The ICE table (Initial, Change, Equilibrium) is a matrix used to solve quantitative problems involving chemical equilibrium. This structure organizes the concentration or partial pressure data of reactants and products as a reversible reaction proceeds. The technique transforms a complex chemical problem into a manageable algebraic equation used to calculate the final, stable concentrations of all species at equilibrium.
The Fundamental Structure of an ICE Table
The foundation of the ICE table is the balanced chemical equation, which is written above the table. The stoichiometric coefficients determine the relative amounts of materials consumed and produced during the reaction. The table is structured with columns dedicated to each reactant and product. The framework consists of three horizontal rows: Initial (I), Change (C), and Equilibrium (E). The Initial row records the starting concentrations or partial pressures before the reaction begins. The Equilibrium row contains the final expressions representing the concentrations of all species at equilibrium.
Step-by-Step Guide to Filling the Table
The process begins by transferring all known starting concentrations or pressures into the Initial (I) row. If a species is not present at the start, its initial value is recorded as zero.
The critical step involves constructing the Change (C) row, which represents the shift in concentration required to reach equilibrium. This unknown extent of change is represented by the variable ‘x’. Reactants are assigned a negative change, and products are assigned a positive change.
The stoichiometric coefficient from the balanced equation must multiply ‘x’ for each species; for example, a reactant with a coefficient of two changes by \(-2x\). The Equilibrium (E) row is mathematically derived by combining the entries from the Initial and Change rows using the vertical summation: \(E = I + C\). The resulting expressions in the ‘E’ row contain ‘x’ and represent the equilibrium concentrations of every species.
Solving the Equilibrium Problem
Solving for the Unknown Variable
Once the ICE table is complete, the expressions from the Equilibrium (E) row are substituted into the equilibrium constant expression (\(K_c\) or \(K_p\)). This expression is structured as the concentration of products divided by the concentration of reactants, both raised to their stoichiometric powers. This substitution creates a single algebraic equation with ‘x’ as the only unknown. Solving for ‘x’ often results in a quadratic equation, requiring the use of the quadratic formula. The mathematically viable value for ‘x’ is then used to find the final equilibrium concentrations by substituting it back into the ‘E’ row expressions.
Using the “x is Small” Approximation
When the equilibrium constant (\(K\)) is very small (typically less than \(10^{-4}\)), the “x is small” approximation can simplify the algebra. This approximation assumes that the change ‘x’ is negligible compared to the initial concentration of the reactants. Under this assumption, the subtraction or addition of ‘x’ to the initial concentration is ignored, simplifying the equation and avoiding the quadratic formula. After solving for ‘x’, the approximation must be verified using the \(5\%\) rule. If the calculated value of ‘x’ is less than \(5\%\) of the initial concentration it was subtracted from, the approximation is valid; otherwise, the original quadratic equation must be solved for an accurate result.
Practical Application: A Worked Example
Consider the decomposition reaction: \(\text{A}(g) \rightleftharpoons 2\text{B}(g)\), with \(K_c = 4.0 \times 10^{-6}\), starting with \(1.00 \text{ M}\) of A. The ICE table setup results in Initial concentrations \([A] = 1.00 \text{ M}\) and \([B] = 0 \text{ M}\), and Change values of \(-x\) for A and \(+2x\) for B. Substituting the resulting Equilibrium expressions (\(1.00 – x\) and \(2x\)) into the equilibrium constant expression yields \(4.0 \times 10^{-6} = \frac{(2x)^2}{(1.00 – x)}\). Given the small \(K_c\), the “x is small” approximation simplifies the denominator to \(1.00\), and solving the resulting equation yields \(x = 0.0010 \text{ M}\). Substituting \(x\) back gives the final concentrations: \([A] = 0.999 \text{ M}\) and \([B] = 0.0020 \text{ M}\). Since \(x\) is \(0.1\%\) of the initial concentration, the approximation is confirmed as valid.