🎯⭐ INTERACTIVE LESSON

ICE Tables and Equilibrium Calculations

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ICE Tables and Equilibrium Calculations - Complete Interactive Lesson

Part 1: Setting Up ICE Tables

🧊 Setting Up ICE Tables

Part 1 of 7 — Initial, Change, Equilibrium

ICE tables are the systematic method for solving equilibrium problems. ICE stands for Initial, Change, Equilibrium — the three rows that track how concentrations evolve from start to finish.

🏗️ The ICE Table Structure

For the reaction: $aA + bB \rightleftharpoons cC + dD$

	$A$	$B$	$C$	$D$
I (Initial)	$[A]_0$

Key Rules

I row: Fill in starting concentrations (often products start at 0)
C row: Use the variable $x$ $x$ with stoichiometric ratios
- Reactants decrease (negative sign)
- Products increase (positive sign)
- Coefficients become multipliers of $x$
E row: I + C for each column
Substitute the E row into the $K$ expression and solve for $x$

Important

The signs in the C row depend on the direction of shift
If the reaction shifts right: reactants lose ( $-$ ), products gain ( $+$ )
If the reaction shifts left: reactants gain ( $+$ ), products lose ( $-$ )

🧪 Worked Example

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ ,

ICE Table Setup 🎯

ICE Table Setup Practice 🧮

For: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$

ICE Table Concepts 🔍

Exit Quiz — ICE Table Setup ✅

Part 2: Solving for x

🧊 Solving for K from Equilibrium Data

Part 2 of 7 — When You Know the Equilibrium Concentrations

The simplest ICE table problem: you're given enough information about the equilibrium state to directly calculate K. No algebra needed — just plug and chug.

⚖️ Method: All Equilibrium Concentrations Given

If you know ALL equilibrium concentrations, just plug them into the K expression.

Example 1

$\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$

Part 3: Small-x Approximation

🧊 Solving for Equilibrium Concentrations Given K

Part 3 of 7 — The Classic ICE Table Problem

This is the most common ICE table scenario: you know K and the initial concentrations, and you need to find the equilibrium concentrations. This requires setting up and solving an algebraic equation.

📌 General Method

Write the balanced equation and $K$ expression
Set up the ICE table with initial concentrations
Determine the direction of shift (usually $Q = 0 < K$ , so shift right)
Express equilibrium concentrations in terms of $x$

Part 4: Quadratic Solutions

🧊 The 5% Approximation

Part 4 of 7 — When x Is Small Enough to Ignore

When K is very small ( $K < 10^{-3}$ ) relative to initial concentrations, the change $x$ is often negligible compared to the initial values. This allows us to simplify the algebra dramatically.

📌 When Can You Use the Approximation?

The Rule of Thumb

If $\frac{[\text{initial}]}{K} > 100$ (or equivalently, ), then is small enough to approximate:

Part 5: ICE Tables with Kp

🧊 When the Approximation Fails — The Quadratic Formula

Part 5 of 7 — Exact Solutions

When K is not small enough for the 5% approximation, or when the 5% test fails, you must solve the full quadratic equation. This part covers the systematic approach.

📌 The Quadratic Formula

For $ax^2 + bx + c = 0$ :

$x = \frac{- b \pm \sqrt{}}{}$

Part 6: Problem-Solving Workshop

🧮 Problem-Solving Workshop: ICE Tables

Part 6 of 7 — Multiple ICE Table Scenarios

This workshop presents varied ICE table problems: finding K, finding equilibrium concentrations, using the approximation, and the quadratic. Practice the decision-making process for each type.

📌 Decision Tree for ICE Table Problems

Step 1: What are you solving for?

K unknown: Use given equilibrium data to find K
Equilibrium concentrations unknown: Use K and initial data to find concentrations

Step 2: Can you use the approximation?

Check: $\frac{[\text{initial}]}{K} > 100$ ?

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — ICE Tables and Equilibrium Calculations

This final part reviews all ICE table techniques: setup, solving for K, solving for concentrations, the 5% approximation, and the quadratic formula. These questions mirror AP Chemistry exam formats.

📋 Complete ICE Table Summary

The ICE Table

	Reactant	Product
I	Initial concentration	Initial concentration (often 0)
C	$-($ coeff $)(x)$	coeff

Initial: $[\text{H}_2] = 1.00$ M, $[\text{I}_2] = 1.00$ M, $[\text{HI}] = 0$ M

Since we start with no products and $K > 0$ , the reaction shifts right.

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

Substitute into K expression:

$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(2x)^2}{(1.00 - x)(1.00 - x)} = \frac{4x^2}{(1.00 - x)^2}$

$50.0 = \frac{4x^2}{(1.00 - x)^2}$

Take the square root of both sides:

$\sqrt{50.0} = \frac{2x}{1.00 - x} \implies 7.07 = \frac{2x}{1.00 - x}$

$7.07(1.00 - x) = 2x \implies 7.07 - 7.07x = 2x$

$7.07 = 9.07x \implies x = 0.780$

Equilibrium concentrations:

$[\text{H}_2] = 1.00 - 0.780 = 0.220$ M
$[\text{I}_2] = 1.00 - 0.780 = 0.220$ M
$[\text{HI}] = 2(0.780) = 1.560$ M

Initial concentrations: $[\text{N}_2\text{O}_4] = 0.50$ M, $[\text{NO}_2] = 0$ M

1) If the change in $[\text{N}_2\text{O}_4]$ is $-x$ , what is the change in $[\text{NO}_2]$ ? (Enter with sign, e.g., "+2x")

2) What is the equilibrium expression for $[\text{N}_2\text{O}_4]$ in terms of x? (Enter, e.g., "0.50 - x")

3) What is the equilibrium expression for $[\text{NO}_2]$ in terms of x? (Enter, e.g., "2x")

At equilibrium: $[\text{N}_2] = 0.50$ , $[\text{H}_2] = 0.30$ , $[\text{NH}_3] = 0.20$ M

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.20)^2}{(0.50)(0.30)^3} = \frac{0.040}{(0.50)(0.027)} = \frac{0.040}{0.0135} = 2.96 \approx 3.0$

⚖️ Method: Initial + One Equilibrium Value Given

When you know initial concentrations and ONE equilibrium concentration, use the ICE table to find x, then calculate all equilibrium concentrations.

Example 2

$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$

Initial: $[\text{PCl}_5] = 1.00$ M, $[\text{PCl}_3] = [\text{Cl}_2] = 0$

At equilibrium: $[\text{PCl}_5] = 0.60$ M

	PCl₅	PCl₃	Cl₂
I	1.00	0	0
C	$-x$	$+x$	$+x$
E	$1.00 - x$

From the equilibrium value: $1.00 - x = 0.60 \implies x = 0.40$

So: $[\text{PCl}_3] = [\text{Cl}_2] = 0.40$ M

$K_c = \frac{(0.40)(0.40)}{0.60} = \frac{0.16}{0.60} = 0.267 \approx 0.27$

Practice Problem 1 🧮

$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$

At equilibrium: $[\text{CO}] = 0.10$ , $[\text{H}_2\text{O}] = 0.10$ , $[\text{CO}_2] = 0.30$ , M

1) Calculate $K_c$ . (Enter as a whole number)

2) Is this reaction product-favored or reactant-favored? (Enter "product-favored" or "reactant-favored")

Practice Problem 2 🧮

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$

Initial: $[\text{N}_2\text{O}_4] = 0.80$ M, $[\text{NO}_2] = 0$ M

At equilibrium: $[\text{NO}_2] = 0.40$ M

1) What is $x$ ? (Enter as a decimal)

2) What is $[\text{N}_2\text{O}_4]$ at equilibrium? (Enter as a decimal)

3) Calculate $K_c$ . (Enter to 3 significant figures)

Solving for K — Concepts 🔍

Exit Quiz — Solving for K ✅

Substitute into the

K

expression

Calculate all equilibrium concentrations

Check: plug values back into K to verify

Perfect-Square Shortcut

When the K expression can be written as a perfect square, take the square root of both sides to avoid the quadratic formula.

This works when: $K = \frac{(\text{something})^2}{(\text{something else})^2}$

🧪 Worked Example: Perfect-Square Case

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ , $K_c = 64.0$

Initial: $[\text{H}_2] = 0.50$ M, $[\text{I}_2] = 0.50$ M,

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

$64.0 = \frac{(2x)^2}{(0.50-x)(0.50-x)} = \frac{4x^2}{(0.50-x)^2}$

Take the square root:

$8.0 = \frac{2x}{0.50-x}$

$8.0(0.50 - x) = 2x \implies 4.0 - 8.0x = 2x \implies 4.0 = 10.0x$

$x = 0.40$

Equilibrium concentrations:

$[\text{H}_2] = [\text{I}_2] = 0.50 - 0.40 = 0.10$ M

Check: $K = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.01} = 64.0$ ✓

🧪 Worked Example: Quadratic Required

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ , $K_c = 0.36$

Initial: $[\text{N}_2\text{O}_4] = 1.00$ M, $[\text{NO}_2] = 0$

	N₂O₄	NO₂
I	1.00	0
C	$-x$	$+2x$
E	$1.00 - x$

$0.36 = \frac{(2x)^2}{1.00 - x} = \frac{4x^2}{1.00 - x}$

$0.36(1.00 - x) = 4x^2$

$0.36 - 0.36x = 4x^2$

$4x^2 + 0.36x - 0.36 = 0$

Using the quadratic formula: $x = \frac{-0.36 \pm \sqrt{(0.36)^2 + 4(4)(0.36)}}{2(4)}$

$x = \frac{-0.36 \pm \sqrt{0.1296 + 5.76}}{8} = \frac{-0.36 \pm \sqrt{5.8896}}{8} = \frac{-0.36 \pm 2.427}{8}$

Taking the positive root: $x = \frac{-0.36 + 2.427}{8} = \frac{2.067}{8} = 0.258$

Equilibrium: $[\text{N}_2\text{O}_4] = 0.742$ M, $[\text{NO}_2] = 0.516$ M

Setting Up the Algebra 🎯

Practice: Solve an ICE Table 🧮

$\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$ , $K_c = 0.25$

Initial: $[\text{A}] = 1.00$ M, $[\text{B}] = [\text{C}] = 0$

	A	B	C
I	1.00	0	0
C	$-x$	$+x$	$+x$
E	$1.00-x$

$0.25 = \frac{x \cdot x}{1.00 - x} = \frac{x^2}{1.00 - x}$

1) Rearrange to standard quadratic form: $x^2 + 0.25x - 0.25 = 0$ . Using the quadratic formula, $x = ?$ (Round to 3 significant figures)

2) What is $[\text{A}]$ at equilibrium? (Round to 3 significant figures)

3) What is $[\text{B}]$ at equilibrium? (Round to 3 significant figures)

Problem-Solving Strategy 🔍

Exit Quiz — Solving for Equilibrium Concentrations ✅

\frac{[ initial ]}{K} > 100

K < 0.01 \times [\text{initial}]

$[\text{initial}] - x \approx [\text{initial}]$

When K is very small, the reaction barely shifts — very little product forms. So $x$ is tiny compared to the initial concentration, and subtracting it doesn't meaningfully change the value.

After solving, check: $\frac{x}{[\text{initial}]} \times 100\% < 5\%$

If the change is less than 5% of the initial concentration, the approximation is valid.

🧪 Worked Example

$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{NO}(g)$ , $K_c = 4.0 \times 10^{-4}$

Initial: $[\text{N}_2] = 0.80$ M, $[\text{O}_2] = 0.20$ M,

Check: $0.20 / (4.0 \times 10^{-4}) = 500 > 100$ ✓ → approximation valid

	N₂	O₂	NO
I	0.80	0.20	0
C	$-x$	$-x$	$+2x$
E

$K_c = \frac{(2x)^2}{(0.80 - x)(0.20 - x)}$

With approximation ( $x \ll 0.20$ ):

$4.0 \times 10^{-4} = \frac{4x^2}{(0.80)(0.20)} = \frac{4x^2}{0.16}$

$4x^2 = (4.0 \times 10^{-4})(0.16) = 6.4 \times 10^{-5}$

$x^2 = 1.6 \times 10^{-5} \implies x = 4.0 \times 10^{-3}$

5% Check: $\frac{x}{0.20} \times 100\% = \frac{4.0 \times 10^{-3}}{0.20} \times 100\% = 2.0\%$ ✓ (< 5%)

Equilibrium:

$[\text{NO}] = 2(4.0 \times 10^{-3}) = 8.0 \times 10^{-3}$ M

The 5% Approximation 🎯

Practice: Using the Approximation 🧮

$\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$ , $K_c = 1.0 \times 10^{-6}$

Initial: $[\text{A}] = 0.50$ M, $[\text{B}] = [\text{C}] = 0$

Using the approximation $0.50 - x \approx 0.50$ :

$1.0 \times 10^{-6} = \frac{x^2}{0.50}$

1) Solve for $x$ . (Enter in scientific notation, e.g. 7.1e-4)

2) What percent of the initial [A] is $x$ ? (Enter as a percentage to 3 significant figures, e.g. 0.14)

3) Is the approximation valid? (Enter "yes" or "no")

Approximation Guidelines 🔍

Exit Quiz — 5% Approximation ✅

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

In Equilibrium Problems

You'll always get two solutions
Only the positive root that gives physically meaningful (non-negative) concentrations is valid
Reject any root that gives a negative concentration

When You Need the Quadratic

The approximation fails when:

$K$ is not very small relative to initial concentrations
$\frac{[\text{initial}]}{K} < 100$
The 5% check yields > 5%

🧪 Worked Example

$\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$ , $K_c = 0.50$

Initial: $[\text{A}] = 1.00$ M, $[\text{B}] = [\text{C}] = 0$

Check approximation: $1.00/0.50 = 2.0 < 100$ → approximation NOT valid

	A	B	C
I	1.00	0	0
C	$-x$	$+x$	$+x$
E	$1.00 - x$

$0.50 = \frac{x^2}{1.00 - x}$

$0.50(1.00 - x) = x^2$

$0.50 - 0.50x = x^2$

$x^2 + 0.50x - 0.50 = 0$

Applying the quadratic formula ( $a = 1, b = 0.50, c = -0.50$ ):

$x = \frac{-0.50 \pm \sqrt{(0.50)^2 - 4(1)(-0.50)}}{2(1)} = \frac{-0.50 \pm \sqrt{0.25 + 2.00}}{2}$

$x = \frac{-0.50 \pm \sqrt{2.25}}{2} = \frac{-0.50 \pm 1.50}{2}$

Two roots:

$x = \frac{-0.50 + 1.50}{2} = \frac{1.00}{2} = 0.50$ ✓

$x = 0.50$

Equilibrium: $[\text{A}] = 0.50$ M, $[\text{B}] = [\text{C}] = 0.50$ M

Check: $K = (0.50)(0.50)/0.50 = 0.50$ ✓

Note: If we had used the approximation: $x = \sqrt{0.50} = 0.71$ . The 5% check: $0.71/1.00 = 71\%$ → fails badly!

Quadratic Approach 🎯

Practice: Full Quadratic 🧮

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ , $K_c = 0.36$

Initial: $[\text{N}_2\text{O}_4] = 0.50$ M, $[\text{NO}_2] = 0$

The equation is: $4x^2 + 0.36x - 0.18 = 0$

Using the quadratic formula:

1) What is the discriminant $b^2 - 4ac$ ? (Enter to 3 significant figures)

2) What is $x$ ? (Round to 3 significant figures)

3) What is $[\text{NO}_2]$ at equilibrium? (Round to 3 significant figures)

Quadratic vs Approximation 🔍

Exit Quiz — Quadratic Formula ✅

Yes → Approximate: $[\text{initial}] - x \approx [\text{initial}]$
No → Full quadratic required

Step 3: Solve and verify

Solve for $x$
Calculate all equilibrium concentrations
Verify: plug back into K expression
If approximation used: check 5% test

Pattern	Example
Perfect square	$\frac{(2x)^2}{(a-x)^2}$ → take square root
Small K with approx	$K = \frac{x^2}{a}$ → $x = \sqrt{Ka}$
Full quadratic	$ax^2 + bx + c = 0$ → quadratic formula

Problem 1: Finding K 🧮

$2\,\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)$

Initial: $[\text{HI}] = 1.00$ M, $[\text{H}_2] = [\text{I}_2] = 0$

At equilibrium: $[\text{HI}] = 0.80$ M

1) What is $x$ ? (Remember: the coefficient of HI is 2)

2) What is $[\text{H}_2]$ at equilibrium?

3) What is $K_c$ ? (Enter to 3 significant figures)

Problem 2: Using the Approximation 🧮

$\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)$ , $K_c = 2.2 \times 10^{-10}$

Initial: $[\text{COCl}_2] = 0.50$ M, $[\text{CO}] = [\text{Cl}_2] = 0$

Using the approximation $0.50 - x \approx 0.50$ :

1) Solve: $x = \sqrt{K_c \times 0.50}$ . What is ? (Enter in scientific notation, e.g. 1.0e-5)

2) Does the 5% test pass? (Enter "yes" or "no")

3) What is $[\text{CO}]$ at equilibrium? (Enter in scientific notation, same as x)

Round all answers to 3 significant figures.

Problem 3: Which Method? 🎯

Problem 4: Non-Zero Initial Products 🧮

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$ , $K_c = 64$

Initial: $[\text{H}_2] = 0.50$ , $[\text{I}_2] = 0.50$ , M

First check: $Q = \frac{(0.20)^2}{(0.50)(0.50)} = \frac{0.04}{0.25} = 0.16$ . Since , shift right.

1) Using the ICE table with shift right, what is $[\text{HI}]$ at equilibrium expressed in terms of $x$ ? (Enter, e.g., "0.20 + 2x")

2) This is a perfect-square case. Taking the square root: $8 = \frac{0.20 + 2x}{0.50 - x}$ . Solve for x. (Round to 3 significant figures)

3) What is $[\text{HI}]$ at equilibrium? (Round to 3 significant figures)

Exit Quiz — ICE Table Workshop ✅

Given	Find	Method
All equilibrium conc.	K	Plug directly into K expression
Initial + one eq. conc.	K	Find x from ICE, then all eq. conc., then K
K + initial conc.	Eq. conc.	Full ICE → solve for x

Condition	Strategy
Perfect square	Take square root
$[\text{init}]/K > 100$	Small-x approximation
$[\text{init}]/K < 100$	Full quadratic
Approx gives > 5%	Switch to quadratic

Always check: Plug equilibrium concentrations back into K expression. The calculated K should match the given K.

AP-Style Multiple Choice — Set 1 🎯

AP Free-Response Style 🧮

$\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g)$ , $K_c = 255$ at 100°C

A 1.00 L flask is charged with 0.400 mol CO and 0.400 mol Cl₂. No COCl₂ is initially present.

1) Write the K expression and set up the ICE table. What is $[\text{COCl}_2]$ at equilibrium in terms of x? (Enter, e.g., "x")

2) The K expression becomes $255 = \frac{x}{(0.400-x)^2}$ . Using the approximation ( $0.400/255$ is small... actually ). Should you use the quadratic? (Enter "yes" or "no")

3) Actually, $[\text{init}]/K = 0.400/255 = 0.00157$ , which is much LESS than 100. This means K is LARGE relative to the initial concentration, meaning the reaction goes nearly to completion. The limiting approach here is to assume the reaction goes to completion, then back-calculate. If the reaction goes to completion, what is the limiting reagent amount of COCl₂ formed? (Enter in mol)

Round all answers to 3 significant figures.

Final Concept Review 🔍

Final Exit Quiz ✅

[\text{H}_2] = 0.30

[\text{HI}] = 2(0.40) = 0.80

[\text{N}_2] \approx 0.80

[\text{O}_2] \approx 0.20

x = \frac{-0.50 - 1.50}{2} = \frac{-2.00}{2} = -1.00

✗ (negative)

[\text{HI}] = 0.20

0.400/255 = 0.00157 < 100

ICE Tables and Equilibrium Calculations

Substitute into K expression:

⚖️ Method: Initial + One Equilibrium Value Given

Example 2

Perfect-Square Shortcut

🧪 Worked Example: Perfect-Square Case

🧪 Worked Example: Quadratic Required

Why This Works

The 5% Test

🧪 Worked Example

In Equilibrium Problems

When You Need the Quadratic

🧪 Worked Example

Step 3: Solve and verify

Common Patterns

Problem Types

Solving Strategies

Verification