ICE Table

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Topic Summary & Highlights
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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).

Practice 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:

      $ax^2 + bx + c = 0 \quad \text{gives} \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}$

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 1: Write the Balanced Equation

  • Ensure the reaction is balanced to correctly relate the stoichiometry of reactants and products.

Step 2: Set Up the ICE Table

  • Organize the species in the reaction.

  • Fill in the known values under the Initial column. Use zeros for species not initially present.

Step 3: Define x

  • Let x represent the change in concentration or pressure, based on stoichiometry.

Step 4: Express Equilibrium Concentrations

  • Use the relationships from the Change column to write expressions for the equilibrium concentrations.

Step 5: Write the Equilibrium Expression

  • Substitute the equilibrium concentrations into the equilibrium constant (K) expression:

    K = $\frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$​

Step 6: Solve for x

  • Solve for x using the equilibrium constant and the given data. This may involve:

    • Direct algebra for simple cases.

    • The quadratic formula if the equation is second-order.

Step 7: Calculate Final Concentrations

  • Plug x back into the equilibrium expressions to find the final concentrations of all species.

Example Problem

Consider the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$

If [H2]0=1.00 M[\text{H}_2]_0 = 1.00 \, \text{M}[H2​]0​=1.00M, [I2]0=1.00 M[\text{I}_2]_0 = 1.00 \, \text{M}[I2​]0​=1.00M, and [HI]0=0[\text{HI}]_0 = 0[HI]0​=0, find the equilibrium concentrations if Kc=50.0K_c = 50.0Kc​=50.0.

Solution:

  1. Write the ICE Table:

    $\begin{array}{c|c|c|c} \text{Species} & \text{Initial (M)} & \text{Change (M)} & \text{Equilibrium (M)} \\ \hline \text{H}_2 & 1.00 & -x & 1.00 - x \\ \text{I}_2 & 1.00 & -x & 1.00 - x \\ \text{HI} & 0 & +2x & 2x \\ \end{array}$

  2. Write the Equilibrium Expression:

    Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}Kc​=[H2​][I2​][HI]2​ 50.0=(2x)2(1.00−x)(1.00−x)50.0 = \frac{(2x)^2}{(1.00 - x)(1.00 - x)}50.0=(1.00−x)(1.00−x)(2x)2​

  3. Simplify:

    50.0=4x2(1.00−x)250.0 = \frac{4x^2}{(1.00 - x)^2}50.0=(1.00−x)24x2​

    Take the square root of both sides:

    50.0=2x1.00−x\sqrt{50.0} = \frac{2x}{1.00 - x}50.0​=1.00−x2x​ 7.07=2x1.00−x7.07 = \frac{2x}{1.00 - x}7.07=1.00−x2x​

  4. Solve for xxx:

    7.07(1.00−x)=2x7.07(1.00 - x) = 2x7.07(1.00−x)=2x 7.07−7.07x=2x7.07 - 7.07x = 2x7.07−7.07x=2x 7.07=9.07x7.07 = 9.07x7.07=9.07x x=7.079.07≈0.779x = \frac{7.07}{9.07} \approx 0.779x=9.077.07​≈0.779

  5. Calculate Equilibrium Concentrations:

    • $[\text{H}_2] = [\text{I}_2] = 1.00 - x = 1.00 - 0.779 = 0.221 \, \text{M}$

    • [HI]=2x=2(0.779)=1.558 M[\text{HI}] = 2x = 2(0.779) = 1.558 \, \text{M}$

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