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

Reaction: 3H2(g) + N2(g) \rightleftharpoons 2NH_3(g), \Delta H = -92 \text{ kJ/mol}
If concentrations of all species are constant between times t1 and t2, 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.

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.

Reaction: H2(g) + Br2(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{reverse} = \frac{1}{K{forward}} = \frac{1}{2.0 \times 10^5} = 0.5 \times 10^{-5} = 5.0 \times 10^{-6}

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.

Reaction: 2H2S(g) + CH4(g) \rightleftharpoons CS2(g) + 4H2(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.

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

Reaction: 2NO2(g) \rightleftharpoons N2O4(g), Kc = 200
Each molecule represents 1 mole in a 100 L container, so each particle corresponds to a concentration of 0.01 M.
When additional N2O4 is added, the system reestablishes equilibrium, keeping K_c constant.
Check answer choices for diagrams with K_c = 200.

Reaction: 3H2(g) + N2(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.

Reaction: N2(g) + 3H2(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.

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^-).

Reaction: CaCO3(s) \rightleftharpoons Ca^{2+}(aq) + CO3^{2-}(aq), K_{sp} = 3.4 \times 10^{-9}
K{sp} = [Ca^{2+}][CO3^{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 CaCO3 in CaCl2 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.

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.

Reaction: CaCO3(s) \rightleftharpoons CaO(s) + CO2(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 CO2 determines Kp and Kp = PCO2 = 1.04 atm, that shows sufficient data obtained to determine the value of the equilibrium constant for the reaction.

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.

	IMFs
H_2	London dispersion forces only
I_2	London dispersion forces only
HI	LDFs and dipole-dipole forces

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