilibrium. In equilibrium problems, we don't need the total pressure. If we know the partial pressures, we can calculate the equilibrium constant. So, the second form of the law is more important. If you have the mole fractions, you can calculate the partial pressures.

Calculating $K_{p}$

When all our reactants and products are gases, we write a new form of the equilibrium constant. It's for partial pressure and denoted with a p-subscript. Instead of concentrations, we use partial pressures.

Let's say we have a generic chemical reaction, $aA+bB\rightarrow cC+dD$. How would we write $K_{p}$? It's similar to our normal equilibrium constant expression. But, instead of concentration, we use partial pressure.

$K_{p}=\frac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}}$

Example Problems

Let's look at the Haber process. Its balanced equation is $N_{2(g)}+3H_{2(g)}\rightleftharpoons 2NH_{3(g)}$.

Problem 1: Let's start with an easy problem. At equilibrium, the partial pressures (in atmospheres) of nitrogen, hydrogen, and ammonia gas are 0.50, 0.30, and 1.40. Calculate the equilibrium constant's value and interpret it.

• Our first step is to write the equilibrium constant expression.

• $K_{p}=\frac{products}{reactants}=\frac{P_{NH_{3}}^{2}}{P_{N_{2}}^{1}\times P_{H_{2}}^{3}}$

• Now we can plug in the partial pressures and solve.

• $K_{p}=\frac{1.40^{2}}{0.50^{1}\times 0.30^{3}}=\frac{1.96}{0.0135}=145.18$

• We can interpret the magnitude of the equilibrium constant. $K_{p}>>1$ so the reaction favors the products. Industrially, we use iron metal as a catalyst to shift the Haber process towards the products.

Problem 2: Let's try a slightly more challenging problem. Let's say that we started with only ammonia gas. Its partial pressure was 0.76 atm. The equilibrium concentration of nitrogen gas is 0.22 atm. Find the other two equilibrium pressures.

• Strategy: For this problem, we'll need an ICE table. We start with only the product. So, we know that the reaction will redirect toward the reactants. In our ICE table, ammonia will lose pressure. Nitrogen and hydrogen will form. So they'll experience an increase in pressure. But, remember we have to use the mole ratios.

• $N_{2(g)}+3H_{2(g)}\rightleftharpoons 2NH_{3(g)}$

$$\begin{array}{|r|c|c|c|} \hline \text{}&N_{2}&H_{2}&NH_{3}\\ \hline \text{Initial:}&0&0&0.76\\ \hline \text{Change:}&+x&+3x&-2x\\ \hline \text{Equilibrium:}&0+x=0.22&0+3x&0.76-2x\\ \hline \end{array}$$

• $P_{N_{2}}=x=0.22atm$
• $P_{H_{2}}=3x=3\times 0.22 = .66atm$
• $P_{NH_{3}}=0.76-2x=0.76-0.44=0.32atm$

Problem 3: Let's do one final problem. This one will be the most challenging. This time we'll start with nitrogen and hydrogen. Their initial partial pressures are 0.22 and .15 atm. The $K_{p}$ at equilibrium is $2.3\times10^{-9}$. What are the partial pressures of each gas?

• Strategy: Like the last problem, we'll need to use an ICE table. But, you can see the question might get tricky. In our equilibrium constant expression $H_{2}$ is raised to the third power. Solving cubic functions is a massive pain. So, we'll need to use the approximation method. The equilibrium constant is minuscule. So, the reaction favors the reactants. The partial pressure of the reactants is not going to change much. So, we can approximate that its final and initial concentrations are equal.

$$\begin{array}{|r|c|c|c|} \hline \text{}&N_{2}&H_{2}&NH_{3}\\ \hline \text{Initial:}&0.22&0.15&0.0\\ \hline \text{Change:}&\cancel{-x}&\cancel{-3x}&+2x\\ \hline \text{Equilibrium:}&0.22&0.15&+2x\\ \hline \end{array}$$

• $K_{p}=\frac{products}{reactants}=\frac{P_{NH_{3}}^{2}}{ P_{N_{2}}^{1} \times P_{H_{2}}^{3}} =\frac{2x^{2}}{0.22\times0.15^{3}}=2.3\times10^{-9}$

• $4x^{2}=2.3\times10^{-9} \times 0.22 \times 0.15^{3} \rightarrow x=6.5\times10^{-7}atm$.

• We can see that $x << P_{H_{2}\,initial}$ and $x << P_{N_{2}\,initial}$. So, our approximation is valid.

A Bit of History of Fritz Haber

Since we looked at the Haber process, I wanted to add some history about Haber himself. Haber is one of the most infamous scientists in history. As we've seen, the Haber process generates most of the world's food supply. But, Haber is also known as the father of chemical warfare. During World War I, he led a team of scientists that developed and implemented chlorine gas. Chlorine gas is a blistering agent. When soldiers breathed in large concentrations, it caused their airways to blister. Many died, unable to breathe. The soldiers that lived often suffered severe long-term effects. Most appallingly, Haber was unapologetic and proud of his accomplishments. He considered himself a patriot.

Maybe this is an opportunity to consider the ethics of chemistry. At dark moments in history, chemistry has caused unbelievable destruction. But, we know that chemistry can help our world. As you study chemistry with us, we hope you learn to love science. And we hope the world becomes a better place because of it.

(Video) Pressure Equilibrium Constants (Kp)

By Mark Lingwood

In this short video, Mark Lingwood explains the pressure equilibrium constant. He uses the ideal gas law to derive the connection between $K_{c}$ and $K_{p}$. Then he shows you an example problem. This video is an excellent and quick explanation of the connection between gas pressure and equilibrium.