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Drill: •List five factors & explain how each affect reaction rates
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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
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)
• Ratef = kf[A]a[B]b
• Rater = kr[P]p[Q]q
• At equilibrium, Ratef = Rater
• kf[A]a[B]b = kr[P]p[Q]q
At equilibrium, Ratef = Rater
kf[A]a[B]b = kr[P]p[Q]q
kf /kr = ([P]p[Q]q)/ ( [A]a[B]b)
kf /kr = Kc = Keq in terms of concentrationKc = ([P]p[Q]q)/ ( [A]a[B]b)
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=
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 & GTriple 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 + D fast
4 C + A 2G fast
2 K 4D + B fast
G + K 2 Q + 2 W fast
Q + W Prod. slow
Drill: Solve Rate LawA + 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
1) Reverse step 3
Drill: Solve Rate LawA + B C + D fast
4 C + A 2G fast
4 D + B 2K fast
G + K 2 Q + 2 W fast
Q + W Prod. slow
1)Reverse Rxn 3Divide Rxn 4 by 2 & cancel
Drill: Solve Rate LawA + B C + D fast
4 C + A 2G fast
4 D + B 2K fast
G/2 + K/2 Q + W fast
Q + W Prod. slow
1) Reverse Rxn 3Divide Rxn 4 by 2 & cancelDivide Rxns 2 & 3 by 4 & cancel
Drill: Solve Rate LawA + B C + D fast
C + A/4 G/2 fast
D + B/4 K/2 fast
G/2 + K/2 Q + W fast
Q + W Prod. slow
1) Reverse Rxn 3Divide Rxn 4 by 2 & cancelDivide Rxns 2 & 3 by 4 & cancelAdd 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.
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• 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
Reaction
Quotient
ba
qpQ
BA
QP
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: NH3 H2 + N2
At a certain temperature at equilibrium Pammonia = 4.0 Atm, Phydrogen = 2.0 Atm, & Pnitrogen = 5.0 Atm.
Calculate Keq:
Review & Collect Drill
& HW
CHM II HW•Review PP 19 & 20
•Complete the attached assignment & turn it in tomorrow
Reaction
Quotient
ba
qpQ
BA
QP
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 + 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
100 -x 50 - x x x
PCO2PH2 x2
PCOPH2O (100-x) (50-x)
Kp = 3.4 x 10-2
=KP=
Send to the next page
Equilibrium Calculations x2
(100 -x)(50 - x)
x2
5000 -150x + x2
x2 = 170 - 5.1x + 0.034x2
0.966x2 + 5.1x - 170 = 0
=
= 3.4 x 10-2
Equilibrium Calculations0.966x2 + 5.1x - 170 = 0
Use the quadratic equation to solve for x
x = 11 x = -16Substitute 11 back into the
originally assigned #
Equilibrium Calculations CO H2O CO2 H2
100 -x 50 - x x x
100-11 50-11 11 11
PCO = 89 kPa PH2O = 39 kPa
PCO2 = 11 kPa PH2 = 11 kPa
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 10-1
Equilibrium Calculations Xe F2 XeF2
50 -x 100 - x x
PXeF2 x
PXePF2 (50-x) (100-x)
Kp = 4.0 x 10-1
=KP=
Send to the next page
Equilibrium Calculations x(50 -x)(100 - x)
x5000 -150x + x2
x = 2000 - 60x + 0.40x2
0.40x2 -61x + 2000 = 0
=
= 4.0 x 10-1
Equilibrium Calculations0.40x2 - 61x + 2000 = 0
Use the quadratic equation to solve for x
x = 48 x = 105Substitute 48 back into the
original assignmented #
Equilibrium Calculations Xe F2 XeF2
50 -x 100 - x x
50-48 100-48 48
PXe = 2 kPa PF2 = 52 kPa
PXeF2 = 48 kPa
SO2 + O2 SO3
• Determine the magnitude of the equilibrium constant if the partial pressure of each gas is 0.667 Atm.
Drill: Write the equilibrium expression & solve its magnitude
when PNO2 & PN2O4
= 50 kPa each at eq:N2O4(g) NO2(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
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
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 - TSG = - RTlnKeq
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
Review & Collect Drill
& HW
Schedule•Lab: Friday
•Test: Next week
CHM II HW•Review PP-20
•Complete the attached assignment & turn it in tomorrow.
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.
Keq = 0.50
Equilibrium CalculationsXe (g) + F2(g) XeF2(g)
Calculate the partial pressure of each portion when 80.0 kPa Xe & 60.0 kPa F2 are combined:
Kp = 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.
Keq = 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
We will return to the lab when the
class has no more field trips.
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
K1
K2
K3
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 CalculationsCuCl6
-4 (aq) + 2 NH3(aq) [Cu(NH3)2Cl4]-2
(aq)
Calculate the molarity of each portion when 0.10 M CuCl6
-4 & 1.0 M NH3 are combined: Kformation = 0.060
Equilibrium CalculationsRn(g) + F2(g) RnF2(g)
Calculate the partial pressure of each portion when 25 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 = 6.0 x 10-8
Review Drill&
Check HW
Next Test
•Monday
Reviewfor theTest
Rate LawaA + bB pP + qQ
k[A]a[B]bRate =
Equilibrium EquationaA + bB pP + qQ
[P]p[Q]q
[A]a[B]bKc =at equilibrium
Reaction QuotientaA + bB pP + qQ
[P]p[Q]q
[A]a[B]bQ =at the other conditions
The data on the next slidewas obtained in lab. Usethat data to solve for the order with respect to each reactant, the reaction order,the rate expression, k, & Ea.
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
Reaction Mechanism• Step 1 A <--> B fast
• Step 2 2 B <--> 3C fast
• Step 3 C <--> 2D fast
• Step 4 D 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 + O2 SO3
Calculate the equilibrium pressures if SO at 80.0 kPa
is combined with O2 at 40.0 kPa. K = 2.00
Equilibrium CalculationsI2 + 2 S2O3
-2 S4O6-2
+ 2 I-
Calculate the equilibrium concentration of each portion when it’s 0.25 M I2 & 0.50 M
S2O3-2 at the start of the rxn.
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
G-S
Go = -RTlnK
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
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
Drill:Calculate the heat of reaction when K =
2.5 x 10-6 at 27oC, &
K = 2.5 x 10-4 at 127oC.
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
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
A + B <---> C + D
C + H <---> M + N
N + T <---> P + Q
•What happens all intermediates if:
1A + 1B 1P + 1Q[Ai] = 0.20 M Calcu-[Bi] = 0.30 Mlate the[Pi] = 0.20 M eq con-[Qi] = 0.30 M centra-Kc = 0.020 tion of ea.
