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

Review Drill & Check HW

CHM II HW Review PP-19 & 20 Complete the attached worksheet & turn it in tomorrow Lab Thursday

Are there any questions on previous material?

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) pP (aq) + qQ (aq) Rate f = k f [A] a [B] b Rate r = k r [P] p [Q] q At equilibrium, Rate f = Rate r k f [A] a [B] b = k r [P] p [Q] q

At equilibrium, Rate f = Rate r k f [A] a [B] b = k r [P] p [Q] q k f / k r = ([P] p [Q] q )/ ( [A] a [B] b ) k f / k r = K c = K eq in terms of concentration K c = ([P] p [Q] q )/ ( [A] a [B] b )

All Aqueous aA + bB pP + qQ

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

Drill: Solve Rate Expr: X + Y M + N fast 3M + N 2G fast 2N K fast 2G + K Prod. slow

Drill: Solve Rate Expr: X + Y M + N fast 3M + N 2G fast 2N K fast 2G + K Prod. slow

Work from the slow step up

Drill: Solve Rate Expr: X + Y M + N fast 3M + N 2G fast 2N K fast 2G + K Prod. slow

1) Cancel K & G

Drill: Solve Rate Expr: X + Y M + N fast 3M + N 2G fast 2N K fast 2G + K Prod. slow

1)Cancel K & G Triple rxn 1 & cancel

Drill: Solve Rate Expr: 3X + 3Y 3M + 3N 3M + N 2G fast 2N K fast 2G + K Prod. slow

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

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

1) Reverse step 3

Drill: Solve Rate Law A + B C + Dfast 4 C + A 2Gfast 4 D + B 2K fast G + K 2 Q + 2 Wfast Q + W Prod.slow

1)Reverse Rxn 3 Divide Rxn 4 by 2 & cancel

Drill: Solve Rate Law A + B C + Dfast 4 C + A 2Gfast 4 D + B 2K fast G/2 + K/2 Q + Wfast Q + W Prod.slow

1)Reverse Rxn 3 Divide Rxn 4 by 2 & cancel Divide Rxns 2 & 3 by 4 & cancel

Drill: Solve Rate Law A + B C + Dfast C + A/4 G/2fast D + B/4 K/2 fast G/2 + K/2 Q + Wfast Q + W Prod.slow

1)Reverse Rxn 3 Divide Rxn 4 by 2 & cancel Divide Rxns 2 & 3 by 4 & cancel Add the rxns

5/4 A + 5/4 B  Product Rate = k[A] 5/4 [B] 5/4

Review & Collect Drill & HW

CHM II HW Review PP-19 & 20 Remember M. S.

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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 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

Reaction Quotient where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium. aA (aq) + bB (aq)  pP (aq) + qQ (aq)

Drill: 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 :

Review & Collect Drill & HW

CHM II HW Review PP 19 & 20 Complete the attached assignment & turn it in tomorrow

Are there any questions on previous material?

Reaction Quotient where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium. aA (aq) + bB (aq)  pP (aq) + qQ (aq)

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

Equilibrium Calculations aA + bB  pP + qQ Stoichiometry is used to calculate the theoretical yield in a one directional rxn

Equilibrium Calculations aA + bB  pP + qQ In equilibrium rxns, no reactant gets used up; so, calculations are different

Equilibrium Calculations Set & balance rxn Assign amounts with x Write eq expression Substitute amounts Solve for x

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

Equilibrium Calculations COH 2 OCO 2 H x 50 - x x x P CO2 P H2 x 2 P CO P H2O (100-x) (50-x) K p = 3.4 x =KP=KP= Send to the next page

Equilibrium Calculations x 2 (100 -x)(50 - x) x x + x 2 x 2 = x x x x = 0 = = 3.4 x 10 -2

Equilibrium Calculations 0.966x x = 0 Use the quadratic equation to solve for x x = 11 x = -16 Substitute 11 back into the originally assigned #

Equilibrium Calculations COH 2 OCO 2 H x 50 - x x x P CO = 89 kPa P H2O = 39 kPa P CO2 = 11 kPa P H2 = 11 kPa

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 -1

Equilibrium Calculations XeF 2 XeF x x x P XeF2 x P Xe P F2 (50-x) (100-x) K p = 4.0 x =KP=KP= Send to the next page

Equilibrium Calculations x (50 -x)(100 - x) x x + x 2 x = x x x 2 -61x = 0 = = 4.0 x 10 -1

Equilibrium Calculations 0.40x x = 0 Use the quadratic equation to solve for x x = 48 x = 105 Substitute 48 back into the original assignmented #

Equilibrium Calculations XeF 2 XeF x x x P Xe = 2 kPa P F2 = 52 kPa P XeF2 = 48 kPa

SO 2 + O 2 SO 3 Determine the magnitude of the equilibrium constant if the partial pressure of each gas is Atm.

Drill: Write the equilibrium expression & solve its magnitude when P NO2 & P N2O4 = 50 kPa each at eq: N 2 O 4(g) NO 2(g)

Review Drill & Check HW

CHM II HW Review PP-20 Complete the attached HW

CHM II Schedule: Lab: Later this week Test Early next week

Are there any questions on previous material?

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

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:

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

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

Review & Collect Drill & HW

Schedule Lab: Friday Test: Next week

CHM II HW Review PP-20 Complete the attached assignment & turn it in tomorrow.

Are there any questions on previous material?

A (aq) + B (aq) AB (aq) Calculate the equilibrium concentration of each species when equal volumes of 0.40 M A & 0.20 M B are combined. K eq = 0.50

Equilibrium Calculations Xe (g) + F 2(g) XeF 2(g) Calculate the partial pressure of each portion when 80.0 kPa Xe & 60.0 kPa F 2 are combined: K p = 4.0 x 10 -2

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. K eq = 5.0 x 10 -6

Review & Collect Drill & HW

CHM II HW Review PPs 19 & 20 Review both for the Test on Monday.

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

Are there any questions on previous material?

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 K1K1 K2K2 K3K3

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 CuCl 6 -4 (aq) + 2 NH 3(aq) [Cu(NH 3 ) 2 Cl 4 ] -2 (aq) Calculate the molarity of each portion when 0.10 M CuCl 6 -4 & 1.0 M NH 3 are combined: K formation = 0.060

Equilibrium Calculations Rn (g) + F 2(g) RnF 2(g) Calculate the partial pressure of each portion when 25 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 = 6.0 x 10 -8

Review Drill & Check HW

Next Test Monday

Review for the Test

Rate Law aA + bB pP + qQ k[A] a [B] b Rate =

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

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

The data on the next slide was obtained in lab. Use that data to solve for the order with respect to each reactant, the reaction order, the rate expression, k, & E a.

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

Reaction Mechanism Step 1A Bfast Step 22 B 3Cfast Step 3C 2Dfast Step 4D  P 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:

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 = 2.00

Equilibrium Calculations I S 2 O 3 -2 S 4 O I - Calculate the equilibrium concentration of each portion when it’s 0.25 M I 2 & 0.50 M S 2 O 3 -2 at the start of the rxn. 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

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

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

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

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

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

A + B C + D C + H M + N N + T P + Q What happens all intermediates if:

1A + 1B1P + 1Q [A i ] = 0.20 MCalcu- [B i ] = 0.30 Mlate the [P i ] = 0.20 Meq con- [Q i ] = 0.30 Mcentra- K c = 0.020tion of ea.