Chemical Equilibrium Practice Flashcards

Introduction to Chemical Equilibrium

Evolution of Chemical Understanding:
- Initially, chemical theory assumed that a reaction arrow ( $\rightarrow$ ) meant the reaction would proceed entirely to completion (e.g., $H_2O(s) \rightarrow H_2O(l)$ ).
- Subsequently, thermodynamics was used to consider conditions of spontaneity, utilizing changes in entropy ( $\Delta S$ ) and Gibbs free energy ( $\Delta G$ ) to predict if a reaction would proceed.
- The current focus is on reversible reactions, which are reactions that reach equilibrium without 100% of the product being formed. This represents the majority of chemical reactions.

Nature of Reversible Reactions and Equilibrium

Definition of Reversible Reaction: A reaction that can easily travel in either direction. For example, the conversion of dinitrogen tetroxide to nitrogen dioxide: $N_2O_4 \rightarrow 2NO_2$ and the reverse $2NO_2 \rightarrow N_2O_4$ . This is written using a double arrow: $N_2O_4 \rightleftharpoons 2NO_2$ .
Defining Chemical Equilibrium: Equilibrium is a state in which there are no observable changes as time passes. It is achieved when:
- The rates (speeds) of the forward and reverse reactions are equal.
- The concentrations of the reactants and products remain constant (though not necessarily equal to each other).
Common Misconceptions (What Equilibrium Isn't):
- It has nothing to do with the speed of the reaction (which falls under kinetics).
- It is not static or unchanging; it is a dynamic process where reactants and products are still being converted, but at equal rates.

The Law of Mass Action

General Expression: For a reversible reaction of the form $aA + bB \rightleftharpoons cC + dD$ , the equilibrium constant ( $K$ ) is defined by the concentrations of the products raised to their stoichiometric coefficients divided by the reactants raised to theirs: $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
Interpreting the Equilibrium Constant (K):
- If $K \gg 1$ : The equilibrium position lies far to the right, meaning products are favored and are much more abundant than reactants.
- If $K \ll 1$ : The equilibrium position lies far to the left, meaning reactants are favored and are much more abundant than products.
Experimental Data (The $NO_2-N_2O_4$ System at $25^\circ C$ ):
- Regardless of whether the reaction starts with only $NO_2$ , only $N_2O_4$ , or a mixture of both, the system reaches an equilibrium where the ratio $\frac{[NO_2]^2}{[N_2O_4]}$ remains constant.
- Observed constant value for this system at $25^\circ C$ : approximately $4.63 \times 10^{-3}$ .

Types of Chemical Equilibrium

Homogeneous Equilibrium: Applies to reactions where all reacting species are in the same phase.
- Example: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ .
- Concentration Constant ( $K_c$ ): $K_c = \frac{[NO_2]^2}{[N_2O_4]}$ .
- Pressure Constant ( $K_p$ ): For gas phase reactions, equilibrium can be expressed via partial pressures: $K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}}$ .
- Relationship between $K_c$ and $K_p$ : $K_p = K_c(RT)^{\Delta n}$ , where $\Delta n$ is the moles of gaseous products minus the moles of gaseous reactants. In most cases, $K_c \neq K_p$ .
- Aqueous Solutions and Solvents: In reactions like acetic acid ionization ( $CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$ ), the concentration of water ( $[H_2O]$ ) is treated as a constant and is omitted from the final $K_c$ expression: $K_c = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ .
Heterogeneous Equilibrium: Applies to reactions in which reactants and products are in different phases.
- Example: Decomposition of calcium carbonate: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ .
- Crucial Rule: The concentration of pure solids and pure liquids are constant and are not included in the equilibrium constant expression.
- For the reaction above: $K_c = [CO_2]$ and $K_p = P_{CO_2}$ .
- Note: $P_{CO_2}$ does not depend on the specific amount of $CaCO_3$ or $CaO$ present, provided some of each is there.

Rules for Writing Equilibrium Expressions

Concentration Units: Reacting species in the condensed phase are expressed in Molarity ( $M$ ). Gaseous species can be expressed in $M$ or atmosphere ( $atm$ ).
Exclusions: Pure solids ( $s$ ), pure liquids ( $l$ ), and solvents are excluded.
Dimensionless Quantity: The equilibrium constant is treated as having no units.
Specification: When quoting $K$ , the balanced chemical equation and the temperature must be specified.
Reaction Sums: If a reaction is the sum of two or more reactions, the overall $K$ is the product of the individual constants ( $K_{overall} = K_1 \times K_2$ ).
Reversing Reactions: If the equation is written in the opposite direction, the new equilibrium constant ( $K'$ ) is the reciprocal of the original constant ( $K' = \frac{1}{K}$ ).

The Reaction Quotient (Q)

Definition: $Q_c$ is calculated using the same formula as $K_c$ but uses initial concentrations rather than equilibrium concentrations.
Predicting Reaction Direction:
- $Q_c > K_c$ : The ratio of products to reactants is too high; the system proceeds from right to left (forming more reactants).
- $Q_c = K_c$ : The system is already at equilibrium.
- $Q_c < K_c$ : The ratio of products to reactants is too low; the system proceeds from left to right (forming more products).

Le Châtelier’s Principle

Definition: If an external stress is applied to a system at equilibrium, the system adjusts to partially offset that stress as it moves toward a new equilibrium position.
Concentration Changes:
- Add Reactant: Shift Right (toward products).
- Add Product: Shift Left (toward reactants).
- Remove Reactant: Shift Left.
- Remove Product: Shift Right.
Pressure and Volume Changes (Gases only):
- Increase Pressure (Decrease Volume): Shift toward the side with the fewest moles of gas.
- Decrease Pressure (Increase Volume): Shift toward the side with the most moles of gas.
Temperature Changes:
- Exothermic Reactions ( $\Delta H < 0$ ): Increase $T$ causes $K$ to decrease (Shift Left); Decrease $T$ causes $K$ to increase (Shift Right).
- Endothermic Reactions ( $\Delta H > 0$ ): Increase $T$ causes $K$ to increase (Shift Right); Decrease $T$ causes $K$ to decrease (Shift Left).
Addition of a Catalyst:
- Lowers activation energy ( $E_a$ ) for both forward and reverse reactions equally.
- Does not change the equilibrium constant ( $K$ ).
- Does not shift the equilibrium position.
- The system simply reaches equilibrium faster.

Thermodynamic Relationships

Spontaneity Scenarios:
1. $\Delta H > 0$ , $\Delta S > 0$ : Spontaneous at high temperatures; nonspontaneous at low temperatures.
2. $\Delta H < 0$ , $\Delta S < 0$ : Spontaneous at low temperatures; nonspontaneous at high temperatures.
3. $\Delta H > 0$ , $\Delta S < 0$ : Nonspontaneous at all temperatures.
4. $\Delta H < 0$ , $\Delta S > 0$ : Spontaneous at all temperatures.
Link between Free Energy and Equilibrium:
- The general relationship is $\Delta G = \Delta G^\circ + RT\ln(Q)$ .
- At equilibrium, $\Delta G = 0$ and $Q = K$ , leading to the formula: $\Delta G^\circ = -RT\ln(K)$ .
- Alternatively, $K = e^{-\frac{\Delta G^\circ}{RT}}$ .
Standard Free Energy ( $\Delta G^\circ$ ) and $K$ :
- If $K > 1$ , $\Delta G^\circ$ is negative (Products more abundant).
- If $K < 1$ , $\Delta G^\circ$ is positive (Reactants more abundant).
- If $K = 1$ , $\Delta G^\circ = 0$ (Reactants and products comparable).

Numerical Practice Problems

1. Calculating Kc and Kp for Carbon Monoxide and Chlorine:

Reaction: $CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$
Conditions: $T = 740^\circ C$ ( $1013.15\,K$ ).
Equilibrium concentrations: $[CO] = 0.012\,M$ , $[Cl_2] = 0.054\,M$ , $[COCl_2] = 0.14\,M$ .
$K_c = \frac{[COCl_2]}{[CO][Cl_2]} = \frac{0.14}{(0.012)(0.054)}$ .

2. Nitrogen Dioxide Equilibrium:

Reaction: $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$
Given: $K_p = 158$ at $1000\,K$ .
Partial Pressures: $P_{NO_2} = 0.400\,atm$ , $P_{NO} = 0.270\,atm$ .
Unknown: Find $P_{O_2}$ .
Expression: $158 = \frac{(0.270)^2(P_{O_2})}{(0.400)^2}$ .

3. Ammonium hydrosulfide decomposition:

Reaction: $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$
Conditions: $T = 295\,K$ , partial pressure of each gas is $0.265\,atm$ .
Calculation: $K_p = (0.265)(0.265) = 0.0702$ . Calculate $K_c$ using $K_p = K_c(RT)^{\Delta n}$ where $\Delta n = 2$ .