Equilibrium · ⁨Cân bằng⁩

Learning Objective 7.1.A: Explain the relationship between the occurrence of a reversible chemical or physical process, and the establishment of equilibrium, to experimental observations.

• 7.1.A.1 Many observable processes are reversible. Examples include evaporation and condensation of water, absorption and desorption of a gas, or dissolution and precipitation of a salt. Some important reversible chemical processes include the transfer of protons in acid-base reactions and the transfer of electrons in redox reactions.
• 7.1.A.2 When equilibrium is reached, no observable changes occur in the system. Reactants and products are simultaneously present, and the concentrations or partial pressures of all species remain constant.
• 7.1.A.3 The equilibrium state is dynamic. The forward and reverse processes continue to occur at equal rates, resulting in no net observable change.
• 7.1.A.4 Graphs of concentration, partial pressure, or rate of reaction versus time for simple chemical reactions can be used to understand the establishment of chemical equilibrium.

Learning Objective 7.2.A: Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.

• 7.2.A.1 If the rate of the forward reaction is greater than the reverse reaction, then there is a net conversion of reactants to products. If the rate of the reverse reaction is greater than that of the forward reaction, then there is a net conversion of products to reactants. An equilibrium state is reached when these rates are equal.

Learning Objective 7.3.A: Represent the reaction quotient $Q_c$ or $Q_p$, for a reversible reaction, and the corresponding equilibrium expressions $K_c = Q_c$ or $K_p = Q_p$.

• 7.3.A.1 The reaction quotient $Q_c$ describes the relative concentrations of reaction species at any time. For gas phase reactions, the reaction quotient may instead be written in terms of partial pressures as $Q_p$. The reaction quotient tends toward the equilibrium constant such that at equilibrium $K_c = Q_c$ and $K_p = Q_p$. As examples, for the reaction

  $$a\,\mathrm{A} + b\,\mathrm{B} \rightleftarrows c\,\mathrm{C} + d\,\mathrm{D}$$
  the law of mass action indicates that the equilibrium expression for $(K_c, Q_c)$ is
  • Equation: $K_c = \dfrac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}$

  and that for $(K_p, Q_p)$ is

  • Equation: $K_p = \dfrac{(P_\mathrm{C})^c (P_\mathrm{D})^d}{(P_\mathrm{A})^a (P_\mathrm{B})^b}$
  • Exclusion statement: Conversion between $K_c$ and $K_p$ will not be assessed on the AP Exam. Students should be aware of the conceptual differences and pay attention to whether $K_c$ or $K_p$ is used in an exam question.
  • Exclusion statement: Equilibrium calculations on systems where a dissolved species is in equilibrium with that species in the gas phase will not be assessed on the AP Exam.

• 7.3.A.2 The reaction quotient does not include substances whose concentrations (or partial pressures) are independent of the amount, such as for solids and pure liquids.

Learning Objective 7.4.A: Calculate $K_c$ or $K_p$ based on experimental observations of concentrations or pressures at equilibrium.

• 7.4.A.1 Equilibrium constants can be determined from experimental measurements of the concentrations or partial pressures of the reactants and products at equilibrium.

Learning Objective 7.5.A: Explain the relationship between very large or very small values of $K$ and the relative concentrations of chemical species at equilibrium.

• 7.5.A.1 Some equilibrium reactions have very large $K$ values and proceed essentially to completion. Others have very small $K$ values and barely proceed at all.

Learning Objective 7.6.A: Represent a multistep process with an overall equilibrium expression, using the constituent $K$ expressions for each individual reaction.

• 7.6.A.1 When a reaction is reversed, $K$ is inverted.
• 7.6.A.2 When the stoichiometric coefficients of a reaction are multiplied by a factor $c$, $K$ is raised to the power $c$.
• 7.6.A.3 When reactions are added together, the $K$ of the resulting overall reaction is the product of the $K$'s for the reactions that were summed.
• 7.6.A.4 Since the expressions for $K$ and $Q$ have identical mathematical forms, all valid algebraic manipulations of $K$ also apply to $Q$.

Learning Objective 7.7.A: Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.

• 7.7.A.1 The concentrations or partial pressures of species at equilibrium can be predicted given the balanced reaction, initial concentrations, and the appropriate $K$.
• 7.7.A.2 When $Q < K$, the reaction will proceed with a net consumption of reactants and generation of products. When $Q > K$, the reaction will proceed with a net consumption of products and generation of reactants. When $Q = K$, the system is at dynamic equilibrium; both forward and reverse reactions proceed at the same rate, and the proportion of reactants and products remains constant.

Learning Objective 7.8.A: Represent a system undergoing a reversible reaction with a particulate model.

• 7.8.A.1 Particulate representations can be used to describe the relative numbers of reactant and product particles present prior to and at equilibrium, and the value of the equilibrium constant.

Learning Objective 7.9.A: Identify the response of a system at equilibrium to an external stress, using Le Châtelier's principle.

• 7.9.A.1 Le Châtelier's principle can be used to predict the response of a system to stresses such as addition or removal of a chemical species, change in temperature, change in volume/pressure of a gas-phase system, or dilution of a reaction system.
• 7.9.A.2 Le Châtelier's principle can be used to predict the effect that a stress will have on experimentally measurable properties such as pH, temperature, and color of a solution.

Learning Objective 7.10.A: Explain the relationships between $Q$, $K$, and the direction in which a reversible reaction will proceed to reach equilibrium.

• 7.10.A.1 A disturbance to a system at equilibrium causes $Q$ to differ from $K$, thereby taking the system out of equilibrium. The system responds by bringing $Q$ back into agreement with $K$, thereby establishing a new equilibrium state.
• 7.10.A.2 Some stresses, such as changes in concentration, cause a change in $Q$ only. A change in temperature causes a change in $K$. In either case, the concentrations or partial pressures of species redistribute to bring $Q$ and $K$ back into equality.

Learning Objective 7.11.A: Calculate the solubility of a salt based on the value of $K_{sp}$ for the salt.

• 7.11.A.1 The dissolution of a salt is a reversible process whose extent can be described by $K_{sp}$, the solubility-product constant.
• 7.11.A.2 The solubility of a substance can be calculated from the $K_{sp}$ for the dissolution process. This relationship can also be used to predict the relative solubility of different substances.
• 7.11.A.3 The solubility rules (see 4.7.A.5) can be quantitatively related to $K_{sp}$, in which $K_{sp}$ values $>1$ correspond to soluble salts.
• 7.11.A.4 The molar solubility of one or more species in a saturated solution can be used to calculate the $K_{sp}$ of a substance.

Learning Objective 7.12.A: Identify the solubility of a salt, and/or the value of $K_{sp}$ for the salt, based on the concentration of a common ion already present in solution.

• 7.12.A.1 The solubility of a salt is reduced when it is dissolved into a solution that already contains one of the ions present in the salt. The impact of this "common-ion effect" on solubility can be understood qualitatively using Le Châtelier's principle or calculated from the $K_{sp}$ for the dissolution process.