drill: list five factors & explain how each affect reaction rates
TRANSCRIPT
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
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)
Equilibrium Expression
•Reactants or products not in the same phase are not included in the equilibrium expression
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
•[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• [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
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
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
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
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
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: 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
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