Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

CHAPTER 15 Chemical Equilibrium Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K) Reaction Quotient (Q) Kc vs Kp ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle van’t Hoff Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Dynamic Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Chemical equilibrium exists when Dynamic Eq Equilibrium Chemical equilibrium exists when Rates of forward and reverse reactions are equal Reaction appears to stop Concentration of reactants and products do not change over time Remain constant Both forward and reverse reaction never cease Equilibrium signified by double arrows ( ) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

N2O4 2 NO2 Dynamic Eq Equilibrium Initially have only N2O4 Only forward reaction As N2O4 reacts NO2 forms As NO2 forms Reverse reaction begins to occur NO2 collide more frequently as concentration of NO2 increases Eventually, equilibrium is reached Concentration of N2O4 does not change Concentration of NO2 does not change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Dynamic Eq Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

N2O4 2NO2 Dynamic Eq Equilibrium Closed system Equilibrium can be reached from either direction Independent of whether it starts with “reactants” or “products” Always have the same composition at equilibrium under same conditions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Reactants Equilibrium Products N2O4 2NO2 Dynamic Eq Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Mass Action Expression Dynamic Eq Mass Action Expression Simple relationship among [reactants] and [products] for any chemical system at equilibrium Called the mass action expression Derived from thermodynamics Forward reaction: A  B Reverse reaction: A  B At equilibrium: A B Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Uses stoichiometric coefficients as exponent for each reactant Dynamic Eq Reaction Quotient Uses stoichiometric coefficients as exponent for each reactant For reaction: aA + bB cC + dD Reaction quotient Numerical value of mass action expression Equals “Q ” at any time, and Equals “K ” only when reaction is known to be at equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C Exp’t Initial Amts Equil’m Amts Equil’m [M] I 1.00 mol H2 0.222 mol H2 0.0222 M H2 10 L 1.00 mol I2 0.222 mol I2 0.0222 M I2 0.00 mol HI 1.56 mol HI 0.156 M HI II 0.00 mol H2 0.350 mol H2 0.0350 M H2 10 L 0.100 mol I2 0.450 mol I2 0.0450 M I2 3.50 mol HI 2.80 mol HI 0.280 M HI

Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C Exp’t Initial Amts Equil’m Amts Equil’m [M] III 0.0150 mol H2 0.150 mol H2 0.0150 M H2 10 L 0.00 mol I2 0.135 mol I2 0.0135 M I2 1.27 mol HI 1.00 mol HI 0.100 M HI IV 0.00 mol H2 0.442 mol H2 0.0442 M H2 10 L 0.00 mol I2 0.442 mol I2 0.0442 M I2 4.00 mol HI 3.11 mol HI 0.311 M HI

Mass Action Expression = same for all data sets at equilibrium Equilibrium Concentrations (M ) Exp’t [H2] [I2] [HI] I 0.0222 0.156 II 0.0350 0.0450 0.280 III 0.0150 0.0135 0.100 IV 0.0442 0.311 Average = 49.5

Write mass action expressions for the following: 2NO2(g) N2O4(g) Group Problem Write mass action expressions for the following: 2NO2(g) N2O4(g) 2CO(g) + O2(g) 2CO2(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)? Group Problem Which of the following is the correct mass action expression for the reaction: Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Equilibrium Laws Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

at equilibrium write the following equilibrium law Equilibrium Laws For reaction H2(g) + I2(g) 2HI(g) at 440 ˚C at equilibrium write the following equilibrium law Equilibrium constant = Kc = constant at given T Use Kc since usually working with concentrations in mol/L For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Predicting Equilibrium Laws For general chemical reaction: dD + eE fF + gG Where D, E, F, and G represent chemical formulas d, e, f, and g are coefficients Mass action expression is Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation. Equilibrium law is: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Predicting Equilibrium Laws Only concentrations that satisfy this equation are equilibrium concentrations Numerator Multiply concentration of products raised to their stoichiometric coefficients Denominator Multiply concentration reactants raised to their stoichiometric coefficients is scientists’ convention Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

What is equilibrium law? Example 3H2(g) + N2(g) 2NH3(g) Kc = 4.26 × 108 at 25 °C What is equilibrium law? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Various operations can be performed on equilibrium expressions 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original A + B C + D C +D A + B Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Operations 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original 3H2(g) + N2(g) 2 NH3(g) at 25˚C 2NH3(g) 3H2(g) + N2(g) at 25 ˚C Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Operations 2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor. A + B C + D 3A + 3B 3C + 3D Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Operations When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor 3H2(g) + N2(g) 2NH3(g) at 25 ˚C Multiply by 3 9H2(g) + 3N2(g) 6NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Operations 3. When chemical equilibria are added, their equilibrium constants are multiplied A + B C + D C + E F + G A + B + E D + F + G Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g) Equilibrium Operations 3. When chemical equilibria are added, their equilibrium constants are multiplied 2 NO2(g) NO3(g) + NO(g) NO3(g) + CO(g) NO2(g) + CO2(g) NO2(g) + CO(g) NO(g) + CO2(g) Therefore Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

For: N2(g) + 3H2(g) 2NH3(g) Kc = 500 at a particular temperature. Group Problem For: N2(g) + 3H2(g) 2NH3(g) Kc = 500 at a particular temperature. What would be Kc for following? 2NH3(g) N2(g) + 3H2(g) 1/2N2(g) + 3/2H2(g) NH3(g) 0.002 22.4 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 15 Chemical Equilibrium Equilibrium Constant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equilibrium Constant Kc Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides Sometimes partial pressures, in atmospheres, may be used in place of concentrations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Based on reactions in which all substances are gaseous Equilibrium Kp Based on reactions in which all substances are gaseous Gas quantities are expressed in atmospheres in mass action expression Use partial pressures for each gas in place of concentrations e.g. N2(g) + 3H2(g) 2NH3(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Relationship between Kp and Kc Equilibrium Relationship between Kp and Kc Start with ideal gas law PV = nRT Rearranging gives Substituting P/RT for molar concentration into Kc results in pressure-based formula ∆n = moles of gas in product – moles of gas in reactant Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Kc = 11.7 Consider the reaction: 2NO2(g) N2O4(g) Group Problem Consider the reaction: 2NO2(g) N2O4(g) If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at same temperature? n = nproducts – nreactants = 1 – 2 = –1 Kc = 11.7 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × (0.082057 × 298.15)1 Kp = 24 Group Problem Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp? 0.99 2.0 24 2400 None of these Δn = (4 – 3) = 1 Kp = Kc(RT)Δn Kp= 0.99 × (0.082057 × 298.15)1 Kp = 24 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Homogeneous and Hetergeneous Equilibrium Homogeneous and Hetergeneous Homogeneous reaction/equilibrium All reactants and products in same phase Can mix freely Heterogeneous reaction/equilibrium Reactants and products in different phases Can’t mix freely Solutions are expressed in M Gases are expressed in M Governed by Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Law = Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Law = Can write in simpler form For any pure liquid or solid, ratio of moles to volume of substance (M ) is constant e.g. 1 mol NaHCO3 occupies 38.9 cm3 2 mol NaHCO3 occupies 77.8 cm3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Equilibrium Heterogeneous 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s) Heterogeneous Write equilibrium laws for the following: Ag+(aq) + Cl–(aq) AgCl(s) H3PO4(aq) + H2O H3O+(aq) + H2PO4–(aq) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) Group Problem Given the reaction: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) What is the mass action expression? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) Group Problem Given the reaction: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) What is mass action expression for the reverse reaction? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E