ICE Tables and Equilibrium Calculations - Complete Interactive Lesson
Part 1: Setting Up ICE Tables
๐ง Setting Up ICE Tables
Part 1 of 7 โ Initial, Change, Equilibrium
ICE tables are the systematic method for solving equilibrium problems. ICE stands for Initial, Change, Equilibrium โ the three rows that track how concentrations evolve from start to finish.
๐๏ธ The ICE Table Structure
For the reaction: aA+bBโcC+dD
A
B
C
D
I (Initial)
Key Rules
I row: Fill in starting concentrations (often products start at 0)
C row: Use the variable 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 (โ), products gain (+)
If the reaction shifts left: reactants gain (+), products lose (โ)
๐งช Worked Example
H2โ(g)+I2โ(g)โ,
ICE Table Setup ๐ฏ
ICE Table Setup Practice ๐งฎ
For: N2โO4โ(g)โ2NO
ICE Table Concepts ๐
Exit Quiz โ ICE Table Setup โ
Part 2: Solving for x
๐ง Solving for K from Equilibrium Data
Part 2 of 7 โ 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 โ just plug and chug.
โ๏ธ Method: All Equilibrium Concentrations Given
If you know ALL equilibrium concentrations, just plug them into the K expression.
Example 1
N2โ(g)+3
Part 3: Small-x Approximation
๐ง Solving for Equilibrium Concentrations Given K
Part 3 of 7 โ 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.
๐ 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
๐ง The 5% Approximation
Part 4 of 7 โ When x Is Small Enough to Ignore
When K is very small (K<10โ3) relative to initial concentrations, the change x is often negligible compared to the initial values. This allows us to simplify the algebra dramatically.
๐ When Can You Use the Approximation?
The Rule of Thumb
If (or equivalently, ), then is small enough to approximate:
Part 5: ICE Tables with Kp
๐ง When the Approximation Fails โ The Quadratic Formula
Part 5 of 7 โ 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.
๐ The Quadratic Formula
For ax2+bx+c=0:
Part 6: Problem-Solving Workshop
๐งฎ Problem-Solving Workshop: ICE Tables
Part 6 of 7 โ 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.
๐ 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: K[initial]โ?
Part 7: Synthesis & AP Review
๐ Synthesis & AP Review
Part 7 of 7 โ 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.
๐ Complete ICE Table Summary
The ICE Table
Reactant
Product
I
Initial concentration
Initial concentration (often 0)
C
โ(coeff)(x)
coeff
[
A
]0โ
[B]0โ
[C]0โ
[D]0โ
C (Change)
โax
โbx
+cx
+dx
E (Equilibrium)
[A]0โโax
[B]0โโbx
[C]0โ+cx
[D]0โ+dx
2
HI
(
g
)
Kcโ=50.0
Initial:[H2โ]=1.00 M, [I2โ]=1.00 M, [HI]=0 M
Since we start with no products and K>0, the reaction shifts right.
โ๏ธ 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
PCl5โ(g)โPCl3โ(g)+Cl2โ(g)
Initial: [PCl5โ]=1.00 M, [PCl3โ]
At equilibrium: [PCl5โ]=0.60 M
PClโ
PClโ
Clโ
I
1.00
0
0
C
โx
+x
+x
E
From the equilibrium value: 1.00โx=0.60โนx=0.40
So: [PCl3โ]=[Cl2โ]=0.40 M
Kcโ=0.60(0.40)(0.40)โ
Finding K ๐ฏ
Practice Problem 1 ๐งฎ
CO(g)+H2โO(g)โCO2โ(g)+H2โ(g)
At equilibrium: [CO]=0.10, [H2โO]=0.10, , M
1) Calculate Kcโ. (Enter as a whole number)
2) Is this reaction product-favored or reactant-favored? (Enter "product-favored" or "reactant-favored")
Practice Problem 2 ๐งฎ
N2โO4โ(g)โ2NO2โ(g)
Initial: [N2โO4โ]=0.80 M, [NO M
At equilibrium: [NO2โ]=0.40 M
1) What is x? (Enter as a decimal)
2) What is [N2โO4โ] at equilibrium? (Enter as a decimal)
3) Calculate Kcโ. (Enter to 3 significant figures)
Solving for K โ Concepts ๐
Exit Quiz โ Solving for K โ
Substitute into the K expression
Solve for x
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=(somethingย else)2(something)2โ
๐งช Worked Example: Perfect-Square Case
H2โ(g)+I2โ(g)โ2HI(g), Kcโ=64.0
Initial: [H2โ]=0.50 M, [I2โ]= M,
Hโ
Iโ
HI
I
0.50
0.50
0
C
โx
โx
+2x
E
64.0=(0.50โx)(0.50โx)(2x)
Take the square root:
8.0=0.50โx2xโ
8.0(0.50โx)=2xโน4.0โ8.0x=2xโน
x=0.40
Equilibrium concentrations:
[H2โ]=[I2โ]=0.50โ0.40 M
Check:K=(0.10)(0.10)(0.80)2โ โ
๐งช Worked Example: Quadratic Required
N2โO4โ(g)โ2NO2โ(g), Kcโ=0.36
Initial: [N2โO4โ]=1.00 M, [NO
NโOโ
NOโ
I
1.00
0
C
โx
+2x
E
1.00โx
0.36=1.00โx(2x)2โ=
0.36(1.00โx)=4x2
0.36โ0.36x=4x2
4x2+0.36xโ0.36=0
Using the quadratic formula: x=2(4)โ0.36ยฑ(0.36)
x=8โ0.36ยฑ0.1296+5.76
Taking the positive root: x=8โ0.36+2.427โ=8
Equilibrium:[N2โO4โ]=0.742 M, [NO M
Setting Up the Algebra ๐ฏ
Practice: Solve an ICE Table ๐งฎ
A(g)โB(g)+C(g), Kcโ=0.25
Initial: [A]=1.00 M, [B]=[C]=0
A
B
C
I
1.00
0
0
C
โx
+x
+x
E
0.25=1.00โxxโ xโ=1.00
1) Rearrange to standard quadratic form: x2+0.25xโ0.25=0. Using the quadratic formula, x=? (Round to 3 significant figures)
2) What is [A] at equilibrium? (Round to 3 significant figures)
3) What is [B] at equilibrium? (Round to 3 significant figures)
Problem-Solving Strategy ๐
Exit Quiz โ Solving for Equilibrium Concentrations โ
K[initial]โ>100
K<0.01ร[initial]
x
[initial]โxโ[initial]
Why This Works
When K is very small, the reaction barely shifts โ very little product forms. So x is tiny compared to the initial concentration, and subtracting it doesn't meaningfully change the value.
The 5% Test
After solving, check: [initial]xโร100%<5%
If the change is less than 5% of the initial concentration, the approximation is valid.
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 ๐
Exit Quiz โ 5% Approximation โ
x=2aโbยฑb2โ4acโโ
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
K[initial]โ<100
The 5% check yields > 5%
๐งช Worked Example
A(g)โB(g)+C(g), Kcโ=0.50
Initial: [A]=1.00 M, [B]=[C]=0
Check approximation:1.00/0.50=2.0<100 โ approximation NOT valid
A
B
C
I
1.00
0
0
C
โx
+x
+x
E
0.50=1.00โxx2โ
0.50(1.00โx)=x2
0.50โ0.50x=x2
x2+0.50xโ0.50=0
Applying the quadratic formula (a=1,b=0.50,c=โ0.50):
x=2(1)โ0.50ยฑ
x=2โ0.50ยฑ2.25
Two roots:
x=2โ0.50+1.50โ=2 โ
x=0.50
Equilibrium:[A]=0.50 M, [B]=[C]=0.50 M
Check:K=(0.50)(0.50)/0.50=0.50 โ
Note: If we had used the approximation: x=0.50โ=0.71. The 5% check: 0.71/1.00 โ fails badly!
Quadratic Approach ๐ฏ
Practice: Full Quadratic ๐งฎ
N2โO4โ(g)โ2NO2โ(g), Kcโ=0.36
Initial: [N2โO4โ]=0.50 M, [NO
The equation is: 4x2+0.36xโ0.18=0
Using the quadratic formula:
1) What is the discriminant b2โ4ac? (Enter to 3 significant figures)
2) What is x? (Round to 3 significant figures)
3) What is [NO2โ] at equilibrium? (Round to 3 significant figures)
Quadratic vs Approximation ๐
Exit Quiz โ Quadratic Formula โ
>
100
Yes โ Approximate: [initial]โxโ[initial]
No โ 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
Common Patterns
Pattern
Example
Perfect square
(aโx)2(2x)2โ โ take square root
Small K with approx
K=ax2โ โ x=
Full quadratic
ax2+bx+c=0 โ quadratic formula
Problem 1: Finding K ๐งฎ
2HI(g)โH2โ(g)+I2โ(g)
Initial: [HI]=1.00 M, [H2โ]=[I
At equilibrium: [HI]=0.80 M
1) What is x? (Remember: the coefficient of HI is 2)
2) What is [H2โ] at equilibrium?
3) What is Kcโ? (Enter to 3 significant figures)
Problem 2: Using the Approximation ๐งฎ
COCl2โ(g)โCO(g)+Cl2โ(g), Kcโ=2.2ร10โ10
Initial: [COCl2โ]=0.50 M, [CO]=[Cl
Using the approximation 0.50โxโ0.50:
1) Solve: x=Kcโร0.50โ. What is ? (Enter in scientific notation, e.g. 1.0e-5)
2) Does the 5% test pass? (Enter "yes" or "no")
3) What is [CO] at equilibrium? (Enter in scientific notation, same as x)
Round all answers to 3 significant figures.
Problem 3: Which Method? ๐ฏ
Problem 4: Non-Zero Initial Products ๐งฎ
H2โ(g)+I2โ(g)โ2HI(g), Kcโ=64
Initial: [H2โ]=0.50, [I2โ]=, M
First check: Q=(0.50)(0.50)(0.20)2โ. Since , shift right.
1) Using the ICE table with shift right, what is [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=0.50โx0.20+2xโ. Solve for x. (Round to 3 significant figures)
3) What is [HI] at equilibrium? (Round to 3 significant figures)
Exit Quiz โ ICE Table Workshop โ
+(
)(x)
E
I + C
I + C
Problem Types
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 โ solve for x
Solving Strategies
Condition
Strategy
Perfect square
Take square root
[init]/K>100
Small-x approximation
[init]/K<100
Full quadratic
Approx gives > 5%
Switch to quadratic
Verification
Always check: Plug equilibrium concentrations back into K expression. The calculated K should match the given K.
AP-Style Multiple Choice โ Set 1 ๐ฏ
AP Free-Response Style ๐งฎ
CO(g)+Cl2โ(g)โCOCl2โ(g), Kcโ=255 at 100ยฐC
A 1.00 L flask is charged with 0.400 mol CO and 0.400 mol Clโ. No COClโ is initially present.
1) Write the K expression and set up the ICE table. What is [COCl2โ] at equilibrium in terms of x? (Enter, e.g., "x")
2) The K expression becomes 255=(0.400โx)2xโ. Using the approximation ( is small... actually ). Should you use the quadratic? (Enter "yes" or "no")
3) Actually, [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 COClโ formed? (Enter in mol)