ICE Table

Core Concept

An ICE table organizes data about the concentrations or pressures of species in a chemical reaction at different stages: Initial, Change, and Equilibrium.

Purpose: To simplify calculations when determining unknown equilibrium concentrations or when solving for the equilibrium constant (K).

Key Tips

Sign Convention:
- Use −x for reactants (consumed) and +x for products (formed).
- Ensure changes are proportional to the stoichiometric coefficients in the balanced equation.
Approximations:
- If K is very small (K \leq 10^{-3}), changes in concentration (x) may be negligible compared to the initial concentration. This allows you to simplify: [A]_0 - x \approx [A]_0
Quadratic Formula:
- Use the quadratic formula if the equilibrium equation is not easily simplified:
  $ax^2 + bx + c = 0 \quad \text{gives} \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}$

Test Yourself

Assorted Multiple Choice

A $1.0 L$ flask is initially filled with $2.0 \text{ moles}$ of pure $PCl_5(g)$ and allowed to decompose according to the equation: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If an ICE table is set up and the equilibrium concentration of $Cl_2$ is determined to be $x$, which of the following expressions correctly represents the equilibrium concentration of $PCl_5$?

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01

Constructing the Change Row with Known Equilibrium Data

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Solving for K using Equilibrium Algebra (x)

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Predicting Equilibrium Concentrations from K

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The "Small K" Approximation

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ICE Tables with Non-Zero Initial Products

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Assorted Multiple Choice

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Structure of an ICE Table

Initial: The starting concentrations (or pressures) of reactants and products.

Change: The amount that each species changes during the reaction.

Use −x for species that are consumed.
Use +x for species that are formed.

Equilibrium: The concentrations (or pressures) of each species at equilibrium, calculated as the initial amount ± change.

Steps for Solving ICE Table Problems

  
      Step & Direction
      Example Application
    
      Step 1: Write the Balanced Equation
Ensure the reaction is balanced to correctly relate the stoichiometry.
      $$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$$
    
      Step 2: Set Up the ICE Table
Organize species and fill in known Initial values. Use zeros for species not initially present.
      
        Initial $[\text{H}_2] = 1.00\text{ M}$

        Initial $[\text{I}_2] = 1.00\text{ M}$

        Initial $[\text{HI}] = 0\text{ M}$
      
      Step 3: Define x
Let x represent the change based on stoichiometry.
      
        Change in $\text{H}_2 = -x$

        Change in $\text{I}_2 = -x$

        Change in $\text{HI} = +2x$ (due to the coefficient 2)
      
      Step 4: Express Equilibrium Concentrations
Use the Change column to write expressions for equilibrium.
      
        Equilibrium $[\text{H}_2] = 1.00 - x$

        Equilibrium $[\text{I}_2] = 1.00 - x$

        Equilibrium $[\text{HI}] = 2x$
      
      Step 5: Write the Equilibrium Expression
Substitute concentrations into the $K$ expression:
 $K = \frac{[\text{products}]^c}{[\text{reactants}]^a}$
      
        $$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$
        $$50.0 = \frac{(2x)^2}{(1.00 - x)(1.00 - x)}$$
      
      Step 6: Solve for x
Solve using algebra or the quadratic formula.
      
        $\sqrt{50.0} = \frac{2x}{1.00 - x}$

        $7.07 = \frac{2x}{1.00 - x}$

        $x \approx 0.779$
      
      Step 7: Calculate Final Concentrations
Plug x back into expressions to find final values.
      
        $[\text{H}_2] = 1.00 - 0.779 = \mathbf{0.221\text{ M}}$

        $[\text{I}_2] = 1.00 - 0.779 = \mathbf{0.221\text{ M}}$

        $[\text{HI}] = 2(0.779) = \mathbf{1.558\text{ M}}$

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