AP Chemistry - Equilibrium

Equilibrium

• Equilibrium is the biggest topic on the AP Chemistry exam.
• Topics covered:
  • Writing equilibrium constant expressions
  • Interpreting the value of the equilibrium constant
  • Applying Le Chatelier's principle to predict the effects of equilibrium disturbances
  • Calculating equilibrium concentrations or pressures using ICE tables
  • Interpreting graphs and particle diagrams related to equilibrium
  • Solubility equilibrium (Ksp) and the common ion effect
• Bonus topics:
  • Using Coulomb's law to explain lattice energy trends
  • Explaining the relationship between IMFs and boiling point

Interpreting Concentration-Time Graphs

• Reaction: $3H<em>2(g) + N</em>2(g) \rightleftharpoons 2NH_3(g)$, $\Delta H = -92 \text{ kJ/mol}$
• If concentrations of all species are constant between times $t<em>1$ and $t</em>2$, the system is at equilibrium.
• At equilibrium:
  • Concentrations are not changing.
  • Temperature is constant.
  • The rates of the forward and reverse reactions are equal.
  • Effective collisions are still happening.

Interpreting K (Equilibrium Constant)

• Reaction: $Fe^{3+} + SCN^- \rightleftharpoons FeSCN^{2+}$ at $25^\circ C$, $K_{eq} = 240$
• If K > 1, the products are favored at equilibrium.
• The size of $K$ does not give information about the reaction rate or whether the reaction is endothermic or exothermic.

Reverse K Calculation

• Reaction: $H<em>2(g) + Br</em>2(g) \rightleftharpoons 2HBr(g)$
• At a certain temperature, $K = 2.0 \times 10^5$
• The $K$ for the reverse reaction is the inverse of the $K$ for the forward reaction.
• $K<em>{reverse} = \frac{1}{K</em>{forward}} = \frac{1}{2.0 \times 10^5} = 0.5 \times 10^{-5} = 5.0 \times 10^{-6}$

Identifying Equilibrium from Particle Diagrams

• Reaction: $X(g) + Y(g) \rightleftharpoons XY(g)$
• At equilibrium, the concentrations of species are not changing.
• Determine equilibrium by finding consecutive boxes where the number of particles of each species remains constant.

Interpreting K - Predicting Concentration

• Reaction: $2H<em>2S(g) + CH</em>4(g) \rightleftharpoons CS<em>2(g) + 4H</em>2(g)$, $K_c = 3.4 \times 10^{-4}$
• Small $K$ means reactants are favored.
• If starting with equimolar concentrations of all species, the system shifts to make more reactants in order to reach the small $K$.
• The system will make them in proportion to their stoichiometric coefficients in the chemical equation.
• If only reactants are present initially, the system will shift to make some products to reach equilibrium; concentrations of reactants will decrease.

Equilibrium from Particle Diagrams with K Calculation

• Reaction: $X(g) + Y(g) \rightleftharpoons XY(g)$
• Each particle represents a concentration of 0.1 M.
• At equilibrium (determined from particle diagram): 2X, 4Y, 3XY
• $K = \frac{[XY]}{[X][Y]} = \frac{0.3}{(0.2)(0.4)} = \frac{0.3}{0.08} = \frac{30}{8} \approx 4$

Predicting Products from K in a Diagram

• Reaction: $2NO<em>2(g) \rightleftharpoons N</em>2O<em>4(g)$, $K</em>c = 200$
• Each molecule represents 1 mole in a 100 L container, so each particle corresponds to a concentration of 0.01 M.
• When additional $N<em>2O</em>4$ is added, the system reestablishes equilibrium, keeping $K_c$ constant.
• Check answer choices for diagrams with $K_c = 200$.

Predicting Effects When Product is Added at Equilibrium

• Reaction: $3H<em>2(g) + N</em>2(g) \rightleftharpoons 2NH_3(g)$, $\Delta H = -92 \text{ kJ/mol}$
• Adding more $NH_3$ (product) shifts the system to the reactants side.
• The value of $K$ does not change when the concentration of a species changes.

Le Chatelier's Principle

• Reaction: $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$, \Delta H < 0 (exothermic)
• Powdered osmium catalyst with a solid cube will only affects the rate of a chemical reaction and don't affect the equilibrium position.
• Increasing temperature shifts to the reactant side.
• Removing reactant shifts to the reactant side.
• Adding reactant shifts to the product side.

Common Ion Effect

• The presence of a common ion reduces the solubility of a slightly soluble salt.
• AgCl solubility in NaCl solution is less than in pure water due to the common ion effect ($Cl^-$).

Free Response Question 1: Calcium Carbonate

• Reaction: $CaCO<em>3(s) \rightleftharpoons Ca^{2+}(aq) + CO</em>3^{2-}(aq)$, $K_{sp} = 3.4 \times 10^{-9}$

• $K<em>{sp} = [Ca^{2+}][CO</em>3^{2-}] = x^2 = 3.4 \times 10^{-9}$

• $x = \sqrt{3.4 \times 10^{-9}} = 5.8 \times 10^{-5} \text{ mol/L}$

• In 50 mL (0.05 L): $(5.8 \times 10^{-5} \text{ mol/L})(0.05 \text{ L}) = 2.9 \times 10^{-6} \text{ mol}$

• $(2.9 \times 10^{-6} \text{ mol})(100.09 \text{ g/mol}) = 2.9 \times 10^{-4} \text{ g}$

• Solubility of $CaCO<em>3$ in $CaCl</em>2$ is less than in pure water due to the common ion effect ($Ca^{2+}$).

• Solubility of $CaCO_3$ in NaCl is equal to that in pure water because $Na^+$ and $Cl^-$ do not interact with the solubility equilibrium.

Lattice Energy and Coulomb's Law

• Coulomb's law: attraction between cation and anion is inversely proportional to the square of the distance between them.
• Magnesium ion is smaller than calcium ion.
• Attractive forces in magnesium carbonate are stronger, and therefore its lattice energy is greater.

Decomposition of Calcium Carbonate

• Reaction: $CaCO<em>3(s) \rightleftharpoons CaO(s) + CO</em>2(g)$
• Using PV = nRT to Calculate Moles of $CO_2$
  • $n = \frac{PV}{RT} = \frac{(1.04 \text{ atm})(2.0 \text{ L})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(1100 \text{ K})} = 0.023 \text{ mol}$
• Based on the data in the experiments, the student's claim is incorrect because:
  • In experiment one, 25 grams of calcium carbonate would be 0.25 moles; but only 0.023 moles of $CO_2$ were produced meaning that a portion of the reactant remains.
  • Experiments starting with 25 and 50 grams of calcium carbonate produced the same constant final pressure meaning that not all of it reacts and reaches equilibrium.
• Since the equilibrium pressure of $CO<em>2$ determines Kp and Kp = PCO2 = 1.04 atm, that shows sufficient data obtained to determine the value of the equilibrium constant for the reaction.

Free Response Question 2: HI Decomposition

• Reaction: $2HI(g) \rightleftharpoons H<em>2(g) + I</em>2(g)$

• Equilibrium constant expression: $K<em>c = \frac{[H</em>2][I_2]}{[HI]^2}$

• Equilibrium concentration of HI from the graph: 0.80 M

• Using ICE table to determine equilibrium concentrations of H2 and I2: 0.1

• Calculating the value of K at 700 Kelvin.
  • $K_c = \frac{[0.1][0.1]}{[0.8]^2} = 0.016$
• The pressure inside the container remains constant because the moles of gas are the same in reactants and products.
• The reaction is endothermic because when K increases and temperature increase the reaction shifts to products.
• Doing a q versus k comparison.
  • $Q = \frac{[0.1][0.5]}{[0.75]^2}= 0.089$, $K_c = 0.026$. When Q > K, the reactants increases and HI will increase.

Intermolecular Forces

• Ordering substances from weakest to strongest IMFs based on boiling points: H2, HI, I2