11.3 Equilibrium Calculations & the Reaction Quotient (Q vs K)

11.3 Equilibrium Calculations & the Reaction Quotient (Q vs K)

Quick Summary: The reaction quotient ($Q$) uses instantaneous concentrations or partial pressures in the equilibrium expression to determine the spontaneous direction of reaction: $Q < K$ shifts forward (right), $Q = K$ indicates dynamic equilibrium, and $Q > K$ shifts reverse (left). ICE tables (Initial, Change, Equilibrium) systematically track stoichiometric transformations to resolve equilibrium concentrations. When starting reactant concentrations exceed $K$ substantially ($[\text{Reactant}]_0 / K > 400$), the 5% approximation rule simplifies calculations by treating reactant consumption as negligible.

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1. The Reaction Quotient ($Q$): The Directional Compass

Before equilibrium is established, the mixture of reactants and products is characterized by the Reaction Quotient ($Q$). Formulated identically to the equilibrium constant expression ($K$), $Q$ is evaluated using instantaneous non-equilibrium concentrations ($Q_c$) or partial pressures ($Q_p$): $\text{For } a\,\text{A} + b\,\text{B} \rightleftharpoons c\,\text{C} + d\,\text{D}, \quad Q_c = \frac{[\text{C}]^c_{\text{inst}} [\text{D}]^d_{\text{inst}}}{[\text{A}]^a_{\text{inst}} [\text{B}]^b_{\text{inst}}}$

Comparing $Q$ to $K$ predicts the net direction the system must shift to attain equilibrium:

• $Q < K$ (Product Deficit): Product ratio is smaller than at equilibrium. Forward reaction rate exceeds reverse rate ($\text{Rate}_{\text{fwd}} > \text{Rate}_{\text{rev}}$). The system shifts right ($\to$), consuming reactants to synthesize products until $Q = K$.
• $Q = K$ (Dynamic Equilibrium): Opposing reaction rates are equal ($\text{Rate}_{\text{fwd}} = \text{Rate}_{\text{rev}}$). The system is in dynamic equilibrium; no net concentration changes occur.
• $Q > K$ (Product Excess): Product ratio exceeds equilibrium values. Reverse rate exceeds forward rate ($\text{Rate}_{\text{rev}} > \text{Rate}_{\text{fwd}}$). The system shifts left ($\leftarrow$), consuming products to regenerate reactants until $Q = K$.

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2. ICE Table Framework & Calculation Categories

Equilibrium problems are structured using ICE Tables:

• I (Initial): Starting concentrations or pressures present before reaction.
• C (Change): Stoichiometric algebraic change, defined using variable $x$. Species consumed carry negative signs ($-ax$); species formed carry positive signs ($+cx$).
• E (Equilibrium): Net algebraic sum of Initial and Change rows ($E = I + C$).

Two Core Problem Types

1. Type 1: Calculating $K$ from Equilibrium Data: When the equilibrium concentration of at least one species is determined experimentally, $x$ is found arithmetically, yielding all equilibrium values to compute $K$ directly.
2. Type 2: Calculating Equilibrium Concentrations from Known $K$: Initial quantities and $K$ are known. Equilibrium algebraic expressions are substituted into the mass-action expression to solve for $x$.

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3. Algebraic Strategies: The 5% Rule vs. Quadratic Formula

Substituting equilibrium expressions into $K$ often yields quadratic forms: $K_c = \frac{x^2}{[\text{Reactant}]_0 - x}$

The Small $K$ Approximation (5% Rule)

When $K$ is very small ($K \ll 1$), very little reactant converts into product ($x \ll [\text{Reactant}]_0$). We can approximate: $[\text{Reactant}]_0 - x \approx [\text{Reactant}]_0$ This eliminates polynomial terms, allowing direct calculation: $x \approx \sqrt{K_c \times [\text{Reactant}]_0}$.

Validation Criteria

• Rule-of-Thumb Ratio: The approximation is generally reliable if: $\frac{[\text{Reactant}]_0}{K} > 400$
• Mandatory 5% Check: The calculated $x$ must satisfy: $\text{Percent Dissociation} = \left(\frac{x}{[\text{Reactant}]_0}\right) \times 100\% < 5.0\%$
  • If $< 5.0\%$, the approximation is verified and retained.
  • If $\ge 5.0\%$, the assumption is invalid, and $x$ must be resolved using the exact quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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4. $Q$ vs $K$ Diagnostic Matrix

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5. Worked Problem: Step-by-Step ICE Table with 5% Rule Validation

Problem: At a certain temperature, phosphorus pentachloride decomposes with the practice value of $K_c$ given below: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \quad K_c = 1.80 \times 10^{-4}$ A $0.250\,\text{mol}$ sample of $\text{PCl}_5$ is placed into a $1.00\,\text{L}$ flask at that temperature. Calculate all equilibrium concentrations.

1. Determine Initial Molarities & $Q_c$: $[\text{PCl}_5]_0 = 0.250\,\text{M}, \quad [\text{PCl}_3]_0 = 0\,\text{M}, \quad [\text{Cl}_2]_0 = 0\,\text{M}$ $Q_c = 0 < K_c \implies \text{Reaction shifts right (}\to\text{)}$

2. Construct ICE Table:

3. Set Up Equilibrium Expression: $K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \implies 1.80 \times 10^{-4} = \frac{x^2}{0.250 - x}$

4. Evaluate Small $K$ Approximation: $[\text{PCl}_5]_0 / K_c = 0.250 / (1.80 \times 10^{-4}) = 1389 > 400$ Approximating $0.250 - x \approx 0.250$: $1.80 \times 10^{-4} \approx \frac{x^2}{0.250} \implies x^2 = 4.50 \times 10^{-5} \implies x = 6.71 \times 10^{-3}\,\text{M}$

5. Check 5% Approximation Rule: $\text{Percent Dissociation} = \left(\frac{6.71 \times 10^{-3}}{0.250}\right) \times 100\% = 2.68\% < 5.0\%$ The approximation is valid ($2.68\% < 5.0\%$).

6. Final Equilibrium Concentrations:

• $[\text{PCl}_5] = 0.250 - 0.00671 = 0.243\,\text{M}$
• $[\text{PCl}_3] = 6.71 \times 10^{-3}\,\text{M}$
• $[\text{Cl}_2] = 6.71 \times 10^{-3}\,\text{M}$