Equilibrium Constant, Kp (Edexcel International A Level (IAL) Chemistry): Revision Note

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

1. N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

2. CaCO₃ (s)  ⇌ CaO (s) + CO₂ (g)

Answer 1

K_p = $\frac{p^2{\mathrm{NH}}_3{\left(\mathrm{g}\right)}_{\mathrm{eqm}}}{p{\mathrm{N}}_2{\left(\mathrm{g}\right)}_{\mathrm{eqm}}\times p^3{\mathrm{H}}_2{\left(\mathrm{g}\right)}_{\mathrm{eqm}}}$

Units = $\frac{{\mathrm{atm}}^2}{\mathrm{atm}\times {\mathrm{atm}}^3}$ = atm^-2

Answer 2

K_p = pCO₂(g)

Units = atm

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 atm

• O₂ = 7.0 atm

• SO₃ = 8.0 atm

Calculate the value of K_p for this reaction.

Answer

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

   $K_p=\frac{p^2{\mathrm{SO}}_3}{p^2{\mathrm{SO}}_2\times p{\mathrm{O}}_2}$

• Step 2: Substitute the equilibrium partial pressures into the expression

   $K_p=\frac{{(8.0)}^2}{{(1.0)}^2\times (7.0)}$

   K_p = 9.1

• Step 3: Deduce the correct units of K_p

   $K_p=\frac{{\mathrm{atm}}^2}{{\mathrm{atm}}^2\times \mathrm{atm}}$

The units of K_p are atm^-1

Therefore, K_p = 9.1 atm^-1

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 atm. 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

   $\begin{array}{l}{\mathrm{H}}_2=\frac{1.71\times {10}^{-3}}{2.112\times {10}^{-2}}=0.0810\\ {\mathrm{I}}_2=\frac{2.91\times {10}^{-3}}{2.112\times {10}^{-2}}=0.1378\\ \mathrm{HI}=\frac{1.65\times {10}^{-2}}{2.112\times {10}^{-2}}=0.7813\end{array}$

• Step 3: Calculate the partial pressure of each gas

H₂ = 0.0810 x 100 = 8.10 atm

I₂ = 0.1378 x 100 = 13.78 atm

HI = 0.7813 x 100 = 78.13 atm

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

   $K_p=\frac{p^2\mathrm{HI}}{p{\mathrm{H}}_2\times p{\mathrm{I}}_2}$

• 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

   $K_p=\frac{{\mathrm{atm}}^2}{\mathrm{atm}\times \mathrm{atm}}$

   All units cancel out

   Therefore, K_p = 54.7