ICE Tables

Overview

ICE tables (Initial-Change-Equilibrium) are a systematic method for solving equilibrium problems. They organize information about concentration changes as a system reaches equilibrium.

ICE Table Structure

	Reactants	Products
Initial	Starting amounts	Starting amounts
Change	How much changes	How much changes
Equilibrium	Final amounts	Final amounts

Setting Up ICE Tables

Step 1: Write the balanced equation

Step 2: Fill in Initial concentrations

Step 3: Express Changes using variable x

• Reactants: $-\text{coefficient} \times x$
• Products: $+\text{coefficient} \times x$

Step 4: Write Equilibrium expressions

• Equilibrium = Initial + Change

Step 5: Substitute into K expression and solve

Example 1: Simple Equilibrium

Problem: For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, $K_c = 50.5$ at 448°C Initial: $[\text{H}_2] = [\text{I}_2] = 0.100$ M, $[\text{HI}] = 0$ Find equilibrium concentrations.

ICE Table:

	H₂	I₂	2HI
I	0.100	0.100	0
C	$-x$	$-x$	$+2x$
E	$0.100-x$	$0.100-x$	$2x$

Solve:

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$ $$50.5 = \frac{(2x)^2}{(0.100-x)(0.100-x)} = \frac{4x^2}{(0.100-x)^2}$$

Take square root:

$$\sqrt{50.5} = \frac{2x}{0.100-x}$$ $$7.11 = \frac{2x}{0.100-x}$$ $$7.11(0.100-x) = 2x$$ $$0.711 - 7.11x = 2x$$ $$x = 0.078 \text{ M}$$ $$[\text{H}_2] = [\text{I}_2] = 0.100 - 0.078 = 0.022 \text{ M}$$ $$[\text{HI}] = 2(0.078) = 0.156 \text{ M}$$

Example 2: Starting with Products

Problem: For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $K_c = 4.63 \times 10^{-3}$ Initial: $[\text{N}_2\text{O}_4] = 0.100$ M Find equilibrium concentrations.

ICE Table:

	N₂O₄	2NO₂
I	0.100	0
C	$-x$	$+2x$
E	$0.100-x$	$2x$

Solve:

$$K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(2x)^2}{0.100-x} = \frac{4x^2}{0.100-x}$$ $$4.63 \times 10^{-3} = \frac{4x^2}{0.100-x}$$

Using quadratic formula: $x = 0.00977$ M

$$[\text{N}_2\text{O}_4] = 0.100 - 0.00977 = 0.090 \text{ M}$$ $$[\text{NO}_2] = 2(0.00977) = 0.0195 \text{ M}$$

Example 3: Using the 5% Approximation

When K is very small, $x \ll$ initial concentration.

Problem: For $\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)$, $K_a = 1.8 \times 10^{-5}$ Initial: $[\text{HA}] = 0.100$ M

ICE Table:

	HA	H⁺	A⁻
I	0.100	0	0
C	$-x$	$+x$	$+x$
E	$0.100-x$	$x$	$x$

Solve with approximation:

$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{0.100-x}$$

Since $K_a$ is small, assume $x \ll 0.100$:

$$1.8 \times 10^{-5} \approx \frac{x^2}{0.100}$$ $$x^2 = 1.8 \times 10^{-6}$$ $$x = 1.34 \times 10^{-3} \text{ M}$$

Check approximation:

$$\frac{1.34 \times 10^{-3}}{0.100} \times 100\% = 1.34\%$$

$1.34\% < 5\%$, so approximation is valid!

When to Use Approximation

Condition	Use Approximation?
$[\text{Initial}]/K > 100$	Yes
$[\text{Initial}]/K < 100$	No, use quadratic
$x/[\text{Initial}] < 5\%$	Approximation valid
$x/[\text{Initial}] > 5\%$	Redo without approximation

Example 4: ICE with Reaction Quotient

Problem: For $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$, $K_c = 5.0$ Initial: All species = 0.10 M Find equilibrium concentrations.

First, find Q:

$$Q = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(0.10)(0.10)}{(0.10)(0.10)} = 1.0$$

$Q < K$, so reaction shifts RIGHT

ICE Table:

	CO	H₂O	CO₂	H₂
I	0.10	0.10	0.10	0.10
C	$-x$	$-x$	$+x$	$+x$
E	$0.10-x$	$0.10-x$	$0.10+x$	$0.10+x$

Solve:

$$5.0 = \frac{(0.10+x)^2}{(0.10-x)^2}$$ $$\sqrt{5.0} = \frac{0.10+x}{0.10-x}$$ $$2.24(0.10-x) = 0.10+x$$ $$0.224 - 2.24x = 0.10 + x$$ $$0.124 = 3.24x$$ $$x = 0.038 \text{ M}$$ $$[\text{CO}] = [\text{H}_2\text{O}] = 0.062 \text{ M}$$ $$[\text{CO}_2] = [\text{H}_2] = 0.138 \text{ M}$$

Common Mistakes to Avoid

1. Wrong signs for change: Reactants decrease ($-$), Products increase ($+$)
2. Forgetting coefficients: Change = coefficient × $x$
3. Invalid approximation: Always check if $x < 5\%$ of initial
4. Wrong direction: Check Q vs K first if not starting from scratch
5. Negative concentrations: $x$ cannot exceed initial reactant concentration

Summary Checklist

1. ☐ Write balanced equation
2. ☐ Set up ICE table
3. ☐ Define $x$ based on stoichiometry
4. ☐ Write equilibrium expressions
5. ☐ Substitute into K expression
6. ☐ Try approximation or use quadratic
7. ☐ Check approximation validity
8. ☐ Calculate final concentrations
9. ☐ Verify by plugging back into K