Drill: List five factors & explain how each affect reaction rates.

Drill:•List five factors & explain how each affect reaction rates

Drill: Solve Rate LawA + B C + D fast

4 C + A 2G fast

2 K 4D + B fast

G + K Q + W fast

2Q + 2W Prod. slow

Chemical Equilibria

Equilibrium•The point at which the

rate of a forward reaction = the rate of its

reverse reaction

Equilibrium•The concentration of all

reactants & products become constant at

equilibrium

Equilibrium• Because concentrations

become constant, equilibrium is sometimes

called steady state

Equilibrium•Reactions do not stop at

equilibrium, forward & reverse reaction rates

become equal

Reaction• aA(aq)+ bB(aq) cC(aq)+ dD(aq)

• Ratef = kf[A]a[B]b

• Rater = kr[C]c[D]d

• At equilibrium, Ratef = Rater

• kf[A]a[B]b = kr[C]c[D]d

At equilibrium, Ratef = Rater

kf[A]a[B]b = kr[C]c[D]d

kf /kr = ([C]c[D]d)/ ( [A]a[B]b)

kf /kr = Kc = Keq in terms of concentrationKc = ([C]c[D]d)/ ( [A]a[B]b)

ReactionaA(g)+ bB(g)<-->cC(g)+ dD(g)

Ratef = kfPAaPB

b

Rater = krPCcPD

d

At equilibrium, Ratef = Rater

kfPAaPB

b = krPCcPD

d

At equilibrium, Ratef = Rater

kfPAaPB

b = krPCcPD

d

kf /kr = (PCcPD

d)/ ( PAaPB

b)

kf /kr = Kp = Keq in terms of pressureKp = (PC

cPDd)/ ( PA

aPBb)

ba

qp

cKBA

QP

All Aqueous

aA + bB pP + qQ

aA + bB(g) pP + qQ

ba

qp

PPP

PPK

BA

QP

Equilibrium Expression

( Products)p

(Reactants)rKeq=

AP CHM HW•Read: Chapter 12

•Work problems: 5, 7, & 12

•Page: 365

CHM II HW•Read: Chapter 17

•Work problems: 17 & 21

•Page: 745

Equilibrium Applications

•When K >1, [p] > [r]

•When K <1, [p] < [r]

Equilibrium Calculations

Kp = Kc(RT)ngas

Equilibrium Expression

•Reactants or products not in the same phase are not included in the equilibrium expression

Equilibrium ExpressionaA(s)+ bB(aq)<--> cC(aq)+ dD(aq)

[C]c [D]d

[B]b Keq=

Reaction Mechanism•Sequence of steps that

make up the total reaction process

Reaction Mechanism•1) A + B <---> C Fast

•2) A + C <---> D Fast

•3) B + D <---> H Fast

•4) H + A -----> P Slow

Reaction Mechanism•The rate determining step is the slowest step

•H + A ----> P Slow

•Rate = k4[H][A]

Reaction Mechanism•Rate = k4[H][A]

•Because H is not one of the original reactants, H cannot be used in a rate expression

Reaction Mechanism

•3) B + D <---> H

•K3 = [H]/([B][D])

•[H] = K3[B][D]

Reaction Mechanism

•[H] = K3[B][D]

•Rate = k4[H][A]

•Rate = k4K3[B][D][A]

Reaction Mechanism

•2) A + C <---> D

•K2 = [D]/([A][C])

•[D] = K2[A][C]

Reaction Mechanism

•[D] = K2[A][C]

•Rate = k4K3[B][D][A]• Rate = k4K3[B]K2[A][C][A]

• Rate = k4K3 K2[B][A]2[C]

Reaction Mechanism

•1) A + B <---> C

•K1 = [C]/([A][B])

•[C] = K1[A][B]

Reaction Mechanism• [C] = K1[A][B]

• Rate = k4K3 K2[B][A]2[C]

• Rate = k4K3 K2[B][A]2K1[A][B]

• Rate = k4K3 K2K1 [B]2[A]3

• Rate = K [B]2[A]3

Solve Rate Expression•1) A + B <---> 2C Fast

•2) A + C <---> D Fast

•3) B + D <---> 2H Fast

•4) 2H + A ----> P Slow

Reaction Mechanism• When one of the

intermediates anywhere in a reaction mechanism is altered, all intermediates are affected

Reaction Mechanism•1) A + B <---> C + D

•2) C + D <---> E + K

•3) E + K <---> H + M

•4) H + M <----> P

Lab Results % 100 80 60 40

RT5.21 8.42 11.9 21.7

WR 2.75 4.23 7.96 11.2

Applications of Equilibrium ConstantsApplications of Equilibrium Constants

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

where [A], [B], [P], and [Q] are molarities at any time.Q = K only at equilibrium.

NH3 H2 + N2

At a certain temperature at equilibrium Pammonia = 4.0 Atm, Phydrogen = 2.0 Atm, & Pnitrogen = 5.0 Atm. Calculate Keq:

Equilibrium Applications

•When K > Q, the reaction goes forward

•When K < Q, the reaction goes in reverse

Drill: SO2 + O2 SO3

• Determine the magnitude of the equilibrium constant the the partial pressure of each gas is 0.667 Atm.

Le Chatelier’s Principle

•If stress is applied to a system at equilibrium, the system will readjust to eliminate the stress

LC Eq Effects•A(aq) +2 B(aq) <--->

C(aq) + D(aq) + heat

•Write equilibrium exp:

•What happens if:

LC Eq Effects•2 A(aq) + B(s) <--->

C(aq) +2 D(aq) + heat

•Write equilibrium exp: What happens if:

LC Eq Effects•2 A(g) + 2 B(g) <--->

3 C(g) + 2 D(l)

•What happens if:

Drill: Write the equilibrium expression & solve when PNO2 & PN2O4 = 50 kPa each:

N2O4(g) 2 NO2(g)

Equilibrium Applications

G = H - TSG = - RTlnKeq

Equilibrium Calculations•aA + bB <--> cC + dD

•Stoichiometry is used to calculate the theoretical yield in a one directional rxn

Equilibrium Calculations•aA + bB <--> cC + dD

•In equilibrium rxns, no reactant gets used up; so, calculations are different

Equilibrium Calculations•CO + H2O CO2 + H2

• Calculate the partial pressure of each portion at eq.when 100.0 kPa CO & 50.0 kPa H2O are combined:

•Kp = 3.4 x 10-2

AP CHM HW•Read: Chapter 12

•Problems: 37 & 39

•Page: 367

CHM II HW•Read: Chapter 17

•Problems: 45

•Page: 747

Equilibrium Calculations• CO + H2O CO2 + H2

• Calculate the partial pressure of each portion when 100 kPa CO & 50 kPa H2O are combined:

• Kp = 3.4 x 10-2

Equilibrium Calculations• CO H2O CO2 H2

• 100 -x 50 - x x x

•Kp = PCO2PH2 = x2

• PCOPH2O (100-x) (50-x)

• Kp = 3.4 x 10-2

Equilibrium Calculations x2 x2

(100 -x)(50 - x) = 5000 -150x + x2

= 3.4 x 10-2

x2 = 170 - 5.1x + 0.034x2

0.966x2 + 5.1x - 170 = 0

Equilibrium CalculationsXe (g) + F2(g) XeF2(g)

Calculate the partial pressure of each portion when 50.0 kPa Xe & 100.0 kPa F2 are combined:

Kp = 4.0 x 10-4

Drill: Solve for K

A(aq)+ 2 B(aq) C(s)+ 2 D(aq)

Calculate Keq if:

[A] = 0.30 M [B] = 0.20 M

C = 5.0 g [D] = 0.30 M

Write the Eq Expression A(aq)+ 3 B(aq) C(g)+ 2 D(aq)

Calculate Keq if:

[A] = 0.40 M [B] = 0.20 M

C = 75 kPa [D] = 0.40 M

Working with Equilibrium Constants

When adding Reactions

Multiply Ks

A B K1

B C K2

A C K3

K3 = (K1)(K2)

Solve K for each:

A + B C + D

C + D P + Q

A + B P + Q

When doubling Reactions

Square Ks

A BK1

2 A 2 B K2

K2 = (K1)2

When a rxn is multiplied by any factor, that factor

becomes the exponent of K

A BK1

1/3 A 1/3 B K2

K2 = (K1)1/3

When reversing Reactions

Take 1/Ks

A B K1

B A K2

K2 = 1/K1

Equilibrium CalculationsKr (g) + F2(g) KrF2(g)

Calculate the partial pressure of each portion when 40.0 kPa of Kr & 80.0 kPa of F2 are combined:

Kp = 4.0 x 10-2

Equilibrium CalculationsXe (g) + 2 F2(g) XeF4(g)

Calculate the partial pressure of each portion when 75 kPa Xe & 20.0 kPa F2 are combined:

Kp = 4.0 x 10-8

Equilibrium CalculationsRn (g) + F2(g) RnF2(g)

Calculate the partial pressure of each portion when 50 kPa Rn & 75 kPa F2 are combined:

Kp = 4.0 x 10-2

Drill: A + B P + Q

Calculate the concentration of each portion at equilibrium when 100.0 mL 0.50 M A is added to 150 mL 0.50 M B:

Kc = 4.0 x 10-2

Equilibrium CalculationsI2 + 2 S2O3

-2 S4O6-2

+ 2 I-

Calculate the concentration of each portion when 100 mL 0.25 M I2 is

added to 150 mL 0.50 M S2O3-2:

Kc = 4.0 x 10-8

Clausius-Claperon Eq

Ea= R ln(T2)(T1) k2

(T2 – T1) k1

Clausius-Claperon Eq

Hv= R ln(T2)(T1) P2

(T2 – T1) P1

Clausius-Claperon Eq

H = R ln(T2)(T1) K2

(T2 – T1) K1

AP CHM HW•Problems: 41 & 43

•Page: 368

CHM II HW•Problems: 47

•Page: 747

Write the Eq Expression AB(aq) A(aq)+ B(aq)

Calculate [A], [B], & [AB] at equilibrium when [AB] = 0.60 M at the start

Keq = 6.0 x 10-5

Drill:Calculate the heat of reaction when K =

2.5 x 10-6 at 27oC, &

K = 2.5 x 10-4 at 127oC.

Drill: 1 A + 1 B 1 Z + 1 Y

Calculate the concentration of each portion at equilibrium

when 1.0 L 0.50 M A is added to 1.5 L 0.50 M B:

Kc = 2.0 x 10-2

Next Test

•Tuesday

Review

Experimental Results• Exp # [A] [B] [C] time

• 1 1.0 1.0 1.0 16• 2 2.0 1.0 1.0 2• 3 1.0 2.0 1.0 8• 4 1.0 1.0 2.0 4

Experimental Results• Exp # [A] [B] Rate

• 127 1.0 1.0 2.0 x 10-2

• 227 2.0 1.0 4.0 x 10-2

• 327 1.0 2.0 8.0 x 10-2

• 477 1.0 1.0 2.0

Write the Eq Expression PQ(aq) P(aq)+ Q(aq)

Calculate [P], [Q], & [PQ] at equilibrium when [PQ] = 0.90 M at the start

Keq = 9.0 x 10-5

Write the Eq Expression AB(aq) A(aq)+ B(aq)

Calculate [A], [B], & [AB] at equilibrium when [AB] = 0.60 M at the start

Keq = 6.0 x 10-5

Experimental Results• Exp # [A] [B] [C] Rate

• 1 0.1 0.1 0.2 2• 2 0.1 0.3 0.2 18• 3 0.1 0.1 0.8 8• 4 0.2 0.1 0.2 64

Reaction Mechanism• Step 1 A <--> B fast

• Step 2 2 B <--> 3C fast

• Step 3 C ---> D slow

LC Eq Effects•2 A(aq) + B(s) <--->

C(aq) +2 D(aq) + heat

•Write equilibrium exp: What happens if:

LC Eq Effects•3 A(g) + B(g) <--->

2 C(g) + 2 D(l)

•Write equilibrium exp:

•What happens if:

A + B <---> C + D

C + H <---> M + N

N + T <---> P + Q

•What happens all intermediates if:

SO + O2 SO3

Calculate the equilibrium pressures if SO at 80.0 kPa

is combined with O2 at 40.0 kPa. K = 0.020