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# Equilibrium Constant Kp

Understanding equilibria is important in chemistry. These equilibria have many real-life applications, particularly in industry. If we know how an equilibrium is set up, how it can be changed, and the factors that influence it, then we can use this to our advantage!

One way of expressing an equilibrium is by using an equilibrium constant. One example is Kp.

Kp is an equilibrium constant based on partial pressures. It tells you the ratio of products to reactants in a reaction at equilibrium.

• We'll start by defining Kp before looking more closely at partial pressures.
• We'll then have a go at calculating Kp.
• After that, we'll look at the properties of Kp.
• Finally, we'll compare Kc and Kp.

## Equilibrium constant Kp for the reaction

If you leave a reversible reaction in a sealed container for long enough, it will eventually reach equilibrium. This is when the concentrations of the reactants and products stay the same, and the rate of the forward reaction equals the rate of the backward reaction. But the interesting thing about each particular reversible reaction is that, provided you keep the external conditions the same, you always end up with the same ratio of products to reactants.

This means that no matter how much of each species you start with, you'll end up with the same relative amounts of products and reactants each time. We can use equilibrium constants to express this ratio of products to reactants in an equilibrium mixture. One such equilibrium constant is Kp.

Kp is very similar to Kc, which you can read more about in Equilibrium Constants. But whilst Kc uses the concentrations of species in equilibrium, Kp uses their partial pressures.

Like Kc, Kp always has the same value for a specific reversible reaction under specific conditions. This means that we can use it to accurately predict the proportions of reactants and products in a reaction at equilibrium. But before we have a go at doing such calculations, we first need to learn about partial pressure.

## Partial Pressure

In a gas equilibrium, the total pressure of the system is the result of each individual gas in the system. Each gas exerts its own pressure, which we call its partial pressure, and the sum of all partial pressures is equal to the total pressure. So, if we have a system with three gaseous components A, B, and C, then the total pressure of the system is given by the partial pressure of A, added to the partial pressure of B, added to the partial pressure of C.

Similar to the way we represent molar concentrations using square brackets, we also have a way of representing partial pressure: ${p}_{\mathrm{x}}$. For example, if we were told that the partial pressure of oxygen gas in the equilibrium is 100 kPa, we would write this as: ${p}_{{\mathrm{O}}_{2}}=100\mathrm{kPa}$.

### Calculating partial pressure

The simplest way of calculating the partial pressure of a gas is using the total pressure of the system and subtracting the partial pressures of all other gases involved. But it won't always be this easy. Instead, we might have to use mole fractions.

Mole fractions represent the number of moles of a species compared to the total number of moles in a system, as a fraction.

Mole fractions are a way of calculating the partial pressure of a gas. You first find the mole fraction of a gas in a system. You then multiply that by the system's total pressure to find the gas' partial pressure. Here are the steps for calculating the partial pressure of gas A:

1. $\text{Mole fraction (A)}=\frac{\text{number of moles of A in the system}}{\text{total number of moles of all gases in the system}}$
2. ${p}_{\mathrm{A}}=\text{Mole fraction (A)}×\text{total pressure of all gases in the system}$

Sounds confusing? Let's look at a real-life example.

The reversible reaction shown in the equation below was allowed to reach equilibrium in a sealed container.

$2{\mathrm{NO}}_{2}\left(\mathrm{g}\right)⇌{\mathrm{N}}_{2}{\mathrm{O}}_{4}\left(\mathrm{g}\right)$

At equilibrium, the mixture contained 2.00 mol NO2 and 1.00 mol N2O4 with a total pressure of 150 kPa. Calculate the partial pressure of each gas in the system.

Partial pressure is found by multiplying the mole fraction of the gas by the total pressure of the system. First, we need to work out each gas' mole fraction. To do this, we divide the number of moles of the gas by the total number of moles of all the gases in the system. Here, the total number of moles of gas in the system is 1 + 2 = 3.

$\mathrm{Mole}\mathrm{fraction}\left({\mathrm{NO}}_{2}\right)=\frac{2}{3}=0.67$

$\mathrm{Mole}\mathrm{fraction}\left({\mathrm{N}}_{2}{\mathrm{O}}_{4}\right)=\frac{1}{3}=0.33$

We then multiply each mole fraction by the total pressure of the system, to find the partial pressure of each gas. Remember that partial pressure is shown using the symbol $p$.

${p}_{{\mathrm{NO}}_{2}}=0.67×150=100\mathrm{kPa}\phantom{\rule{0ex}{0ex}}{p}_{{\mathrm{N}}_{2}{\mathrm{O}}_{4}}=0.33×150=50\mathrm{kPa}$

Now let's move on to the equation for Kp.

## Equation for equilibrium constant Kp

To reiterate: Kp expresses the ratio of products to reactants in a gaseous reversible reaction at equilibrium. It uses the gases' partial pressures, which is the pressure each gas exerts on the system. We'll now look at the equation for Kp that links the partial pressures of each gaseous species to find a ratio of products to reactants at equilibrium.

Let's say that you have the following reaction:

$\mathrm{aA}\left(\mathrm{g}\right)+\mathrm{bB}\left(\mathrm{g}\right)⇌\mathrm{cC}\left(\mathrm{g}\right)+\mathrm{dD}\left(\mathrm{g}\right)$

We use these values to find Kp:

$\mathrm{Kp}=\frac{{{p}_{\mathrm{C}}}^{\mathrm{c}}×{{p}_{\mathrm{D}}}^{\mathrm{d}}}{{{p}_{\mathrm{A}}}^{\mathrm{a}}×{{p}_{\mathrm{B}}}^{\mathrm{b}}}$

What does this mean? Well, ${{p}_{\mathrm{A}}}^{\mathrm{a}}$ means the partial pressure of A, raised to the power of the number of moles given in the equation. You calculate a similar value for each of the gaseous species involved in the reaction. You then multiply the values given by the products together and place them in the numerator (at the top of the fraction). You multiply the values given by the reactants together and place them in the denominator (at the bottom of the fraction).

When working with Kp, all partial pressures are taken at equilibrium. You could represent this using ${{{p}_{\mathrm{A}}}^{\mathrm{a}}}_{\mathrm{eqm}}$, but you don't need to - just remember this fact when carrying out any calculations.

Don't worry if this seems complicated. We'll have a go at an example in a second, but we'll quickly take a look at the units of Kp first.

### Units of Kp

To find the units of Kp, you need to use the equation for Kp that you just worked out. You take each term in the equation and insert its units, and then cancel all of the units down.

Sulphur dioxide reacts with oxygen to form sulphur trioxide according to the following equation:

$2{\mathrm{SO}}_{2}\left(\mathrm{g}\right)+{\mathrm{O}}_{2}\left(\mathrm{g}\right)⇌2{\mathrm{SO}}_{3}\left(\mathrm{g}\right)$

The total pressure is 54 kPa. Write an equation for the equilibrium constant Kp for this reaction, and calculate its units.

We know our product is sulphur trioxide, SO3. The equation tells us that 2 moles of sulfur trioxide are produced. Therefore, we have ${{p}_{{\mathrm{SO}}_{3}}}^{2}$ in the numerator. This means that we take the partial pressure of sulphur trioxide and raise it to the power of 2.

Likewise, our reactants are sulphur dioxide and oxygen, SO2 and O2. The equation tells us that there are 2 moles and 1 mole respectively. Therefore, we have ${{p}_{{\mathrm{SO}}_{2}}}^{2}×{p}_{{\mathrm{O}}_{2}}$ in the numerator. This means that we take the partial pressure of sulphur dioxide and raise it to the power of 2 and then we multiply it by the partial pressure of oxygen. Overall, we get the following equation for Kp:

$\mathrm{Kp}=\frac{{{p}_{{\mathrm{SO}}_{3}}}^{2}}{{{p}_{{\mathrm{SO}}_{2}}}^{2}×{p}_{{\mathrm{O}}_{2}}}$

We now need to calculate the units of Kp. The question gives us the total pressure in kilopascals, kPa, so we would measure partial pressures in kPa too. If we substitute the units of partial pressure into each term in the equation for Kp, we get the following:

$\mathrm{units}=\frac{{\mathrm{kPa}}^{2}}{{\mathrm{kPa}}^{2}×\mathrm{kPa}}$

Cancelling the units down, we are left with just $\frac{1}{\mathrm{kPa}}$, which equals ${\mathrm{kPa}}^{-1}$. These are the units of Kp for this particular reaction.

## How do we calculate Kp?

Next, we can try working out an actual value for Kp using the equation we've just found. To calculate Kp, you first need to know the partial pressures of each gas in a mixture, then you simply put the partial pressure values into the equation you worked out earlier.

Here's an example.

An equilibrium mixture contains 2 moles of sulphur dioxide, 1 mole of oxygen and 6 moles of sulphur trioxide. The total pressure is 54 kPa. The reversible reaction is given below:

$2{\mathrm{SO}}_{2}\left(\mathrm{g}\right)+{\mathrm{O}}_{2}\left(\mathrm{g}\right)⇌2{\mathrm{SO}}_{3}\left(\mathrm{g}\right)$

This is the same reaction that we looked at earlier. We already know the equation for Kp - we simply need to find the partial pressures of all of the gases involved and substitute them into the equation.

Remember that to find partial pressures, you multiply the gases' mole fraction by the total pressure of the system. Although we're not given mole fractions in the question, we are given the molar quantities of each species at equilibrium. We use this to calculate mole fractions by dividing the number of moles of each gas by the total number of moles of gas in the system. Here, the total number of moles is $2+ 1+6=9$. We can then multiply this by the total pressure, 54 kPa, to find each gases' partial pressure.

Let's draw a table to show all these values.

 Species SO2 O2 SO3 Moles at equilibrium 2 1 6 Mole fraction $\frac{2}{9}$ $\frac{1}{9}$ $\frac{6}{9}$ Partial pressure (kPa) 12 6 36
Now let's substitute them into our equation for Kp: $\mathrm{Kp}=\frac{{{p}_{{\mathrm{SO}}_{3}}}^{2}}{{{p}_{{\mathrm{SO}}_{2}}}^{2}×{p}_{{\mathrm{O}}_{2}}}=\frac{{36}^{2}}{{12}^{2}×6}=1.5$

Add on the units that we worked out earlier, which are kPa-1, and we are left with our final answer:

$\mathrm{Kp}=1.5{\mathrm{kPa}}^{-1}$

## Properties of Kp

Remember how at the beginning we said that equilibrium constants are, well, constant, provided you keep the external conditions the same? Some conditions can in fact change Kp. We'll now focus on Kp's properties and how it is affected by these changes.

### Kp and temperature

Changing the temperature of a gaseous system at equilibrium changes the value of Kp and shifts the position of the equilibrium. As with Kc, this is the only environmental condition that affects Kp.

Le Chatelier's Principle helps us account for the shifted position of the equilibrium. It states that changes in the conditions of an equilibrium cause the equilibrium position to shift in order to counteract the change. When it comes to temperature, we can say the following:

• Increasing the temperature shifts the position of the equilibrium in the direction of the endothermic reaction. This 'takes in' excess heat, counteracting the change in conditions.
• Decreasing the temperature shifts the position of the equilibrium in the direction of the exothermic reaction. This releases extra heat, counteracting the change in conditions.

But what does this mean for Kp? If the forward reaction in an equilibrium is endothermic, then increasing the temperature favours the forward reaction. We produce more of the products and have less of the reactants. This means that the partial pressure of the products increases, and the partial pressure of the reactants decreases. Therefore, if we look at our equation for Kp, the numerator increases whilst the denominator decreases. This increases Kp.

If the forward reaction in an equilibrium is exothermic, then increasing the temperature has the opposite effect. The backward reaction is favoured and more of the reactants are produced. This decreases the numerator and increases the denominator, lowering the value of Kp.

### Kp and pressure

Changing the pressure of a gaseous system does not change the value of Kp. However, it does shift the position of the equilibrium in order to counteract the change.

• Increasing the pressure shifts the position of the equilibrium to favour the side that produces less moles of gas. This decreases the pressure in the system, counteracting the change in conditions.
• Decreasing the pressure shifts the position of the equilibrium to favour the side that produces the greater number of moles of gas. This increases the pressure in the system, counteracting the change in conditions.

### Kp and concentration

Changing the concentration of a gaseous system at equilibrium does not change the value of Kp. However, as with pressure, it does shift the position of the equilibrium.

• Increasing the concentration of the reactants shifts the position of the equilibrium to favour the forward reaction. This uses up some of the reactants, counteracting the change in conditions.
• Increasing the concentration of the products shifts the position of the equilibrium to favour the backward reaction. This uses up some of the products, counteracting the change in conditions.

### Kp and catalysts

Adding a catalyst to a gaseous system at equilibrium neither changes the value of Kp, nor shifts the position of the equilibrium. Catalysts simply increase the rate of reaction. They increase the rate of both reactions equally, so the position of the equilibrium stays the same.

How does changing the pressure or concentration of a gaseous system at equilibrium shift the position of the equilibrium constant without changing the value of Kp?

Let's start with concentration. Imagine that you take a reaction at equilibrium and remove some of the products. This decreases the products' partial pressure. If we look at the equation for Kp, we can see that this would reduce the numerator (the number on top of the fraction), and thus reduce Kp. However, Le Chatelier's principle tells us that the equilibrium shifts to oppose the change caused by reducing the concentration of the products. It does this by producing more of the products. This increases the numerator, bringing it back to the value it was at before we made any changes. Overall, the fraction in the equation for Kp has stayed the same. This means that Kp has also stayed the same.

Now let's look at pressure. This is a little more complicated. Remember how the equation for Kp features partial pressures? They're calculated using mole fractions and total pressure. Let's take an example reaction, find its equation for Kp, and replace all the partial pressures with their respective mole fractions and total pressure.

For example, the reaction $2{\mathrm{SO}}_{2}+ {\mathrm{O}}_{2}⇌2{\mathrm{SO}}_{3}$ has the following equation for Kp:

$\frac{{{p}_{{\mathrm{SO}}_{3}}}^{2}}{{{p}_{{\mathrm{SO}}_{2}}}^{2}×{p}_{{\mathrm{O}}_{2}}}$

If we replace all of the partial pressures with mole fractions, which we'll represent using M, and the total pressure, which we'll represent using P, we get the following:

$\frac{{\left({\mathrm{M}}_{{\mathrm{SO}}_{3}}\mathrm{P}\right)}^{2}}{{\left({\mathrm{M}}_{{\mathrm{SO}}_{2}}\mathrm{P}\right)}^{2}×\left({\mathrm{M}}_{{\mathrm{O}}_{2}}\mathrm{P}\right)}$

P cancels out, until we are left with an equation featuring just mole fractions and one P at the bottom of the fraction. If we were to increase the total pressure of the system, this means that P increases, so the denominator increases. This would decrease Kp. However, Le Chatelier's principle tells us that the equilibrium shifts to oppose the change caused by changing the pressure. It does this by favouring the reaction that produces the fewest moles of gas. Here, it favours the forward reaction, so we produce more of the products. This increases the value of the numerator and increases the value of Kp, back to what it was before we made any changes. Overall, the fraction stays the same. This means that Kp also stays the same.

You can apply a similar process to any equation showing a gaseous equilibrium. However, you'll find that if both the forward and backward reactions produce the same number of moles of gas, P will cancel out completely on the top and the bottom of the fraction. In this case, changing the pressure has no effect on the position of the equilibrium.

## Kp vs Kc

Kc and Kp are both equilibrium constants, so why don't we use Kc when working with gas equilibria?

Well, we can use Kc, but when working with gases, it is more useful to think about them in terms of their pressures, instead of their molar concentrations. That's the primary difference between the two equilibrium constants.

• Kc is calculated using the concentrations of aqueous, liquid or gaseous species in an equilibrium.
• Kp is calculated using the partial pressures of gaseous species in an equilibrium.

## Equilibrium Constant Kp - Key takeaways

• Kp is an equilibrium constant based on partial pressures. It tells you the ratio of products to reactants in a reaction at equilibrium.
• Partial pressure is the pressure that a single constituent gas exerts on a closed system. The partial pressure of a gas can be worked out using the mole fraction of the gas and the total pressure of the system.
• The expression for Kp depends on the reactants and products involved and their number of moles. For the general reaction $\mathrm{aA}\left(\mathrm{g}\right)+\mathrm{bB}\left(\mathrm{g}\right)⇌\mathrm{cC}\left(\mathrm{g}\right)+\mathrm{dD}\left(\mathrm{g}\right)$, Kp = $\frac{{{p}_{\mathrm{C}}}^{\mathrm{c}}×{{p}_{\mathrm{D}}}^{\mathrm{d}}}{{{p}_{\mathrm{A}}}^{\mathrm{a}}×{{p}_{\mathrm{B}}}^{\mathrm{b}}}$
• Changing the temperature of a gaseous equilibrium changes the value of Kp and shifts the position of the equilibrium.
• Changing the concentration or pressure has no effect on the value of Kp, but does shift the position of the equilibrium.
• Adding a catalyst neither affects the value of Kp, nor shifts the position of the equilibrium.
• Kp is very similar to Kc. Whilst Kc uses molar concentrations, Kp uses partial pressures.

#### Flashcards in Equilibrium Constant Kp 3

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What is the equilibrium constant Kp of a reaction?

The equilibrium constant, Kp, is found using the partial pressures of the reactants and products of the reaction.

How do you find the equilibrium constant Kp?

To find the equilibrium constant Kp you need to know the partial pressure that each reactant and each product exerts on the reaction mixture. Once you know each partial pressure, you input them into the expression for Kp, which is different for different gas equilibria, to find the value of the constant.

What are equilibrium constants Kp and Kc?

The equilibrium constants Kc and Kp are both values that tell you about the rates of the forward and backward reactions involved in a dynamic equilibrium. The difference between Kp and Kc is that Kp is only applicable to gas equilibria.

Write an expression for the equilibrium constant Kp?

For the gas equilibrium:

aA + bB -> dD + eE

Where all the reactants and products are in their gaseous state, the expression for Kp is:

(p(D)d • p(E)e) ÷ (p(A)a • p(B)b)

What are the properties of Kp?

The properties of Kp are based on how the value of Kp changes depending on reaction conditions. Kp is only constant at fixed temperatures, but is unaffected by pressure, or the addition of a catalyst.

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