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

Drill: Solve Rate Law A + B C + D fast 4 C + A 2G fast 2 K 4D + B fast G + K 2 Q + 2 W fast Q + W 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

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 concentration Kc = ([C]c[D]d)/ ( [A]a[B]b)

aA(g)+ bB(g)<-->cC(g)+ dD(g) Reaction aA(g)+ bB(g)<-->cC(g)+ dD(g) Ratef = kfPAaPBb Rater = krPCcPDd At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd

At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd kf /kr = (PCcPDd)/ ( PAaPBb) kf /kr = Kp = Keq in terms of pressure Kp = (PCcPDd)/ ( PAaPBb)

All Aqueous aA + bB pP + qQ

Equilibrium Expression ( Products)p (Reactants)r Keq=

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)Dngas

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

Keq= [C]c [D]d [B]b Equilibrium Expression aA(s)+ bB(aq)<--> cC(aq)+ dD(aq) [C]c [D]d [B]b Keq=

Sequence of steps that make up the total reaction process 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

K3 = [H]/([B][D]) [H] = K3[B][D] Reaction Mechanism

Reaction Mechanism [H] = K3[B][D] Rate = k4[H][A] Rate = k4K3[B][D][A]

K2 = [D]/([A][C]) [D] = K2[A][C] Reaction Mechanism

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]

K1 = [C]/([A][B]) [C] = K1[A][B] Reaction Mechanism

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 RT 5.21 8.42 11.9 21.7 WR 2.75 4.23 7.96 11.2

Applications of Equilibrium Constants 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 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 DG = DH - TDS DG = - 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 Set & balance rxn Assign amounts Write eq expression Substitute amounts Solve for x

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 Calculations Xe (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

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

A(aq)+ B(aq) D(aq) Drill: Calculate the equilibrium concentration of each species when equal volumes of 0.40 M A & 0.20 M B are combined. Keq = 0.50

Equilibrium Calculations Xe (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

Drill: A + B C + D Calculate the equilibrium concentration of each species when equal volumes of 0.60 M A & 0.80 M B are combined. Keq = 0.50

The Test on Rxn Rates & Chemical Equilibria will be on Tuesday

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 B K1 2 A 2 B K2 K2 = (K1)2

When a rxn is multiplied by any factor, that factor becomes the exponent of K

A B K1 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 Calculations Kr (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 Calculations Rn (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 Calculations I2 + 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 (T2)(T1) k2 (T2 – T1) k1 Ea= R ln

Clausius-Claperon Eq (T2)(T1) P2 (T2 – T1) P1 Hv= R ln

Clausius-Claperon Eq (T2)(T1) K2 (T2 – T1) K1 DH = R ln

DG = DH - TDS DGo = -RTlnK

All aqueous aA + bB pP + qQ [P]p[Q]q [A]a[B]b Kc = at equilibrium

Q = [P]p[Q]q [A]a[B]b All aqueous aA + bB pP + qQ at the other conditions

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

Calculate the heat of reaction when K = 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

Drill: 1 A + 1 B 1 Z Calculate the concentration of each portion at equilibrium when the original solution has 0.50 M A & 0.50 M B: Kc = 2.0 x 10-2

Review

Experimental Results Exp # [A] [B] [C] time 1 1.0 1.0 1.0 16 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 10-2 477 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 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 SO + O2 SO3 Calculate the equilibrium pressures if SO at 80.0 kPa is combined with O2 at 40.0 kPa. K = 0.020

1A + 1B. 1P + 1Q [Ai] = 0. 20 M. Calcu- [Bi] = 0. 30 M 1A + 1B 1P + 1Q [Ai] = 0.20 M Calcu- [Bi] = 0.30 M late the [Pi] = 0.20 M eq con- [Qi] = 0.30 M centra- Kc = 0.020 tion of ea.