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:

$\text{mole fraction of a gas }=\frac{\text{ number of moles of gas}}{\text{total number of moles in the mixture}}$

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 $\frac{2}{3}$.

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 is 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)

Give your answer to 3 significant figures.

Step 1: Calculate total moles

Total moles = 1.75 + 1.75 + 1.25 = 4.75 mol

Step 2: Calculate PCl₃ mole fraction

χ$(PC{l}_3)=\frac{{\text{ Moles of PCl}}_3}{\text{total moles}}=\frac{1.75}{4.75}=0.368$

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 Kp

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 Kp expressions

For the general equilibrium reaction:

aA_(g) + bB_(g) ⇌ dD_(g) + eE_(g)

The K_p expression for this equilibrium is:

$K_p=\frac{{(pD)}^d{(pE)}^e}{{(pA)}^a{(pB)}^b}$

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:

$K_p=\frac{p(PC{l}_3)p(C{l}_2)}{p(PC{l}_5)}$

Calculating values for Kp

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 2 - Calculating Kp 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

$K_p=\frac{p(PC{l}_3)p(C{l}_2)}{p(PC{l}_5)}$

Step 2: Substitution and correct evaluation

$K_p=\frac{(150\times 200)}{100}=300\text{ kPa}$

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

Worked example 3 - Calculating partial pressure from Kp

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). Give your answer to 2 significant figures.

Step 1: Write the equilibrium constant (K_p) expression

$K_p=\frac{p{(N{H}_3)}^2}{p(N_2)p{(H_2)}^3}$

Step 2: Rearrange the equilibrium constant (K_p) expression

$p(N{H}_3)=\sqrt{{\text{ K}}_{\text{p }}\times p(N_2)p{(H_2)}^3}$

Step 3: Substitution and correct evaluation

$p(N{H}_3)=\sqrt{0.50\times 0.30\times {(0.60)}^3}=\sqrt{0.0324}=0.18\text{ atm}$

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 4 - Calculating Kp 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)

Give your answer to 2 significant figures.

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

$p(PC{l}_5)=\frac{0.18}{(0.18+0.12+0.12)}\times 200=86$ kPa

$p(PC{l}_3)=\frac{0.12}{(0.18+0.12+0.12)}\times 200=57$ kPa

$p(C{l}_2)=\frac{0.12}{(0.18+0.12+0.12)}\times 200=57$ kPa

Step 4: Write the K_p expression

$K_p=\frac{p(PC{l}_3)p(C{l}_2)}{p(PC{l}_5)}$

Step 5: Substitution and correct evaluation

$K_p=\frac{(57\times 57)}{86}=38$ kPa

Kp 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 shifting 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: