ICE = Initial, Change, Equilibrium

Organized method to track concentrations:

Key points:

• Reactants decrease (negative change)
• Products increase (positive change)
• Changes related by stoichiometry
• Use variable x for unknown change

Setting Up ICE Table

Steps:

1. Write balanced equation with ⇌
2. Initial row: Given concentrations (often some are 0)
3. Change row: Use coefficients
   • Reactants: -coefficient × x
   • Products: +coefficient × x
4. Equilibrium row: I + C for each species
5. Write K expression using E row
6. Solve for x

Stoichiometric Relationships

For reaction: aA + bB ⇌ cC + dD

If A changes by amount ax:

• B changes by bx (same x, different coefficient)
• C changes by +cx
• D changes by +dx

Ratio of changes = ratio of coefficients

Example: N₂ + 3H₂ ⇌ 2NH₃

If N₂ changes by -x:

• H₂ changes by -3x (1:3 ratio)
• NH₃ changes by +2x

Types of Equilibrium Problems

Type 1: Calculate K from equilibrium concentrations

Given: All equilibrium concentrations Find: K

Method:

• Plug directly into K expression
• No ICE table needed (already at equilibrium)

Type 2: Calculate equilibrium concentrations from K

Given: Initial concentrations and K Find: Equilibrium concentrations

Method:

1. Set up ICE table
2. Write K expression
3. Solve for x
4. Find equilibrium concentrations

Type 3: Two initial concentrations given

Given: Some reactants, some products initially Find: Equilibrium concentrations

Method:

1. Calculate Q to find direction
2. Set up ICE table
3. Solve for x

Small x Approximation

When K is very small (< 10⁻³):

If [initial] >> K:

• Assume x is negligible
• Simplify: [initial] - x ≈ [initial]
• Easier algebra

Check validity:

• Calculate x
• If x/[initial] < 5%, approximation valid
• If x/[initial] > 5%, must use quadratic

Example:

K = 1.0 × 10⁻⁵, [initial] = 0.10 M

With approximation: 0.10 - x ≈ 0.10

Check: x/0.10 < 5%? If yes, good!

Quadratic Equation

When approximation invalid:

Standard form: ax² + bx + c = 0

Quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Use (+) or (-) based on physical meaning:

• Concentrations must be positive
• Changes must make sense

Example ICE Table Setup

Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)

Given: [H₂]₀ = 0.50 M, [I₂]₀ = 0.50 M, [HI]₀ = 0

ICE Table:

K expression:

$K = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.50-x)(0.50-x)}$

Problem-Solving Strategy

Step 1: Write balanced equation

Step 2: Organize data

• List all initial concentrations
• Identify what you're finding

Step 3: Set up ICE table

• Use stoichiometry for changes
• Express equilibrium in terms of x

Step 4: Write K expression

• Use equilibrium row

Step 5: Solve for x

• Try small x approximation if K small
• Use quadratic if needed
• Take physically meaningful root

Step 6: Calculate final answer

• Substitute x back
• Check: concentrations positive?
• Verify with K expression

Common Mistakes to Avoid

❌ Wrong stoichiometry in change row ✓ Use coefficients from balanced equation

❌ Forgetting to square/cube in K expression ✓ Exponents = coefficients

❌ Using initial concentrations in K ✓ Use equilibrium (E row)

❌ Taking wrong quadratic root ✓ Concentrations must be positive

❌ Invalid small x approximation ✓ Check x < 5% of initial

Note stoichiometry:

• NO has coefficient 2 → change is -2x
• Br₂ has coefficient 1 → change is -x
• NOBr has coefficient 2 → change is +2x

Write K expression:

$K_c = \frac{[NOBr]^2}{[NO]^2[Br_2]} = \frac{(2x)^2}{(0.020-2x)^2(0.025-x)}$

$1.3 \times 10^4 = \frac{4x^2}{(0.020-2x)^2(0.025-x)}$

Large K analysis:

K = 1.3 × 10⁴ is VERY large → reaction goes nearly to completion

Limiting reactant:

• NO: 0.020 M ÷ 2 = 0.010 M available
• Br₂: 0.025 M available
• NO is limiting

Assume reaction goes to completion:

• All NO consumed
• 0.020 M NO → 0.010 M Br₂ consumed
• 0.020 M NOBr formed

Then small amount comes back:

With small y:

$1.3 \times 10^4 = \frac{(0.020-2y)^2}{(2y)^2(0.015+y)}$

Assume y << 0.015:

$1.3 \times 10^4 \approx \frac{(0.020)^2}{(2y)^2(0.015)}$

$1.3 \times 10^4 = \frac{4.0 \times 10^{-4}}{4y^2(0.015)}$

$1.3 \times 10^4 = \frac{4.0 \times 10^{-4}}{0.060y^2}$

$y^2 = \frac{4.0 \times 10^{-4}}{(1.3 \times 10^4)(0.060)}$

$y^2 = 5.13 \times 10^{-7}$

$y = 7.2 \times 10^{-4}$

Equilibrium concentrations:

[NO]: 2y = 2(7.2 × 10⁻⁴) = 1.4 × 10⁻³ M

[Br₂]: 0.015 + y ≈ 0.015 M = 0.015 M

[NOBr]: 0.020 - 2y = 0.020 - 0.0014 = 0.019 M

Answers:

• [NO] = 1.4 × 10⁻³ M
• [Br₂] = 0.015 M
• [NOBr] = 0.019 M

Write K expression:

$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.050-x}$

$4.6 \times 10^{-3} = \frac{4x^2}{0.050-x}$

Try small x approximation:

Since K is small (4.6 × 10⁻³):

• Little dissociation occurs
• Try: 0.050 - x ≈ 0.050

$4.6 \times 10^{-3} = \frac{4x^2}{0.050}$

$4x^2 = (4.6 \times 10^{-3})(0.050)$

$4x^2 = 2.3 \times 10^{-4}$

$x^2 = 5.75 \times 10^{-5}$

$x = 7.58 \times 10^{-3}$

Check approximation:

$\frac{x}{[N_2O_4]_0} = \frac{7.58 \times 10^{-3}}{0.050} = 0.152 = 15.2\%$

15.2% > 5% → approximation INVALID!

Must use quadratic equation.

Quadratic solution:

$4.6 \times 10^{-3} = \frac{4x^2}{0.050-x}$

$(4.6 \times 10^{-3})(0.050-x) = 4x^2$

$2.3 \times 10^{-4} - 4.6 \times 10^{-3}x = 4x^2$

$4x^2 + 4.6 \times 10^{-3}x - 2.3 \times 10^{-4} = 0$

Quadratic formula:

• a = 4
• b = 4.6 × 10⁻³
• c = -2.3 × 10⁻⁴

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-4.6 \times 10^{-3} \pm \sqrt{(4.6 \times 10^{-3})^2 - 4(4)(-2.3 \times 10^{-4})}}{2(4)}$

$x = \frac{-4.6 \times 10^{-3} \pm \sqrt{2.1 \times 10^{-5} + 3.68 \times 10^{-3}}}{8}$

$x = \frac{-4.6 \times 10^{-3} \pm \sqrt{3.70 \times 10^{-3}}}{8}$

$x = \frac{-4.6 \times 10^{-3} \pm 0.0608}{8}$

Take positive root:

$x = \frac{-0.0046 + 0.0608}{8} = \frac{0.0562}{8} = 7.03 \times 10^{-3}$

Calculate equilibrium concentrations:

[N₂O₄]: 0.050 - x = 0.050 - 0.00703 = 0.043 M

[NO₂]: 2x = 2(0.00703) = 0.014 M

Verify:

$K_c = \frac{(0.014)^2}{0.043} = \frac{1.96 \times 10^{-4}}{0.043} = 4.6 \times 10^{-3}$ ✓

Answers:

• [N₂O₄] = 0.043 M
• [NO₂] = 0.014 M