# Chapter 15. Overview Equilibrium Reactions Equilibrium Constants K c & K p Equilibrium Expression product/reactant Reaction Quotient Q Calculations Le.

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Chapter 15

Overview Equilibrium Reactions Equilibrium Constants K c & K p Equilibrium Expression product/reactant Reaction Quotient Q Calculations Le Chatelier’s Principle –disturbing the equilibrium

Overview, cont’d All reactions are reversible Dynamic Equilibrium –When the rates of the forward and –reverse reactions are equal Reactions do not “go to completion” Cannot use stoichiometric methods to calculate amount of products formed

A B k f [A] = k r [B] forward rate = reverse rate [B ] = k f = constant [A] k r Equilibrium and Rates

Ratio of Products to Reactant –raised to each coefficient for example, –N 2 + 3H 2 2NH 3 –K = [NH 3 ] 2 [N 2 ][H 2 ] 3 Equilibrium Constant Expression

N 2 + 3H 2 2NH 3 Large K, more product Product Favored Small K, more reactant Reactant Favored cont’d

aA + bB cC + dD K c = [C] c [D] d [A] a [B] b products reactants concentration In General:

Forms of Eq. Constant Expression CO (g) + H 2 O (g) CO 2(g) + H 2(g) K c = [CO 2 ][H 2 ] [CO][H 2 O] 4HCl (g) + O 2(g) 2H 2 O (g) + 2Cl 2(g) K c = [Cl 2 ] 2 [H 2 O] 2 [O 2 ][HCl] 4

Cont’d 2HI H 2 + I 2 K f = [H 2 ][I 2 ] [HI] 2 H 2 + I 2 2HI K r = [HI] 2 = 1 [H 2 ][I 2 ] K f

Cont’d 2{ 2HI H 2 + I 2 } K f = [H 2 ][I 2 ] [HI] 2 4HI 2H 2 + 2I 2 K = K f 2 = [H 2 ] 2 [I 2 ] 2 [HI] 4

Example N 2 O 4 2NO 2 Initial Initial Equilibrium Equilibrium Kc N 2 O 4 NO 2 N 2 O 4 NO 2 0.00.020.0014 0.0172 0.211 0.00.030.0028 0.0243 0.211 0.00.040.0045 0.0310 0.213 0.020.00.0045 0.0310 0.213 Kc = [NO 2 ] 2 [N 2 O 4 ] generally unitless

N 2 + 3H 2 2NH 3 Large K, more product K > > 1 Product Favored N 2 + 3H 2 2NH 3 Small K, more reactant Reactant Favored K < < 1 Review values of K:

Example: N 2 + 3H 2 2NH 3 K c = 4.34 x 10 -3 at 300°C = [NH 3 ] 2 [H 2 ] 2 [N 2 ] What is K for reverse reaction? What is K for 2N 2 + 6H 2 4NH 3 ? What is K for 4NH 3 6H 2 + 2N 2 ? K c reverse = 230 K c = 1.88 x 10 -5 K c = 5.31 x 10 4

Heterogeneous Equilibria: When pure solid or liquid is involved –Pure solids & liquids do not appear in the equilibrium constant expression When H 2 O is a reactant or product and is the solvent –H 2 O does not appear in the equilibrium constant expression

Examples: CaCO 3(s) CaO (s) + CO 2(g) –conc. = mol = g/cm 3 = density cm 3 g/mol MM K c = [CaO] [CO 2 ] = (constant 1) [CO 2 ] [CaCO 3 ] (constant 2) K c ’ = K c (constant 2) = [CO 2 ] (constant 1) both are constant

Multi-Step Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K 1 = [Ag + ][Cl - ] Ag + (aq) + 2NH 3(aq) Ag(NH 3 ) 2 + (aq) K 2 = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 AgCl (s) + 2NH 3(aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) K tot = K 1 K 2 = [Ag(NH 3 ) 2 + ][Cl - ] [NH 3 ] 2

Problem: 1.00 mole of H 2 & 1.00 mole of I 2 are placed in a 1.0 L container at 520 K and allowed to come to equilibrium. Analysis reveals 0.12 mol of HI present at equilibrium. Calculate K c. H 2(g) + I 2(g) 2HI (g) initial 1.00 1.00 0 change -0.06 -0.06 +0.12 equil. 0.94 mol 0.94 mol +0.12 Kc = (0.12) 2 = 1.6 x 10 -2 (0.94)(0.94)

Conversion between K p and K c Kc –Equilibrium constant using concentrations Kp –Equilibrium constant using partial pressures Kp = Kc (RT)  n P = n RT V R = 0.0821 L atm/mol K T = Temperature in K  n = tot. mol product - tot. mol reactant

Problem: For 2SO 3(g) 2SO 2(g) + O 2(g) K c = 4.08 x 10 -3 at 1000 K. Calculate K p. K p = K c (RT)  n = 4.08 x 10 -3 (0.0821 x 1000) 1 K p = 0.0335

Problem: For 3H 2(g) + N 2(g) 2NH 3(g) K c = 0.105 at 472°C. Calculate K p. K p = K c (RT)  n = 0.105 (0.0821 x 745) -2 K p = 2.81 x 10 -5

Applications of Eq. Constants Reaction Quotient –Non-equilibrium concentrations used in the equilibrium constant expression Q = KReaction is at equilibrium Q > KReaction will shift left to equilibrium Q < KReaction will shift right to equilibrium

Problem: K c = 5.6 x 10 -12 at 500 K for I 2(g) 2I (g) [I 2 ] = 0.020 M & [I] = 2.0 x 10 -8 M. Is the reaction at equilibrium? Which direction will it shift to reach equilibrium? Q = [I ] 2 = (2.0 x 10 -8 ) 2 = 2.0 x 10 -14 [I 2 ] (0.020) Q < K (2.0 x 10 -14 ) < (5.6 x 10 -12 ) Reaction Shifts Right to get to equilibrium not at equilibrium because Q  K

Calculation of Eq. Concentrations use the stoichiometry of reaction initial concentration of all species change that occurs to all species equilibrium concentration of all species reaction will occur to reach the equilibrium point no matter the direction of reaction

Problem: Cyclohexane, C 6 H 12(g), can isomerize to form methylcyclopentane, C 5 H 9 CH 3(g). The equilibrium constant at 25°C is 0.12. If the original amount was 0.045 mol cyclohexane in a 2.8 L flask, what are the concentrations at equilibrium? C 6 H 12 C 5 H 9 CH 3 initial 0.045 mol 0 change -x +x equil. 0.045 - x x

Cont’d 0.12 = (x) (0.045 - x) Solve for x which is the equilibrium concentration of methylcyclopentane or the product x = 4.8 x 10 -3 mol C 5 H 9 CH 3 in 2.8 L flask [C 5 H 9 CH 3 ] = 1.7 x 10 -3 M [C 6 H 12 ] = 1.4 x 10 -2 M

Problem: For the reaction H 2 + I 2 2HI the K c = 55.64. You start with 1.00 mol H 2 and 1.00 mol I 2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species? H 2 + I 2 2HI initial 2.00 M 2.00 M 0 change -x -x +2x equil. 2.00 - x 2.00 - x 2x

Cont’d 55.64 = (2x) 2 (2.00 -x) 2 (55.64) ½ = (2x) 2 (2.00 -x) 2 x = 1.58 M [HI] = 3.16 M [H 2 ] = 0.42 M [I 2 ] = 0.42 M = 2x= 2.00 - x = 2.00 - x ½ perfect square

Problem: For the reaction H 2 + I 2 2HI the K c = 55.64. You start with 1.00 mol H 2 and 0.50 mol I 2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species? H 2 + I 2 2HI initial 2.00 M 1.00 M0 change -x-x +2x equil. 2.00 - x 1.00 - x 2x

Cont’d K c = [HI] 2 [H 2 ][I 2 ] 55.64 = (2x) 2 (2.00 -x)(1.00 -x) reduces to a quadratic equation: x 2 - 3.232 x + 2.155 = 0 not a perfect square

Quadratic Equation x = - b ± b 2 - 4ac 2a for ax 2 + bx + c = 0 ½

Cont’d x = +3.232 ± 10.446 - 8.62 2 x = 1.616 ± 0.6755 [HI] = 1.88 M [H 2 ] = 1.06 M [I 2 ] = 0.060 M = 2x = 2.00 - x = 1.00 - x ½

Le Chatelier’s Principle When a stress is applied to an equilibrium reaction, the equilibrium shifts to reduce the stress Types of Stress –Addition or removal of reactant –Addition or removal or product –Increase or decrease of temperature –Change in pressure or volume

2NOCl (g) Cl 2(g) + 2NO (g)  H  = +77 kJ temp. NOCl reaction shifts temp. NOCl reaction shifts pressure volume reaction shifts Cl 2 NO

[C 6 H 12 ][C 5 H 9 CH 3 ] initial 1.4 x 10 -2 + 1.0 x 10 -2 M1.7 x 10 -3 M change -x+x equil. 2.4 x 10 -2 - x1.7 x 10 -3 + x 0.12 = (1.7 x 10 -3 + x) x = 1.05 x 10 -3 M (2.4 x 10 -2 - x) [C 6 H 12 ] = 0.023 M[C 5 H 9 CH 3 ] = 0.0028 M Addition of Reactant or Product

Changes in Temperature will change K for an endothermic reaction –increasing T increases K for an exothermic reaction –increasing T decreases K

A B Energy Reactions Path without catalyst with catalyst rfrfr E a (f) E a (r) Effect of a Catalyst

2O 3(g) 3O 2(g) overall rxn O 3(g) O 2(g) + O (g) fast equil.rate1 = k 1 [O 3 ] rate2 = k 2 [O 2 ][O] O (g) + O 3(g) 2O 2(g) slow rate3 = k 3 [O][O 3 ] rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities Reaction Mechanisms & Equilibria k1k1 k3k3 k2k2

Substitution Method at equilibriumk 1 [O 3 ] = k 2 [O 2 ][O] rate3 =k 3 [O][O 3 ] [O] = k 1 [O 3 ] k 2 [O 2 ] rate3 = k 3 k 1 [O 3 ] 2 or k 2 [O 2 ] overall rate = k’ [O 3 ] 2 [O 2 ] substitute

Problem: Derive the rate law for the following reaction given the mechanism step below: OCl - (aq) + I - (aq) OI - (aq) + Cl - (aq) OCl - + H 2 O HOCl + OH - fast I - + HOCl HOI + Cl - slow HOI + OH - H 2 O + OI - fast k1k1 k2k2 k3k3 k4k4

Cont’d rate1 = k 1 [OCl - ][H 2 O] = rate 2 = k 2 [HOCl][OH - ] [HOCl] = k 1 [OCl - ][H 2 O] k 2 [OH - ] rate 3 = k 3 [HOCl][I - ] rate 3 = k 3 k 1 [OCl - ][H 2 O][I - ] k 2 [OH - ] overall rate = k’ [OCl - ][I - ] [OH - ] solvent

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