ICE Tables and Equilibrium Calculations

Master ICE tables to solve equilibrium problems and calculate equilibrium concentrations from initial conditions.

ICE Tables and Equilibrium Calculations

What is an ICE Table?

ICE = Initial, Change, Equilibrium

Organized method to track concentrations:

| | A | + | B | ⇌ | C | + | D | |-|---|---|---|---|---|---|---| | Initial | [A]₀ | | [B]₀ | | [C]₀ | | [D]₀ | | Change | -ax | | -bx | | +cx | | +dx | | Equilibrium | [A]₀-ax | | [B]₀-bx | | [C]₀+cx | | [D]₀+dx |

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=−b±b2−4ac2ax = \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:

| | H₂ | + | I₂ | ⇌ | 2HI | |-|-------|---|-------|---|---------| | I | 0.50 | | 0.50 | | 0 | | C | -x | | -x | | +2x | | E | 0.50-x | | 0.50-x | | 2x |

K expression:

K=[HI]2[H2][I2]=(2x)2(0.50−x)(0.50−x)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

📚 Practice Problems

1Problem 1easy

❓ Question:

For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50.0 at 500 K. If 1.00 mol H₂ and 1.00 mol I₂ are placed in a 1.00 L container, calculate equilibrium concentrations.

💡 Show Solution

Given:

  • Reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
  • K_c = 50.0
  • Initial: 1.00 mol each in 1.00 L → [H₂]₀ = [I₂]₀ = 1.00 M
  • [HI]₀ = 0

Set up ICE table:

| | H₂ | I₂ | 2HI | |-|---------|---------|---------| | I | 1.00 | 1.00 | 0 | | C | -x | -x | +2x | | E | 1.00-x | 1.00-x | 2x |

Write K expression:

Kc=[HI]2[H2][I2]=(2x)2(1.00−x)(1.00−x)K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(1.00-x)(1.00-x)}

50.0=4x2(1.00−x)250.0 = \frac{4x^2}{(1.00-x)^2}

Solve for x:

Take square root of both sides:

50.0=2x1.00−x\sqrt{50.0} = \frac{2x}{1.00-x}

7.07=2x1.00−x7.07 = \frac{2x}{1.00-x}

Cross multiply:

7.07(1.00−x)=2x7.07(1.00-x) = 2x

7.07−7.07x=2x7.07 - 7.07x = 2x

7.07=2x+7.07x7.07 = 2x + 7.07x

7.07=9.07x7.07 = 9.07x

x=7.079.07=0.780x = \frac{7.07}{9.07} = 0.780

Calculate equilibrium concentrations:

[H₂]: 1.00 - x = 1.00 - 0.780 = 0.22 M

[I₂]: 1.00 - x = 1.00 - 0.780 = 0.22 M

[HI]: 2x = 2(0.780) = 1.56 M

Check answer:

Kc=(1.56)2(0.22)(0.22)=2.430.048=50.6≈50.0K_c = \frac{(1.56)^2}{(0.22)(0.22)} = \frac{2.43}{0.048} = 50.6 \approx 50.0 ✓

Answers:

  • [H₂] = 0.22 M
  • [I₂] = 0.22 M
  • [HI] = 1.56 M

2Problem 2medium

❓ Question:

For the reaction: 2NO(g) + Br₂(g) ⇌ 2NOBr(g), K_c = 1.3 × 10⁴ at 300 K. If [NO]₀ = 0.020 M and [Br₂]₀ = 0.025 M, find equilibrium concentrations.

💡 Show Solution

Given:

  • Reaction: 2NO(g) + Br₂(g) ⇌ 2NOBr(g)
  • K_c = 1.3 × 10⁴ (very large!)
  • [NO]₀ = 0.020 M
  • [Br₂]₀ = 0.025 M
  • [NOBr]₀ = 0

Set up ICE table:

| | 2NO | Br₂ | 2NOBr | |-|-----------|-----------|-----------| | I | 0.020 | 0.025 | 0 | | C | -2x | -x | +2x | | E | 0.020-2x | 0.025-x | 2x |

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:

Kc=[NOBr]2[NO]2[Br2]=(2x)2(0.020−2x)2(0.025−x)K_c = \frac{[NOBr]^2}{[NO]^2[Br_2]} = \frac{(2x)^2}{(0.020-2x)^2(0.025-x)}

1.3×104=4x2(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:

| | 2NO | Br₂ | 2NOBr | |-|------|------------|---------| | After completion | 0 | 0.015 | 0.020 | | C | +2y | +y | -2y | | E | 2y | 0.015+y | 0.020-2y |

With small y:

1.3×104=(0.020−2y)2(2y)2(0.015+y)1.3 \times 10^4 = \frac{(0.020-2y)^2}{(2y)^2(0.015+y)}

Assume y << 0.015:

1.3×104≈(0.020)2(2y)2(0.015)1.3 \times 10^4 \approx \frac{(0.020)^2}{(2y)^2(0.015)}

1.3×104=4.0×10−44y2(0.015)1.3 \times 10^4 = \frac{4.0 \times 10^{-4}}{4y^2(0.015)}

1.3×104=4.0×10−40.060y21.3 \times 10^4 = \frac{4.0 \times 10^{-4}}{0.060y^2}

y2=4.0×10−4(1.3×104)(0.060)y^2 = \frac{4.0 \times 10^{-4}}{(1.3 \times 10^4)(0.060)}

y2=5.13×10−7y^2 = 5.13 \times 10^{-7}

y=7.2×10−4y = 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

3Problem 3hard

❓ Question:

For N₂O₄(g) ⇌ 2NO₂(g), K_c = 4.6 × 10⁻³ at 298 K. Initially, [N₂O₄] = 0.050 M and [NO₂] = 0. Calculate equilibrium concentrations.

💡 Show Solution

Given:

  • Reaction: N₂O₄(g) ⇌ 2NO₂(g)
  • K_c = 4.6 × 10⁻³ (small K)
  • [N₂O₄]₀ = 0.050 M
  • [NO₂]₀ = 0

Set up ICE table:

| | N₂O₄ | 2NO₂ | |-|-----------|---------| | I | 0.050 | 0 | | C | -x | +2x | | E | 0.050-x | 2x |

Write K expression:

Kc=[NO2]2[N2O4]=(2x)20.050−xK_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.050-x}

4.6×10−3=4x20.050−x4.6 \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

4.6×10−3=4x20.0504.6 \times 10^{-3} = \frac{4x^2}{0.050}

4x2=(4.6×10−3)(0.050)4x^2 = (4.6 \times 10^{-3})(0.050)

4x2=2.3×10−44x^2 = 2.3 \times 10^{-4}

x2=5.75×10−5x^2 = 5.75 \times 10^{-5}

x=7.58×10−3x = 7.58 \times 10^{-3}

Check approximation:

x[N2O4]0=7.58×10−30.050=0.152=15.2%\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:

4.6×10−3=4x20.050−x4.6 \times 10^{-3} = \frac{4x^2}{0.050-x}

(4.6×10−3)(0.050−x)=4x2(4.6 \times 10^{-3})(0.050-x) = 4x^2

2.3×10−4−4.6×10−3x=4x22.3 \times 10^{-4} - 4.6 \times 10^{-3}x = 4x^2

4x2+4.6×10−3x−2.3×10−4=04x^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=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=−4.6×10−3±(4.6×10−3)2−4(4)(−2.3×10−4)2(4)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=−4.6×10−3±2.1×10−5+3.68×10−38x = \frac{-4.6 \times 10^{-3} \pm \sqrt{2.1 \times 10^{-5} + 3.68 \times 10^{-3}}}{8}

x=−4.6×10−3±3.70×10−38x = \frac{-4.6 \times 10^{-3} \pm \sqrt{3.70 \times 10^{-3}}}{8}

x=−4.6×10−3±0.06088x = \frac{-4.6 \times 10^{-3} \pm 0.0608}{8}

Take positive root:

x=−0.0046+0.06088=0.05628=7.03×10−3x = \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:

Kc=(0.014)20.043=1.96×10−40.043=4.6×10−3K_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
🎴

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