Chemical Equilibrium
CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop
2
CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Learning Objectives:
Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium
Constant (K) Reaction Quotient (Q) Kc vs Kp
ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle
3
CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Lecture Road Map:
① Dynamic Equilibrium
② Equilibrium Laws
③ Equilibrium Constant
④ Le Chatelier’s Principle
⑤ Calculating Equilibrium
4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Dynamic Equilibrium
CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
5
Dynamic Eq Equilibrium
• Chemical equilibrium exists when– Rates of forward and reverse reactions are equal– Reaction appears to stop – Concentration of reactants and products do not
change over time• Remain constant• Both forward and reverse reaction never cease
• Equilibrium signified by double arrows ( )
Dynamic Eq Equilibrium
N2O4 2 NO2
• Initially have only N2O4
– Only forward reaction
– As N2O4 reacts NO2 forms
• As NO2 forms
– Reverse reaction begins to occur
– NO2 collide more frequently as concentration of NO2 increases
• Eventually, equilibrium is reached
– Concentration of N2O4 does not change
– Concentration of NO2 does not change
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6
Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7
Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8
N2O4 2NO2
Closed system• Equilibrium can be
reached from either direction
• Independent of whether it starts with “reactants” or “products”
• Always have the same composition at equilibrium under same conditions
Dynamic Eq Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9
N2O4 2NO2
Reactants ProductsEquilibrium
Dynamic Eq Mass Action Expression
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10
• Simple relationship among [reactants] and [products] for any chemical system at equilibrium
• Called the mass action expression– Derived from thermodynamics
• Forward reaction: A B • Reverse reaction: A B • At equilibrium: A B
Dynamic Eq Reaction Quotient
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11
• Uses stoichiometric coefficients as exponent for each reactant
• For reaction: aA + bB cC + dD
Reaction quotient– Numerical value of mass action expression– Equals “Q ” at any time, and– Equals “K ” only when reaction is known to be at
equilibrium
12
Ex. 1 H2(g) + I2(g) 2HI(g) 440˚C
Exp’t Initial Amts
Equil’m Amts
Equil’m [M]
I 1.00 mol H2
0.222 mol H2
0.0222 M H2
10 L
1.00 mol I2
0.222 mol I2
0.0222 M I2
0.00 mol HI
1.56 mol HI 0.156 M HIII 0.00 mol H2
0.350 mol H2
0.0350 M H2
10 L 0.100 mol I2
0.450 mol I2
0.0450 M I2
3.50 mol HI
2.80 mol HI 0.280 M HI
13
Ex. 1 H2(g) + I2(g) 2HI(g) 440 ˚C
Exp’t Initial Amts
Equil’m Amts
Equil’m [M]
III 0.0150 mol H2
0.150 mol H2
0.0150 M H2
10 L 0.00 mol I2 0.135 mol I2 0.0135 M I21.27 mol HI 1.00 mol HI 0.100 M HI
IV 0.00 mol H2 0.442 mol H2
0.0442 M H2
10 L 0.00 mol I2 0.442 mol I2
0.0442 M I2
4.00 mol HI 3.11 mol HI 0.311 M HI
Equilibrium Concentrations (M )
Exp’t [H2] [I2] [HI]
I 0.0222 0.0222 0.156
II 0.0350 0.0450 0.280
III 0.0150 0.0135 0.100
IV 0.0442 0.0442 0.311
14
Mass Action Expression
= same for all data sets at equilibrium
4.49)0222.0)(0222.0(
)156.0( 2
8.49)0450.0)(0350.0(
)280.0( 2
4.49)0135.0)(0150.0(
)100.0( 2
5.49)0442.0)(0442.0(
)311.0( 2
Average = 49.5
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
15
GroupProblem
Write mass action expressions for the following:
• 2NO2(g) N2O4(g)
• 2CO(g) + O2(g) 2CO2(g)
16
GroupProblem
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Which of the following is the correct mass action expression for the reaction:
Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?
17Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Equilibrium Laws
CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
18
Equilibrium Equilibrium Laws
• For reaction
H2(g) + I2(g) 2HI(g) at 440 ˚C
at equilibrium write the following equilibrium law
• Equilibrium constant = Kc = constant at given T
• Use Kc since usually working with concentrations in mol/L
• For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to equilibrium constant, Kc
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
19
Equilibrium Predicting Equilibrium Laws
For general chemical reaction:• dD + eE fF + gG
– Where D, E, F, and G represent chemical formulas– d, e, f, and g are coefficients
• Mass action expression is
• Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation.
• Equilibrium law is:
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
20
Equilibrium Predicting Equilibrium Laws
• Only concentrations that satisfy this equation are equilibrium concentrations
• Numerator– Multiply concentration of products raised to their
stoichiometric coefficients• Denominator
– Multiply concentration reactants raised to their stoichiometric coefficients
is scientists’ convention
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
21
Equilibrium Example
3H2(g) + N2(g) 2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
22
Equilibrium Operations
Various operations can be performed on equilibrium expressions
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original
A + B C + D
C +D A + B
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
23
Equilibrium Operations
1. When direction of equation is reversed, new equilibrium constant is reciprocal of original
3H2(g) + N2(g) 2 NH3(g) at 25˚C
2NH3(g) 3H2(g) + N2(g) at 25 ˚C
8
23
2
23 1026.4
][N][H
][NHcK
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
24
Equilibrium Operations
2. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor.
A + B C + D
3A + 3B 3C + 3D
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
25
Equilibrium Operations
2. When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor
3H2(g) + N2(g) 2NH3(g) at 25 ˚C
Multiply by 3
9H2(g) + 3N2(g) 6NH3(g)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
26
Equilibrium Operations
3. When chemical equilibria are added, their equilibrium constants are multiplied
A + B C + D
C + E F + G
A + B + E D + F + G
27
Equilibrium Operations
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
3. When chemical equilibria are added, their equilibrium constants are multiplied
][CO][NO]][CO[NO
][NO
][NO][NO
3
222
2
3
2 NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g) NO2(g) + CO2(g)
NO2(g) + CO(g) NO(g) + CO2(g)
Therefore
][CO][NO][NO][CO
2
2
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
28
GroupProblem For: N2(g) + 3H2(g) 2NH3(g)
Kc = 500 at a particular temperature.
What would be Kc for following?
• 2NH3(g) N2(g) + 3H2(g)
• 1/2N2(g) + 3/2H2(g) NH3(g)
22.4
0.002
29Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Equilibrium Constant
CHAPTER 15 Chemical Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
30
Equilibrium Constant Kc
• Most often Kc is expressed in terms of a ratio of concentrations of products and reactants as shown on previous slides
• Sometimes partial pressures, in atmospheres, may be used in place of concentrations
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
31
Equilibrium Kp
• Based on reactions in which all substances are gaseous
• Gas quantities are expressed in atmospheres in mass action expression
• Use partial pressures for each gas in place of concentrations
e.g. N2(g) + 3H2(g) 2NH3(g)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
32
Equilibrium Relationship between Kp and Kc
• Start with ideal gas law
PV = nRT• Rearranging gives
• Substituting P/RT for molar concentration into Kc results in pressure-based formula
• ∆n = moles of gas in product – moles of gas in reactant
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
33
GroupProblem
Consider the reaction: 2NO2(g) N2O4(g)
If Kp = 0.480 for the reaction at 25 ˚C, what is value of Kc at same temperature?
n = nproducts – nreactants = 1 – 2 = –1
Kc = 11.7
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
34
GroupProblem
Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25 ˚C, what would be the Kp?
A. 0.99
B. 2.0
C. 24
D. 2400
E. None of these
Δn = (4 – 3) = 1
Kp = Kc(RT)Δn
Kp= 0.99 × (0.082057 × 298.15)1
Kp = 24
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
35
Equilibrium Homogeneous and Hetergeneous
Homogeneous reaction/equilibrium– All reactants and products in same phase– Can mix freely
Heterogeneous reaction/equilibrium– Reactants and products in different phases– Can’t mix freely– Solutions are expressed in M– Gases are expressed in M
– Governed by Kc
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
36
Equilibrium Heterogeneous
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
• Equilibrium Law =
• Can write in simpler form• For any pure liquid or solid, ratio of moles to
volume of substance (M ) is constant– e.g. 1 mol NaHCO3 occupies 38.9 cm3
2 mol NaHCO3 occupies 77.8 cm3
37
Equilibrium Heterogeneous
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
– Ratio (n/V ) or M of NaHCO3 is constant (25.7 mol/L) regardless of sample size
– Likewise can show that molar concentration of Na2CO3 solid is constant regardless of sample size
• So concentrations of pure solids and liquids can be incorporated into equilibrium constant, Kc
• Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
38
Equilibrium Heterogeneous
Write equilibrium laws for the following:
Ag+(aq) + Cl–(aq) AgCl(s)
H3PO4(aq) + H2O H3O+(aq) + H2PO4
–(aq)
39
Interpreting KC
• Large K (K >>1)–Means product rich
mixture–Reaction goes far
toward completione.g.2SO2(g) + O2(g)
2SO3(g)
Kc = 7.0 1025 at 25 °C
40
Interpreting KC
• Small K (K << 1)–Means reactant rich
mixture–Only very small
amounts of product formed
e.g. H2(g) + Br2(g)
2HBr(g) Kc = 1.4 10–21 at 25 °C
41
Interpreting KC
• K 1–Means product and
reactant concentrations close to equal
–Reaction goes only about halfway
42
• Size of K gives measure of how reaction proceeds
• K >> 1 [products] >> [reactants]
• K = 1 [products] = [reactants]• K << 1 [products] <<
[reactants]
43Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier’s Principle
CHAPTER 15 Chemical Equilibrium
44
Le Chatelier Definition
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
• Equilibrium positions – Combination of concentrations that allow Q = K– Infinite number of possible equilibrium positions
• Le Châtelier’s principle– System at equilibrium (Q = K) when upset by
disturbance (Q ≠ K) will shift to offset stress• System said to “shift to right” when forward reaction
is dominant (Q < K) • System said to “shift to left” when reverse direction
is dominant (Q > K)
45Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
• Q = K reaction at equilibrium
• Q < K reactants go to products– Too many reactants– Must convert some reactant to product to
move reaction toward equilibrium
• Q > K products go to reactants– Too many products– Must convert some product to reactant to
move reaction toward equilibrium
Le Chatelier Q & K Relationships
46Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl4
2–(aq) + 4H2O
blue yellow• Equilibrium mixture is blue-green
• Add excess Cl– (conc. HCl)– Equilibrium shifts to products
– Makes more yellow CuCl42–
– Solution becomes green
47Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
• Add Ag+ – Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s)– Equilibrium shifts to reactants
– Makes more blue Cu(H2O)42+
– Solution becomes increasingly more blue
• Add H2O?
48Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Concentration: Example
For the reaction 2SO2(g) + O2(g)
2SO3(g)
Kc = 2.4 × 10–3 at 700 °C
Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture?
A.Towards the products
B.Towards the reactants
C.No change will occur
49Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Concentration
• When changing concentrations of reactants or products– Equilibrium shifts to remove reactants or products that
have been added– Equilibrium shifts to replace reactants or products that
have been removed
50Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Pressure or Volume
• Consider gaseous system at constant T and n
3H2(g) + N2(g) 2NH3(g)
• If volume is reduced– Expect pressure to increase– To reduce pressure, look at each side of reaction– Which has less moles of gas– Reactants = 3 mol + 1 mol = 4 mol gas– Products = 2 mol gas– Reaction favors products (shifts to right)
51Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Pressure or Volume
Consider gaseous system at constant T and n
H2(g) + I2(g) 2HI(g)
• If pressure is increased, what is the effect on equilibrium?
– nreactant = 1 + 1 = 2
– nproduct = 2
– Predict no change or shift in equilibrium
52Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Pressure or Volume
2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g)
• If you decrease volume of reaction, what is the effect on equilibrium?– Reactants: All solids, no moles gas– Products: 2 moles gas– Decrease in V, causes an increase in P– Reaction shifts to left (reactants), as this has fewer
moles of gas
53Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Pressure or Volume
• Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can– Increasing pressure
• Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids– Substances are already almost incompressible
• Changes in V, P and [X ] effect position of equilibrium (Q), but not K
54Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Temperature
Ice water
Boiling water
Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow– Reaction endothermic– Adding heat shifts equilibrium toward products– Cooling shifts equilibrium toward reactants
55Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Temperature
Hf°=+6 kJ (at 0 °C)
– Energy + H2O(s) H2O(l )
– Energy is reactant– Add heat energy, shift reaction right
3H2(g) + N2(g) 2NH3(g) Hrxn= –47.19 kJ
– 3 H2(g) + N2(g) 2 NH3(g) + energy
– Energy is product– Add heat, shift reaction left
H2O(s) H2O(l)
56Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Temperature
• Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change
• Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change
57Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change in Temperature
• Changes in T change value of mass action expression at equilibrium, so K changed– K depends on T– Increase in temperature of exothermic reaction
makes K smaller• More heat (product) forces equilibrium to
reactants– Increase in temperature of endothermic reaction
makes K larger• More heat (reactant) forces equilibrium to
products
58Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Change with Catalyst
• Catalyst lowers Ea for both forward and reverse reaction
• Change in Ea affects rates k r and k f equally
• Catalysts have no effect on equilibrium
59Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier Addition of Inert Gas at Constant Volume
Inert gas – One that does not react with components of reaction
e.g. argon, helium, neon, usually N2
• Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products
• Since it doesn’t react with anything– No change in concentrations of reactants or products– No net effect on reaction
60Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Le Chatelier How To Use Le Chatelier’s Principle
1. Write mass action expression for reaction
2. Examine relationship between affected concentration and Q (direct or indirect)
3. Compare Q to K– If change makes Q > K, shifts left– If change makes Q < K, shifts right– If change has no effect on Q, no shift expected
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 61
GroupProblem
Consider:
H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq)
What will happen if PO43– is removed?
Q is proportional to [PO43–]
Decrease [PO43–], decrease in Q
Q < K equilibrium shifts to right
62Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
GroupProblem
The reaction
H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq)
is exothermic.
What will happen if system is cooled?
Since reaction is exothermic, heat is product Heat is directly proportional to Q Decrease in T, decrease in Q Q < K equilibrium shifts to right
heat
63Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
GroupProblem
The equilibrium between aqueous cobalt ion and the chlorine ion is shown:
[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O
pink blue
It is noted that heating a pink sample causes it to turn violet.
The reaction is:
A. endothermic
B. exothermic
C. cannot tell from the given information
64Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
GroupProblem
The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products?
A. Ka = 2.2 × 10–3
B. Ka = 1.8 × 10–5
C. Ka = 4.0 × 10–10
D. Ka = 6.3 × 10–3
65Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculating Equilibrium
CHAPTER 15 Chemical Equilibrium
66Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Overview
• For gaseous reactions, use either KP or KC
• For solution reactions, must use KC
• Either way, two basic categories of calculations
1. Calculate K from known equilibrium concentrations or partial pressures
2. Calculate one or more equilibrium concentrations or partial pressures using known KP or KC
67Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Kc with Known Equilibrium Concentrations
• When all concentrations at equilibrium are known– Use mass action expression to relate
concentrations to KC
• Two common types of calculationsA. Given equilibrium concentrations, calculate KB. Given initial concentrations and one final
concentration• Calculate equilibrium concentration of
all other species• Then calculate K
68Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Ex. 3 N2O4(g) 2NO2(g)
• If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC?
• [N2O4]eq = 0.0292 M
• [NO2]eq = 0.0116 M
KC = 4.61 10–3
Kc with Known Equilibrium Concentrations
69Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature?
A. 14
B. 0.15
C. 1.5
D. 6.75
GroupProblem
70Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Ex. 4 2SO2(g) + O2(g) 2SO3(g)
At 1000 K, 1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask. At equilibrium 0.925 mol SO3 has formed. Calculate K C for this reaction.
• First calculate concentrations of each– Initial
– Equilibrium
Kc with Known Equilibrium Concentrations
71Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Example Continued
• Set up concentration table– Based on the following:
• Changes in concentration must be in same ratio as coefficients of balanced equation
• Set up table under balanced chemical equation– Initial concentrations
• Controlled by person running experiment– Changes in concentrations
• Controlled by stoichiometry of reaction– Equilibrium concentrations
EquilibriumConcentration = Initial
Concentration– Change in
Concentration
72Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Example Continued
2SO2(g)
+ O2(g) 2SO3(g)
Initial Conc. (M) 1.000 1.000
0.000
Changes in Conc. (M)
Equilibrium Conc. (M)
[SO2] consumed = amount of SO3 formed
= [SO3] at equilibrium = 0.925 M
[O2] consumed = ½ amount SO3 formed = 0.925/2 = 0.462 M
[SO2] at equilibrium = 1.000 – 0.975 = 0.075
[O2] at equilibrium = 1.00 – 0.462 = 0.538 M
–0.925 –0.462 +0.925
0.075 0.538 0.925
73Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Example Continued
• Finally calculate KC at 1000 K
][O][SO
][SO
22
2
23
c K
]538.0[]075.0[
]925.0[2
2cK
Kc = 2.8 × 102 = 280
74Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations ICE Table Summary
ICE tables used for most equilibrium calculations:1. Equilibrium concentrations are only values used in mass
action expression Values in last row of table
2. Initial value in table must be in units of mol/L (M) [X]initial = those present when reaction prepared
No reaction occurs until everything is mixed
3. Changes in concentrations always occur in same ratio as coefficients in balanced equation
4. In “change” row be sure all [reactants] change in same directions and all [products] change in opposite direction. If [reactant]initial = 0, its change must be an increase (+) because
[reactant]final cannot be negative
If [reactants] decreases, all entries for reactants in change row should have minus sign and all entries for products should be positive
75Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X ]equilibrium from Kc and [X ]initial
• When all concentrations but one are known
– Use mass action expression to relate Kc and known concentrations to obtain missing concentrations
Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g)
• At 1500 °C, Kc = 5.67. An equilibrium mixture of gases had the following concentrations: [CH4] = 0.400 M and [H2] = 0.800 M and [CO] = 0.300 M.
What is [H2O] at equilibrium ?
76Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 5.67
[CH4] = 0.400 M; [H2] = 0.800 M; [CO] =0.300 M
• What is [H2O] at equilibrium?
• First, set up equilibrium
• Next, plug in equilibrium concentrations and Kc
[H2O] = 0.0678 M
27.2154.0
5.67)([0.400]800][0.300][0.
O][H3
2
Calculate [X ]equilibrium from Kc and [X ]initial
77Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculating [X ]Equilibrium from Kc
When Initial Concentrations Are Given
• Write equilibrium law/mass action expression• Set up Concentration table
– Allow reaction to proceed as expected, using “x” to represent change in concentration
• Substitute equilibrium terms from table into mass action expression and solve
78Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Ex. 6 H2(g) + I2(g) 2HI(g) at 425 ˚C
KC = 55.64
If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?
Step 1. Write Equilibrium Law
79Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Step 2: Construct an ICE table
• Initial [H2] = [I2] = 1.00 mol/0.500 L =2.00 M
• Amt of H2 consumed = Amt of I2 consumed = x
• Amount of HI formed = 2x
Conc (M) H2(g) + I2(g) 2HI (g)
Initial 2.00 2.00 0.000
Change
Equilibrium
– x +2x– x
+2x2.00 – x 2.00 – x
Calculate [X]equilibrium from [X]initial and KC
80Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Step 3. Solve for x• Both sides are squared so we can take square root of both
sides to simplify
)00.2(2
459.7x
x
xx 2)00.2(459.7
xx 2459.7918.14
58.1459.9918.14 x
x459.9918.14
Calculate [X]equilibrium from [X]initial and KC
81Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Step 4. Equilibrium Concentrations
• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M
• [HI]equil = 2x = 2(1.58) = 3.16
Conc (M) H2(g) + I2(g) 2HI (g)
Initial 2.00 2.00 0.00
Change
Equilibrium
– 1.58 +3.16– 1.58
+3.160.42 0.42
Calculate [X]equilibrium from [X]initial and KC
82Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Ex. 7 H2(g) + I2(g) 2HI(g) at 425 ˚C
KC = 55.64
• If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 ˚C, what are the equilibrium concentrations of H2, I2 and HI?
• Now have product as well as reactants initially
Step 1. Write Equilibrium Law
83Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations
Conc (M) H2(g) + I2(g) 2HI (g)
Initial 2.00 2.00 2.00
Change
Equil’m
– x +2x– x
2.00 + 2x2.00 – x 2.00 – x
2
22
)00.2(
)200.2()00.2)(00.2(
)200.2(64.55
x
xxx
x
2
2
)00.2(
)200.2(64.55
x
xK
Step 2. Concentration Table
Calculate [X]equilibrium from [X]initial and KC
84Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
)00.2(200.2
459.7xx
xx 200.2)00.2(459.7
xx 200.2459.7918.14
37.1459.9918.12
x
x459.9918.12
Step 3. Solve for x
[H2]equil = [I2]equil = 2.00 – x
= 2.00 – 1.37 = 0.63 M
[HI]equil = 2.00 + 2x = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M
85Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
N2(g) + O2(g) 2NO(g)
Kc = 0.0123 at 3900 ˚C
If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species?
A. 0.0526 M, 0.947 M, 0.105 M
B. 0.947 M, 0.947 M, 0.105 M
C. 0.947 M, 0.105 M, 0.0526 M
D. 0.105 M, 0.105 M, 0.947 M
GroupProblem
86Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Conc (M) N2(g) + O2(g) 2NO (g)
• Initial 1.00 1.00 0.00• Change – x – x + 2x• Equil 1.00 – x 1.00 – x + 2x
GroupProblem
87Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Ex. 8
CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O(l)acetic acid ethanol ethyl acetate
KC = 0.11
An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?
88Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 1. Write equilibrium law
• Need to find equilibrium values that satisfy this
Step 2: Set up concentration table using “x” for unknown– Initial concentrations– Change in concentrations– Equilibrium concentrations
11.0H]COOH][CHH[C
]HCCO[CH
2352
5223 cK
89Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 2 Concentration Table
• Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x
• Amt of CH3CO2C2H5 formed = + x
• [CH3CO2H]eq and [C2H5OH ] = 0.810 – x
• [CH3CO2C2H5] = x
(M) CH3CO2H(aq) +
C2H5OH(aq)
CH3CO2C2H5(aq) + H2O(l)
I 0.810 0.810 0.000
C
E
–x +x– x
+x0.810 – x 0.810 – x
90Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 3. Solve for x• Rearranging gives
• Then put in form of quadratic equation
ax2 + bx + c = 0
• Solve for the quadratic equation using
xxx )62.16561.0(11.0 2
011.01782.007217.0 2 xxx
007217.01782.111.0 2 xx
aacbb
x2
42
91Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation Step 3. Solve for x
• This gives two roots: x = 10.6 and x = 0.064• Only x = 0.064 is possible
– x = 10.6 is >> 0.810 initial concentrations – 0.810 – 10.6 = negative concentration,
which is impossible
)11.0(2)07217.0)(11.0(4)1782.1()1782.1( 2
x
22.0164.11782.1
22.0)032.0()388.1(1782.1
x
92Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 4. Equilibrium Concentrations
[CH3CO2C2H5]equil = x = 0.064 M
[CH3CO2H]equil = [C2H5OH]equil = 0.810 M – x = 0.810 M – 0.064 M = 0.746 M
CH3CO2H(aq) +
C2H5OH(aq)
CH3CO2C2H5(aq) + H2O
I 0.810 0.810 0.000
C
E
–0.064 +0.064– 0.064
+0.0640.746 0.746
93Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
When KC is very smallEx. 9 2H2O(g) 2H2(g) + O2(g)
• At 1000 °C, KC = 7.3 10–18
• If the initial H2O concentration is 0.100 M, what will the H2 concentration be at equilibrium?
Step 1. Write Equilibrium Law
182
2
22
2 103.7O][H
][O][H cK
94Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: CubicStep 2. Concentration Table
• Cubic equation – tough to solve• Make approximation
– KC very small, so x will be very small– Assume we can neglect x – Must prove valid later
Conc (M ) 2H2O(g)
2H2(g)
+O2(g)
Initial 0.100 0.00 0.00
Change
Equil’m
– 2x +x+2x
+x+2x 0.100 – 2x
2
3
2
218
)2100.0(
4
)2100.0(
)2(103.7
x
x
x
xx
95Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Step 3. Solve for x• Assume (0.100 – 2x) 0.100
• Now our equilibrium expression simplifies to
Conc (M) 2H2O (g) 2H2 (g)
+ O2 (g)
Initial 0.100 0.00 0.00
Change
Equil’m
– 2x +x+2x+x+2x 0.100
010.04
)100.0(
)2(103.7
3
2
218 xxx
)103.7(010.04 183 x = 7.3 × 10–20
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
96
Calculations Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Step 3. Solve for x
• Now take cube root
• x is very small • 0.100 – 2(2.6 10–7) = 0.09999948 • Which rounds to 0.100 (3 decimal places)
• [H2] = 2x = 2(2.6 10–7) = 5.2 10–7 M
2020
3 108.14103.7
x
73 20 106.2108.1 x
97Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
Calculations Simplifications: When Can You Ignore x In Binomial (Ci – x)?
• If equilibrium law gives very complicated mathematical problems and if K is small– Then the change (x term) will also be small and we can
assume it can be ignored when added or subtracted from the initial concentration, Ci.
• How do we check that the assumption is correct?– If the calculated x is so small it does not change the
initial concentration
(e.g. 0.10 Minitial – 0.003 Mx-calc = 0.10)
– Or if the answer achieved by using the assumption differs from the true value by less than five percent. This often occurs when Ci > 100 x Kc
98Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
For the reaction 2A(g) B(g)
given that Kp = 3.5 × 10–16 at 25 ˚C, and we place 0.2 atm A into the container, what will be the pressure of B at equilibrium?
2A B
I 0.2 0 atm
C –2x +x
E 0.2 – 2x x ≈0.2 x = 1.4 × 10–17
[B]= 1.4 × 10–17 atm
Proof: 0.2 - 1.4 × 10–17 = 0.2
GroupProblem