When do I need to use the quadratic formula?

Use the quadratic formula when the equilibrium expression expands to the form $ax^2 + bx + c = 0$ and the equation cannot be simplified by taking the square root of both sides.

What if the initial concentration of products is not zero?

Calculate the reaction quotient $Q$ using the initial concentrations. If $Q K$, the reaction shifts left, so products lose $x$ and reactants gain $x$.

Can I ignore $x$ in the denominator to simplify the math?

You can use the 'small $x$ approximation' ($[A]_0 - x \approx [A]_0$) if $K$ is very small, typically $10^{-4}$ or smaller, meaning the change in concentration is negligible compared to the initial amount. Always verify that $x$ is less than 5% of the initial concentration.

How to solve an equilibrium problem with an ICE table

An ICE table organizes the initial concentrations, changes in concentrations, and equilibrium concentrations of a reversible chemical reaction. It applies when you are given initial amounts of reactants or products alongside the equilibrium constant ( $K_c$ or $K_p$ ), and need to calculate the final equilibrium state of the system.

The setup

Write the balanced chemical equation. Define the initial concentrations (in Molarity) or partial pressures (in atmospheres) of all species. Ensure all units are consistent and omit pure solids or liquids, as they do not appear in the equilibrium expression.

The steps

Create a table with rows for Initial, Change, and Equilibrium directly under the balanced chemical equation. 2. Fill in the given initial values. 3. Define the change row using a variable $x$ , multiplied by the stoichiometric coefficients. Use negative signs for the side decreasing in amount and positive signs for the side increasing. 4. Add the initial and change rows to write the equilibrium expressions in terms of $x$ . 5. Substitute these equilibrium expressions into the equilibrium constant expression $K = \frac{[Products]^b}{[Reactants]^a}$ . 6. Solve the resulting algebraic equation for $x$ . 7. Substitute $x$ back into the equilibrium row expressions to find the final concentrations.

Checking the result

Substitute your final calculated equilibrium concentrations back into the equilibrium constant expression. The calculated $K$ should match the given $K$ value, allowing for minor rounding differences.

Common errors

Failing to square or cube concentrations according to their stoichiometric coefficients in the $K$ expression. Incorrectly assigning the signs in the change row without checking the reaction quotient $Q$ . Including pure liquids (like water in aqueous reactions) or solids in the ICE table and $K$ expression.

Worked example

Find the equilibrium concentrations for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at a temperature where $K_c = 54.3$ . The initial concentrations are $[H_2] = 0.10 ext{ M}$ , $[I_2] = 0.10 ext{ M}$ , and $[HI] = 0 ext{ M}$ .

Initial: $[H_2] = 0.10$ , $[I_2] = 0.10$ , $[HI] = 0$ . Change: $[H_2] = -x$ , $[I_2] = -x$ , $[HI] = +2x$ . Equilibrium: $[H_2] = 0.10 - x$ , $[I_2] = 0.10 - x$ , $[HI] = 2x$ . Substitute into the equilibrium expression: $K_c = \frac{[HI]^2}{[H_2][I_2]}$ . This gives $54.3 = \frac{(2x)^2}{(0.10 - x)(0.10 - x)}$ . Take the square root of both sides: $7.369 = \frac{2x}{0.10 - x}$ . Multiply out: $0.7369 - 7.369x = 2x$ . Rearrange to solve for $x$ : $9.369x = 0.7369$ , which gives $x = 0.0787$ . Calculate final concentrations: $[H_2] = 0.10 - 0.0787 = 0.0213 ext{ M}$ , $[I_2] = 0.10 - 0.0787 = 0.0213 ext{ M}$ , and $[HI] = 2(0.0787) = 0.157 ext{ M}$ .

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References: OpenStax Chemistry 2e, Chapter 13.4: Equilibrium Calculations · Khan Academy: Solving equilibrium problems