Chemical Equilibrium and Solubility: Study Notes for General Chemistry

Chemical Equilibrium

Equilibrium Constant Expressions

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The equilibrium constant quantifies the ratio of product and reactant concentrations at equilibrium.

• Kc (Equilibrium Constant in Terms of Concentration): For a general reaction aA + bB → cC + dD, the equilibrium constant expression is: $K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Kp (Equilibrium Constant in Terms of Partial Pressure): For gases, the equilibrium constant can be expressed using partial pressures: $K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$

• Relationship between Kc and Kp: $K_p = K_c(RT)^{\Delta n}$ where $\Delta n$ is the change in moles of gas ($\text{moles of gaseous products} - \text{moles of gaseous reactants}$), $R$ is the gas constant (0.08206 L·atm/mol·K), and $T$ is temperature in Kelvin.

Example: For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2 NO(g)$, $K_c$ and $K_p$ can be related using the above formula.

Equilibrium for Reverse Reactions

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

• Key Point: If $K$ is the equilibrium constant for the forward reaction, then for the reverse reaction: $K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}$

Equilibrium Concentration and Rate Plots

Understanding how concentrations and rates change over time helps visualize the approach to equilibrium.

• Concentration vs. Time Plot: When only reactants are present initially, their concentrations decrease while product concentrations increase until equilibrium is reached.

• Rate vs. Time Plot: The rates of the forward and reverse reactions change over time and become equal at equilibrium.

• Equilibrium Condition: At equilibrium, the concentrations of reactants and products no longer change.

Le Châtelier's Principle

Le Châtelier's Principle predicts how a system at equilibrium responds to disturbances.

• Adding Heat: For endothermic reactions, adding heat shifts equilibrium toward products; for exothermic, toward reactants.

• Removing/Adding Reactants or Products: Removing a reactant or product shifts equilibrium to replace what was removed.

• Changing Pressure: Increasing pressure favors the side with fewer moles of gas.

• Adding a Catalyst: A catalyst speeds up both forward and reverse reactions equally, allowing equilibrium to be reached faster but does not change the position of equilibrium.

Example: Sketches of [Br2] vs. time for catalyzed and uncatalyzed reactions show faster approach to equilibrium with a catalyst.

Interpreting Kc Values

The magnitude of the equilibrium constant indicates the relative amounts of products and reactants at equilibrium.

• If $K_c \gg 1$: Products dominate at equilibrium.

• If $K_c \ll 1$: Reactants dominate at equilibrium.

Equilibrium Calculations Using Tables

Tabular data is often used to organize initial and equilibrium concentrations or pressures for equilibrium calculations.

Initial PNOBr	Initial PNO	Initial PBr2	Eq. PNOBr	Eq. PNO	Eq. PBr2	Kc
0.200	0.100	0.100	--	--	--	1.14×10-2
0.150	0.250	--	--	--	--	1.14×10-2
--	--	--	1.20	3.45×10-1	--	1.14×10-2

Additional info: The table is used to calculate equilibrium partial pressures and apply the equilibrium constant expression.

Solubility Equilibria

Solubility Product Constant (Ksp)

The solubility product constant, $K_{sp}$, describes the equilibrium between a solid and its ions in solution.

• Definition: For a salt $AB$ dissolving as $AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$: $K_{sp} = [A^+][B^-]$

• Magnitude of $K_{sp}$: The larger the $K_{sp}$, the more soluble the ionic solid.

Calculating Molar Solubility

Molar solubility is the number of moles of solute that dissolve per liter of solution to reach saturation.

• Example: For $Ca(OH)_2$ with $K_{sp} = 6.5 \times 10^{-6}$: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$ Let $s$ = molar solubility of $Ca(OH)_2$: $K_{sp} = [Ca^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$ Solve for $s$ to find molar solubility.

• For $La(OH)_3$: If molar solubility of $La^{3+}$ is given, $K_{sp}$ can be calculated using: $La(OH)_3(s) \rightleftharpoons La^{3+}(aq) + 3OH^-(aq)$ $K_{sp} = [La^{3+}][OH^-]^3$

Applications of Solubility Product

• Predicting Precipitation: If the ion product exceeds $K_{sp}$, precipitation occurs.

• Calculating Ion Concentrations: Use $K_{sp}$ to determine concentrations of ions in saturated solutions.

Worked Example: Equilibrium Partial Pressure Calculation

Given initial pressures and $K_c$, calculate the equilibrium partial pressure of a gas.

• Example: For $N_2(g) + O_2(g) \rightleftharpoons 2 NO(g)$, with $K_c = 1.5 \times 10^{-10}$ at 25°C, initial $P_{N_2} = 0.200$ atm, $P_{O_2} = 0.800$ atm, calculate $P_{O_2}$ at equilibrium.

• Steps:

  1. Write the equilibrium expression.

  2. Set up an ICE (Initial, Change, Equilibrium) table.

  3. Solve for the change in pressure using $K_c$.

Summary Table: Key Equilibrium Concepts

Concept	Definition/Formula	Application
Kc	$K_c = \frac{[\text{products}]}{[\text{reactants}]}$	Concentration-based equilibrium
Kp	$K_p = \frac{(P_{\text{products}})}{(P_{\text{reactants}})}$	Partial pressure-based equilibrium (gases)
Ksp	$K_{sp} = [\text{cation}][\text{anion}]^n$	Solubility of ionic solids
Le Châtelier's Principle	System shifts to counteract disturbance	Predicts direction of shift

Additional info: These notes expand on the original questions by providing definitions, formulas, and context for equilibrium and solubility concepts, suitable for exam preparation in General Chemistry.