Chemical%20Equilibrium

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Chapter 6

• Chemical Equilibrium

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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 LeChateliers Principle 6.9
Equilibria Involving Real Gases
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Nitrogen dioxide shown immediately after
expanding
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Figure 6.1 Reaction of 2NO2(g) and N2O4(g) over
time in a closed vessel
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Reddish brown nitrogen dioxide, NO2 (g)
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Reaching Equilibrium on the Macroscopic and
Molecular Level
N2O4 (g) 2 NO2
(g) Colorless Brown
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The State of Equilibrium
For the Nitrogen dioxide - dinitrogen tetroxide
equilibrium
N2O4 (g, colorless) 2 NO2 (g, brown)
At equilibrium ratefwd raterev
ratefwd kfwdN2O4 raterev
krevNO22
kfwdN2O4 krevNO22
1) Small k
N2 (g) O2 (g) 2 NO(g)
K 1 x 10 -30
2) Large k 2 CO(g) O2 (g)
2 CO2 (g) K 2.2 x 1022
3) Intermediate k 2 BrCl(g)
Br2 (g) Cl2 (g) K 5
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Reaction Direction and the Relative Sizes of Q
and K
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Initial and Equilibrium Concentrations for
theN2O4-NO2 System at 100C
Initial Equilibrium
Ratio
N2O4
N2O4
N2O4
NO2
NO2
NO22
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
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Figure 6.2 Changes in concentration with time
for the reaction
H2O(g) CO(g) H2 (g)
CO2 (g)
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Molecular model When equilibrium is reached,
how many molecules of H2O, CO, H2, and CO2 are
present?
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Figure 6.3 H2O and CO are mixed in equal
numbers
H2O(g) CO(g) H2 (g)
CO2 (g)
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Figure 6.4 Changes with time in the rates of
forward and reverse reactions
H2O(g) CO(g) H2
(g) CO2 (g)
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Figure 6.5 Concentration profile for the
reaction
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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 CH4
  0.387 mol/L
• H2 1.839 mol/L
  H2O 0.387 mol/L
• Calculate the value of K at 927oC for this
  reaction.
• Calculate the value of the equilibrium constant
  at 927oC for
• H2O(g) CH4 (g)
  CO(g) 3 H2 (g)
• 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
______L2/mol2
CO
H23
(0.613 mol/L)
(1.839 mol/L)3
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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
H23
(0.613 mol/L)
(1.839 mol/L)3
K
25.45 mol2/L2
CH4
H2O
(0.387 mol/L)
(0.387 mol/L)
This is the reciprocal of K
1
1 K
25.45 mol2/L2
0.0393 L2/mol2

• Calculate the value of the equilibrium constant
  at 927oC for
• 1/3 CO(g) H2 (g)
  1/3 CH4 (g) 1/3 H2O(g)

H2O1/3
CH41/3
(0.387mol/L)1/3
(0.387 mol/L)1/3
K
CO1/3
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)
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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.
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Expressing K with Pressure Units
n V
For gases, PVnRT can be rearranged to give P
RT
n P V RT
n V
or
Since Molarity, and R is a constant if
we 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 2 NO(g) O2 (g) 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
P 2NO2
Qp
P 2NO x PO2
Kp Kc(RT) ngas
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Figure 6.6 Position of the equilibrium
CaCO3 (s) CaO(s) CO2 (g)
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Writing the Reaction Quotient or Mass-Action
Expression
Q mass-action expression or reaction quotient
Product of the Product Concentrations
Q
Product of the Reactant Concentrations
For the general reaction
Example The Haber process for ammonia
production
N2 (g) 3 H2 (g) 2 NH3 (g)
NH32
Q
N2H23
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Reaction Direction and the Relative Sizes of Q
and K
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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)
Qc
KCl2O23
KClO32
(b) 2 C4H10 (g) 13 O2 (g) 8 CO2
(g) 10 H2O(g)
CO28 H2O10
Qc
C4H102 O213
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Writing the Reaction Quotient for an Overall
ReactionI
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 NO(g)
Kc1 4.3 x 10-25 (2) 2 NO(g) O2 (g)
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 Qcs
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 Kcs 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)
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Writing the Reaction Quotient for an Overall
ReactionII
(a) cont.
NO2
Qc (overall)
N2O22
For the individual steps
NO2
(1) N2 (g) O2 (g) 2 NO(g)
Qc1
N2 O2
NO22
(2) 2 NO(g) O2 (g) 2 NO2 (g)
Qc2
NO2 O2
NO2
NO22
NO22
Qc1 x Qc2 x

The same!
N2 O2
NO2 O2
N2O22
(b) K Kc1 x Kc2 (4.3 x 10-25)(6.4 x 109)
______________
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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)
SO32
Qc(fwd)
SO22O2
2 SO3 (g) 2 SO2 (g) O2 (g)
For the reverse reaction
at 1000K Kc(fwd) 261
1
1
and Kc(fwd)
_____________
Kc(rev)
261
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Ways of Expressing the Reaction Quotient, Q
Form of Chemical Equation Form of Q
Value of K
Beq
B
Reference reaction A B
Q(ref) K(ref) Reverse
reaction B A Q
K Reaction as sum of two
steps
A
Aeq
1 A
1 K(ref)
Q(ref) B
(1) A C
Qoverall Q1 x Q2 Q(ref) Koverall K1 x K2
x
(2) C B
C B B
K(ref)
A C A
Coefficients multiplied by n Q
Qn(ref) K Kn(ref)
Reaction with pure solid or Q Q(ref)A
B K K(ref)A B liquid component,
such as A(s)
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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.
• NH30 1.0 x 10-3 M N20 1.0 x 10-5 M
  H202.0 x 10-3 M
• NH30 2.00 x 10-4 M N20 1.50 x 10-5 M
  H20 3.54 x 10-1 M
• NH30 1.0 x 10-4 M N20 5.0 M H20 1.0 x
  10-2 M
• Solution
• a) First we calculate the Q

(1.0 x 10-3 mol/L)2
Q
____________________ L2/mol2
(1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3
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
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Example 6.2 (P 202) - II
b) We calculate the value of Q
(2.00 x 10-4 mol/L)2
Q 6.01 x 10-2
L2/mol2
(1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3)
In this case Q K, so the system is at
equilibrium. No shift will occur.
c) The value of Q is
(1.0 x 10-4 mol/L)2
Q
_________________ L2/mol2
(5.0 mol/L) (1.0 x 10-2 mol/L)3
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!
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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) K 2.76 x 102
Cl2 H2 4.000mol/2.000L 2.000M

HCl 2.000 mol/2.000L 1.000M Initial
Concentration Change
Equilibrium Conc. (mol/L)
(mol/L)
(mol/L) H2o 2.000M
-x H2
2.000-x Cl2o 2.000M
-x Cl2
2.000-x HClo 1.000M
2x HCl 1.000
2x
HCl2 H2 Cl2
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Like Example 6.3 (P203-5) - II
HCl2 H2 Cl2
(1.000 2x)2 (2.000 x)(2.000 x)
(1.000 2x)2 (2.000 x)2
K 2.76 x 102

Take the square root of each side
16.61
(1.000 2x) (2.000 x)
33.22 16.61x 1.000 2x Therefore
H2 0.269 M 32.22 18.61x
Cl2 0.269 M
x 1.731
HCl 4.462 M Check

276 OK!
HCl2 H2 Cl2
(4.462)2 (0.269)(0.269)
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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.
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Determining Equilibrium Concentrations from KI
Problem One laboratory method of making methane
is from carbon disulfide reacting with hydrogen
gas, and K this reaction at 900C 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)
CH4 H2S2
K 27.8
CS2 H24
CS2 ____________ mol/L
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Determining Equilibrium Concentrations from KII
Solution cont.
1.10 mol
0.450 mol
H2 0.23404 mol/L
H2S 0.095745 mol/L
4.70 L
4.70 L
Kc CS2 H24
(27.8)(0.05319)(0.23404)4
CH4
H2S2
(0.095745)2
0.004436
CH4 0.485547 mol/L
0.486 M
0.009167
Check Substitute the concentrations back into
the equation for K and make sure
that you get the correct value of K
(0.485547 M)(0.095745 M)2
CH4 H2S2
K
27.81875
CS2 H24
(0.05319 M)(0.23404 M)4
OK!
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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
3.000 mol
H2 2.000 M
1.500 L
3.000 mol
F2 2.000 M
1.500 L
3.000 mol
HF 2.000 M
1.500 L
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Determining Equilibrium Concentrations from
Initial Concentrations and KII
Concentration (M) H2 F2
HF Initial 2.000
2.000 2.000 Change
-x -x
2x Final 2.000-x
2.000-x 2.0002x
(2.000 2x)2
(2.000 2x)2
(2.000 - x)
(2.000 - x)
(2.000 - x)2
Taking the square root of each side we get
(2.000 2x)
(115)1/2 10.7238
x 1.528
(2.000 - x)
(5.056 M)2
H2 2.000 - 1.528 0.472 M
(0.472 M)(0.472 M)
F2 2.000 - 1.528 0.472 M
K 115
check
HF 2.000 2(1.528) 5.056 M
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Calculating K from Concentration DataI
Problem Hydrogen iodide decomposes at moderate
temperatures by the reaction below When 4.00
mol HI was placed in a 5.00 L vessel at 458C,
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 its value. Solution To calculate the
concentrations of HI and I2, we divide the
amounts of these compounds by the volume of the
vessel.
2 HI(g) H2 (g) I2 (g)
4.00 mol 5.00 L
Starting conc. of HI 0.800 M
0.442 mol 5.00 L
Equilibrium conc. of I2
0.0884 M
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
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Calculating K from Concentration DataII
HI M (0.800 - 2 x 0.0884) M 0.623 M
H2 x 0.0884 M I2
H2 I2
( 0.0884)(0.0884)
Kc
____________
(0.623)2
HI2
Therefore the equilibrium constant for the
decomposition of Hydrogen Iodide at 458C 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.
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Using the Quadratic Formula to Solve for the
Unknown
Given the Reaction between CO and H2O
Concentration (M) CO(g) H2O(g)
CO2(g) H2(g)
Initial 2.00
1.00 0
0 Change -x
-x x
x Equilibrium 2.00-x
1.00-x x x
CO2H2
(x) (x)
x2
Qc
1.56
COH2O
(2.00-x)(1.00-x)
x2 - 3.00x 2.00
We rearrange the equation 0.56x2 - 4.68x
3.12 0
ax2 bx c 0
quadratic equation
CO 1.27 M H2O 0.27 M CO2 0.73 M
H2 0.73 M
4.68 (-4.68)2 - 4(0.56)(3.12)
x
7.6 M
and 0.73 M
2(0.56)
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Predicting Reaction Direction and Calculating
Equilibrium Concentrations I
Problem Two components of natural gas can react
according to the following chemical equation
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 960C. 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
1.00 mol 0.250 L
CH4 4.00 M
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Predicting Reaction Direction and Calculating
Equilibrium Concentrations II
CS2 H24
4.00 x (8.00)4
Q
64.0
CH4 H2S2
4.00 x (8.00)2
Comparing Q and K Q gt K (64.0 gt 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 4.00
8.00 4.00 8.00 Change
x 2x
-x - 4x Equilibrium
4.00 x 8.00 2x 4.00
- x 8.00 -4x
Solving for x at equilibrium CH4 5.56 M
4.00 M x
x ____________ M
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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
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Le Chateliers 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.
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Henri Louis Le Chatelier
Source Photo Researchers
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Blue Anhydrous cobalt(II) chloride
CoCl2 (s) 6 H2O(g)
CoCl2 6 H2O(s)
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Figure 6.7 Equilibrium mixture
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The Effect of a Change in ConcentrationI
Given an equilibrium equation such as
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.
Forces equilibrium to produce more product.
Remove NH3
Forces the reaction equilibrium to go back to
the left and produce more of the reactants.
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The Effect of a Change in ConcentrationII
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.
Forces equilibrium to go toward the reactant
direction.
Add H2
Remove HCN
Forces equilibrium to make more produce and
replace the lost HCN.
51
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.
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Figure 6.8 A mixture of NH3(g), N2(g), and
H2(g) at equilibrium
N2 (g) 3 H2 (g) 2 NH3 (g)
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Figure 6.9 Brown NO2(g) and colorless N2O4(g)
at equilibrium in a syringe
2 NO2 (g) N2O4 (g)
Brown Colorless
Source Ken ODonoghue
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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
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Shifting the N2O4(g) and 2NO2(g) equilibrium by
changing the temperature
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Table 6.4 (P 216)
Shifts in the Equilibrium Position for the
Reaction N2O4(g)
2 NO2 (g)
Change
Shift
Addition of N2O4 (g)
More
Products Addition of NO2 (g)
More Reactants Removal
of N2O4 (g)
More Reactants Removal of NO2 (g)
More
Products Addition of He(g)
None Decrease in
Container Volume More
Reactants Increase in Container Volume
More Products Increase in
Temperature
More Products Decrease in Temperature
More Reactants
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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
Kpobs (atm)
(atm-2)
10
4.4 x 10-5 50
4.6 x 10-5 100
5.2 x
10-5 300
7.7 x 10-5 600
1.7 x 10-4 1000
5.3
x 10-4
66
Percent Yield of Ammonia vs. Temperature (C)
at five different operating pressures.
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Key Stages in the Haber Synthesis of Ammonia
68
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.
69
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.
70
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 H0rxngt 0

Decreases if
H0rxnlt 0 Decrease T Toward release
of heat Increases if H0rxnlt 0

Decreases if H0rxngt
0 Catalyst added None rates of forward
and reverse
reactions increase equally None