Equilibrium Constant, Kp | Edexcel International A Level Chemistry Revision Notes 2017

Kp Expressions - Deduction

We have seen previously that equilibrium reactions can be quantified by reference to an equilibrium expression and equilibrium constant
The equilibrium expression links the equilibrium constant, K_c, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account
So, for a given reaction:

aA + bB ⇌ cC + dD

K_c is defined as follows:

Equilibria Equilibrium Expression, downloadable AS & A Level Chemistry revision notes

Equilibrium expression linking the equilibrium concentration of reactants and products at equilibrium

Solids are ignored in equilibrium expressions
The K_c of a reaction is constant and only changes if the temperature of the reaction changes

Homogeneous reactions

In the generic example above, if all the substances are gases, we can show the equation with that state symbol

aA (g) + bB (g) ⇌ cC (g) + dD (g)

We can write a different equilibrium expression in terms of the partial pressure of the gases
This equilibrium constant is called K_pand is defined as follows

Kp Expressions, downloadable AS & A Level Chemistry revision notes

Equilibrium expression linking the partial pressures of reactants and products at equilibrium

Heterogeneous reactions

For heterogenous reactions, solids and liquids are ignored in K_pequilibrium expressions
The K_p of a reaction is constant and only changes if the temperature of the reaction changes

Worked example

Write a K_p expression for the following equilibria and deduce the units of K_p :

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
CaCO₃ (s) ⇌ CaO(s) + CO₂ (g)

Answer 1

Worked Example Answer 1, downloadable AS & A Level Chemistry revision notes

Answer 2

K_p = pCO₂(g)

Kp Expressions - Calculations

Calculations involving K_p

In the equilibrium expression, the p represents the partial pressure of the reactants and products in Pa
The units of K_p therefore depend on the form of the equilibrium expression

Worked example

Calculating K_p of a gaseous reaction

Sulfur dioxide and oxygen react according to the following equation:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

The equilibrium partial pressures at constant temperature are:

SO₂ =1.0 x 10⁶ Pa
O₂ = 7.0 x 10⁶ Pa
SO₃ = 8.0 x 10⁶ Pa

Calculate the value of K_p for this reaction.

Answer

- Step 1: Write the equilibrium constant for the reaction in terms of partial pressures

$K_{p} = \frac{p^{2} {SO}_{3}}{p^{2} {SO}_{2} \times p O_{2}}$

- Step 2: Substitute the equilibrium concentrations into the expression

$K_{p} = \frac{{(8.0 \times 10^{6})}^{2}}{{(1.0 \times 10^{6})}^{2} \times (7.0 \times 10^{6})}$

K_p= 9.1 x 10^-6

- Step 3: Deduce the correct units of K_p

_{$K_{p} = \frac{P a^{2}}{P a^{2} \times P a}$}

The units of K_p are Pa^-1

Therefore, K_p = 9.1 x 10^-6 Pa^-1

Some questions only give the number of moles of gases present and the total pressure
The number of moles of each gas should be used to first calculate the mole fractions
The mole fractions are then used to calculate the partial pressures
The values of the partial pressures are then substituted in the equilibrium expression

Worked example

Calculating K_p of hydrogen iodide equilibrium reaction

The equilibrium between hydrogen, iodine and hydrogen iodide, at 600 K, is as follows:

H₂ (g) + I₂ (g) ⇌ 2HI (g)

At equilibrium, the number of moles present are:

H₂ = 1.71 x 10^-3
I₂ = 2.91 x 10^-3
HI = 1.65 x 10^-2

The total pressure is 100 kPa. Calculate the value of K_p for this reaction.

Answer

- Step 1: Calculate the total number of moles

Total number of moles = 1.71 x 10^-3+ 2.91 x 10^-3+ 1.65 x 10^-2

Total number of moles = 2.112 x 10^-2

- Step 2: Calculate the mole fraction of each gas

$H_{2} = \frac{1.71 \times 10^{- 3}}{2.112 \times 10^{- 2}} = 0.0810 I_{2} = \frac{2.91 \times 10^{- 3}}{2.112 \times 10^{- 2}} = 0.1378 HI = \frac{1.65 \times 10^{- 2}}{2.112 \times 10^{- 2}} = 0.7813$

- Step 3: Calculate the partial pressure of each gas

H₂ = 0.0810 x 100 = 8.10 kPa

I₂ = 0.1378 x 100 = 13.78 kPa

HI = 0.7813 x 100 = 78.13 kPa

- Step 4: Write the equilibrium constant in terms of partial pressure

$K_{p} = \frac{p^{2} HI}{p H_{2} \times p I_{2}}$

- Step 5: Substitute the values into the equilibrium expression

$K_{p} = \frac{78 . 13^{2}}{8.10 \times 13.78}$

K_p= 54.7

- Step 6: Deduce the correct units for K_p

$K_{p} = \frac{P a^{2}}{P a \times P a}$

All units cancel out

Therefore, K_p = 54.7

Other questions related to equilibrium expressions may involve calculating quantities present at equilibrium given appropriate data

Equilibrium Constant, Kp (Edexcel International A Level Chemistry)

Revision Note

Kp Expressions - Deduction

Homogeneous reactions

Heterogeneous reactions

Worked example

Kp Expressions - Calculations

Calculations involving K_p

Worked example

Calculating K_p of a gaseous reaction

Worked example

Calculating K_p of hydrogen iodide equilibrium reaction

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Sonny

Equilibrium Constant, Kp (Edexcel International A Level Chemistry)

Revision Note

Homogeneous reactions

Heterogeneous reactions

Calculations involving Kp

Calculating Kp of a gaseous reaction

Calculating Kp of hydrogen iodide equilibrium reaction

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Sonny

Calculations involving K_p

Calculating K_p of a gaseous reaction

Calculating K_p of hydrogen iodide equilibrium reaction