Kc and Kp Calculations

Learning Objectives (OCR Spec 5.1.2)

By the end of this lesson you should be able to:

• Use ICE tables (Initial, Change, Equilibrium) to work out equilibrium amounts
• Calculate Kc from equilibrium concentrations
• Calculate Kp from equilibrium partial pressures
• Handle calculations in which only some equilibrium quantities are given and the rest must be deduced from stoichiometry

The ICE Table Method

An ICE table is a structured way to keep track of moles (or concentrations) during an equilibrium calculation. The rows are:

• Initial — amounts at the start, before any reaction has occurred
• Change — the change during approach to equilibrium, consistent with stoichiometry
• Equilibrium — the amounts at equilibrium (I + C)

Example — Simple 1:1 Reaction

Consider A(g) <-> B(g) with 1.00 mol A initially and no B. At equilibrium, 0.60 mol A remain.

	A	B
Initial	1.00	0
Change	-0.40	+0.40
Equilibrium	0.60	0.40

The change -0.40 in A must be matched by +0.40 in B because of 1:1 stoichiometry.

Example — 1:2 Stoichiometry

For A(g) <-> 2B(g), starting with 1.00 mol A and 0 B. At equilibrium, 0.70 mol A.

	A	2B
Initial	1.00	0
Change	-0.30	+0.60
Equilibrium	0.70	0.60

Note the 1:2 ratio: 0.30 mol A gives 0.60 mol B.

Worked Example 1 — Finding Kc from Initial Moles

For H2(g) + I2(g) <-> 2HI(g), 2.0 mol H2 and 2.0 mol I2 are mixed in a 2.0 dm^3 flask. At equilibrium, 3.2 mol HI is present. Calculate Kc.

Step 1 — ICE in moles

	H2	I2	2HI
Initial	2.0	2.0	0
Change	-1.6	-1.6	+3.2
Equilibrium	0.4	0.4	3.2

Change in HI = +3.2, so H2 and I2 each decrease by 3.2/2 = 1.6 (stoichiometry). Equilibrium: H2 = 0.4, I2 = 0.4, HI = 3.2.

Step 2 — Convert moles to concentrations (divide by V = 2.0 dm^3)

• [H2] = 0.4/2.0 = 0.20 mol dm^-3
• [I2] = 0.4/2.0 = 0.20 mol dm^-3
• [HI] = 3.2/2.0 = 1.6 mol dm^-3

Step 3 — Substitute into Kc expression

Kc = [HI]^2 / ([H2][I2])
   = (1.6)^2 / (0.20 x 0.20)
   = 2.56 / 0.040
   = 64

Step 4 — Units

Top power = 2, bottom power = 2. Net = 0. Kc is dimensionless.

Kc = 64.

Worked Example 2 — Kc with Units

N2O4(g) <-> 2NO2(g). A 2.0 dm^3 flask initially contains 1.0 mol N2O4 and no NO2. At equilibrium, 0.20 mol N2O4 is present. Calculate Kc.

ICE table (moles)

	N2O4	2NO2
Initial	1.0	0
Change	-0.80	+1.60
Equilibrium	0.20	1.60

Change in N2O4 = -0.80 (decreased from 1.0 to 0.20). Therefore change in NO2 = +1.60 (stoichiometric factor of 2).

Concentrations

• [N2O4] = 0.20/2.0 = 0.10 mol dm^-3
• [NO2] = 1.60/2.0 = 0.80 mol dm^-3

Kc

Kc = [NO2]^2 / [N2O4]
   = (0.80)^2 / 0.10
   = 0.64 / 0.10
   = 6.4 mol dm^-3

Net power = 2 - 1 = 1. Units = mol dm^-3.

Worked Example 3 — Kp with ICE Table

For 2SO2(g) + O2(g) <-> 2SO3(g), a flask initially contains 2.0 mol SO2, 1.0 mol O2 and no SO3. The total pressure at equilibrium is 150 kPa. At equilibrium, 1.6 mol SO3 is present.

Step 1 — ICE table

	2SO2	O2	2SO3
Initial	2.0	1.0	0
Change	-1.6	-0.80	+1.6
Equilibrium	0.40	0.20	1.60

Change in SO3 = +1.60. Stoichiometrically, SO2 decreases by 1.60 (2:2 ratio) and O2 decreases by 0.80 (2:1 ratio).

Step 2 — Mole fractions