The Equilibrium Constant, Kp

This lesson covers: 

1. The concept of mole fractions and partial pressure for gas mixtures
2. What the equilibrium constant (K_p) is
3. How to write an expression for K_p
4. Calculations involving K_p
5. How temperature changes affect K_p values

Mole fraction (χ)

The mole fraction (χ) of a gas is the proportion of the total gas mixture that is made up of that particular gas.

This can be represented by the equation:

mole fraction of a gas =total number of moles in the mixture number of moles of gas​

For example, if there are 3 moles of gas in total and two of them are gas A, the mole fraction of gas A is 2/3.

The mole fraction (χ) has no units because it is a ratio.

Partial pressure (p)

Partial pressure (p) is the pressure exerted by each individual gas component in a gaseous mixture. The total pressure exerted by the gas mixture equals the sum of the partial pressures.

This can be represented by the equation:

p(total) = ∑p(components)

where ∑ = sum of

For example, if the partial pressure of gas A is 50 kPa and the partial pressure of gas B is 70 kPa, the total pressure of a mixture containing gases A and B would be 120 kPa.

Linking partial pressures to mole fractions

We can use mole fractions to determine the partial pressures of gases in a mixture if the total pressure if known.

The equation for the partial pressure of a gas is:

Partial pressure of a gas = mole fraction of that gas x total pressure

Worked example 1 - Calculating partial pressures from mole fractions

The equilibrium mixture below contains 1.75 moles each of PCl₃ and Cl₂, and 1.25 moles of PCl₅ at equilibrium. If the total pressure of 714 kPa, calculate the partial pressure of PCl₃.

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

Step 1: Calculate total moles

Total moles = 1.75 + 1.75 + 1.25 = 4.75 mol

Step 2: Calculate PCl₃ mole fraction

χ(PCl3​)=total moles Moles of PCl3​​=4.751.75​=0.368

Step 3: Equation

p(PCl₃) = χ(PCl₃) x total pressure

Step 4: Substitution and correct evaluation

p(PCl₃) = 0.368 x 714 = 263 kPa

The equilibrium constant K_p

Unlike liquids, it's easier to measure gas amounts using pressure rather than concentration.

So for gas phase equilibria, we use K_p instead of K_c.

• K_p is calculated in the same way as K_c, but using partial pressures instead of concentrations
• Units for K_p are pressure units e.g. kPa or atm
• Solids and liquids do NOT get included in K_p expressions

Just like with K_c, a large K_p value indicates the equilibrium favours products, while a small K_p indicates equilibrium favours reactants.

Writing K_p expressions

For the general equilibrium reaction:

aA_(g) + bB_(g) ⇌ dD_(g) + eE_(g)

The K_p expression for this equilibrium is:

Kp​=(pA)a(pB)b(pD)d(pE)e​

Where p represents partial pressure and the lower case letters represent the coefficients in the balanced chemical equation.

Note: Round brackets are used instead of square brackets in the K_p expression. Square brackets mean concentration, but partial pressures do not have concentration units.

For example, for the reaction:

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

The K_p expression would be:

Kp​=p(PCl5​)p(PCl3​)p(Cl2​)​

Calculating values for K_p

If we know the equilibrium partial pressures of all gases, we can substitute them into the K_p expression to calculate K_p.

The units for K_p vary, so we must determine the units after each calculation.

Worked example 1 - Calculating K_p from known partial pressures

For the reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) at 500 K, the equilibrium partial pressures are:

p(PCl₃) = 150 kPa

p(Cl₂) = 200 kPa

p(PCl₅) = 100 kPa

Determine the equilibrium constant K_p.

Step 1: Write the equilibrium constant (K_p) expression

Kp​=p(PCl5​)p(PCl3​)p(Cl2​)​

Step 2: Substitution and correct evaluation

Kp​=100(150×200)​=300 kPa

If the equilibrium constant K_p and some equilibrium partial pressures are known, an unknown partial pressure can be calculated.

Worked example 2 - Calculating partial pressure from K_p

The reaction N_2(g) + 3H_2(g) ⇌ 2NH_3(g) has an equilibrium constant K_p = 0.50 atm at 25°C.

At equilibrium:

p(N₂) = 0.30 atm

p(H₂) = 0.60 atm

Determine p(NH₃).

Step 1: Write the equilibrium constant (K_p) expression

Kp​=p(N2​)p(H2​)3p(NH3​)2​

Step 2: Rearrange the equilibrium constant (K_p) expression

p(NH3​)=√ Kp ​× p(N2​)p(H2​)3​

Step 3: Substitution and correct evaluation

p(NH3​)=√0.50×0.30×(0.60)3​=√0.0324=0.18 atm

If the total pressure of an equilibrium mixture is given, along with some equilibrium information, the partial pressures can be derived to calculate K_p.

Worked example 3 - Calculating K_p from starting numbers of moles

0.30 mol PCl₅ decomposes at 575 K in a 5.0 dm³ vessel at a pressure of 200 kPa. At equilibrium 0.12 mol Cl₂ is present. Determine K_p for:

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

Step 1: Deduce moles of PCl₃ formed

Moles of PCl₃ = moles of Cl₂ = 0.12 mol

Step 2: Calculate remaining moles of PCl₅ at equilibrium

Moles of PCl₅ decomposed = 0.12 mol

Moles of PCl₅ at equilibrium = 0.30 - 0.12 = 0.18 mol

Step 3: Determine equilibrium partial pressures

p(PCl5​)=(0.18+0.12+0.12)0.18​×200=86 kPa

p(PCl3​)=(0.18+0.12+0.12)0.12​×200=57 kPa

p(Cl2​)=(0.18+0.12+0.12)0.12​×200=57 kPa

Step 4: Write the K_p expression

Kp​=p(PCl5​)p(PCl3​)p(Cl2​)​

Step 5: Substitution and correct evaluation

Kp​=86(57×57)​=38 kPa

K_p changes with temperature

The equilibrium constant K_p changes with temperature in the same way as the equilibrium constant K_c.

• Increasing temperature causes K_p to increase for endothermic reactions and decrease for exothermic reactions.
• Decreasing temperature has the opposite effects on K_p.

Changes in pressure and adding a catalyst do not affect the value of K_p, only the equilibrium position. The value of K_p itself depends only on temperature, just like K_c.

The effects of temperature changes on exothermic and endothermic reactions can be summarised in the following table: