Chemical Equilibrium Practice Flashcards

Definition and Characteristics of Equilibrium

• General Definition of Equilibrium: A state in which there are no observable changes as time goes by.

• Definition of Chemical Equilibrium: A condition reached in a chemical reaction when:

  • The rates of the forward and reverse reactions become equal.

  • The concentrations of the reactants and products remain constant over time.

• Physical Equilibrium: An equilibrium involving the change of physical state (phase change).

  • Example: The equilibrium between liquid water and water vapor in a closed system:

    • $H_2O(l) \rightleftharpoons H_2O(g)$

• Chemical Equilibrium Example: The reversible reaction of dinitrogen tetroxide and nitrogen dioxide:

  • $N_2O_4(g) \rightleftharpoons 2NO_2(g)$

The Nitrogen Dioxide-Dinitrogen Tetroxide System

• Starting Conditions and Reversibility: Equilibrium for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ can be reached starting from different initial concentrations:

  • Starting with pure $NO_2$.

  • Starting with pure $N_2O_4$.

  • Starting with a mixture of both $NO_2$ and $N_2O_4$.

• Concentration-Time Dynamics:

  • When starting with $NO_2$, its concentration decreases while $N_2O_4$ increases until they both reach a constant level.

  • When starting with $N_2O_4$, its concentration decreases while $NO_2$ increases until they both reach a constant level.

  • At equilibrium, the concentrations do not necessarily equal each other, but their ratio defined by the law of mass action remains constant.

• Experimental Data at $25^{\circ}C$ (Table 14.1):

  • Initial Concentrations ($M$):

    • Case 1: $[NO_2] = 0.000$, $[N_2O_4] = 0.670$

    • Case 2: $[NO_2] = 0.0500$, $[N_2O_4] = 0.446$

    • Case 3: $[NO_2] = 0.0300$, $[N_2O_4] = 0.500$

    • Case 4: $[NO_2] = 0.0400$, $[N_2O_4] = 0.600$

    • Case 5: $[NO_2] = 0.200$, $[N_2O_4] = 0.000$

  • Equilibrium Concentrations ($M$) and Ratios ($\frac{[NO_2]^2}{[N_2O_4]}$):

    • Case 1: $[NO_2] = 0.0547$, $[N_2O_4] = 0.643$\; Ratio = $4.65 \times 10^{-3}$

    • Case 2: $[NO_2] = 0.0457$, $[N_2O_4] = 0.448$\; Ratio = $4.66 \times 10^{-3} \text{ (Note: transcript lists as } 4.66 \times 10^3 \text{ in one column and } 10^{-3} \text{ in header)}$

    • Case 3: $[NO_2] = 0.0475$, $[N_2O_4] = 0.491$\; Ratio = $4.60 \times 10^{-3}$

    • Case 4: $[NO_2] = 0.0523$, $[N_2O_4] = 0.594$\; Ratio = $4.60 \times 10^{-3}$

    • Case 5: $[NO_2] = 0.0204$, $[N_2O_4] = 0.0898$\; Ratio = $4.63 \times 10^{-3}$

The Law of Mass Action and Equilibrium Constant

• The Law of Mass Action: For any reversible reaction at a constant temperature, a specific ratio of reactant and product concentrations has a constant value ($K$).

• General Reaction Equation: $aA + bB \rightleftharpoons cC + dD$

• Equilibrium Constant Expression ($K$):

  • $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Significance of the Magnitude of $K$:

  • If $K \gg 1$: The equilibrium lies to the right, favoring the formation of products.

  • If $K \ll 1$: The equilibrium lies to the left, favoring the reactants.

Homogeneous Equilibrium

• Definition: An equilibrium in which all reacting species are in the same phase.

• Gaseous Homogeneous Equilibrium:

  • Example: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$

  • Concentration scale ($K_c$): $K_c = \frac{[NO_2]^2}{[N_2O_4]}$

  • Pressure scale ($K_p$): $K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}}$

• Relationship between $K_c$ and $K_p$:

  • Based on the Ideal Gas Law: $PV = nRT \implies P = \frac{n}{V}RT = CRT$

  • General formula: $K_p = K_c(RT)^{\Delta n}$

  • Variables:

    • $R = 0.0821$

    • $T = \text{Absolute temperature in Kelvin (K)}$

    • $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = (c + d) - (a + b)$

  • Note: In most cases, $K_c \neq K_p$ unless $\Delta n = 0$.

• Homogeneous Equilibrium in Solution:

  • Example: Ionization of acetic acid in water:

    • $CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$

  • The concentration of water ($[H_2O]$) is treated as a constant.

  • The equilibrium expression is simplified to: $K_c = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$

• Conventions for Equilibrium Constants:

  • It is general practice not to include units for the equilibrium constant.

Calculations and Example Problems

• Example 1: Calculating $K_c$ and $K_p$ for CO + $Cl_2$:

  • Reaction: $CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$

  • Data at $74^{\circ}C$: $[CO] = 0.012\,M$, $[Cl_2] = 0.054\,M$, $[COCl_2] = 0.14\,M$

  • Calculation for $K_c$:

    • $K_c = \frac{0.14}{0.012 \times 0.054} = 220$

  • Calculation for $K_p$:

    • $\Delta n = 1 - 2 = -1$

    • $T = 273 + 74 = 347\,K$

    • $K_p = 220 \times (0.0821 \times 347)^{-1} = 7.7$

• Example 2: Oxygen Equilibrium Pressure:

  • Reaction: $2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$

  • Data: $K_p = 158$ at $1000\,K$; $P_{NO} = 0.400\,atm$, $P_{NO_2} = 0.270\,atm$

  • Formula: $K_p = \frac{P_{NO}^2 P_{O_2}}{P_{NO_2}^2}$

  • Rearranging: $P_{O_2} = K_p \frac{P_{NO_2}^2}{P_{NO}^2}$

  • Calculation: $P_{O_2} = 158 \times \frac{(0.400)^2}{(0.270)^2} = 347\,atm$

Heterogeneous Equilibrium

• Definition: An equilibrium involving reactants and products that are in different phases.

• Key Principle: The concentrations of pure solids and pure liquids are not included in the equilibrium constant expression because their concentrations remain constant regardless of the amount present.

• Example: Decomposition of Calcium Carbonate:

  • Equation: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

  • Concentrations of $CaCO_3$ and $CaO$ are constant.

  • $K_c = [CO_2]$

  • $K_p = P_{CO_2}$

  • The equilibrium pressure of $CO_2$ does not depend on the amount of $CaCO_3$ or $CaO$ present.

• Example 3: $NH_4HS$ Decomposition:

  • Reaction: $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$

  • Partial pressures: Each gas is $0.265\,atm$ at $295\,K$.

  • $K_p = P_{NH_3} P_{H_2S} = 0.265 \times 0.265 = 0.0702$

  • $K_c = K_p(RT)^{-\Delta n}$

  • $\Delta n = 2 - 0 = 2$

  • $K_c = 0.0702 \times (0.0821 \times 295)^{-2} = 1.20 \times 10^{-4}$

Multiple Equilibria and Directional Changes

• Multiple Equilibria (Sum of Reactions): If a reaction is the sum of two or more individual reactions, the overall equilibrium constant is the product of the individual constant values.

  • $Reaction 1: A + B \rightleftharpoons C + D \implies K_c'$

  • $Reaction 2: C + D \rightleftharpoons E + F \implies K_c''$

  • $Overall: A + B \rightleftharpoons E + F \implies K_c = K_c' \times K_c''$

• Reversing a Reaction: When a reversible reaction equation is written in the opposite direction, the new equilibrium constant ($K'$) is the reciprocal of the original constant ($K$).

  • For $N_2O_4 \rightleftharpoons 2NO_2$, $K = 4.63 \times 10^{-3}$.

  • For $2NO_2 \rightleftharpoons N_2O_4$, $K' = \frac{1}{K} = 216$.

Rules for Writing Equilibrium Constant Expressions

1. Concentrations in the condensed phase (liquid/solution) are expressed in Molarity ($M$). In the gas phase, they can be expressed in $M$ or $atm$.

2. Pure solids, pure liquids, and solvents do not appear in the expression.

3. The equilibrium constant is a dimensionless quantity.

4. Reporting a $K$ value requires specifying the balanced chemical equation and the temperature.

5. The product rule applies to multi-step reactions.

Chemical Kinetics and Equilibrium

• For a elementary step reaction $A + 2B \rightleftharpoons AB_2$:

  • Forward rate = $k_f [A][B]^2$

  • Reverse rate = $k_r [AB_2]$

• At equilibrium, Forward rate = Reverse rate:

  • $k_f [A][B]^2 = k_r [AB_2]$

  • $\frac{k_f}{k_r} = \frac{[AB_2]}{[A][B]^2} = K_c$

Predicting the Direction of Reaction

• Reaction Quotient ($Q_c$): Calculated by substituting initial concentrations into the equilibrium constant expression.

• Comparison of $Q_c$ and $K_c$:

  • Q_c > K_c: System shifts left (towards reactants) to reach equilibrium.

  • $Q_c = K_c$: System is already at equilibrium.

  • Q_c < K_c: System shifts right (towards products) to reach equilibrium.

Calculating Equilibrium Concentrations (ICE Method)

1. Express equilibrium concentrations in terms of initial concentrations and a single unknown $x$ (change in concentration).

2. Write the $K_c$ expression in terms of these equilibrium values and solve for $x$.

3. Calculate final equilibrium concentrations.

• Exhaustive Problem Example (Bromine dissociation):

  • Reaction: $Br_2(g) \rightleftharpoons 2Br(g)$

  • Data: $K_c = 1.1 \times 10^{-3}$ at $1280^{\circ}C$. Initial $[Br_2] = 0.063\,M$, $[Br] = 0.012\,M$.

  • Set up ICE table:

    • Initial ($M$): $[Br_2] = 0.063$, $[Br] = 0.012$

    • Change ($M$): $[Br_2] = -x$, $[Br] = +2x$

    • Equilibrium ($M$): $[Br_2] = 0.063 - x$, $[Br] = 0.012 + 2x$

  • Expression: $1.1 \times 10^{-3} = \frac{(0.012 + 2x)^2}{0.063 - x}$

  • Quadratic Equation: $4x^2 + 0.0491x + 0.0000747 = 0$

  • Roots: $x = -0.0105$ or $x = -0.00178$

  • Choosing Physical Answer: If $x = -0.0105$, $[Br]$ becomes negative ($0.012 + 2(-0.0105) = -0.009\,M$), which is impossible.

  • Final concentrations: $[Br] = 0.00844\,M$; $[Br_2] = 0.0648\,M$.

Le Châtelier’s Principle

• General Law: If an external stress is applied to a system at equilibrium, the system adjusts to partially offset the stress as it reaches a new equilibrium.

• Changes in Concentration:

  • Increase product: Shift Left.

  • Decrease product: Shift Right.

  • Increase reactant: Shift Right.

  • Decrease reactant: Shift Left.

• Changes in Volume and Pressure:

  • Increase pressure (Decrease volume): Shift to the side with fewest moles of gas.

  • Decrease pressure (Increase volume): Shift to the side with most moles of gas.

• Changes in Temperature:

  • Exothermic Reaction (\Delta H < 0):

    • Increase Temp ($T$): $K$ decreases, shifts Left.

    • Decrease Temp ($T$): $K$ increases, shifts Right.

  • Endothermic Reaction (\Delta H > 0):

    • Increase Temp ($T$): $K$ increases, shifts Right.

    • Decrease Temp ($T$): $K$ decreases, shifts Left.

• Adding a Catalyst:

  • Lowers activation energy ($E_a$) for both forward and reverse reactions.

  • Does not change $K$.

  • Does not shift the equilibrium position.

  • System simply reaches equilibrium sooner.

Chemistry in Action

• Life at High Altitudes:

  • Equilibrium: $Hb(aq) + O_2(aq) \rightleftharpoons HbO_2(aq)$

  • $K_c = \frac{[HbO_2]}{[Hb][O_2]}$

  • Lower $O_2$ at high altitudes causes shift left, decreasing oxygenated hemoglobin ($HbO_2$).

• The Haber Process:

  • Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

  • $\Delta H^0 = -92.6\,kJ/mol$ (Exothermic).

Summary of Le Châtelier’s Principle