ICE tables Knowledge-Based Study

Study notes

1. What is an ICE Table?

   An ICE table (also known as a RICE table, where 'R' stands for Reaction) is a structured tabular framework used in chemical equilibrium calculations to track how concentrations or partial pressures change as a dynamic system proceeds from a non-equilibrium state to equilibrium.

   The acronym ICE represents the three key stages of the process:

   • I (Initial): The concentrations or partial pressures of all reactants and products present before the reaction begins or shifts.
   • C (Change): The algebraic quantities by which concentrations or pressures increase or decrease as the system moves toward equilibrium, directly determined by reaction stoichiometry.
   • E (Equilibrium): The final concentrations or partial pressures when the forward and reverse reaction rates become equal, expressed as the algebraic sum of the initial and change rows: $E = I + C$.

   Step-by-Step Construction

   To successfully set up an ICE table, follow these fundamental rules:

   1. Write the balanced chemical equation: Identify the stoichiometric coefficients, as they determine the relative mole ratios in the Change row.
   2. Enter Initial quantities (Row I): Use molarity ($\text{M}$) for solutions or concentration equilibrium constants ($K_c$), and partial pressures ($\text{atm}$ or $\text{bar}$) for gas-phase systems using $K_p$. If a substance is not added initially, its entry is $0$.
   3. Define the Change row (Row C): Let $x$ represent the extent of reaction. Reactants consumed are assigned negative changes ($-ax$), while products formed are assigned positive changes ($+bx$), where $a$ and $b$ are stoichiometric coefficients. If both reactants and products are present initially, calculate the reaction quotient ($Q$) to determine the direction of the shift.
   4. Determine Equilibrium values (Row E): Add the Initial and Change rows together for each species: $\text{Equilibrium} = \text{Initial} + \text{Change}$.

   Connecting the ICE Table to the Equilibrium Expression

   Once the equilibrium row is expressed in terms of the variable $x$, substitute these expressions into the equilibrium constant equation:

   $$K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}$$

   Mathematical Strategies

   • Direct Linear Equations: Occurs when terms simplify or when one of the species has a known equilibrium value, allowing immediate determination of $x$.
   • Perfect Squares: If both the numerator and the denominator share the same algebraic power (such as $(2x)^2 / (1.0 - x)^2$), take the square root of both sides to avoid the quadratic formula.
   • Quadratic Formula: Rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$ and solve using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Discard physically impossible values (e.g., negative concentrations or concentrations greater than initial amounts).
   • The Small $K$ Approximation: If $K$ is very small (typically $10^{-4}$ or smaller) and initial concentrations are relatively large (the ratio $\frac{[\text{Initial}]}{K} > 400$), assume that $x$ is negligible compared to the initial concentration: $[\text{Initial}] - x \approx [\text{Initial}]$. Always verify that $x$ is less than 5% of the initial value.

   Problem

   Consider the gaseous reaction:

   $$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \quad K_c = 64.0 \text{ at } 450^\circ\text{C}$$

   If $1.00\text{ M}$ $\text{H}_2$ and $1.00\text{ M}$ $\text{I}_2$ are placed in an evacuated container, calculate the equilibrium concentrations of all species.

   ICE Table Setup

   • Initial (I): $[\text{H}_2]_0 = 1.00\text{ M}$, $[\text{I}_2]_0 = 1.00\text{ M}$, $[\text{HI}]_0 = 0\text{ M}$
   • Change (C): $[\text{H}_2] = -x$, $[\text{I}_2] = -x$, $[\text{HI}] = +2x$
   • Equilibrium (E): $[\text{H}_2] = 1.00 - x$, $[\text{I}_2] = 1.00 - x$, $[\text{HI}] = 2x$

   Calculation

   $$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(2x)^2}{(1.00 - x)(1.00 - x)} = \frac{(2x)^2}{(1.00 - x)^2} = 64.0$$

   Taking the square root of both sides:

   $$\frac{2x}{1.00 - x} = 8.00 \implies 2x = 8.00 - 8.00x \implies 10.00x = 8.00 \implies x = 0.800\text{ M}$$

   Final Equilibrium Concentrations

   • $[\text{H}_2] = 1.00 - 0.800 = 0.20\text{ M}$
   • $[\text{I}_2] = 1.00 - 0.800 = 0.20\text{ M}$
   • $[\text{HI}] = 2(0.800) = 1.60\text{ M}$