#Equilibrium Calculations: Mastering ICE Tables

Hey there, future AP Chem master! 🧪 Let's break down ICE tables – your secret weapon for equilibrium problems. It might seem like a lot, but by the end of this review, you’ll be solving these like a pro.

#What are ICE Tables?

At equilibrium, the forward and reverse reaction rates are equal, and concentrations remain constant. But what if we need to find those equilibrium concentrations? That’s where ICE tables come in!

• Initial: Concentrations at the start of the reaction.
• Change: Change in concentration as the reaction reaches equilibrium (use 'x' to represent unknown changes).
• Equilibrium: Final concentrations at equilibrium.
• Reaction: (Optional) The balanced chemical reaction.

#Setting Up an ICE Table

Let's use the reaction: CH₃COOH ⇌ CH₃COO⁻ + H⁺ (K = 1.8 * 10⁻⁵)

#Solving for Equilibrium Concentrations

1. Write the equilibrium expression:

   K = [CH₃COO⁻][H⁺] / [CH₃COOH]

2. Plug in equilibrium concentrations from your ICE table:

   1.8 * 10⁻⁵ = (x)(x) / (1 - x) = x² / (1 - x)

3. Use the 5% approximation: 💡

   • If K is small (usually less than 10⁻⁴), assume 'x' is negligible compared to initial concentrations. This simplifies calculations by making (1-x) ≈ 1. - Important: Do not approximate 'x' when it's multiplied by a coefficient (e.g., 2x, 3x) or when it is the only term (e.g. x).

4. Solve for x:

   1.8 * 10⁻⁵ ≈ x² / 1 x = √(1.8 * 10⁻⁵) = 0.0042

5. Find equilibrium concentrations:

   • [CH₃COO⁻] = [H⁺] = x = 0.0042 M
   • [CH₃COOH] = 1 - x = 0.9958 M

#The 5% Approximation: Your Best Friend

#Why does it work?

• When K is very small, the reaction doesn't proceed much, so 'x' is tiny compared to initial concentrations.

• If we have a number like 3 + 0.0000001, it's basically 3. We can approximate 3 + x ≈ 3 if x is small enough.

• The 5% rule says that if the change in concentration is less than 5% of the initial concentration, the approximation is valid.

#When to Use ICE Tables

You can use ICE tables when you have:

1. The equilibrium constant (K).

2. Initial concentrations. (If not given, assume 0 for products)

#Practice Time!

Let's try another example:

H₂CO₃ ⇌ HCO₃⁻ + H⁺ (K = 4.3 x 10⁻⁷)

If the initial [H₂CO₃] = 1.2 M, what is [HCO₃⁻] at equilibrium?

1. Set up the equilibrium expression:

   K = [HCO₃⁻][H⁺] / [H₂CO₃]

2. Plug in equilibrium concentrations:

   4.3 * 10⁻⁷ = (x)(x) / (1.2 - x)

3. Use the 5% approximation:

   4.3 * 10⁻⁷ ≈ x² / 1.2

4. Solve for x:

   x² = 1.2 * (4.3 * 10⁻⁷) = 5.16 * 10⁻⁷ x = √(5.16 * 10⁻⁷) = 0.0007

5. Answer:

   [HCO₃⁻] = x = 0.0007 M

#Final Exam Focus

Key Topics:

• Mastering ICE table setup and solving for equilibrium concentrations.

• Understanding and applying the 5% approximation.

• Connecting equilibrium concepts to acid-base chemistry (Unit 8).

• Always check if the 5% approximation is valid (if K is small).

• Pay attention to stoichiometric coefficients when setting up the 'Change' row.

• Practice, practice, practice! The more you use ICE tables, the easier they become.

• Forgetting to include stoichiometric coefficients in the 'Change' row.
• Approximating 'x' when it is not valid (e.g., when it's the only term).
• Not checking if the 5% approximation is valid.

#

Practice Question

Practice Questions

#Multiple Choice Questions

1. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant K is 4.0. If the initial concentrations are [N₂] = 2.0 M and [H₂] = 1.0 M, what is the equilibrium concentration of NH₃?

   (A) 0.5 M (B) 1.0 M (C) 1.5 M (D) 2.0 M

2. The equilibrium constant for the reaction A(g) ⇌ 2B(g) is 0.25. If the initial concentration of A is 1.0 M, what is the equilibrium concentration of B?

   (A) 0.25 M (B) 0.50 M (C) 0.75 M (D) 1.0 M

#Free Response Question

Consider the following reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

The equilibrium constant, K, for this reaction is 5.10 at 800 K. Initially, a reaction vessel contains 1.00 M of CO and 1.00 M of H₂O.

(a) Set up an ICE table for this reaction. (b) Calculate the equilibrium concentrations of all species. (c) Calculate the partial pressure of each gas at equilibrium, given that the total pressure is 2.0 atm.

Answer Key:

(a) ICE Table:

(b) Equilibrium Concentrations:

K = [CO₂][H₂] / [CO][H₂O] 5.10 = x² / (1.00 - x)² √5.10 = x / (1.00 - x) 2.26 = x / (1.00 - x) 2.26 - 2.26x = x 2.26 = 3.26x x = 0.693

[CO] = [H₂O] = 1.00 - 0.693 = 0.307 M [CO₂] = [H₂] = 0.693 M

(c) Partial Pressures:

Total pressure = 2.0 atm

Since the total number of moles is constant (2 initial moles and 2 equilibrium moles), the mole fraction of each gas is equal to its concentration divided by the total concentration. The partial pressure of each gas is then the mole fraction times the total pressure:

Partial pressure of CO = (0.307 / 2) * 2.0 atm = 0.307 atm Partial pressure of H₂O = (0.307 / 2) * 2.0 atm = 0.307 atm Partial pressure of CO₂ = (0.693 / 2) * 2.0 atm = 0.693 atm Partial pressure of H₂ = (0.693 / 2) * 2.0 atm = 0.693 atm

#Answers to Multiple Choice Questions

1. D
2. B