E. Chemical Systems and Equilibrium · Chemical Equilibrium

ICE Tables I: Calculating K from Experimental Data

E2.4 E3.2 E2.1 A1.12 A1.13About 70 minutes

Learning goals

I can set up an ICE (Initial, Change, Equilibrium) table in mol/L.
I can use the mole ratio and one measured equilibrium value to complete an ICE table.
I can calculate K from initial amounts and one equilibrium measurement.
I can handle starting mixtures that already contain some product.

Words to know

ICE table
A table with rows for Initial, Change and Equilibrium concentrations of each species, used to organize equilibrium calculations.

change row
The middle row of an ICE table. Its entries are in the same ratio as the coefficients: reactants decrease and products increase when the reaction shifts forward.

x (in an ICE table)
The unknown amount (in mol/L) by which a substance with coefficient 1 changes as the system moves to equilibrium.

initial concentration
The concentration of each substance when it is mixed, before the system starts moving toward equilibrium.

Measuring just one thing

In the lab, it is often hard to measure every concentration in an equilibrium mixture. But it is usually possible to measure one: the intensity of a colour, the pressure of a gas, or the pH. If you also know what you put in at the start, the mole ratio in the balanced equation lets you find everything else. An ICE table keeps all of this organized, and it will be your main tool for the rest of the unit.

Building an ICE table

Write the balanced equation across the top.
Initial (I): the concentrations when mixed, in mol/L. Convert moles to concentration first: c = n ÷ V.
Change (C): use x for the change. The changes must be in the same ratio as the coefficients. If the reaction moves forward, reactants get −, products get +.
Equilibrium (E): I + C for each column.
Use the measured equilibrium value to find x, then fill in every E entry.
Substitute the E values into the K expression.

Only the C row follows the mole ratio. The I and E rows can be any values.

An ICE table for 1.00 mol of H₂ and 1.00 mol of I₂ in a 2.00 L flask. The change row uses the 1 : 1 : 2 coefficients; each equilibrium entry is initial + change.

Worked example

Example 1: K for the hydrogen iodide equilibrium

1.00 mol of H₂ and 1.00 mol of I₂ are placed in a sealed 2.00 L flask at about 450 °C. At equilibrium the flask contains 1.56 mol of HI. Calculate K for H₂(g) + I₂(g) ⇌ 2 HI(g).

Convert to mol/L: [H₂]ᵢ = [I₂]ᵢ = 1.00 ÷ 2.00 = 0.500 mol/L; [HI]ₑ = 1.56 ÷ 2.00 = 0.780 mol/L.
ICE table: I: 0.500, 0.500, 0. C: −x, −x, +2x. E: 0.500 − x, 0.500 − x, 2x.
From the measurement, 2x = 0.780, so x = 0.390.
[H₂] = [I₂] = 0.500 − 0.390 = 0.110 mol/L.
K = [HI]² ÷ ([H₂][I₂]) = (0.780)² ÷ (0.110)² = 0.6084 ÷ 0.0121 = 50.3.

K = 50.3, the same value found in the last lesson for a 1.00 L flask. K does not depend on the volume or starting amounts.

Initial and equilibrium concentrations from Example 1. The changes are in the 1 : 1 : 2 ratio of the coefficients, even though the final amounts are not.

Worked example

Example 2: A reaction with different coefficients

1.00 mol of N₂ and 3.00 mol of H₂ are sealed in a 5.00 L vessel at a high temperature. At equilibrium, [NH₃] = 0.0800 mol/L. Find K for N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g).

Initial: [N₂] = 1.00 ÷ 5.00 = 0.200 mol/L; [H₂] = 3.00 ÷ 5.00 = 0.600 mol/L; [NH₃] = 0.
Change: −x, −3x, +2x. Equilibrium: 0.200 − x, 0.600 − 3x, 2x.
2x = 0.0800, so x = 0.0400.
[N₂] = 0.200 − 0.0400 = 0.160 mol/L; [H₂] = 0.600 − 3(0.0400) = 0.480 mol/L.
K = [NH₃]² ÷ ([N₂][H₂]³) = (0.0800)² ÷ [(0.160)(0.480)³] = 0.00640 ÷ (0.160 × 0.110592) = 0.00640 ÷ 0.017695 = 0.362.

K = 0.362. Watch the H₂ column: its change is −3x because of the coefficient 3, and it is cubed in K.

Worked example

Example 3: When some product is there at the start

A flask initially contains [PCl₅] = 0.0500 mol/L, [PCl₃] = 0 and [Cl₂] = 0.0200 mol/L. At equilibrium, [PCl₃] = 0.0150 mol/L. Find K for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

I: 0.0500, 0, 0.0200. C: −x, +x, +x. E: 0.0500 − x, x, 0.0200 + x.
x = [PCl₃] = 0.0150.
[PCl₅] = 0.0500 − 0.0150 = 0.0350 mol/L; [Cl₂] = 0.0200 + 0.0150 = 0.0350 mol/L.
K = [PCl₃][Cl₂] ÷ [PCl₅] = (0.0150)(0.0350) ÷ 0.0350 = 0.0150.

K = 0.0150. The Cl₂ that was there at the start stays in the I row; the change row still uses the 1 : 1 : 1 ratio.

Five ICE-table habits

Always work in mol/L, not moles (except when the volume is exactly 1 L, where they are the same number).
Put the coefficients into the C row (−x, −3x, +2x), not into the I row.
Use the E row, never the I row, in the K expression.
Keep unrounded values in your calculator and round only the final K.
Check: does every E value make sense (none negative)?

Just for fun

Canadian chemistry students are the only people who look forward to ICE season in every unit test. (No skates required, just a pencil.)

Hands-on activity

Model an equilibrium with 60 beans

You need: 60 dried beans or buttons, two bowls labelled A and B, paper and a pencil.

Model the reaction A ⇌ B in which, each round, one third of the A particles turn into B, and one sixth of the B particles turn back into A.

Start with all 60 beans in bowl A (round 0: A = 60, B = 0).
Each round, work out both moves from the counts at the start of the round, rounding to the nearest whole bean (round halves up): move A ÷ 3 beans from A to B, and B ÷ 6 beans from B to A. Then do both moves.
Keep going until the counts stop changing. Record each round in a table.
Make an ICE table using bean counts (pretend the bowl volume is 1 L) and calculate K = [B] ÷ [A].

What to look for: the rounds go A 60 → 40 → 30 → 25 → 23 → 21, and the counts then stay at A = 21, B = 39. ICE: I = 60, 0; C = −39, +39; E = 21, 39. K = 39 ÷ 21 ≈ 1.9, close to the exact value of 2 (one third ÷ one sixth). The small difference comes from rounding to whole beans. Then try starting with all 60 beans in B: you should finish near the same counts.

Your turn

What do the letters in an ICE table stand for?

For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), write the change row if the reaction moves forward.

0.200 mol/L of NO₂ is sealed in a flask: 2 NO₂(g) ⇌ N₂O₄(g). At equilibrium [N₂O₄] = 0.0500 mol/L. Find [NO₂] at equilibrium and K.

0.500 mol/L of HI is placed in a flask: 2 HI(g) ⇌ H₂(g) + I₂(g). At equilibrium [H₂] = 0.0550 mol/L. Find K.

2.00 mol of CO and 2.00 mol of H₂O(g) are placed in a 20.0 L flask: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). At equilibrium there are 1.33 mol of CO₂. Find K.

Why must you use the E row, not the I row, in the K expression?

More practice: warm-up

Which row of an ICE table follows the coefficients of the balanced equation?

Convert 0.60 mol in a 3.0 L flask to mol/L.

In A ⇌ 2 B, if A changes by −x, what is the change in B?

If [A]ᵢ = 0.80 mol/L and the change is −0.25 mol/L, what is [A]ₑ?

If a product's initial concentration is 0, what is the sign of its change?

More practice: on your own

0.400 mol/L of SO₂ and 0.200 mol/L of O₂ react: 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g). At equilibrium [SO₃] = 0.300 mol/L. Find K.

In a lab, 2.00 × 10⁻³ mol/L Fe³⁺ and 1.00 × 10⁻³ mol/L SCN⁻ (concentrations after mixing) form FeSCN²⁺: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). A colorimeter shows [FeSCN²⁺] = 2.4 × 10⁻⁴ mol/L at equilibrium. Find K.

4.00 mol of NOCl is placed in a 2.00 L flask: 2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g). At equilibrium, [Cl₂] = 0.200 mol/L. Find K.

A flask starts with [H₂] = 0.100, [I₂] = 0.100 and [HI] = 0.200 mol/L. At equilibrium [HI] = 0.311 mol/L. Find K.

A student forgot to convert moles to mol/L in Example 2 and used 1.00 mol, 3.00 mol and 0.400 mol of NH₃. Would they get the right K? Explain.

Challenge

A 1.00 L flask starts with 0.500 mol of N₂O₄ and 0.100 mol of NO₂: N₂O₄(g) ⇌ 2 NO₂(g). At equilibrium the total moles of gas is 0.650. Find the equilibrium concentrations and K.

In a sealed 2.00 L flask, 0.80 mol of NH₃ partly decomposes: 2 NH₃(g) ⇌ N₂(g) + 3 H₂(g). At equilibrium, 40.0 % of the NH₃ has decomposed. Find K.

For A(g) + B(g) ⇌ C(g), a flask starts with [A] = [B] = 1.00 mol/L. At equilibrium, [A] is 4 times [C]. Find K.

Check your understanding

Quiz

In an ICE table for 2 A ⇌ 3 B, if A changes by −2x, B changes by:

0.50 mol of a gas is placed in a 2.0 L flask. Its initial concentration is:

For A ⇌ B, [A]ᵢ = 1.0 mol/L and [B]ᵢ = 0. At equilibrium [B] = 0.75 mol/L. K is:

Which values go into the K expression?

For H₂ + I₂ ⇌ 2 HI, [H₂]ᵢ = [I₂]ᵢ = 0.30 mol/L and [HI]ₑ = 0.40 mol/L. What is [H₂]ₑ?

For H₂ + I₂ ⇌ 2 HI, the equilibrium concentrations are [H₂] = [I₂] = 0.10 mol/L and [HI] = 0.40 mol/L. K is:

For parents and teachers

Big idea (E2.4, E3.2): an ICE table turns initial amounts plus one equilibrium measurement into all the equilibrium concentrations, and then K. The change row follows the coefficients.

How to help: ask your teen to talk through each row out loud: "What did I start with? How did it change, and in what ratio? What's left?" The bean activity makes the idea concrete and gives a real K to calculate.

Common difficulties:

Using moles instead of mol/L when the volume is not 1 L.
Putting coefficients in the I row instead of the C row, or forgetting them (−x instead of −3x).
Substituting initial values into K.
Rounding x too early. Keep full calculator values until the end.

Looking ahead: next lesson runs the ICE table the other way: given K, find the equilibrium concentrations.