Chapter 17: Equilibrium: The Extent of Chemical Reactions Study Guide

Chapter 17 Overview: Equilibrium: The Extent of Chemical Reactions

• Chapter Sections:

  • 17.1 The Equilibrium State and the Equilibrium Constant.

  • 17.2 The Reaction Quotient and the Equilibrium Constant.

  • 17.3 Expressing Equilibria with Pressure Terms: Relation Between $K_c$ and $K_p$.

  • 17.4 Comparing $Q$ and $K$ to Determine Reaction Direction.

  • 17.5 How to Solve Equilibrium Problems.

  • 17.6 Reaction Conditions and Equilibrium: Le Châtelier’s Principle.

The Equilibrium State

• Core Principles:

  • All reactions are reversible and, under suitable conditions, will reach a state of equilibrium.

  • At equilibrium, the concentrations of products and reactants no longer change. This occurs because the rates of the forward and reverse reactions are exactly equal ($rate_{fwd} = rate_{rev}$).

  • Dynamic Nature: Chemical equilibrium is a dynamic state. Reactions continue to occur at the molecular level, but because they occur at the same rate, no net change is observed on the macroscopic level.

The Equilibrium Constant ($K$)

• Derivation from Kinetics:

  • Consider the general reaction: $aA + bB \rightleftharpoons cC + dD$

  • Forward rate: $rate_{fwd} = k_{fwd}[A]^a[B]^b$

  • Reverse rate: $rate_{rev} = k_{rev}[C]^c[D]^d$

  • At equilibrium ($rate_{fwd} = rate_{rev}$): $k_{fwd}[A]^a[B]^b = k_{rev}[C]^c[D]^d$

  • The ratio of rate constants gives the equilibrium constant ($K$):

  • Equation 17.2: $K = \frac{k_{fwd}}{k_{rev}} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Extent of Reaction:

  • The value of $K$ reflects the ratio of product equilibrium concentrations to reactant equilibrium concentrations at a specific temperature.

  • Small Value for $K$: Indicates the reaction yields very little product before reaching equilibrium. The equilibrium position favors the reactants.

  • Intermediate Value for $K$: (e.g., $K \text{ approx. } 1$) Significant amounts of both reactants and products are present at equilibrium.

  • Large Value for $K$: Indicates the reaction reaches equilibrium with very little reactant remaining. The equilibrium position favors the products.

The Reaction Quotient ($Q$)

• Definition:

  • For the general reaction $aA + bB \rightleftharpoons cC + dD$, the reaction quotient is:

  • Equation 17.4: $Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

  • $Q$ provides the ratio of product concentrations to reactant concentrations at any point in a reaction.

• Relationship to $K$:

  • Equation 17.3: At equilibrium, $Q = K$.

  • The value of $Q$ indicates how close a reaction is to equilibrium and the direction it must proceed to reach it.

  • Regardless of starting concentrations, a specific system at a constant temperature will always attain the same equilibrium state (the same $K$ value).

Heterogeneous Equilibrium

• Definition: A heterogeneous equilibrium involves reactants and/or products in different phases (e.g., solids, liquids, and gases).

• Concentration of Pure Solids and Liquids: A pure solid or liquid always has the same "concentration," defined as its density divided by molar mass (moles per liter). Since this value does not change, it is incorporated into the constant $K$.

• Exclusion Rule: The expressions for $Q$ and $K$ include only species whose concentrations change as the reaction approaches equilibrium. Therefore, pure solids and pure liquids are omitted from the expression for $Q$ or $K$.

• Example: For the reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, the reaction quotient is simply $Q = [CO_2]$.

Sample Problem 17.1: Writing Reaction Quotients

• Problem: Write the reaction quotient ($Q_c$) for:

  • (a) $N_2O_5(g) \rightleftharpoons NO_2(g) + O_2(g)$ (unbalanced)

  • (b) $H_2O(l) \rightleftharpoons H_2(g) + O_2(g)$ (unbalanced)

  • (c) $NH_3(g) + O_2(g) \rightleftharpoons NO(g) + H_2O(g)$ (unbalanced)

• Solutions:

  • (a) Balanced: $2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g)$; $Q_c = \frac{[NO_2]^4[O_2]}{[N_2O_5]^2}$

  • (b) Balanced: $2H_2O(l) \rightleftharpoons 2H_2(g) + O_2(g)$; $Q_c = [H_2]^2[O_2]$ (liquid $H_2O$ is omitted).

  • (c) Balanced: $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$; $Q_c = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}$

Forms of $K$ and $Q$

• Direction of Equation (Equation 17.5): The reaction quotient/equilibrium constant for a forward reaction is the reciprocal of that for the reverse reaction.

  • $Q_{fwd} = \frac{1}{Q_{rev}}$; $K_{fwd} = \frac{1}{K_{rev}}$

• Multiplied Coefficients (Equation 17.6): If the coefficients of a balanced equation are multiplied by a common factor $n$, that factor becomes the exponent for the new constant.

  • $Q' = Q^n$; $K' = K^n$

• Sum of Reactions (Equation 17.7): If an overall reaction is the sum of two or more step reactions, the overall equilibrium constant is the product of the constants for the individual steps.

  • $Q_{overall} = Q_1 \times Q_2 \times ... \times Q_n$

  • $K_{overall} = K_1 \times K_2 \times ... \times K_n$

Sample Problem 17.2: Manipulating $K$

• Problem: Find $K$ at $700.\text{ K}$ for: $2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$

• Given Data ($700.\text{ K}$):

  1. $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$; $K_1 = 4.3 \times 10^{-4}$

  2. $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g)$; $K_2 = 4.1 \times 10^{-13}$

  3. $H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g)$; $K_3 = 1.3 \times 10^{10}$

• Plan:

  1. Reverse reaction (1): $K_{new1} = \frac{1}{K_1} = 2.3 \times 10^3$

  2. Multiply reaction (2) by 2: $K_{new2} = (K_2)^2 = 1.7 \times 10^{-25}$

  3. Multiply reaction (3) by 3: $K_{new3} = (K_3)^3 = 2.2 \times 10^{30}$

• Result: $K_{overall} = (2.3 \times 10^3)(1.7 \times 10^{-25})(2.2 \times 10^{30}) = 8.6 \times 10^8$

Expressing Equilibria with Pressure Terms ($K_p$)

• Equation 17.8: For gaseous reactions, $K$ can be expressed via partial pressures.

  • $P = \frac{n}{V}RT = [M]RT$

  • Relationship: $K_p = K_c(RT)^{\triangle n_{gas}}$

  • $\triangle n_{gas}$ = (mols of gaseous product) - (mols of gaseous reactant).

• Note: If the amount (mol) of gas does not change ($\triangle n_{gas} = 0$), then $K_p = K_c$.

Sample Problem 17.3: Converting $K_c$ to $K_p$

• Problem: Find $K_p$ for $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$ at $25.^\text{o}C$ if $K_c = 1.6 \times 10^{-4}$.

• Plan/Solution:

  • $\triangle n_{gas} = (1 + 1) - 0 = 2$

  • $T = 25.^\text{o}C + 273.15 = 298.15\text{ K}$

  • $R = 0.0821\text{ L}\times\text{atm}\times\text{mol}^{-1}\times\text{K}^{-1}$

  • $K_p = K_c(RT)^2 = (1.6 \times 10^{-4})(0.0821 \times 298.15)^2 = 0.096$

Determining Reaction Direction

• Comparing the relative sizes of $Q$ and $K$:

  • Case 1: Q < K: The ratio of products to reactants is less than at equilibrium. Reactants decrease, products increase. Reaction proceeds to the right (reactants $\rightarrow$ products).

  • Case 2: Q > K: The ratio of products to reactants is greater than at equilibrium. Products decrease, reactants increase. Reaction proceeds to the left (reactants $\text{←}$ products).

  • Case 3: $Q = K$: The system is at equilibrium. No net change occurs.

Sample Problem 17.4: Molecular Scenes

• Problem: Reaction $A(g) + B(g) \rightleftharpoons AB(g)$ where $K_c = 2$ at $175.^\text{o}C$. Given four scenes, determine direction.

• Calculation Method: Count red (A) and blue (B) spheres and purple (AB) molecules to find $Q = \frac{[AB]}{[A][B]}$.

• Scenario Results:

  1. Q = 15 > K: Proceeds left.

  2. $Q = 2 = K$: Equilibrium.

  3. Q = 6 > K: Proceeds left.

  4. Q = 0.33 < K: Proceeds right.

Sample Problem 17.5: Using Concentrations for Direction

• Problem: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$; $K_c = 0.21$ at $100.^\text{o}C$. Given: $[N_2O_4] = 0.12\text{ M}$ and $[NO_2] = 0.55\text{ M}$. Is it at equilibrium?

• Solution:

  • $Q_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.55)^2}{0.12} = 2.5$

  • Q_c (2.5) > K_c (0.21), therefore the reaction is not at equilibrium and will proceed to the left (from products to reactants).

Solving Equilibrium Problems: Reaction Tables (ICE)

• Definition: A reaction table (Initial, Change, Equilibrium) tracks concentrations or pressures.

• Structure:

  1. Balanced Equation.

  2. Initial quantities.

  3. Changes in quantities (defined by $x$ and stoichiometric coefficients).

  4. Equilibrium quantities (Initial + Change).

Sample Problem 17.6: Finding $K_c$

• Problem: $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$. A $2.00\text{-L}$ flask is filled with $0.200\text{ mol}$ of $HI$ at $453.^\text{o}C$. At equilibrium, $[HI]_{eq} = 0.078\text{ M}$. Calculate $K_c$.

• Step 1: Calculate Initial [HI]:

  • $[HI]_{init} = \frac{0.200\text{ mol}}{2.00\text{ L}} = 0.100\text{ M}$

• Step 2: Table:

  • Initial: $[HI]=0.100$, $[H_2]=0$, $[I_2]=0$

  • Change: $[HI]=-2x$, $[H_2]=+x$, $[I_2]=+x$

  • Equilibrium: $[HI]=0.100-2x$, $[H_2]=x$, $[I_2]=x$

• Step 3: Solve for $x$:

  • $0.100 - 2x = 0.078 \rightarrow 2x = 0.022 \rightarrow x = 0.011\text{ M}$

• Step 4: Calculate $K_c$:

  • $K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{(0.011)(0.011)}{(0.078)^2} = 0.020$

Sample Problem 17.8: Perfect Squares

• Problem: $I_2(g) + Cl_2(g) \rightleftharpoons 2ICl(g)$; $K_c = 81.9$. Initial $[I_2] = [Cl_2] = 0.100\text{ M}$.

• ICE Setup:

  • Equilibrium: $[I_2]=0.100-x$, $[Cl_2]=0.100-x$, $[ICl]=2x$

• Math: $K_c = \frac{(2x)^2}{(0.100-x)^2} = 81.9$

• Taking square root: $\frac{2x}{0.100-x} = \text{sqrt}(81.9) = 9.05$

• $2x = 0.905 - 9.05x \rightarrow 11.05x = 0.905 \rightarrow x = 0.0819\text{ M}$

• Equilibrium concentrations: $[I_2] = [Cl_2] = 0.018\text{ M}$, $[ICl] = 0.164\text{ M}$.

The Simplifying Assumption

• Condition: If $K$ is very small and the initial concentration is relatively large, the change ($x$) may be negligible.

• The 5% Rule: If the assumption results in a change that is less than 5.\text{%} of the initial concentration, the error is insignificant.

• Guidelines:

  • If \frac{[A]_{init}}{K_c} > 400, the assumption is generally justified.

  • If \frac{[A]_{init}}{K_c} < 400, the assumption is not justified; use the quadratic formula.

Sample Problem 17.9: Simplifying Assumption (Phosgene)

• Reaction: $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$; $K_c = 8.3 \times 10^{-4}$

• (a) 5.00 mol in 10.0 L ($0.500\text{ M}$):

  • $K_c = \frac{x^2}{0.500-x} \text{ approx. } \frac{x^2}{0.500} = 8.3 \times 10^{-4}$

  • $x = 0.020\text{ M}$. Check: \frac{0.020}{0.500} \times 100 = 4.0\text{%}. Justified.

• (b) 0.100 mol in 10.0 L ($0.0100\text{ M}$):

  • Assumed $x = 0.0029\text{ M}$. Check: \frac{0.0029}{0.0100} \times 100 = 29\text{%}. Not justified.

  • Must solve via quadratic formula: $x = 0.0025\text{ M}$.

Le Châtelier’s Principle

• Definition: When a chemical system at equilibrium is disturbed, it reattains equilibrium by undergoing a net reaction that reduces the effect of the disturbance.

Concentration Disturbances

• Add reactant: Equilibrium shifts right (toward products) to consume reactant.

• Remove reactant: Equilibrium shifts left (toward reactants) to produce more.

• Add product: Equilibrium shifts left.

• Remove product: Equilibrium shifts right.

• Note: Adding pure solids or liquids has no effect. Concentration changes do not change the value of $K$.

Pressure and Volume Disturbances

• Decrease Volume (Increase Pressure): Equilibrium shifts toward the side with fewer moles of gas to reduce pressure.

• Increase Volume (Decrease Pressure): Equilibrium shifts toward the side with more moles of gas.

• Add Inert Gas (at constant V): No effect on equilibrium position because concentrations and partial pressures of reactive gases remain unchanged.

• Note: Pressure/volume changes do not change the value of $K$.

Temperature Disturbances

• Exothermic Reaction (\triangle H < 0): Heat is a product ($A + B \rightleftharpoons C + D + \text{heat}$).

  • Increase $T$: Shifts left, $K$ decreases.

  • Decrease $T$: Shifts right, $K$ increases.

• Endothermic Reaction (\triangle H > 0): Heat is a reactant ($\text{heat} + A + B \rightleftharpoons C + D$).

  • Increase $T$: Shifts right, $K$ increases.

  • Decrease $T$: Shifts left, $K$ decreases.

• Critical Fact: Temperature is the only factor that affects the value of $K$.

The van’t Hoff Equation

• Relates equilibrium constants at two different temperatures:

• Equation 17.10: $\text{ln}\frac{K_2}{K_1} = -\frac{\triangle H^\text{o}_{rxn}}{R} \times (\frac{1}{T_2} - \frac{1}{T_1})$

• Where $K_1$ is at $T_1$ and $K_2$ is at $T_2$.

Catalysts and Equilibrium

• A catalyst speeds up a reaction by lowering the activation energy ($E_a$).

• It speeds up both forward and reverse reactions to the same extent.

• Result: A catalyst causes a reaction to reach equilibrium faster but has no effect on the equilibrium position or the value of $K$.

Industrial Application: The Haber Process

• Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$; $\triangle H^\text{o}_{rxn} = -91.8\text{ kJ}$

• Maximizing Yield:

  1. Remove $NH_3$: Continually liquefy and remove products to pull equilibrium right.

  2. High Pressure: Decreasing volume shifts equilibrium to the side with fewer gas molecules ($4 \rightarrow 2$), favoring $NH_3$. Industrial conditions: $200\text{--}300\text{ atm}$.

  3. Optimum Temperature: High $T$ increases rate but decreases $K$ (exothermic). Low $T$ increases yield but slows rate. Industrial compromise: approx. $400.^\text{o}C$ with a catalyst.