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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. - PowerPoint PPT Presentation
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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
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)
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=
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•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
Reaction Mechanism
•3) B + D <---> H
•K3 = [H]/([B][D])
•[H] = K3[B][D]
Reaction Mechanism
•[H] = K3[B][D]
•Rate = k4[H][A]
•Rate = k4K3[B][D][A]
Reaction Mechanism
•2) A + C <---> D
•K2 = [D]/([A][C])
•[D] = K2[A][C]
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
•1) A + B <---> C
•K1 = [C]/([A][B])
•[C] = K1[A][B]
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
RT5.21 8.42 11.9 21.7
WR 2.75 4.23 7.96 11.2
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
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
& slove for:
N2O4(g) 2 NO2(g)
Equilibrium Applications
G = H - TSG = - 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•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
Write the Eq Expression 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 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 10-6
Equilibrium CalculationsXe (g) + F2(g) XeF2(g)
Calculate the partial pressure of each portion when 40.0 kPa Xe & 80.0 kPa F2 are combined:
Kp = 4.0 x 10-6
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 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
Equilibrium Constants
A + B C + D
C + D P + Q
A + B P + Q
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
Drill:Calculate the heat of reaction when K =
2.5 x 10-4 at 27oC, &
K = 2.5 x 10-6 at 127oC.
Review
Experimental Results• Exp # [A] [B] [C] Rate
• 1 1.0 1.0 1.0 2• 2 2.0 1.0 1.0 32• 3 1.0 2.0 1.0 8• 4 1.0 1.0 2.0 16
Write the Eq Expression AB(aq) A(aq)+ B(aq)
Calculate [A], [B], & [AB] at equilibrium when [AB] = 0.40 M at the start
Keq = 4.0 x 10-7
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
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 80.0 kPa is combined with O2 at 40.0 kPa.
K = 1.0
