The Reading is for the next class. Problems are for that day’s class. Problems for each week (MWF) are due the following Monday.

Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = or [C] c [D] d [A] a [B] b

Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = P A, P B, etc. not at equilibrium

Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = Q will go to the value of K as the partial pressures go to equilibrium

Q vs K can predict where the reaction is in respect to equilibrium.

PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K aA + bB cC + dD

PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K aA + bB cC + dD Too much reactant, not enough product.

PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K Q > K aA + bB cC + dD

PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K Q > K aA + bB cC + dD Too much product, not enough reactant.

PCPDPCPD PAPBPAPB ab c d K = Q = Q > K aA + bB cC + dD Q < K Large vs c d PCPDPCPD c d PCPDPCPD ab PAPBPAPB ab PAPBPAPB

Q = c d PCPDPCPD ab PAPBPAPB

Exercise page 291 P 4 2 P 2 K = 400 o C 1.40 mol P mol P 2 Volume = 25.0 L

P 4 2 P 2 Q = (P P ) 2 2 P 4

P 4 2 P 2 Q = (P P ) 2 2 P 4 P = nRT V T = 673 K V = 25.0 L R = L atm mol -1 K -1

P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = nRT V 4 = (1.4)( )(673) 25.0 = 3.09 atm

P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = nRT V 2 = (1.25)( )(673) 25.0 = 2.76 atm P P = nRT V 4 = (1.4)( )(673) 25.0 = 3.09 atm

P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = atm P P = atm = (2.76) =

P 4 2 P 2 (P P ) 2 2 P 4 P P = atm P P = atm = (2.76) = 2.46 Q = K = 1.39

P 4 2 P 2 (P P ) 2 2 P 4 P P = atm P P = atm = (2.76) = 2.46 Q = K = 1.39 P 4 2 P 2

Converting between partial pressures and concentrations.

Converting between partial pressures and concentrations. Concentration = moles V

Converting between partial pressures and concentrations. Concentration = moles V PV = nRT

Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V PV = nRT

Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V = PAPA RT PV = nRT

Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V = PAPA RT P A = RT[A] PV = nRT

P A = RT[A] 2 NO 2(g) N 2 O 4(g)

P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm

P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm P N O 24 /P ref (P NO 2 /P ref ) 2 = K

P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm 4 /P ref (P NO 2 /P ref ) 2 = K P N O = RT[N 2 O 4 ]

P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm 4 /P ref (P NO 2 /P ref ) 2 = K P N O = RT[N 2 O 4 ] P NO 2 = RT[NO 2 ]

2 NO 2(g) N 2 O 4(g) 4 /P ref (P NO 2 /P ref ) 2 = K P N O = RT[N 2 O 4 ] P NO 2 = RT[NO 2 ] [N 2 O 4 ](RT/P ref ) [NO 2 ] 2 (RT/P ref ) = [N 2 O 4 ] [NO 2 ] 2 x RT P ref ) ( K =

2 NO 2(g) N 2 O 4(g) [N 2 O 4 ](RT/P ref ) [NO 2 ] 2 (RT/P ref ) = x K = [N 2 O 4 ] [NO 2 ] 2 [N 2 O 4 ] [NO 2 ] 2 = K RT P ref ) ( RT P ref ) (

[N 2 O 4 ] [NO 2 ] 2 = K RT P ref ) ( aA + bB cC + dD c = K [C] c [D] d PCPDPCPD PAPBPAPB ab d [A] a [B] b = K ( RT P ref ) a+b-c-d ?

[C] c [D] d [A] a [B] b = K ( RT P ref ) a+b-c-d Exercise page 297 CH 4 + H 2 O CO + 3 H 2 K = [H 2 ]=[CO]=[H 2 O]= mol L -1

[C] c [D] d [A] a [B] b = K ( RT P ref ) a+b-c-d CH 4 + H 2 O CO + 3 H 2 K = [H 2 ]=[CO]=[H 2 O]= mol L -1 ( ) 4 [CH 4 ]( ) = ( RT P ref ) -2

[CH 4 ] = ( RT P ref ) -2 CH 4 + H 2 O CO + 3 H 2 [CH 4 ] = K(RT) -2 = x ( ) x = 8.39 x mol L -1

Le Chatelier’s Principle

A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress.

Le Chatelier’s Principle Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation of the factor considered in the opposite direction.

Le Chatelier’s Principle Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation of the factor considered in the opposite direction.

Le Chatelier’s Principle A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. Stress = a factor affecting equilibrium

K = [C] c [D] d [A] a [B] b Anything causing a change in the concentration (or partial pressure) of a reactant or product is a stress.

Factors affecting equilibrium:

temperature

Factors affecting equilibrium: temperature pressure

Factors affecting equilibrium: temperature pressure volume

Factors affecting equilibrium: temperature pressure removal or addition of product volume

Factors affecting equilibrium: temperature pressure removal or addition of reactant volume removal or addition of product

Removal or addition of a reactant or product.

Removal or addition of a reactant or product. Increases or decreases concentration.

PAPA [A] PBPB [B] Removal or addition of a reactant or product. Increases or decreases concentration.

K = [B] [A]

I 2 + H 2 2 HI

P HI = atm PIPI 2 = atm PHPH 2 = atm

I 2 + H 2 2 HI = atm = atm = atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K

I 2 + H 2 2 HI = atm = atm = atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K (3.009) 2 (0.4756)(0.2056) =

I 2 + H 2 2 HI = atm = atm = atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K (3.009) 2 (0.4756)(0.2056) = 92.60

I 2 + H 2 2 HI Increase P I 2 to 2.00 atm.

I 2 + H 2 2 HI K = Start P PP 2 nd eq P Increase P I 2 to atm

I 2 + H 2 2 HI K = Start P PP 2 nd eq P Increase P I 2 to atm -x -x +2x

I 2 + H 2 2 HI K = Start P PP 2 nd eq P Increase P I 2 to atm -x -x +2x x x x

I 2 + H 2 2 HI K = Start P PP 2 nd eq P Increase P I 2 to atm -x -x +2x x x x 92.6 = ( x) 2 (2.000-x)( x) =

92.6 = ( x) 2 (2.000-x)( x) =

92.6 = ( x) 2 (2.000-x)( x) = ( x + 4x 2 ) ( x + x 2 ) = 92.6

92.6 = ( x) 2 (2.000-x)( x) = ( x + 4x 2 ) = x + 4x 2 = 92.6( x + x 2 ) ( x + x 2 )

92.6 = ( x) 2 (2.000-x)( x) = ( x + 4x 2 ) = 92.6 ( x + x 2 ) ( x + x 2 ) x + 4x 2 = x x x + 4x 2 = 92.6

x + 4x 2 = x x x x = 0 x = -b  b ac 2a abc =

x + 4x 2 = x x x x = 0 x = -b  b ac 2a abc = or 2.30

Start P PP 2 nd eq P -x -x +2x x x x or 2.30x = I 2 + H 2 2 HI

Start P PP 2 nd eq P -x -x +2x x x x or 2.30x = I 2 + H 2 2 HI x = 2.30 would give a negative pressure for I 2

Start P PP 2 nd eq P or 2.30x = I 2 + H 2 2 HI

Calculate K with new partial pressures.

Calculate K with new partial pressures. (3.293) 2 (1.858)(0.0636) =

Calculate K with new partial pressures. (3.293) 2 (1.858)(0.0636) = K given = 92.60

Changing volume of system

V  1 P

Changing volume of system V  1 P If V is reduced, P increases. PV = nRT

Changing volume of system V  1 P If V is reduced, P increases. Le Chatlier’s principle requires that the equilibrium shift so that P decreases.

2 NO 2(g) N 2 O 4(g)

2 NO 2(g) N 2 O 4(g) An increase in pressure should favor an increase in the product, N 2 O 4.

2 NO 2(g) N 2 O 4(g) An increase in pressure should favor an increase in the product, N 2 O 4. Each N 2 O 4 produced means a loss of two NO 2 molecules, a net loss of one molecule and a lower pressure.

2 NO 2(g) N 2 O 4(g) A decrease in pressure favors a shift to NO 2.

Raising or lowering the temperature

Some reactions liberate heat to form products.

Raising or lowering the temperature Some reactions liberate heat to form products. This is an exothermic reaction.

Raising or lowering the temperature This is an endothermic reaction. Other reactions absorb heat to produce products.

Raising the temperature of an endothermic reaction will favor the formation of more product. R P

Raising the temperature of an exothermic reaction will favor the formation of more reactant. R P

NaOH (s) + H 2 O (l) Na + (aq) + OH - (aq) + H 2 O (l)

The solution of solid sodium hydroxide into water is exothermic.

K = [P] [R]

K = [P] [R] If a reaction is exothermic and the temperature is raised, K will decrease.

K = [P] [R] If a reaction is endothermic and the temperature is raised, K will increase.

Driving reactions to completion

Completion = 100% yield of product

Cl - (aq) + Ag + (aq) AgCl (s)

AgCl precipitates from the solution.

Cl - (aq) + Ag + (aq) AgCl (s) AgCl precipitates from the solution. As the AgCl precipitates, product is removed from solution.

Cl - (aq) + Ag + (aq) AgCl (s)

N H 2 2 NH 3 All gases

N H 2 2 NH 3 All gases exothermic

N H 2 2 NH 3 All gases exothermic cool

N H 2 2 NH 3 All gases exothermic cool

N H 2 2 NH 3 Although a lower temperature favors more NH 3 formed, the lower temperature also leads to a very slow reaction.

N H 2 2 NH 3 An increase in pressure should favor product.

N H 2 2 NH 3 An increase in pressure should favor product.

N H 2 2 NH 3 Ultimate solution: react at high Temperature to speed up reaction, cool until NH 3 becomes liquid. Remove from reaction vessel and repeat.

(time) CONCENTRATIONCONCENTRATION

Heterogeneous equilibrium

Involves at least two phases.

Heterogeneous equilibrium Involves at least two phases. What is the concentration of a pure liquid or a pure solid?

Concentrations are not a valid way to define a pure liquid or solid.

Concentrations are not a valid way to define a pure liquid or solid. Moles water Liters solvent =?

The concentration of a pure liquid or solid is defined as 1.

Law of Mass Action

1. Gases enter equilibrium expressions as partial pressures in atmospheres.

Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L -1.

Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L Pure solids and liquids are represented by 1 at equilibrium, a dilute solvent is 1.

Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L Pure solids and liquids are represented by 1 at equilibrium, a dilute solvent is Partial pressures or concentrations of products appear in the numerator, reactants in the denominator. Each is raised to the power of its coefficient.

hemoglobin

Fe

heme Fe

Fe heme + O 2 Fe heme O 2

4 heme Fe sites to bond O 2

Increase P O more Fe heme O 2 Fe heme + O 2 Fe heme O 2 2

4 heme Fe sites to bond O 2 Fe heme O 2 Fe heme + O 2 Fe heme O 2 Increase P O by changing local atmosphere. 2 Increase P O more 2

4 heme Fe sites to bond O 2 Decrease P O less Fe heme O 2 Fe heme + O 2 Fe heme O 2 2

4 heme Fe sites to bond O 2 Decrease P O less Fe heme O 2 Fe heme + O 2 Fe heme O 2 2 Decrease P O 2 by going to higher altitude.

Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2

Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm Affinity for CO over O 2 factor of 200+.

Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm Affinity for CO over O 2 factor of CO 1 x 10 3 ppm < 1% P CO < 0.01 atm

Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm CO 1 x 10 3 ppm < 1% P CO < 0.01 atm K heme-CO >> K heme-O 2