# The Equilibrium Constant Kp

A LevelAQA

## The Equilibrium Constant $\text{K}_p$

In reactions that contain gases, the equilibrium constant, $\text{K}_p$, can be used instead of $\text{K}_c$. $\text{K}_p$ is calculated using partial pressure instead of concentration.

## What is $\text{K}_p$?

$\text{K}_p$ is a value that shows the relationship between the partial pressure of the reactants and the partial pressure of the products. $\text{K}_p$ is calculated by dividing the partial pressures of the products by the partial pressures of the reactants.

$2\text{A} + \text{B}\rarr 4\text{C}$

$\text{K}_p=\frac{_p\text{(C)}^4}{_p\text{(A)}^2\text{ }_p\text{(B)}}$

The unit for pressure is usually $\text{kPa}$, however, some questions may give you pressures in $\text{MPa}$ and that is what would be used instead.

For the equation:

$2\text{A} + \text{B} \rightleftharpoons4\text{C}$

The units can be calculated using the following method

$\text{Kp}=\frac{\text{(kPa)}^4}{\text{(kPa)}^2}\text{{(kPa)}}$

We can cancel down the units to find out what the unit for $\text{K}_p$ is

$\text{K}_p= \text{kPa}$

A LevelAQA

## Mole Fraction and Partial Pressure

The mole fraction of a gas is the ratio of the number of moles of that gas to the total amount of gaseous moles in the mixture.

$\text{Mole Fraction} = \frac{\text{Number of Moles of Gas at Equilibrium}}{\text{Total Number of Moles of all Gases at Equilibrium}}$

For example, if $6\text{ mol}$ of oxygen pumped into a container containing $7\text{ mol}$ of nitrogen and $4\text{ mol}$ of argon, the mole fraction of oxygen is given my:

$_p\left(\text{O}_2\right)=\frac{6}{6+8+4}=0.33\text{ mol}$

The partial pressure of a gas is the pressure that a gas in a mixture would exert if it occupied the given volume on its own.

$\text{Partial Pressure} = \text{Mole Fraction}\times\text{Total Pressure of a Gas}$

For example, if $12\text{ kPa}$ of oxygen pumped into a container containing $17\text{ kPa}$ of nitrogen and $20\text{ kPa}$ of argon, the mole fraction of oxygen is given my:

$\text{n}_{\text{O}_2}=\frac{12}{12+17+20}=0.24\text{ kPa}$

A LevelAQA

## Effect of Conditions on $\text{K}_p$

The value of $\text{K}_P$, similar to the equilibrium constant, $\text{K}_c$,  will tell us how far to which side of the equilibrium a reaction lies.  A large $\text{K}_p$ means that the equilibrium favours the products while a smaller $\text{K}_p$ similarly shows that the equilibrium favours the reactants.

Changing the concentration, pressure or adding a catalyst would have no effect on $\text{K}_p$.

Effect of Temperature

Like the equilibrium constant $\text{K}_c$, the $\text{K}_p$ value only changes if the temperature changes.

The affect of changing the temperature on the value of $\text{K}_p$ will depend on nature of the reaction. If the equilibrium is exothermic in the forward direction then increasing the temperature would shift the equilibrium towards the reactants to oppose this change. This will lead to a higher partial pressure in the reactants and a lower partial pressure in the products, leading to a lower value of $\text{K}_p$. The opposite is true for an A. In this case, and increasing temperature increases the value of $\text{K}_p$.

A LevelAQA

## Example: Calculating $\text{K}_p$

An experiment is set up to investigate the reaction shown below.

$\text{CO}_{2\text{(g)}} + 3\text{H}_{2\text{(g)}}\rightleftharpoons\text{CH}_3\text{OH}_\text{(g)} + \text{H}_2\text{O}_\text{(g)}$

$\textcolor{#00bfa8}{1.3\text{ mol}}$ of carbon dioxide and $\textcolor{#f21cc2}{3.9\text{ mol}}$ of hydrogen were sealed in a container. After the mixture reached equilibrium, at a pressure of $\textcolor{#a233ff}{500\text{ kPa}}$, the yield of the methanol was $\textcolor{#eb6517}{1.1\text{ mol}}$.

Calculate the $\text{K}p$ value and the units. Give your answer to an appropriate amount of significant figures.

[6 marks]

Step 1: Find the moles of the gases at equilibrium. We can use the table to work out the moles.

 $\text{CO}_2$ $3\text{H}_2$ $\text{CH}_3\text{OH}$ $\text{H}_2\text{O}$ Initial Moles $\textcolor{#00bfa8}{1.3}$ $\textcolor{#f21cc2}{3.9}$ $0$ $0$ Equilibrium Moles $0.2$ $0.6$ $\textcolor{#eb6517}{1.1}$ $1.1$

Step 2: Calculate the partial pressures.

$\text{Partial Pressure} = \text{Mole Fraction}\times\text{Total Pressure of a Gas}$

$\text{Total Moles of Gas} = 0.2+0.6+1.1+1.1 =\textcolor{#008d65}{3\text{ mol}}$

$_p\text{CO}_2=\frac{0.2}{3}\times\textcolor{#a233ff}{500}=\textcolor{#008d65}{33.33\text{ kPa}}$

$_p\text{H}_2=\frac{0.6}{3}\times\textcolor{#a233ff}{500}=\textcolor{#008d65}{100\text{ kPa}}$

$_p\text{CH}_3\text{OH}=\frac{\textcolor{#eb6517}{1.1}}{3}\times\textcolor{#a233ff}{500}=\textcolor{#008d65}{183.33\text{kPa}}$

$_p\text{H}_2\text{O}=\frac{1.1}{3}\times\textcolor{#a233ff}{500}=\textcolor{#008d65}{183.33\text{ kPa}}$

Step 3: Using the partial pressures, calculate $\text{K}_p$.

\begin{aligned}\text{K}_p&=\frac{_p\text{(CH}_3\text{OH)} _p\text{(H2O)}}{_p\text{(CO}_2\text{)}_p\text{(H}_2\text{)}^3}\\\text{}\\ &=\frac{183.33\times 183.33}{33.33\times 100^3}\\\text{}\\ &=\textcolor{#008d65}{0.034}\end{aligned}

$\text{Units} = \textcolor{#008d65}{\text{kPa}^{-2}}$

A LevelAQA

## The Equilibrium Constant Kp Example Questions

$\text{K}_p=\frac{_p\text{(Z)}^2}{_p\text{(X)}_ p\text{(Y)}^3}$

$_p\left(\text{X}\right)=\frac{1}{1+3+2}\times17=2.8\text{ MPa}$

$_p\left(\text{Y}\right)=\frac{3}{1+3+2}\times17=8.5\text{ MPa}$

Increase

Increase in temperature means the equilibrium will move in the endothermic direction and favours the products

No effect

Step 1: Deduce the equation for $\text{K}_p$.

$\text{K}_p=\frac{_p\text{ (CH}_3\text{OH)}}{_p\text{(CO)}_ p\text{(H}2\text{)}^2}$

Step 2: Calculate the eqm moles of each carbon monoxide and hydrogen.

$\text{CO Eqm Moles} = 60.3-23.4=36.9$

Step 3: calculate the total moles at equilibrium.

$\text{H}_2\text{ Eqm Moles} = 141-(2\times23.4)=94.2$

$\text{Total no.of Eqm Moles} = 23.4+94.2+36.9=154.5$

Step 4: calculate the value of $\text{K}_p$.

\begin{aligned}\text{K}_p&=\frac{0.151\times8.9}{(0.239\times8.9)(0.610\times)^2}\\ &=\frac{1.348}{2.130\times5.426^2}\\ &=0.022\text{ MPa}^{-2}\end{aligned}

(One mark for step 1. Two marks for step 2, one mark per correct eqm moles. One mark for step 3, one mark for correct total moles. Three marks for step 4, one mark per correct calculation of $\text{K}_p$.)

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