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

Drill: Solve Rate Law A + B C + Dfast 4 C + A 2Gfast 2 K 4D + B fast G + K 2 Q + 2 Wfast 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) Rate f = k f [A] a [B] b Rate r = k r [C] c [D] d At equilibrium, Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d

At equilibrium, Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d k f / k r = ([C] c [D] d )/ ( [A] a [B] b ) k f / k r = K c = K eq in terms of concentration K c = ([C] c [D] d )/ ( [A] a [B] b )

Reaction aA (g) + bB (g) cC (g) + dD (g) Rate f = k f P A a P B b Rate r = k r P C c P D d At equilibrium, Rate f = Rate r k f P A a P B b = k r P C c P D d

At equilibrium, Rate f = Rate r k f P A a P B b = k r P C c P D d k f / k r = (P C c P D d )/ ( P A a P B b ) k f / k r = K p = K eq in terms of pressure K p = (P C c P D d )/ ( P A a P B b )

All Aqueous aA + bB pP + qQ

Equilibrium Expression ( Products) p (Reactants) r K eq =

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 K p = K c (RT)  n gas

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

Equilibrium Expression aA (s) + bB (aq) cC (aq) + dD (aq) [C] c [D] d [B] b K eq =

Reaction Mechanism Sequence of steps that make up the total reaction process

Reaction Mechanism 1) A + B CFast 2) A + C DFast 3) B + D HFast 4) H + A -----> PSlow

Reaction Mechanism The rate determining step is the slowest step H + A ----> PSlow Rate = k 4 [H][A]

Reaction Mechanism Rate = k 4 [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 K 3 = [H]/([B][D]) [H] = K 3 [B][D]

Reaction Mechanism [H] = K 3 [B][D] Rate = k 4 [H][A] Rate = k 4 K 3 [B][D][A]

Reaction Mechanism 2) A + C D K 2 = [D]/([A][C]) [D] = K 2 [A][C]

Reaction Mechanism [D] = K 2 [A][C] Rate = k 4 K 3 [B][D][A] Rate = k 4 K 3 [B]K 2 [A][C][A] Rate = k 4 K 3 K 2 [B][A] 2 [C]

Reaction Mechanism 1) A + B C K 1 = [C]/([A][B]) [C] = K 1 [A][B]

Reaction Mechanism [C] = K 1 [A][B] Rate = k 4 K 3 K 2 [B][A] 2 [C] Rate = k 4 K 3 K 2 [B][A] 2 K 1 [A][B] Rate = k 4 K 3 K 2 K 1 [B] 2 [A] 3 Rate = K [B] 2 [A] 3

Solve Rate Expression 1) A + B 2CFast 2) A + C DFast 3) B + D 2HFast 4) 2H + A ----> PSlow

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 % RT WR

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

NH 3 H 2 + N 2 At a certain temperature at equilibrium P ammonia = 4.0 Atm, P hydrogen = 2.0 Atm, & P nitrogen = 5.0 Atm. Calculate K eq :

Equilibrium Applications When K > Q, the reaction goes forward When K < Q, the reaction goes in reverse

Drill: SO 2 + O 2 SO 3 Determine the magnitude of the equilibrium constant the the partial pressure of each gas is 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 P NO2 & P N2O4 = 50 kPa each: N 2 O 4(g) 2 NO 2(g)

Equilibrium Applications  G =  H - T  S  G = - RTlnK eq

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 + H 2 O CO 2 + H 2 Calculate the partial pressure of each portion at eq.when kPa CO & 50.0 kPa H 2 O are combined: K p = 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 + H 2 O CO 2 + H 2 Calculate the partial pressure of each portion when 100 kPa CO & 50 kPa H 2 O are combined: K p = 3.4 x 10 -2

Equilibrium Calculations COH 2 OCO 2 H x50 - x x x K p = P CO2 P H2 = x 2 P CO P H2O (100-x) (50-x) K p = 3.4 x 10 -2

Equilibrium Calculations x 2 x 2 (100 -x)(50 - x) = x + x 2 = 3.4 x x 2 = x x x x = 0

Equilibrium Calculations Xe (g) + F 2(g) XeF 2(g) Calculate the partial pressure of each portion when 50.0 kPa Xe & kPa F 2 are combined: K p = 4.0 x 10 -4

Equilibrium Calculations Xe (g) + 2 F 2(g) XeF 4(g) Calculate the partial pressure of each portion when 75 kPa Xe & 20.0 kPa F 2 are combined: K p = 4.0 x 10 -8

Drill: Solve for K A (aq) + 2 B (aq) C (s) + 2 D (aq) Calculate K eq if: [A] = 0.30 M [B] = 0.20 M C = 5.0 g [D] = 0.30 M

Drill: A (aq) + B (aq) D (aq) Calculate the equilibrium concentration of each species when equal volumes of 0.40 M A & 0.20 M B are combined. K eq = 0.050

Working with Equilibrium Constants

When adding Reactions Multiply Ks

ABK 1 B CK 2 ACK 3 K 3 = (K 1 )(K 2 )

Solve K for each: A + B C + D C + D P + Q A + B P + Q

When doubling Reactions Square Ks

ABK 1 2 A2 B K 2 K 2 = (K 1 ) 2

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

ABK 1 1/3 A1/3 B K 2 K 2 = (K 1 ) 1/3

When reversing Reactions Take 1/Ks

ABK 1 B AK 2 K 2 = 1/K 1

Equilibrium Calculations Kr (g) + F 2(g) KrF 2(g) Calculate the partial pressure of each portion when 40.0 kPa of Kr & 80.0 kPa of F 2 are combined: K p = 4.0 x 10 -2

Equilibrium Calculations Rn (g) + F 2(g) RnF 2(g) Calculate the partial pressure of each portion when 50 kPa Rn & 75 kPa F 2 are combined: K p = 4.0 x 10 -2

Drill: A + B P + Q Calculate the concentration of each portion at equilibrium when mL 0.50 M A is added to 150 mL 0.50 M B: K c = 4.0 x 10 -2

Equilibrium Calculations I S 2 O 3 -2 S 4 O I - Calculate the concentration of each portion when 100 mL 0.25 M I 2 is added to 150 mL 0.50 M S 2 O 3 -2 : K c = 4.0 x 10 -8

Clausius-Claperon Eq E a = R ln (T 2 )(T 1 ) k 2 (T 2 – T 1 ) k 1

Clausius-Claperon Eq H v = R ln (T 2 )(T 1 ) P 2 (T 2 – T 1 ) P 1

Clausius-Claperon Eq  H = R ln (T 2 )(T 1 ) K 2 (T 2 – T 1 ) K 1

 G  -  S  G o = -RTlnK

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

All aqueous aA + bB pP + qQ [P] p [Q] q [A] a [B] b Q = 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 K eq = 6.0 x 10 -5

Drill: Calculate the heat of reaction when K = 2.5 x at 27 o C, & K = 2.5 x at 127 o C.

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: K c = 2.0 x 10 -2

Next Test Tuesday

Review

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

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

Write the Eq Expression PQ (aq) P (aq) + Q (aq) Calculate [P], [Q], & [PQ] at equilibrium when [PQ] = 0.90 M at the start K eq = 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 K eq = 6.0 x 10 -5

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

Reaction Mechanism Step 1A Bfast Step 22 B 3Cfast Step 3C ---> Dslow

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 + O 2 SO 3 Calculate the equilibrium pressures if SO at 80.0 kPa is combined with O 2 at 40.0 kPa. K = 0.020