Transcript Chapter 14

Chapter 14
Chemical Equilibrium
Chemistry, 7e Thinkwell vs. Chang
CH 14:
14-30(even), 38-46(even),
52-60(even), 66, 67, 72,
74, 75, 80, 84, 90, 102
Thinkwell vs. Chang
14.1 The Concept of Equilibrium
and the Equilibrium Constant
• Most reactions are reversible
– this is denoted by… reactant qe products
• With reversible reactions, a state of chemical
equilibrium is achieved
• Chemical equilibrium - the rates of the forward
and reverse reactions are equal and the
concentrations of the reactants and products
remain constant.
Introduction to CPS Remote Use
Question 1
Question 1
I am present and my remote is working properly.
(a) True
(b) False (if so tell me now)
Introduction to CPS Remote Use
Question 1 Answer
Question 1
I am present and my remote is working properly.
(a) True
Correct Answer
(b) False (if so tell me now)
N2O4
colorless gas
NO2
dark brown
• No matter how much of the substances you start
with, eventually equilibrium will be reached.
• Although there doesn’t appear to be a connection
between the concentrations of the three above...
...THERE IS!
For the reaction:
aA + bB qe cC + dD

C  D
K a
b
A B
c
Equilibrium constant (a
dimensionless quantity)
d
[ ] represents the equilibrium
concentration in mol/L

NO2 
K
N 2O4 
2
N2O4 qe 2NO2
0.670M
0.2000M
0.446M
0.643M
0.448M
0.0500M
0.0547M
0.0898M
0.0457M
0.000M
0.0204M
0.02042
 4.63103
0.0898
0.000M
0.05472
 4.65103
0.643
0.04572
 4.66103
0.448

NO2 
K
N 2O4 
2
N2O4 qe 2NO2
The value obtained for K = 4.63 x 10-3
• What does the size of K tell us about the equilibrium?
• If K«1, the denominator is much larger than the numerator.
Therefore the reactants are favored.
• If K»1, the numerator is much larger than the denominator.
Therefore the products are favored.
• Which case do we have for the above reaction?
• The equilibrium favors the reactants. We say that “it lies to
the left.” Look at the graphs once again.
Question 2
Question 2
(Feel free to discuss with a neighbor)
When K = 4.63 x 10-3 for the reaction above, which is true?
(a) Products are favored
(b) Reactants are favored
(c) Neither is favored
Question 2 Answers
Question 2
(Feel free to discuss with a neighbor)
When K = 4.63 x 10-3 for the reaction above, which is true?
(a) Products are favored
(b) Reactants are favored
(c) Neither is favored
Correct Answer
Is the reactant of N2O4 favored in each case?
Yes
14.2 Writing Equilibrium
Constant Expressions
• We must distinguish between two types of
equilibria:
1 Homogeneous equilibria - all reacting species are
in the same phase. Examples:
• all are gasses
• all are aqueous
2 Heterogeneous equilibria - reacting species exist in
different phases. Examples:
• solids and gases
• solids and liquids
HOMOGENEOUS EQUILIBRIA
• Include all species in the reaction.
• For the reaction:
aA qe bB
• Can be expressed in terms of concentrations as in
our previous example or
• Can be expressed in terms of partial pressure if
gases.

B
Kc  a
A
b
PBb
Kp  a
PA
partial pressure is in atm
HOMOGENEOUS EQUILIBRIA
What is the relationship between Kc and Kp?
See derivation pg 564.
R=0.08206 Latm/mol K
Temperature in?
Kelvin
Kp=Kc(RT) n
n = moles of gaseous products - moles of gaseous reactants
When n = 0
Kp = Kc
Question 3
An equilibrium mixture of the reaction
2H2S(g) qe 2H2(g) + S2(g)
on analysis,was found to contain 1.0 mol H2S, 4.0
mol H2 and 0.80 mol S2 in a 4.0 L vessel. Calculate
the equilibrium constants Kc and Kp for this
reaction. (Temp = 25oC)
Discuss with neighbor
(a) Kc = 78.3 and Kp = 3.2
(b) Kc = 3.2 and Kp = 78.3
(c) Kc = 12.8 and Kp = 19.6
HOMOGENEOUS EQUILIBRIA
An equilibrium mixture of the reaction
2H2S(g) qe 2H2(g) + S2(g)
on analysis,was found to contain 1.0 mol H2S, 4.0
mol H2 and 0.80 mol S2 in a 4.0 L vessel. Calculate
the equilibrium constants Kc and Kp for this
reaction. (Temp = 25oC)
T = 25ºC
V = 4.0 L
(a) Kc = 78.3 and Kp = 3.2
(b) Kc = 12.8 and Kp = 19.6
(c)
Kc = 3.2 and Kp = 78.3
Correct Answer
2H2S(g) qe 2H2(g) + S2(g)
Kc =
[H2]2[S2]1
[H2S]2
(4/4)2(0.8/4)1
(1/4)2
=
= 3.2
the units of M are
omitted by
convention
Since PV = nRT
P = nRT/V = partial pressure of a gas
at constant V and T
Kp =
(P H2 )2(P S2 )1
(PH S )2
2
=
=
2
[n H ]2[n S2 ]1 (RT)3
V
2
[n H S ]2 • V3 • (RT)2
2
(16)(0.8)(0.08206 L·atm·mol-1·K)(298 K)
(1)(4.0 L)
= 78.3 (no units)
HOMOGENEOUS EQUILIBRIA
Consider the reaction 2NO(g) + O2(g) qe 2NO2(g)
At 430oC, an equilibrium mixture consists of
0.020 mol of O2, 0.040 mol of NO, and 0.96 mol
of NO2. Calculate Kp for the reaction, given that
the total pressure is 0.20 atm.
T = 430ºC
PT = 0.20 atm
Ans: Kp=1.5 x 105
Solving for Kp
Given: 2NO(g) + O2(g) qe 2NO2(g)
Via Dalton’s Law of partial pressures (CH 5):



NO
O2
NO2
=
=
=
n
n
n
NO
NO
NO
n NO
+ n
+ n
O2
nO
2
+ n
O2
+ n
n NO
2
+ n
+ n
O2
=
0.040
1.02
= 3.92 x 10-2
=
0.020
1.02
= 1.96 x 10-2
=
0.96
1.02
=
NO2
NO2
NO2
0.941
Solving for Kp
P
P
P
Kp =
NO
O2
NO2
=

=

=

(PNO2)2
(P O )1(P NO )2
2
NO
PT = 7.84 x 10-3 atm
PT
O2
= 3.92 x 10-3 atm
PT = 0.188 atm
NO2
=
(0.188)2
(3.92 x 10-3)(7.84 x 10-3)2
= 1.5 x 105
Why Assumptions Can be Bad
From last time………recall that…….
Kp =
(PNO2)2
(P O )1(P NO )2
2
=
(0.188)2
(3.92 x 10-3)(7.84 x 10-3)2
= 1.5 x 105
PTV = nTRT
V =
nTRT
PT
=
(1.02 mol)(0.08206 L • atm • mol-1 • K-1)(273 + 430)K
(0.20 atm)
V = 294.2 L (very different from 1 L)
Another way to solve last time’s example
Given: 2NO(g) + O2(g) qe 2NO2(g)
P =
Kp =
nRT
V
Insert V = 294.2 L, n’s, and T provided
(P NO2
(P NO
)2
Kp
=
)2
n2 NO
2
=
(P O )
2
1.5 x 105
n2
NO • n O2
(RT)2
•
V3
•
V2
(RT)3
What about Kc ?
Given: 2NO(g) + O2(g) qe 2NO2(g)
KP = Kc(RT)n
Kc
=
KP
(RT)n
= 8.7 x 106
=
(1.5 x 105)
[(0.08206)(430 + 273)](2 – 3)
=
Another way to solve for Kc
Kc =
[NO2]2
n2 NO
2
=
[NO]2[O2]
n2
NO • n O2
Kc = 8.7 x 106
V2
1
•
V2
•
V
•
1
1
HOMOGENEOUS EQUILIBRIA
Ammonia is placed in a reactor, and the temperature is
increased to 745oC, where some will decompose to
nitrogen and hydrogen. The initial concentration of
ammonia was 0.0240 M. After equilibrium is attained,
the concentration of ammonia is 0.0040 M. Calculate
Kc at 745oC for
2NH3(g) qeN2(g) + 3H2(g)
Via ICE table (a.k.a. Initial…Change…..Equilibrium)
Conc. (M)
Initial
Change
NH3
0.0240
-0.020
N2
0
+?
H2
0
+?
Equilib.
0.0040
?
?
ans: 1.7 x 10-2
NH3 (g)
qe
N2 (g) + 3H2 (g
0.02 mol NH3
1
1 mol N2
• 2 mol NH
3
= 0.01 mol (0.01 M if 1 L)
0.02 mol NH3
1
3 mol H2
• 2 mol NH
3
=
Kc
=
(0.01) (0.03)3
(0.004)2
0.03 mol (0.03 M if 1 L)
= 1.7 x 10-2
HETEROGENEOUS EQUILIBRIA
• Let’s look at the following example:
CaCO3(s) qe CaO(s) + CO2(g)
• What would the equilibrium constant be?

CaOCO2 
Kc 
CaCO3 

I’ll call this constant...
• Concentration of a solid is “intensive” - it is a
constant for that substance.
• Because [CaO] and [CaCO3] are constants, I can
join them with the other constant, giving...
HETEROGENEOUS EQUILIBRIA

CaOCO2 
Kc 
CaCO3 

cancel...
giving...
K c CaCO3  CaOCO2 CaCO3 

CaO
CaCO3 CaO
K c CaCO3 
 CO2 
CaO
What is a constant, divided and/or multiplied by
a constant?
Kc  CO2 
HETEROGENEOUS EQUILIBRIA
Also, for the above example...
K p  PCO2
Heterogeneous Equilibria Expressions
(1)
2SO2 (g) + O2 (g) qe 2SO3 (g)
Kc =
(2)
[ SO3 ]2
[O2][SO2]2
(PSO3)2
(P O )1(P SO )2
2
CaCO3 (s) qe CaO + CO2 (g)
Kc = [CO2]
(3)
Kp =
Kp =
P
CO2
H2O (l) CO2 (g) qe H2CO3 (aq)
Kc = [ H2CO3 ]
[CO2]
Kp = (P CO )-1
2
(4) S (s) = H2SO3 (aq) qe H2S2O3 (aq)
Kc = [ H2S2O3]
[H2SO3]
Kp =
undefined
(no gases)
2
HETEROGENEOUS EQUILIBRIA
What is the bottom line?
Do not include solids and liquids in the
equilibrium constant expression.
HETEROGENEOUS EQUILIBRIA
At equilibrium, the pressure of the reacting mixture
CaCO3(s) qe CaO(s) + CO2(g)
is 0.105 atm at 350oC. Calculate Kp and Kc for this
reaction.
Kc = [CO2]
N =
V
mol
L
=
Kp =
P
P = Kc
RT
Kp = 0.105 and Kc = 2.05 x 10-3
CO2
MULTIPLE EQUILIBRIA
Let us consider what to do when the product molecules
of one equilibrium system are involved in a second
equilibrium.
overall reaction:
A + B qe C + D
C + D qe E + F
(1)
(2)
A + B qe E + F
(3)

C D
K c (1) 
AB

E F 
K c (2) 
C D

C D E F 

E F 
 K c (1)  K c (2) 

K c (3) 
AB C D
AB
MULTIPLE EQUILIBRIA
Therefore:
If a reaction can be expressed as the
sum of two or more reactions, the
equilibrium constant for the overall
reaction is given by the product of the
equilibrium constants of the individual
reactions.
HOW MANIPULATIONS TO THE
EQUATION CHANGES K
1. Reversing the direction of the equilibrium:
•
Take the reciprocal of K
2. How the equation is balanced will determine the
value for K. The K must be accompanied by a
balanced equation!
•
For example, if you double the coefficients, what
will happen to the K expression
Let’s look at examples of the above on the overhead.
14.3 The Relationship between
Chemical Kinetics and Chemical
Equilibrium
The following reaction takes place as a single elementary
step:
k
A + B qwef C
kr
ratef = kf[A][B]
rater = kr[C]
At equilibrium, ratef = rater. Therefore
kf[A][B] = kr[C], giving…

kf
C
 Kc 
AB
kr

kf
C
 Kc 
AB
kr
Since we know that the rate constant if affected
by temperature, we can see that the value of the
equilibrium constant will be affected by
temperature.
14.4 What does the Equilibrium
Constant Tell Us?
We can use the equilibrium constant to:
1.
Predict the direction of a reaction.
2.
Calculate equilibrium concentrations.
PREDICTING THE
DIRECTION OF A RXN
• We know that in order to calculate the
equilibrium constant, we simply must plug in
equilibrium concentrations into the expression.
• We can obtain the reaction quotient (Qc) by
substituting the initial concentrations into the
equilibrium constant expression.
• Once you have a value of Qc, a judgement of
the direction can be made according to the
following:
• Qc>Kc
system proceeds from rt to left.
• Qc=Kc
system is at equilibrium
• Qc<Kc
system proceeds from left to right
PREDICTING THE
DIRECTION OF A RXN
PREDICTING THE
DIRECTION OF A RXN
At 700 K, the reaction
2SO2(g) + O2(g) qe 2SO3(g)
has Kc = 4.3 x 106, and the following
concentrations are present:
[SO2] = 0.10 M; [SO3] = 10 M; [O2] = 0.10 M
Is the mixture at equilibrium, if not, in which
direction will the reaction proceed?
Q < Kc
And……?
CALCULATING EQUILIBRIM
CONCENTRATIONS
You will be given the equilibrium constant
value and the initial concentrations. The
following is an outline of the procedure to
follow.
1. Use an “ICE table” to expression the
equilibrium concentrations in terms of the
initial concentration and a variable x.
2. Write the equilibrium constant expression in
terms of the equilibrium concentrations.
3. Solve for x.
4. Relate the value for x to the concentrations
of all species.
CALCULATING EQUILIBRIM
CONCENTRATIONS
At 250o C, the equilibrium constant, Kp for the
reaction below is 1.80.
PCl5(g) qe PCl3(g) + Cl2(g)
Sufficient PCl5 is put into a reaction vessel to give an
initial pressure of 2.74 atm at 250oC. Calculate the
pressure of PCl5 after the system has reached
equilibrium.
Kp =
Ans. 1.24 atm
P
Cl2
P
P
PCl3
PCl5
Via ICE table (a.k.a. Initial…Change…..Equilibrium)
P (atm)
Initial
Change
PCl5
2.74
-x
Equilib.
2.74 - x
PCl3
0
+x
Cl2
0
+x
x
x
Kp =
P
Cl2
P
P
PCl3
=
x2
(2.74 – x)
PCl5
0 = (1)x2 + Kp(x) - (2.74)Kp
and x
And x = 1.50 atm
Pf (PCl5) = 2.74 – 1.50 = 1.24 atm
=
-b ± [ b2 – 4ac]0.5
2(a)
CALCULATING EQUILIBRIM
CONCENTRATIONS
Hydrogen iodide decomposes according to the
equation:
2HI(g) qe H2(g) + I2(g)
Kc= 0.0156 at 400oC
0.550 mol HI was injected into a 2.0 L reaction
vessel at 400oC. Calculate the concentration of HI
at equilibrium.
Solve for next time
Ans. 0.220M
Solution to Last Time’s Problem
Kc= 0.0156 at 400oC
2HI(g) qe H2(g) + I2(g)
Kc =
[H2][I2]
[HI]2
Kc =
x2
(0.275 – 2x)2
=
=
(x)(x)
( [HI]o – 2x )
=
x2
(0.0756 – 1.1x + 4x2)
, where [HI]o = 0.550 mol
2.0 L
x2
(0.275 – 2x)(0.275 – 2x)
and with rearrangement…….
0 = -x2 + Kc[0.0756 – 1.1x + 4x2] = -(0.938)x2 – (0.0172)x + 0.00118
Still Working….
Via quadratic equation and solving for x…Where
x =
x =
-b ± [b2 – 4ac]0.5
2a
(0.0172) ± [(0.0172)2 – 4(-0.938)(0.00118)]0.5
2(-0.938)
= 0.0172 ± [(2.94 x 10-4) + (4.43 x 10-3)]0.5
-1.88
= 2.74 x 10-2 M = [H2] = [I2]
=
(at equilibrium)
Since [HI] = [HI]o – 2x = [0.274 – 2(2.74 x 10-2)]M
[HI] = 0.22 M (Finally)
0.0172 – 6.87 x 10-2
-1.88
14.5 Factors that Affect Chemical
Equilibrium
• The balance between the forward and reverse reaction,
once in equilibrium, is quite delicate.
• When conditions change, the equilibrium will shift to
the left or right to reestablish equilibrium.
• Le Chatelier’s Principle - If an external stress is
applied to a system at equilibrium, the system adjusts
in such a way that the stress is partially offset.
• We will look at how the equilibrium shifts
with…
• changes in concentration.
• changes in pressure and volume.
• changes in temperature.
• addition of a catalyst.
CHANGES IN CONCENTRATION
• When a substance in the reaction is added to a
system at equilibrium, the reaction will shift to
use up some of the added substance.
• When a substance in the reaction is removed
from a system at equilibrium, the reaction will
shift to form some more of that substance.
• Example:
2NOBr(g) qe 2NO(g) + Br2(g)
• Addition of NO will cause the equilibrium to
shift to the __________.
left
• Removal of Br2 will cause the equilibrium to
shift to the __________.
right
•Consider the equilibrium:
3H2+ N2 qe 2NH3
•NH3 is added.
• Which direction will
the equilibrium shift?
• To the left.
CHANGES IN VOLUME AND PRESSURE
• When dealing with gases; as volume goes
down, what happens to the pressure?
• P goes up.
• When P goes up or V goes down, the
equilibrium will shift in the direction of less
gas moles in order to relieve the extra
pressure.
• And visa versa
CHANGES IN VOLUME AND PRESSURE
• If the reaction has no net change in the number
of gas moles (i.e. n = 0) a change in volume
and pressure has no effect on the position of the
equilibrium.
• Because the equilibrium constant expression
has partial pressure of the gases in it, the
change of the total pressure by addition of an
inert gas has no effect on the equilibrium.
CHANGES IN VOLUME AND PRESSURE
Consider the following reaction at equilibrium
2SO2(g) + O2(g) qe 2SO3(g)
If the volume of the system is compressed at
constant temperature, what changes will this
cause in the system?
A shift to produce more SO3
CHANGES IN TEMPERATURE
• To determine the effect of temperature, you must
know if the reaction, as written, is exothermic or
endothermic.
• A temperature increase favors an endothermic
reaction, and a temperature decrease favors an
exothermic reaction.
• To help remember this, write heat as a reactant
for endothermic reactions and a product for
exothermic reactions.
• Think of raising the temperature as adding heat
and lowering the temperature as removing heat.
Then, using Le Chatelier’s principle, determine
which way the reaction will shift.
CHANGES IN TEMPERATURE
For the chemical reaction:
2NO(g) qe N2(g) + O2(g) H=-180 kJ.
Which way will the equilibrium shift if the
temperature is increased?
to the left
THE EFFECT OF A CATALYST
• A catalyst lowers the activation energy for the
forward and the reverse reaction to the same
extent.
• Therefore, a catalyst does not alter the
equilibrium constant, nor shifts the
equilibrium.
The End.