Transcript lecture CH15 chem162pikul partC

Chemical Equilibrium
CHAPTER 15
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 15 Chemical Equilibrium
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
 van’t Hoff Equation
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CHAPTER 15 Chemical Equilibrium
Lecture Road Map:
① Dynamic Equilibrium
② Equilibrium Laws
③ Equilibrium Constant
④ Le Chatelier’s Principle
⑤ Calculating Equilibrium
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CHAPTER 15 Chemical Equilibrium
Calculating
Equilibrium
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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
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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 calculations
A. Given equilibrium concentrations, calculate K
B. Given initial concentrations and one final
concentration
• Calculate equilibrium concentration of
all other species
• Then calculate K
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Kc with Known Equilibrium Concentrations
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
2
K c=
[NO2 ]
2
[0.0116]
K c=
[0.0292]
[N2O 4 ]
KC = 4.61  10–3
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Group
Problem
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
3
[C ]
C. 1.5
Kc =
2
[
A
]
[B ]
D. 6.75
Kc =
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[3.0]
2
[2.0] [1.0]
8
Kc with Known Equilibrium Concentrations
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
1.00 mol
[SO2 ] = [O2 ] =
= 1.00 M
1.00 L
– Equilibrium
0.925 mol
[SO3 ] =
= 0.925 M
1.00 L
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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
Equilibrium
Change in
Initial
=
–
Concentration
Concentration
Concentration
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Calculations
Example Continued
2SO2(g) + O2(g)
Initial Conc. (M)
1.000
Changes in Conc. (M) –0.925
Equilibrium Conc. (M) 0.075
1.000
–0.462
0.538
2SO3(g)
0.000
+0.925
0.925
[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
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Calculations
Overview
• Finally calculate KC at 1000 K
Kc 
Kc 
[SO 3 ]2
[SO 2 ]2 [O 2 ]
2
[0.925]
2
[0.075] [0.538]
Kc = 2.8 × 102 = 280
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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
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Calculations
ICE Table Summary
ICE tables used for most equilibrium calculations:
1.
Equilibrium concentrations are only values used in mass
action expression

2.
Initial value in table must be in units of mol/L (M)


3.
4.
Values in last row of table
[X]initial = those present when reaction prepared
No reaction occurs until everything is
mixed
Changes in concentrations always occur in same ratio as
coefficients in balanced equation
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
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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 ?
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Calculate [X ]equilibrium from Kc and [X ]initial
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
Kc =
[CO][H2 ]3
[CH4 ][H2O]
[H2O] =
[CO][H2 ]3
[CH4 ]K c
• Next, plug in equilibrium concentrations and Kc
[0.300][0. 800]3 0.154
[H2O] 

[0.400](5.67)
2.27
[H2O] = 0.0678 M
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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
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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
[HI]2
Kc =
= 55.64
[H2 ][I2 ]
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 2: Construct an ICE table
Conc (M)
H2(g) + I2(g)
Initial
2.00
–x
Change
Equilibrium 2.00 – x
2HI (g)
2.00
0.000
–x
+2x
2.00 – x
+2x
• 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
(2x )
(2x )
55.64 =
=
(2.00 - x )(2.00 - x ) (2.00 - x )2
2
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 3. Solve for x
• Both sides are squared so we can take square root of both
sides to simplify
(2x)2
K = 55.64 =
2
(2.00 - x)
2x
7.459 
7.459(2.00  x )  2x
(2.00  x )
14.918  7.459x  2x
14.918  9.459x
14.918
x 
 1.58
9.459
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Step 4. Equilibrium Concentrations
Conc (M)
H2(g) +
Initial
2.00
– 1.58
Change
Equilibrium 0.42
I2(g)
2.00
2HI (g)
0.00
– 1.58
+3.16
0.42
+3.16
• [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M
• [HI]equil = 2x = 2(1.58) = 3.16
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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
2
[HI]
Kc =
= 55.64
[H2 ][I2 ]
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Calculate [X]equilibrium from [X]initial and KC
Calculations
Step 2. Concentration Table
Conc (M)
H2(g) +
I2(g)
2HI (g)
Initial
Change
Equil’m
2.00
2.00
2.00
–x
–x
+2x
2.00 – x
2.00 – x
2.00 + 2x
(2.00  2x ) 2
(2.00  2x ) 2
55.64 

(2.00  x )(2.00  x )
(2.00  x ) 2
K  55.64 
(2.00  2x ) 2
(2.00  x ) 2
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Calculate [X]equilibrium from [X]initial and KC
Calculations
Step 3. Solve for x
2.00  2x
7.459 
(2.00  x )
7.459(2.00  x )  2.00  2x
14.918  7.459x  2.00  2x
 [H2]equil = [I2]equil = 2.00 – x
= 2.00 – 1.37 = 0.63 M
12.918  9.459 x
12.918
x 
 1.37
9.459
 [HI]equil = 2.00 + 2x
= 2.00 + 2(1.37)
= 2.00 + 2.74
= 4.74 M
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Group
Problem
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.
B.
C.
D.
0.0526 M, 0.947 M, 0.105 M
0.947 M, 0.947 M, 0.105 M
0.947 M, 0.105 M, 0.0526 M
0.105 M, 0.105 M, 0.947 M
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Group
Problem
Conc (M) N2(g) + O2(g)
• Initial
1.00
1.00
• Change
–x
–x
• Equil
1.00 – x 1.00 – x
2NO (g)
0.00
+ 2x
+ 2x
0.250 mol
[N2 ] = [O2 ] =
= 1.00 M
0.250 L
(2x )2
2x
0.0123 =
0.0123 =
2
1- x
(1 - x )
x = 0.0526 M [NO] = 2x = 0.105 M
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Ex. 8
CH3CO2H(aq) + C2H5OH(aq)
acetic acid ethanol
KC = 0.11
CH3CO2C2H5(aq) + H2O(l)
ethyl acetate
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?
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 1. Write equilibrium law
[CH3CO2C2H5 ]
Kc 
 0.11
[C 2H5OH][CH3CO2H]
• 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
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 2 Concentration Table
(M) CH3CO2H(aq) + C2H5OH(aq)
I
C
E
•
•
•
•
0.810
–x
0.810 – x
CH3CO2C2H5(aq) + H2O(l)
0.810
0.000
0.810 – x
+x
–x
+x
Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x
Amt of CH3CO2C2H5 formed = + x
[CH3CO2H]eq and [C2H5OH ] = 0.810 – x
[CH3CO2C2H5] = x
x
0.11 =
(0.810 - x )(0.810 - x )
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 3. Solve for x
• Rearranging gives
0.11  (0.6561  1.62x  x 2 )  x
• Then put in form of quadratic equation
ax2 + bx + c = 0
2
0.07217  0.1782 x  0.11x
0.11x
2
x 0
 1.1782x  0.07217  0
• Solve for the quadratic equation using
 b  b 2  4ac
x 
2a
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 3. Solve for x
 (1.1782)  (1.1782) 2  4(0.11)(0.07217)
x 
2(0.11)
1.1782  (1.388)  (0.032) 1.1782  1.164
x 

0.22
0.22
• 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
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Quadratic Equation
Step 4. Equilibrium Concentrations
CH3CO2H(aq) + C2H5OH(aq)
I 0.810
C –0.064
E 0.746
0.810
– 0.064
0.746
CH3CO2C2H5(aq) + H2O
0.000
+0.064
+0.064
[CH3CO2C2H5]equil = x = 0.064 M
[CH3CO2H]equil = [C2H5OH]equil = 0.810 M – x
= 0.810 M – 0.064 M
= 0.746 M
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
When KC is very small
Ex. 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
Kc 
2
[H2 ] [O 2 ]
[H2 O]2
 7.3  10
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Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Calculations
Step 2. Concentration Table
Conc (M ) 2H2O(g)
Initial
0.100
– 2x
Change
Equil’m 0.100 – 2x
7.3  10 18 
2H2(g) + O2(g)
0.00
0.00
+2x
+x
+2x
+x
(2x )2 x
2

4x 3
(0.100  2x )
(0.100  2x )2
• Cubic equation – tough to solve
• Make approximation
– KC very small, so x will be very small
– Assume we can neglect x
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Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Calculations
Step 3. Solve for x
• Assume (0.100 – 2x)  0.100
Conc (M)
2H2O (g)
2H2 (g) +
Initial
Change
Equil’m
0.100
0.00
+2x
– 2x
0.100
O2 (g)
0.00
+2x
+x
+x
• Now our equilibrium expression simplifies to
2
3
(2x ) x
4x
18
7.3  10


2
0.010
4x 3
(0.100)
 0.010(7.3  10 18 ) = 7.3 × 10–20
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Calculations
Calculate [X]equilibrium from [X]initial and KC
Example: Cubic
Step 3. Solve for x
x
3
7.3  10

4
20
 1.8  10 20
• Now take cube root
3
x  1.8  10
•
•
•
•
 20
 2.6  10
7
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
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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
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Group
Problem
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?
PB
Q = KP = 2
2A  B
PA
I
0.2
0 atm
x
-16
3.5 ´ 10 =
C
–2x
+x
2
(0.2)
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
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