Transcript Chapter 6 Chemical Equilibrium Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of.

Chapter 6
Chemical Equilibrium
Chapter 6: Chemical Equilibrium
6.1 The Equilibrium Condition
6.2 The Equilibrium Constant
6.3 Equilibrium Expressions Involving Pressures
6.4 The Concept of Activity
6.5 Heterogeneous Equilibria
6.6 Applications of the Equilibrium Constant
6.7 Solving Equilibrium Problems
6.8 LeChatelier’s Principle
6.9 Equilibria Involving Real Gases
Nitrogen dioxide shown immediately
after expanding
Figure 6.1: Reaction of 2NO2(g) and
N2O4(g) over time in a closed vessel
Reddish brown nitrogen dioxide, NO2 (g)
Reaching Equilibrium on the
Macroscopic and Molecular Level
N2O4 (g)
2 NO2 (g)
Colorless
Brown
The State of Equilibrium
For the Nitrogen dioxide - dinitrogen tetroxide equilibrium:
N2O4 (g, colorless) = 2 NO2 (g, brown)
At equilibrium: ratefwd = raterev
raterev = krev[NO2]2
ratefwd = kfwd[N2O4]
kfwd[N2O4] = krev[NO2
]2
1) Small k
N2 (g) + O2 (g)
2) Large k
2 CO(g) + O2 (g)
3) Intermediate k
2 BrCl(g)
kfwd = [NO2]2 = K
eq
krev
[N2O4]
2 NO(g)
2 CO2 (g)
K = 1 x 10 -30
K = 2.2 x 1022
Br2 (g) + Cl2 (g)
K=5
Reaction Direction and the
Relative Sizes of Q and K
Initial and Equilibrium Concentrations for the
N2O4-NO2 System at 100°C
Initial
Equilibrium
Ratio
[N2O4]
[NO2]
[N2O4]
[NO2]
[NO2]2 [N2O4]
0.1000
0.0000
0.0491
0.1018
0.211
0.0000
0.1000
0.0185
0.0627
0.212
0.0500
0.0500
0.0332
0.0837
0.211
0.0750
0.0250
0.0411
0.0930
0.210
Figure 6.2: Changes in concentration with
time for the reaction
H2O(g) + CO(g)
H2 (g) + CO2 (g)
Molecular model: When equilibrium is
reached, how many molecules of
H2O, CO, H2, and CO2 are present?
Figure 6.3: H2O and CO are mixed
in equal numbers
H2O(g) + CO(g)
H2 (g) + CO2 (g)
Figure 6.4: Changes with time in the
rates of forward and reverse reactions
H2O(g) + CO(g)
H2 (g) + CO2 (g)
Figure 6.5: Concentration profile
for the reaction
Like Example 6.1 (P 195) - I
The following equilibrium concentrations were observed for the
Reaction between CO and H2 to form CH4 and H2O at 927oC.
CO(g) + 3 H2 (g) = CH4 (g) + H2O(g)
[CO] = 0.613 mol/L
[H2] = 1.839 mol/L
[CH4] = 0.387 mol/L
[H2O] = 0.387 mol/L
a) Calculate the value of K at 927oC for this reaction.
b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)
Solution:
a) Given the equation above:
[CH4] [H2O] (0.387 mol/L) (0.387 mol/L)
K = [CO] [H ]3 = (0.613 mol/L) (1.839 mol/L)3 = ______L2/mol2
2
Like Example 6.1 (P 195) - II
b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) = CO(g) + 3 H2 (g)
[CO] [H2]3
(0.613 mol/L) (1.839 mol/L)3
K=
=
= 25.45 mol2/L2
[H2O] [CH4]
(0.387 mol/L) (0.387 mol/L)
This is the reciprocal of K:
1
1 =
= 25.45 mol2/L2
K
0.0393 L2/mol2
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)
[H2O]1/3 [CH4]1/3
(0.387mol/L)1/3 (0.387 mol/L)1/3
K=
=
1/3
[CO] [H2]
(0.613 mol/L)1/3 (1.839 mol/L)
(0.729) (0.729)
K=
= 0.340 L2/3/mol2/3 = (0.0393L2/mol2)1/3
(0.850)(1.839)
Summary:
Some Characteristics of the
Equilibrium Expression
The equilibrium expression for a reaction written in reverse is the
reciprocal of that for the original reaction.
When the balanced equation for a reaction is multiplied by a
factor n, the equilibrium expression for the new reaction is the
original expression raised to the nth power. Thus Knew = (Koriginal)n
The apparent units for K are determined by the powers of the
various concentration terms. The (apparent) units for K therefore
depend on the reaction being considered. We will have more to say
about the units for K in section 6.4.
Expressing K with Pressure Units
For gases, PV=nRT can be rearranged to give: P = n RT
V
or: n = P Since n = Molarity, and R is a constant if we
V RT
V keep the temperature constant then
the molar concentration is directly
proportional to the pressure.
Therefore for an equilibrium between gaseous compounds we can
express the reaction quotient in terms of partial pressures.
For:
Qp =
2 NO(g) + O2 (g)
P2
NO2
P 2NO x PO2
2 NO2 (g)
If there is no change in the number of moles of
reactants and products then
n = 0 then
Kc = Kp , or if there is a change in the number of
moles of reactants or products then:
Kp = Kc(RT)
ngas
Figure 6.6: Position of the equilibrium
CaCO3 (s)
CaO(s) + CO2 (g)
Writing the Reaction Quotient or Mass-Action Expression
Q = mass-action expression or reaction quotient
Q=
Product of the Product Concentrations
Product of the Reactant Concentrations
For the general reaction:
a A + bB
cC + dD
[C]c [D]d
[A]a [B]b
Q=
Example: The Haber process for ammonia production:
N2 (g) + 3 H2 (g)
2 NH3 (g)
[NH3]2
Q=
[N2][H2]3
Reaction Direction and the
Relative Sizes of Q and K
Writing the Reaction Quotient from the Balanced Equation
Problem: Write the reaction quotient for each of the following reactions:
(a) The thermal decomposition of potassium chlorate:
KClO3 (s) = KCl(s) + O2 (g)
(b) The combustion of butane in oxygen:
C4H10 (g) + O2 (g) = CO2 (g) + H2O(g)
Plan: We first balance the equations, then construct the reaction quotient
as described by equation 17.4.
Solution:
(a) 2 KClO3 (s)
2 KCl(s) + 3 O2 (g)
(b) 2 C4H10 (g) + 13 O2 (g)
Qc =
Qc =
[KCl]2[O2]3
8 CO2 (g) + 10 H2O(g)
[CO2]8 [H2O]10
[C4H10]2 [O2]13
[KClO3]2
Writing the Reaction Quotient for an Overall
Reaction–I
Problem: Oxygen gas combines with nitrogen gas in the internal
combustion engine to produce nitric oxide, which when out in the
atmosphere combines with additional oxygen to form nitrogen dioxide.
(1) N2 (g) + O2 (g)
(2) 2 NO(g) + O2 (g)
2 NO(g)
Kc1 = 4.3 x 10-25
2 NO2 (g) Kc2 = 6.4 x 109
(a) Show that the overall Qc for this reaction sequence is the same as the
product of the Qc’s for the individual reactions.
(b) Calculate Kc for the overall reaction.
Plan: We first write the overall reaction by adding the two reactions
together and write the Qc. We then multiply the individual Kc’s for the
total K.
(1)
N2 (g) + O2 (g)
2 NO(g)
(2)
2 NO(g) + O2 (g)
2 NO2 (g)
overall: N2 (g) + 2 O2 (g)
2 NO2 (g)
Writing the Reaction Quotient for an Overall
Reaction–II
(a) cont.
2
[NO]
Qc (overall) =
[N2][O2]2
For the individual steps:
[NO]2
(1) N2 (g) + O2 (g)
2 NO(g)
Qc1 =
[N2] [O2]
2
[NO
]
2
(2) 2 NO(g) + O2 (g)
2 NO2 (g)
Qc2 =
[NO]2 [O2]
2
2
2
[NO]
[NO
]
[NO
]
2
2
Qc1 x Qc2 =
x
=
The same!
2
2
[N2] [O2]
[NO] [O2]
[N2][O2]
(b) K = Kc1 x Kc2 = (4.3 x 10-25)(6.4 x 109) = ______________
The Form of Q for a Forward and Reverse Reaction
The production of sulfuric acid depends upon the conversion of sulfur
dioxide to sulfuric trioxide before the sulfur trioxide is reacted with
water to make the sulfuric acid.
2 SO2 (g) + O2 (g)
2 SO3 (g)
2
[SO
]
3
Qc(fwd) =
[SO2]2[O2]
For the reverse reaction:
Qc(rev) =
2 SO3 (g)
[SO2]2[O2]
[SO3]2
and: Kc(fwd) =
1
Kc(rev)
=
=
1
Qc(fwd)
1
261
2 SO2 (g) + O2 (g)
at 1000K Kc(fwd) = 261
= _____________
Ways of Expressing the Reaction Quotient, Q
Form of Chemical Equation
Reference reaction: A
Reverse reaction: B
B
A
Form of Q
[B]
Q(ref) =
[A]
Q = 1 = [A]
Q(ref) [B]
Value of K
K(ref) =
K=
[B]eq
[A]eq
1
K(ref)
Reaction as sum of two steps:
(1) A
(2) C
C
B
Q1 = [C] ; Q2 = [B]
[A]
[C]
Qoverall = Q1 x Q2 = Q(ref) Koverall = K1 x K2
Coefficients multiplied by n
Reaction with pure solid or
liquid component, such as A(s)
[C] [B] [B]
= K(ref)
=
x
=
[A] [C] [A]
Q = Qn(ref)
K = Kn(ref)
Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B]
Example 6.2 (P 202) - I
For the synthesis of ammonia at 500oC, the equilibrium constant is
6.0 x 10-2 L2/mol2. Predict the direction in which the system will
shift to reach equilibrium in each of the following cases.
a) [NH3]0 = 1.0 x 10-3 M; [N2]0= 1.0 x 10-5 M; [H2]0=2.0 x 10-3 M
b) [NH3]0 = 2.00 x 10-4 M; [N2]0= 1.50 x 10-5 M; [H2]0= 3.54 x 10-1 M
c) [NH3]0 = 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 M
Solution
a) First we calculate the Q:
[NH3]02
(1.0 x 10-3 mol/L)2
Q=
=
3
[N2]0[H2]0
(1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3
= ____________________ L2/mol2
Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For the
system to attain equilibrium, the concentrations of the products
must be decreased and the concentrations of the reactants
increased. The system will shift to the left:
Example 6.2 (P 202) - II
b) We calculate the value of Q:
[NH3]02
Q=
[N2]0[H2]03
(2.00 x 10-4 mol/L)2
=
(1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3)
= 6.01 x 10-2 L2/mol2
In this case Q = K, so the system is at equilibrium. No shift will occur.
c) The value of Q is:
[NH3]02
Q=
[N2]0[H2]03
=
(1.0 x 10-4 mol/L)2
(5.0 mol/L) (1.0 x 10-2 mol/L)3
= _________________ L2/mol2
Here Q is less than K, so the system will shift to the right, attaining
equilibrium by increasing the concentration of the product and
decreasing the concentrations of the reactants. More Ammonia!
Like Example 6.3 (P203-5) - I
Look at the equilibrium example for the formation of Hydrogen Chloride
gas from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and
4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter
flask.
H2 (g) + Cl2 (g)
2 HCl(g)
2
[HCl]
K = 2.76 x 102 =
[H2] [Cl2]
Initial Concentration
(mol/L)
[Cl2] = [H2] = 4.000mol/2.000L = 2.000M
[HCl] = 2.000 mol/2.000L = 1.000M
Change
(mol/L)
Equilibrium Conc.
(mol/L)
[H2]o = 2.000M
-x
[H2] = 2.000-x
[Cl2]o = 2.000M
-x
[Cl2] = 2.000-x
[HCl]o = 1.000M
+2x
[HCl] = 1.000 + 2x
Like Example 6.3 (P203-5) - II
2
2
2
[HCl]
(1.000
+
2x)
(1.000
+
2x)
K = 2.76 x 102 = [H ] [Cl ] = (2.000 –x)(2.000 – x) = (2.000 – x)2
2
2
Take the square root of each side:
(1.000 + 2x)
16.61 = (2.000 – x)
33.22 – 16.61x = 1.000 + 2x
32.22 = 18.61x
x = 1.731
Therefore:
[H2] = 0.269 M
[Cl2] = 0.269 M
[HCl] = 4.462 M
Check:
[HCl]2
[H2] [Cl2]
(4.462)2
= (0.269)(0.269)
= 276
OK!
Summary: Solving Equilibrium Problems
Write the balanced equation for the reaction.
Write the equilibrium expression using the law of mass action.
List the initial concentrations.
Calculate Q and determine the direction of the shift to equilibrium.
Determine the change needed to reach equilibrium, and define the
equilibrium concentrations by applying the change to the
initial concentrations.
Substitute the equilibrium concentrations into the equilibrium
expression, and solve for the unknown.
Check your calculated equilibrium concentrations by making sure that
they give the correct value of K.
Determining Equilibrium Concentrations from K–I
Problem: One laboratory method of making methane is from carbon
disulfide reacting with hydrogen gas, and K this reaction at 900°C is 27.8.
CS2 (g) + 4 H2 (g)
CH4 (g) + 2 H2 S(g)
At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol
CS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was
formed?
Plan: Write the reaction quotient, and calculate the equilibrium
concentrations from the moles given and the volume of the container.
Use the reaction quotient and solve for the concentration of methane.
Solution:
CS2 (g) + 4 H2 (g)
CH4 (g) + 2 H2 S(g)
K=
[CH4] [H2S]2
[CS2] [H2]4
= 27.8
[CS2] = 0.250 mol
4.70 L
[CS2] = ____________ mol/L
Determining Equilibrium Concentrations from K–II
Solution cont.
[H2] = 1.10 mol = 0.23404 mol/L [H2S] = 0.450 mol = 0.095745 mol/L
4.70 L
4.70 L
[CH4] =
Kc [CS2] [H2]4
[H2S]2
[CH4] = 0.004436
0.009167
=
(27.8)(0.05319)(0.23404)4
(0.095745)2
= 0.485547 mol/L = 0.486 M
Check: Substitute the concentrations back into the equation for K and
make sure that you get the correct value of K
K=
S]2
[CH4] [H2
[CS2] [H2]4
(0.485547 M)(0.095745 M)2
=
= 27.81875
(0.05319 M)(0.23404 M)4
OK!
Determining Equilibrium Concentrations from Initial
Concentrations and K –I
Problem: Given the that the reaction to form HF from molecular
hydrogen and fluorine has a reaction quotient of 115 at a certain
temperature. If 3.000 mol of each component is added to a 1.500 L flask,
calculate the equilibrium concentrations of each species.
H2 (g) + F2 (g)
2 HF(g)
Plan: Calculate the concentrations of each component, and then figure
the changes, and solve the equilibrium equation to find the resultant
concentrations.
Solution:
[H2] =3.000 mol = 2.000 M
1.500 L
2
K = [HF] = 115
3.000 mol = 2.000 M
[F
]
=
2
[H2] [F2]
1.500 L
[HF] = 3.000 mol = 2.000 M
1.500 L
Determining Equilibrium Concentrations from
Initial Concentrations and K–II
Concentration (M)
Initial
Change
Final
H2
F2
HF
2.000
-x
2.000-x
2.000
-x
2.000-x
2.000
+2x
2.000+2x
2
(2.000 + 2x)2
(2.000 + 2x)2
[HF]
K=
= 115 =
=
(2.000 - x) (2.000 - x)
(2.000 - x)2
[H2][F2]
Taking the square root of each side we get:
(115)1/2 =(2.000 + 2x) =10.7238
(2.000 - x)
[H2] = 2.000 - 1.528 = 0.472 M
[F2] = 2.000 - 1.528 = 0.472 M
[HF] = 2.000 + 2(1.528) = 5.056 M
x = 1.528
2
(5.056 M)2
[HF]
K=
=
[H2][F2] (0.472 M)(0.472 M)
check:
K = 115
Calculating K from Concentration Data–I
Problem: Hydrogen iodide decomposes at moderate temperatures by the
reaction below:
2 HI
H
+I
(g)
2 (g)
2 (g)
When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium
mixture was found to contain 0.442 mol I2. What is the value of Kc ?
Plan: First we calculate the molar concentrations, and then put them into
the equilibrium expression to find it’s value.
Solution: To calculate the concentrations of HI and I2, we divide the
amounts of these compounds by the volume of the vessel.
Starting conc. of HI =4.00 mol = 0.800 M
5.00 L
Equilibrium conc. of I2 = 0.442 mol = 0.0884 M
5.00 L
Conc. (M)
2HI(g)
H2 (g)
I2 (g)
Starting
0.800
0
0
Change
- 2x
x
x
Equilibrium
0.800 - 2x
x
x = 0.0884
Calculating K from Concentration Data–II
[HI] = M = (0.800 - 2 x 0.0884) M = 0.623 M
[H2] = x = 0.0884 M = [I2]
Kc =
[H2] [I2]
[HI]2
=
( 0.0884)(0.0884)
(0.623)2
= ____________
Therefore the equilibrium constant for the decomposition of Hydrogen
Iodide at 458°C is only 0.0201 meaning that the decomposition does not
proceed very far under these temperature conditions. We were given the
initial concentrations, and that of one at equilibrium, and found the
others that were needed to calculate the equilibrium constant.
Using the Quadratic Formula to Solve for the Unknown
Given the Reaction between CO and H2O:
Concentration (M)
CO(g) + H2O(g)
CO2(g)
Initial
Change
Equilibrium
0
+x
x
[CO2][H2]
2.00
-x
2.00-x
1.00
-x
1.00-x
+
H2(g)
0
+x
x
2
(x)
(x)
x
Qc =
=
=
= 1.56
2
[CO][H2O]
(2.00-x)(1.00-x)
x - 3.00x + 2.00
We rearrange the equation:
0.56x2 - 4.68x + 3.12 = 0
ax2 + bx + c = 0
quadratic equation:
2 - 4ac
b
+
b
x=
[CO] = 1.27 M
2a
[H2O] = 0.27 M
2
[CO2] = 0.73 M
x = 4.68 + (-4.68) - 4(0.56)(3.12) = 7.6 M
2(0.56)
[H2] = 0.73 M
and 0.73 M
Predicting Reaction Direction and Calculating
Equilibrium Concentrations –I
Problem: Two components of natural gas can react according to the
following chemical equation:
CH4(g) + 2 H2S(g)
CS2(g) + 4 H2(g)
In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and
2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature,
K = 0.036. (a) In which direction will the reaction go?
(b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of the
other substances?
Plan: The find the direction, we calculate Qc using the calculated
concentrations from the data given, and compare it with Kc. (b) Based
upon (a), we determine the sign of each component for the reaction table
and then use the given [CH4] at equilibrium to determine the others.
Solution:
[H2S] = 8.00 M, [CS2] = 4.00 M
and [H2 ] = 8.00 M
[CH4] = 1.00 mol = 4.00 M
0.250 L
Predicting Reaction Direction and Calculating
Equilibrium Concentrations –II
Q=
[CS2] [H2]4
=
2
[CH4] [H2S]
4.00 x (8.00)4 = 64.0
4.00 x (8.00)2
Comparing Q and K: Q > K (64.0 > 0.036, so the reaction goes to
the left. Therefore, reactants increase and products decrease their
concentrations.
(b) Setting up the reaction table, with x = [CS2] that reacts, which equals
the [CH4] that forms.
Concentration (M)
CH4 (g) + 2 H2S(g)
CS2(g) + 4 H2(g)
Initial
Change
Equilibrium
4.00
+x
4.00 + x
8.00
+2x
8.00 + 2x
4.00
-x
4.00 - x
Solving for x at equilibrium: [CH4] = 5.56 M = 4.00 M + x
x = ____________ M
8.00
- 4x
8.00 -4x
Predicting Reaction Direction and Calculating
Equilibrium Concentrations –III
x = 1.56 M = [CH4]
Therefore:
[H2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = _________ M
[CS2] = 4.00 M - x = 4.00 M - 1.56 M = __________ M
[H2] = 8.00 M - 4x = 8.00 M - 4(1.56 M) = __________ M
[CH4] = __________ M
Le Chatelier’s Principle
“If a change in conditions (a “stress”) is imposed on a system
at equilibrium, the equilibrium position will shift in a direction
that tends to reduce that change in conditions.”
A + B
C + D + Energy
For example: In the reaction above, if more A or B is added you
will force the reaction to produce more product, if they are removed,
it will force the equilibrium to form more reactants. If C or D is
added you will force the reaction to form more reactants, if they are
Removed from the reaction mixture, it will force the equilibrium to
Form more products. If it is heated, you will get more reactants,
and if cooled, more products.
Henri Louis Le Chatelier
Source: Photo Researchers
Blue Anhydrous cobalt(II) chloride
CoCl2 (s) + 6 H2O(g)
CoCl2 6 H2O(s)
Figure 6.7: Equilibrium mixture
The Effect of a Change in Concentration–I
Given an equilibrium equation such as :
CH4 (g) + NH3 (g)
HCN(g) + 3 H2 (g)
If one adds ammonia to the reaction mixture at equilibrium, it will force
the reaction to go to the right producing more product. Likewise, if one
takes ammonia from the equilibrium mixture, it will force the reaction
back to produce more reactants by recombining H2 and HCN to give
more of the initial reactants, CH4 and NH3.
CH4 (g) + NH3 (g)
HCN(g) + 3 H2 (g)
Forces equilibrium to
Add NH3
produce more product.
CH4 (g) + NH3 (g)
Remove NH3
HCN(g) + 3 H2 (g)
Forces the reaction equilibrium to go back
to the left and produce more of the reactants.
The Effect of a Change in Concentration–II
CH4 (g) + NH3 (g)
HCN(g) + 3 H2 (g)
If to this same equilibrium mixture one decides to add one of the
products to the equilibrium mixture, it will force the equilibrium back
toward the reactant side and increase the concentrations of reactants.
Likewise, if one takes away some of the hydrogen or hydrogen cyanide
from the product side, it will force the equilibrium to replace it.
CH4 (g) + NH3 (g)
Forces equilibrium to go
toward the reactant direction.
CH4 (g) + NH3 (g)
Forces equilibrium to make more
produce and replace the lost HCN.
HCN(g) + 3 H2 (g)
Add H2
HCN(g) + 3 H2 (g)
Remove HCN
The Effect of a Change in Pressure (Volume)
Pressure changes are mainly involving gases as liquids and solids
are nearly incompressible. For gases, pressure changes can occur in
three ways:
Changing the concentration of a gaseous component
Adding an inert gas (one that does not take part in the reaction)
Changing the volume of the reaction vessel
When a system at equilibrium that contains a gas undergoes a change
in pressure as a result of a change in volume, the equilibrium position
shifts to reduce the effect of the change.
If the volume is lower (pressure is higher), the total number of gas
molecules decrease.
If the volume is higher (pressure is lower), the total number of gas
molecules increases.
Figure 6.8: A mixture of NH3(g), N2(g),
and H2(g) at equilibrium
N2 (g) + 3 H2 (g)
2 NH3 (g)
Figure 6.9: Brown NO2(g) and colorless
N2O4(g) at equilibrium in a syringe
2 NO2 (g)
N2O4 (g)
Brown
Colorless
Source: Ken O’Donoghue
The Effect of a Change in Temperature
Only temperature changes will alter the equilibrium constant, and that is
why we always specify the temperature when giving the value of Kc.
The best way to look at temperature effects is to realize that temperature
is a component of the equation, the same as a reactant, or product. For
example, if you have an exothermic reaction, heat (energy) is on the
product side of the equation, but if it is an endothermic reaction, it will
be on the reactant side of the equation.
O2 (g) + 2 H2 (g)
2 H2O(g) + Energy = Exothermic
Electrical energy + 2 H2O(g)
2 H2 (g) + O2 (g) = Endothermic
A temperature increase favors the endothermic direction and a
temperature decrease favors the exothermic direction.
A temperature rise will increase Kc for a system with a positive H0rxn
A temperature rise will decrease Kc for a system with a negative H0rxn
Shifting the N2O4(g) and 2NO2(g)
equilibrium by changing the temperature
Table 6.4 (P 216)
Shifts in the Equilibrium Position for the Reaction
N2O4(g)
2 NO2 (g)
Change
Addition of N2O4 (g)
Addition of NO2 (g)
Removal of N2O4 (g)
Removal of NO2 (g)
Addition of He(g)
Decrease in Container Volume
Increase in Container Volume
Increase in Temperature
Decrease in Temperature
Shift
More Products
More Reactants
More Reactants
More Products
None
More Reactants
More Products
More Products
More Reactants
Table 6.5 (P 217)
Values of Kpobs at 723 K for the Reaction
N2 (g) + 3 H2 (g)
2 NH3 (g)
as a Function of Total Pressure (at equilibrium)
Total Pressure
(atm)
10
50
100
300
600
1000
Kpobs
(atm-2)
4.4 x 10-5
4.6 x 10-5
5.2 x 10-5
7.7 x 10-5
1.7 x 10-4
5.3 x 10-4
Percent Yield of Ammonia vs. Temperature (°C)
at five different operating pressures.
Key Stages in the Haber Synthesis
of Ammonia
Predicting the Effect of a Change in Concentration
on the Position of the Equilibrium
Problem: Carbon will react with water to yield carbon monoxide and
and hydrogen, in a reaction called the water gas reaction that was used
to convert coal into a fuel that can be used by industry.
C(s) + H2O (g)
CO(g) + H2 (g)
What happens to:
(a) [CO] if C is added?
(c) [H2O] if H2 is added?
(b) [CO] if H2O is added?
(d) [H2O] if CO is removed?
Plan: We either write the reaction quotient to see how equilibrium will
be effected, or look at the equation, and predict the change in direction
of the reaction, and the effect of the material desired.
Solution: (a) No change, as carbon is a solid, and not involved in the
equilibrium, as long as some carbon is present to allow the reaction.
(b) The reaction moves to the product side, and [CO] increases.
(c) The reaction moves to the reactant side, and [H2O] increases.
(d) The reaction moves to the product side, and [H2O] decreases.
Predicting the Effect of Temperature and Pressure
Problem: How would you change the volume (pressure) or temperature
in the following reactions to increase the chemical yield of the products?
(a) 2 SO2 (g) + O2 (g)
2 SO3 (g); H0 = 197 kJ
(b) CO(g) + 2 H2 (g)
CH3OH(g); H0 = -90.7 kJ
(c) C(s) + CO2 (g)
2 CO(g); H0 = 172.5 kJ
(d) N2(g) + 3 H2(g)
2 NH3(g); H0 = -91.8 kJ
Plan: For the impact of volume (pressure), we examine the reaction for
the side with the most gaseous molecules formed. For temperature, we
see if the reaction is exothermic, or endothermic. An increase in volume
(pressure) will force a reaction toward fewer gas molecules.
Solution: To get a higher yield of the products you should:
(a) Increase the pressure, and increase the temperature.
(b) Increase the pressure, and decrease the temperature.
(c) A pressure change will not change the yield, an increase in the
temperature will increase the product yield.
(d) Increase the pressure, and decrease the temperature.
Effect of Various Disturbances on an
Equilibrium System
Disturbance
Net Direction of Reaction Effect on Value of K
Concentration
Increase [reactant] Toward formation of product
None
Decrease [reactant] Toward formation of reactant
None
Pressure (volume)
Increase P
Toward formation of lower
amount (mol) of gas
None
Decrease P
Toward formation of higher
amount (mol) of gas
None
Temperature
Increase T
Toward absorption of heat Increases if H0rxn> 0
Decreases if H0rxn< 0
Decrease T
Toward release of heat
Increases if H0rxn< 0
Decreases if H0rxn> 0
Catalyst added
None; rates of forward and reverse
reactions increase equally
None