PP 20: Chemical Equilibria
Drill: List five factors & explain how each affect reaction rates
Drill: Solve Rate Expr:
Reaction Mechanism
X + Y  M + N
3 M + N  3 G
2 N  3 K
2 G + 2 K  Prod.
Rxn Rate
fast
fast
fast
slow
Chemical Equilibria
Equilibrium:
• The point at which the rate of a forward reaction equals the rate of its reverse reaction
• The concentration of all reactants & products become constant at equilibrium
• Because concentrations become constant, equilibrium is sometimes called steady state
• Reactions do not stop at equilibrium, forward & reverse reaction rates become equal
General Reaction:
aA + bB  pP + qQ
Finding the equilibrium constant for aqueous equilibria:
aA(aq) + bB(aq)  pP(aq) + qQ(aq)
Forward Rxn Rate Law:
Ratef = kf[A]a[B]b
Reverse Rxn Rate Law:
Rater = kr[P]p[Q]p
At equilibrium:
Ratef = Rater
Thus:
kf[A]a[B]b = kr[P]p[Q]p
Isolate the k’s:
kf / kr = [P]p[Q]p / [A]a[B]b
kf / kr is a new constant (K): (K or Keq or Kc for concentration)
Kc = [P]p[Q]p
[A]a[B]b
Equilibrium constants in term of pressure are the same except for the units:
KP = PPp PQq
PAa PBb
Equilibrium Constants: Keq= (Products)p / (Reactants)r
Equilibrium Applications:
• When K >1, [P] > [R]
• When K <1, [P] < [R]
What do you know when:
• When K = 2.3 x 103
• When K = 3.4 x 10-2
Relationship between Kc & Kp:
Kp = Kc(RT)ngas
Equilibrium Expression:
•
Reactants or products not in the same phase are not included in the equilibrium expression
aA(s )+ bB(aq)  cC(aq) + dD(aq)
Kc = [C]c[D]d / [B]b
Problem: Solve for the magnitude of Keq when A = 5.0 g, [B] = 0.20 M, [C] = 0.10 M, & [D] = 0.40 M
for the following reaction: 2 A(s) + 3 B(aq)  2 C(aq ) + 2 D(aq)
Drill: Calculate the value of Kp using the given amounts in the following reaction:
__ PH3(g)  __ H2(g) + __ P4(s)
(PPH3 = 4.0 Atm & PH2 = 2.0 Atm & mPP4 = 5.0 g)
Calculations to determine the amounts of each substance at equilibrium: (Equilibrium Calculations)
•
•
Stoichiometry is used to calculate the theoretical yield in a one directional rxn
In equilibrium rxns, no reactant gets used up; so, calculations are different
•
The steps for solving equilibrium are as follows:
1. Set & balance rxn
2. Assign amounts with x
3. Write eq formula
4. Substitute amounts
5. Solve for x
Demonstration of an Equilibrium Calculation:
Problem: Use the following reaction to determine the partial pressure of each portion at equilibrium
when 100.0 kPa CO & 50.0 kPa H2O are combined: (Kp for the rxn is 3.4 x 10-2)
__ CO + __ H2O  __ CO2 + __ H2
1. Set up & balance rxn:
1 CO + 1 H2O  1 CO2 + 1 H2
2. Assign eq. amounts in terms of x
100 – x 50 - x
3. Write formula for Keq
Kp = PCO2 PH2 / PCO PH2O
4. Substitute Eq. amounts into Keq
KP = (x)(x) / (100-x)(50-x)
5. Solve for x
KP = x2 / (5000 – 150 x + x2) = 0.034
x
Cross multiply:
x2 = 170 - 5.1 x + 0.34 x2
Move all term to the left side:
0.966 x2 + 5.1 x - 170 = 0
Solve for x using quadratic equation:
a
b
c
x = 11
&
x = -16
Throw out the x that will not work:
x = 11 kPa
Assign all of the amounts:
• PCO = 100 - x = 100 - 11 =
• PH2O = 50 - x = 50 - 11 =
• PCO2 = PH2 = x =
PCO = 89 kPa
PH2O = 39 kPa
PCO2 = PH2 = 11 lPa
x
Problem: Calculate the partial pressure of each portion when 50.0 kPa Xe & 100.0 kPa F2 are combined
in the following reaction:
__ Xe (g) + __F2(g)  __XeF2(g)
Problem: At equilibrium, Pammonia = 4.0 Atm, Phydrogen = 2.0 Atm, & Pnitrogen = 5.0 Atm in the following rxn.
__NH3  __H2 + __N2
Calculate the value of the equilibrium constant:
Problem: Determine the magnitude of the equilibrium constant of the following reaction if the partial pressure
of each gas in the system is 0.667 Atm at equilibrium:
__SO2 + __O2  __SO3
Drill: Write the equilibrium expression & solve its magnitude when PNO2 & PN2O4 = 50 kPa each
at equilibrium in the following reaction:
__ N2O4  __NO2
Reaction Quotient: for the following rxn:
aA(aq ) + bB(aq)  pP(aq) + qQ(aq)
Q = [P]p[Q]q / [A]a[B]b
The formula for the reaction quotient (Q) & the equilibrium constant (K) are the same.
• When using starting concentrations, use Q
• When using equilibrium concentrations, use K
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 for the following rxn:
A(aq) + 2 B(aq)  C(aq) + D(aq) + heat
• Write the formula for the equilibrium constant
• What happens to all others if each of the following is changed in the system at equilibrium?
o Extra A is added
o B is somehow removed
o C is added
o The temperature in increased
LC Eq Effects for the following rxn:
2 A(aq) + B(s)  C(aq) + 2 D(aq) + heat
• Write the formula for the equilibrium constant
• What happens to all others if each of the following is changed in the system at equilibrium?
o Extra A is added
o B is added
o The temperature in increased
LC Eq Effects for the following rxn:
2 A(g) + B(g)  3 C(g) + 2 D(l)
• Write the formula for the equilibrium constant
• What happens to all others if each of the following is changed in the system at equilibrium?
o Extra A is added
o B is somehow removed
o D is added
o The temperature in increased
Equilibrium Applications:
•
•
G = H - TS
G = - RTlnKeq
Problem: Calculate the equilibrium concentration of each species when the following reaction is started
by combining A & B making 0.40 M A & 0.20 M B: A(aq) + B(aq)  AB(aq)
(KC = 0.50)
Problem: Calculate the partial pressure of each portion when 75 kPa Xe & 20.0 kPa F2 are combined
in the following reaction:
Xe (g) + 2 F2(g)  XeF4(g)
(Kp = 4.0 x 10-8)
Drill: Solve for value of K when the following reaction is at the following amounts:
A(aq)+ 2 B(aq) C(s)+ 2 D(aq)
[A] = 0.30 M [B] = 0.20 M
C = 5.0 g
[D] = 0.30 M
Working with Equilibrium Constants:
When adding Reactions: Multiply K’s
A  B
K1
B  C
K2
A  C
K3 = (K1)(K2)
Solve K for each of the following reactions:
A + B  C + D Reaction 1, solve K1
C + D  P + Q Reaction 2, solve K2
A + B  P + Q Reaction 3 which is the addition of reactions 1 & 2, solve K3
When doubling Reactions: Square K’s
A  B
K1
2 A  2 B
K2 = (K1)2
When a reaction is multiplied by any factor, that factor becomes the exponent of K:
A  B
K1
1/3 A  1/3 B
K2 = (K1)1/3
When reversing Reactions: Take 1/K’s
A  B
K1
B  A
K2 = 1/K1
Problem: Calculate the equilibrium concentration of each species when initially the mixture is
0.60 M A & 0.80 M B in the following rxn: 1 A + 2 B  1 C + 1 D
(Keq = 5.0 x 10-8)
Problem: Calculate the partial pressure of each portion when 80.0 kPa Xe & 60.0 kPa F2 are combined
in the following reaction:
Xe (g) + F2(g)  XeF2(g)
(Kp = 4.0 x 10-2)
Drill: Solve for K:
A(aq) + 3 B(aq)  C(s) + 2 D(aq)
Calculate Kc if:
[A] = 0.30 M, [B] = 0.20 M, C = 5.0 g, & [D] = 0.30 M at equilibrium
A + B  C + D
C + H  M + N
N + T  P + Q
Problem: In the reaction mechanism to the
right, determine what happens to all species when
the following changes are made at equilibriom
•
What happens to all other species when:
o A is added at equilibrium
o C is removed at equilibrium
o H is added at equilibrium
o P is added at equilibrium
Problem: Using to following problem, CuCl6-4 (aq) + 2 NH3(aq)  [Cu(NH3)2Cl4]-2(aq) + 2 Cl-(aq)
• Calculate the molarity of each portion when 0.10 M CuCl6-4 & 1.0 M NH3 are combined:
• Kformation = 3.2 x 10-10
Problem: Calculate the partial pressure of each portion when 15 kPa Rn & 85 kPa F2 are combined in the
following rxn:
Rn(g) + F2(g)  RnF2(g)
(Kp = 4.0 x 10-2)
Drill: Rxn: 1A + 1B  1P + 2Q
• Calculate the concentration of each portion at equilibrium when before mixing the solution
contains 0.50 M A & 0.50 M B:
Kc = 1.6 x 10-8
Test Review:
General Reaction:
aA + bB  pP + qQ
Rate Law:
Rate = k[A]a[B]b
Equilibrium Constant:
Kc = [P]p[Q]q / [A]a[B]b
Reaction Quotient
Q = [P]p[Q]q / [A]a[B]b
:
or
KP = PPp PQq / PAa PBb
Clausius-Claperon Eq:
Ea = R(T2T1) / (T2 – T1) ln (k2 / k1)
Clausius-Claperon Eq:
HV = R(T2T1) / (T2 – T1) ln (P2 / P1)
Problem: Use the reaction below &
Exp # Temp
[A]
[B]
Rate
the data from the experimental
1
27oC
0.10
0.10
0.020
data table to the right to determine
2
27oC
0.20
0.10
0.040
the rate law, the reaction order, k, & Ea:
3
27oC
0.10
0.20
0.080
4
77oC
0.10
0.10
2.0
aA + bB  Products
Problem: Determine the Rate Expression for the following Reaction mechanism:
Reaction Mechanism
Rxn Rate
A  B
fast
2 B  3 C
fast
C  2 D
fast
D  Prod.
slow
LC Eq Effects for the following rxn:
•
•
2 A(aq) + B(s)  C(aq) + 2 D(aq) + heat
Write the formula for the equilibrium constant
What happens to all others if each of the following is changed in the system at equilibrium?
o Extra A is added
o B is added
o The temperature in increased
LC Eq Effects for the following rxn:
3 A(g) + B(g)  2 C(g) + 2 D(l)
• Write the formula for the equilibrium constant
• What happens to all others if each of the following is changed in the system at equilibrium?
o Extra A is added
o B is somehow removed
o D is added
o The temperature in increased
Problem: Using the following reaction, (__SO + __O2  __SO3) Calculate the equilibrium pressures
if SO at 80.0 kPa is combined with O2 at 40.0 kPa
(K = 2.00)
Problem: Using the following reaction, (I2 + 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. (K = 4.0 x 10-8)
Problem: Use the reaction below &
[B]
[C]
Rate
the data from the experimental
1
0.10
0.10
0.10
16
data table to the right to determine
2
0.20
0.10
0.10
2
the rate law, the reaction order, k, & Ea:
3
0.10
0.20
0.10
8
4
0.10
0.10
0.20
4
aA + bB + cC  Products
:
Exp # [A]