Chapter 17
Equilibrium: The Extent of Chemical Reactions
17-1
Equilibrium: The Extent of Chemical Reactions
17.1 The dynamic nature of the equilibrium state
17.2 The reaction quotient and the equilibrium constant
17.3 Expressing equilibria with pressure terms: Relation
between Kc and Kp
17.4 Reaction direction: Comparing Q and K
17.5 How to solve equilibrium problems
17.6 Reaction conditions and the equilibrium state:
LeChatelier’s principle
17-2
Kinetics addresses the speed of a reaction, the change in product
concentration (or reactant concentration) per unit time.
Equilibrium addresses the extent of a reaction, the concentration of
product that results given unlimited reaction time.
At equilibrium:
rateforward = ratereverse
A system at equilibrium is dynamic on the molecular level; no further net
change in reactant and product concentrations is observed because
changes in one direction are balanced by changes in the opposite direction.
17-3
Example:
N2O4(g)
At equilibrium:
2NO2(g)
rateforward = ratereverse
kfwd[N2O4]eq = krev[NO2]2eq
(we assume here
that the forward and
reverse reactions are
elementary steps)
kfwd/krev = [NO2]2eq / N2O4]eq = K
K = the equilibrium constant
The magnitude of K indicates how far a reaction proceeds
toward product at a given temperature.
17-4
Reaching Equilibrium on the Macroscopic
and Molecular Levels
at equilibrium
N2O4(g)
Figure 17.1
17-5
2NO2(g)
N2O4(g) is colorless and NO2(g) is brown
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If rateforward = ratereverse
then
kforward[reactants]m = kreverse[products]n
kforward
kreverse
=
[products]n
[reactants]m
= Keq
(the equilibrium
constant)
This is also known as the Law of Mass Action.
The exponents, m and n, are equal to the coefficients in the balanced
chemical equation. Note that this is equilibrium, not kinetics. The rates of
the forward and reverse reactions are equal, not the concentrations of
reactants and products.
17-6
The range of equilibrium constants
small K
nearly all reactant
large K
nearly all product
Figure 17.2
17-7
intermediate K
mix of reactant and
product
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The reaction quotient, Q
At any time, t, the system can be sampled to determine the amounts of
reactants and products present. A ratio of products to reactants,
calculated in the same manner as K, tells us whether the system has
reached equilibrium (Q = K) or whether the reaction has to proceed further
from reactants to products (Q < K) or in the reverse direction from
products to reactants (Q > K).
Molar concentrations of the substances in the reaction are used. This is
symbolized using square brackets - [ ].
For a general reaction
aA + bB
cC + dD
where a, b, c and
d are the numerical coefficients in the balanced equation, Q (and K) can
be calculated as:
[C] c [D
________
Q=
]d a
[A] [B
]b
17-8
Table 17.1 Initial and equilibrium concentration ratios for
the N2O4-NO2 system at 100 oC
ratio (Q)
ratio (K)
equilibrium
initial
[NO2]2
[N2O4] [NO2]
[N2O4]
[N2O4]eq [NO2]eq
[N2O4]eq
1
0.1000
0.0000
0.0000
0.0491 0.1018
0.211
2
0.0000
0.1000
∞
0.0185 0.0627
0.212
3
0.0500
0.0500
0.0500
0.0332 0.0837
0.231
4
0.0750
0.0250
0.00833
0.0411 0.0930
0.210
experiment
[NO2]eq2
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17-9
The change in Q
during the N2O4NO2 reaction
For experiment 1 in
Table 17.1
Figure 17.3
17-10
Q<K
Writing the Reaction Quotient
aA + bB
cC + dD
[C] c [D
________
Qc=
]d a
[A] [B
]b
(the reaction quotient
based on concentrations)
To construct the reaction quotient for a reaction,
you must know the balanced chemical equation
for the reaction!
17-11
Sample Problem 17.1
PROBLEM:
Writing the reaction quotient from the balanced equation
Write the reaction quotient, Qc, for each of the following reactions:
(a) The decomposition of dinitrogen pentoxide, N2O5(g)
NO2(g) + O2(g)
(b) The combustion of propane gas, C3H8(g) + O2(g)
CO2(g) + H2O(g)
PLAN: Balance the equations before writing the Qc expression.
SOLUTION:
(a) 2 N2O5(g)
(b)
C3H8(g) +
[NO2]4[O2]
4 NO2(g) + O2(g)
5 O2(g)
3CO2(g) +
Qc =
[N2O5]2
4H2O(g)
[CO2]3[H2O]
Qc =
17-12
[C3H8][O2]5
4
Determining Qc for an Overall Reaction
If an overall reaction is the sum of two or more reactions,
the overall reaction quotient (or K) is equal to the product
of the reaction quotients (or Ks) for the individual steps.
Thus....
Qoverall = Q1 x Q2 x Q3........
and
Koverall = K1 x K2 x K3.......
17-13
Sample Problem 17.2
Writing the reaction quotient for an overall reaction
PROBLEM: At the very high temperatures reached during the explosive
combustion of gasoline within the cylinders of a car engine, some of
the N2 and O2 present form nitric oxide which combines with more
O2 to form nitrogen dioxide, a toxic pollutant that contributes to
photochemical smog.
(1) N2(g) + O2(g)
(2) 2NO(g) + O2(g)
2NO(g)
2NO2(g)
Kc1 = 4.3 x 10-25
Kc2 = 6.4 x 109
(a) Show that Qc for the overall reaction sequence is the same as the
product of the Qcs of the individual reactions.
(b) Calculate Kc for the overall reaction.
PLAN:
Sum the equations to give the overall reaction and write its Qc. Write
the Qcs for the individual reactions and multiply the expressions.
The Kcs for the individual reactions are multiplied to give the equilibrium
constants for the overall reaction.
17-14
Sample Problem 17.2 (continued)
SOLUTION:
(a)
(1) N2(g) + O2(g)
(2) 2NO(g) + O2(g)
2NO(g)
Qc1 =
[N2][O2]
2NO2(g)
Qc2 =
[NO2]2
Qc =
N2(g) + 2O2(g)
[N2][O2]2
Qc1 x Qc2 =
(b)
17-15
Kc =
2NO2(g)
[NO]2 x [NO2]
=
2
[N2][O2] [NO]2[O2
]
[NO]2
[NO2]
2
[NO]2[O2
]
[NO2]2
[N2] [O2]2
Kc1 x Kc2 = (4.3 x 10-25) x (6.4 x 109) = 2.8 x 10-15
Q for a forward and reverse reaction
2SO2(g) + O2(g)
2SO3(g)
Qc(fwd) = [SO3]2/[SO2]2[O2]
2SO3(g)
2SO2(g) + O2(g)
Qc(rev) = [SO2]2[O2]/[SO3]2 = 1/Qc(fwd)
Thus: Qc(fwd) = (Qc(rev))-1 and Kc(fwd) = (Kc(rev))-1
Kc(fwd) = 261 and Kc(rev) = 1/261 = 3.83 x 10-3 (at 1000 K)
17-16
Q for a reaction with coefficients multiplied
by a common factor
(multiplying by 1/2)
SO2(g) + 1/2O2(g)
SO3(g)
Q’c(fwd) = [SO3]/[SO2][O2]0.5
Thus, Q’c(fwd) = (Qc(fwd))0.5
Also, K’c(fwd) = (Kc(fwd))0.5 = (261)0.5 = 16.2
A particular K has meaning only in relation to a particular
balanced equation.
17-17
General Relationships
n(aA + bB
cC + dD)
Q’ = Qn = ([C]c[D]d/[A]a[B]b)n and K’ = Kn
17-18
Sample Problem 17.3
PROBLEM:
Determining the equilibrium constant for an equation
multiplied by a common factor
For the ammonia formation reaction, N2(g) + 3H2(g)
2NH3(g)
the equilibrium constant, Kc, is 2.4 x 10-3 at 1000 K. If we change the
coefficients of the equation, which we’ll call the reference (ref) equation, what
are the values of Kc for the following balanced equations?
(a) 1/3N2(g) + H2(g)
2/3NH3(g)
(b) NH3(g)
1/2N2(g) + 3/2H2(g)
PLAN: Compare each equation to the reference. Keep in mind that
changing the coefficients will be reflected in a power change in Kc
and a reversal of the equation will show up as an inversion of Kc.
SOLUTION:
(a) The reference equation is multiplied by 1/3, so Kc(ref) will be to the 1/3 power.
Kc = (2.4 x 10-3)1/3 = 0.13
(b) The reference equation is reversed and halved, so Kc(ref) is to the -1/2 power.
Kc = (2.4 x 10-3)-1/2 =
20.
17-19
Q for a reaction involving pure liquids and solids
heterogeneous
equilibria
CaCO3(s)
CaO(s) + CO2(g)
Qc = [CaO][CO2]/[CaCO3]
But a pure solid always has the same concentration at
a given temperature!
Thus: Terms for pure solids (and liquids) are eliminated from
the reaction quotient.
Q’c = Qc x [CaCO3]/[CaO] = [CO2]
As long as some CaCO3 and CaO are present, the
reaction quotient equals the CO2 concentration!
17-20
The reaction quotient for a heterogeneous system
solids do not
change their
concentrations!
Figure 17.4
17-21
17-22
Equilibria containing pressure terms (Kc and Kp)
PV = nRT or P/RT = n/V
n/V = molar concentration
Thus, pressure is directly proportional to molar concentration.
(at constant T)
2NO(g) + O2(g)
2NO2(g)
Qp = P2(NO2)/P2(NO) x P(O2)
Kp = equilibrium constant based on pressures
Note that in the above equation, ∆ngas = -1
17-23
But Qc = [NO2]2/[NO]2[O2]
What is the mathematical relationship between Qp and Qc?
It can be shown that: Qc = Qp(RT)
Likewise: Kc = Kp(RT) or Kp = Kc(RT)-1
Note that the exponent = ∆ngas
Generalizing:
Kp = Kc(RT)∆n(gas)
17-24
Sample Problem 17.4
PROBLEM:
Calculate Kc for the following reaction if CO2 pressure is given in
atmospheres.
CaCO3(s)
PLAN:
CaO(s) + CO2(g)
Kp = 2.1 x 10-4 at 1000. K
We know Kp and can calculate Kc after finding Dngas (R = 0.0821
L.atm/mol.K).
SOLUTION:
Dngas = 1 - 0, since there is only one gaseous product and
no gaseous reactants.
Kp = Kc(RT)1
17-25
Converting between Kc and Kp
Kc = Kp(RT)-1 = (2.1 x 10-4)(0.0821 x 1000.)-1 = 2.6 x 10-6
Reaction direction and the relative sizes of Q and K
reaction
progress
reaction
progress
reactants
products
equilibrium:
no net change
reactants
Figure 17.5
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17-26
products
Summarizing:
Q < K: reaction progresses to the right (reactants are
converted to product until equilibrium is achieved)
Q > K: reaction proceeds to the left (products are
converted to reactants until equilibrium is achieved)
Q = K: the system is at equilibrium
17-27
Sample Problem 17.5
PROBLEM:
PLAN:
Comparing Q and K to determine reaction direction
For the reaction N2O4(g)
2NO2(g), Kc = 0.21 at 100 oC. At
a point during the reaction, [N2O4] = 0.12 M and [NO2] = 0.55 M.
Has the reaction reached equilibrium? If not, in which direction
is it progressing?
Write an expression for Qc, substitute with the values given, and
compare the Qc with the given Kc.
SOLUTION:
Qc =
[NO2]
2
=
(0.55)2
= 2.5
[N2O4
(0.12)
]
Qc > Kc, therefore the reaction is not at equilibrium and will
proceed from right to left, from products to reactants, until Qc = Kc.
17-28
Solving Equilibrium Problems
Equilibrium quantities are given (concentrations or partial pressures)
and we solve for K
K and initial quantities are given and we solve for the equilibrium
concentrations
Use of reaction tables to perform calculations
17-29
Sample Problem 17.6
PROBLEM:
Calculating Kc from concentration data
In a study of hydrogen halide decomposition, a researcher fills an
evacuated 2.00 L flask with 0.200 mol of HI gas and allows the
reaction to proceed at 453 oC.
2HI(g)
H2(g) + I2(g)
At equilibrium, [HI]eq = 0.078 M. Calculate Kc.
PLAN:
Find the molar concentration of the starting material and then find the
amount of reactants and products at equilibrium.
SOLUTION: [HI] =
0.200 mol
= 0.100 M
2.00 L
Let x be the amount of [H2] at equilibrium. Then x will also be the
concentration of [I2] and the amount of [HI] is 2x or 0.078 M.
We can summarize these data in a reaction table.
17-30
Sample Problem 17.6
(continued)
A Reaction Table
concentration (M)
initial
change
equilibrium
2HI(g)
H2(g)
17-31
I2(g)
0.100
0
0
-2x
x
x
0.100 - 2x
x
x
[HI]eq = 0.078 M = 0.100 - 2x ;
[H2]
Qc = [I2]
[HI]2
+
x = 0.011 M
[0.011][0.011]
=
[0.078]2
= 0.020 = Kc
Sample Problem 17.7
PROBLEM:
PLAN:
Determining equilibrium concentrations from Kc
In a study concerning the conversion of methane to other fuels, a
chemical engineer mixes gaseous CH4 and H2O in a 0.32 L flask at
1200 K. At equilibrium the flask contains 0.26 mol of CO, 0.091 mol
of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium? Kc =
0.26 for the equation:
CH4(g) + H2O(g)
CO(g) + 3H2(g)
Use the balanced equation to write the Kc expression, and then
substitute values for each component.
[CO][H2]3
CH4(g) + H2O(g)
CO(g) + 3H2(g)
SOLUTION:
Kc =
[CH4][H2O]
0.041 mol
[CO]eq[H2]eq3
[CH4]eq =
= 0.13 M
[H2O]eq =
0.32 L
[CH4]eq Kc
0.26mol
[CO]eq =
= 0.81 M
(0.81)(0.28]3
0.32 L
=
= 0.53 M
(0.13)(0.26]
0.091 mol
[H2]eq =
= 0.28 M
0.32 L
17-32
Determining equilibrium concentrations from initial
concentrations and Kc
Sample Problem 17.8
PROBLEM:
Fuel engineers use the extent of the change from CO and H2O to
CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If
0.250 mol of CO and 0.250 mol of H2O are placed in a 125 mL flask
at 900 K, what is the composition of the equilibrium mixture? At this
temperature, Kc is 1.56 for the equation,
CO(g) + H2O(g)
PLAN:
CO2(g) + H2(g)
Find the concentrations of all species at equilibrium and then
substitute into a Kc expression.
SOLUTION: Initial concentrations must be calculated as M, so [CO] = [H2O] =
0.250/0.125L.
CO(g) + H2O(g)
CO2(g) + H2(g)
concentration
_________________________________________________
initial
change
equilibrium
17-33
2.00
-x
2.00 -x
2.00
0
0
-x
x
x
2.00 -x
x
x
Sample Problem 17.8
Qc = Kc =
(continued)
[CO2][H2]
[CO][H2O]
x
1.56 =
(x)
=
(x)
=
(x)2
(2.00 - x) (2.00 - x) (2.00 - x)2
= +/- 1.25
(negative result is ignored)
2.00 - x
17-34
x = 1.11 M
[CO] = [H2O] = 0.89 M
2.00 - x = 0.89 M
[CO2] = [H2] = 1.11 M
= 1.56
Sample Problem 17.9
PROBLEM:
Calculating equilibrium concentration with
simplifying assumptions
Phosgene is a potent chemical warfare agent that is now outlawed
by international agreement. It decomposes by the reaction,
COCl2(g)
CO(g) + Cl2(g)
Kc = 8.3 x 10-4 (at 360 oC)
Calculate [CO], [Cl2], and [COCl2] when the following amounts of phosgene
decompose and reach equilibrium in a 10.0 L flask.
(a) 5.00 mol COCl2
PLAN:
(b) 0.100 mol COCl2
After finding the concentration of starting material, write the expressions
for the equilibrium concentrations. When solving for the remaining
amount of reactant, try to make an assumption about the initial and final
concentrations that could simplify the calculations by eliminating the need
to solve a quadratic equation.
SOLUTION:
(a) 5.00 mol/10.0 L = 0.500 M
(b) 0.100 mol/10.0 L = 0.0100 M
Let x = [CO]eq = [Cl2]eq and 0.500 - x and 0.0100 - x = [COCl2]eq,
respectively, for (a) and (b).
17-35
Sample Problem 17.9
Kc =
[CO][Cl2]
[COCl2]
(continued)
(a)
Kc = 8.3 x 10-4 =
(x) (x)
(0.500 - x)
Assume x is << 0.500 so that we can drop x in the denominator.
8.3 x 10-4 =
(x) (x)
(0.500)
4.15 x 10-4 = x2
x ≈ 2.0 x 10-2 M
(0.500 - x) = 4.80 x 10-1 M
CHECK: 0.020/0.500 = 0.04 or 4% percent error (simplification is justified)
(b)
Kc = 8.3 x 10-4 =
(x) (x)
(0.010 - x)
Dropping the x gives a value of x = 2.9 x 10-3 M. (0.010 - x) ≈ 0.0071 M
CHECK: 0.0029/0.010 = 0.29 or 29% percent error (simplification not justified)
Must use the quadratic formula to solve: yields x = 2.5 x 10-3 M
and 0.0100 - x = 7.5 x 10-3 M.
17-36
Sample Problem 17.10
Predicting reaction direction and calculating
equilibrium concentrations
PROBLEM: The research and development unit of a chemical company is studying
the reaction of CH4 and H2S, two components of natural gas.
CH4(g) + 2H2S(g)
CS2(g) + 4H2(g)
In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S and 2.00 mol
of H2 are mixed in a 250 mL vessel at 960 oC. At this temperature, Kc = 0.036.
(a) In which direction will the reaction proceed to reach equilibrium?
(b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations
of the other three substances?
PLAN:
Find the initial molar concentrations of all components and use these to
calculate Qc. Compare Qc to Kc, determine in which direction the reaction
will progress, and draw up expressions for the equilibrium concentrations.
SOLUTION:
[CH4]initial = 1.00 mol/0.25 L = 4.00 M [CS2]initial = 1.00 mol/0.25 L = 4.00 M
[H2S]initial = 2.00 mol/0.25 L = 8.00 M [H2]initial = 2.00 mol/0.25 L = 8.00 M
17-37
Sample Problem 17.10
Qc =
[CS2][H2]4
[CH4][H2S]2
(continued)
=
[4.0][8.0]4
Qc of 64 is >> than Kc = 0.036
= 64.0
[4.0][8.0]2
The reaction will
progress to the left.
concentrations
CH4(g) + 2H2S(g)
CS2(g) + 4H2(g)
______________________________________________________
initial
4.00
8.00
4.00
8.00
change
+x
+ 2x
-x
- 4x
4.00 + x
8.00 + 2x
4.00 - x
8.00 - 4x
equilibrium
At equilibrium [CH4] = 5.56 M, so 5.56 = 4.00 + x; thus, x = 1.56 M
Therefore:
[H2S] = 8.00 + 2x = 11.12 M
[CS2] = 4.00 - x = 2.44 M
17-38
[H2] = 8.00 - 4x = 1.76 M
General procedure to solve equilibrium problems
PRELIMINARY SET UP
1.
2.
3.
Write balanced equation.
Write reaction quotient, Q.
Convert all amounts into the
correct units (M or atm).
WORKING ON A REACTION TABLE
4.
5.
When reaction direction is not
known, compare Q with K.
Construct a reaction table.
Check the sign of x, the change in
the quantity.
SOLVING FOR x AND EQUILIBRIUM QUANTITIES
6.
7.
8.
9.
10.
Substitute the quantities into Q.
To simplify the math, assume that x is
negligible.
[A]init - x = [A]eq ≈ [A]init
Solve for x.
Find the equilibrium quantities.
Check that assumption is
justified (< 5% error). If not,
solve quadratic equation for x.
Check that calculated values
give the known K.
Figure 17.6
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17-39
Le Chatelier’s Principle
When a chemical system at equilibrium is perturbed, it re-attains equilibrium
by undergoing a net reaction that reduces the effect of the perturbant.
Perturbants include (a) change in concentration, (b) change in
pressure, (c) change in temperature, or (d) the presence of a catalyst.
PCl3(g) + Cl2(g)
PCl5(g)
What happens when we add Cl2 to the system at equilibrium?
17-40
Table 17.3
Effect of Added Cl2 on the PCl3-Cl2-PCl5 system
concentration (M)
PCl3(g)
original equilibrium
0.200
disturbance
+
Cl2(g)
PCl5(g)
0.125
0.600
+0.075
new initial
change
new equilibrium
0.200
0.200
0.600
-x
-x
+x
0.200 - x
0.200 - x
0.600 + x
(0.637)*
*Experimentally determined value.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
17-41
If the new [PCl5] = 0.637 M, then x = 0.037 M
We can thus calculate Kc for the original and new conditions: in
both cases, Kc = 24.0.
Thus, at a given temperature, Kc does not change with a
change in concentration.
17-42
The effect of added
Cl2 on the
PCl3-Cl2-PCl5 system
[PCl5] increases
[PCl3] decreases
K is unaffected!
Figure 17.7
17-43
Sample Problem 17.11
PROBLEM:
Predicting the effect of a change in concentration on
the equilibrium position
To improve air quality and obtain a useful product, sulfur is often
removed from coal and natural gas by treating the fuel
contaminant hydrogen sulfide with O2:
2H2S(g) + O2(g)
2S(s) + 2H2O(g)
What happens to:
(a) [H2O] if O2 is added?
(b) [H2S] if O2 is added?
(c) [O2] if H2S is removed?
(d) [H2S] if sulfur is added?
PLAN: Write an expression for Q and compare it to K when the system is
perturbed to see in which direction the reaction will progress.
SOLUTION: Q =
[H2O]2
[H2S]2[O2]
(note that [S] is absent from this expression)
(a) When O2 is added, Q decreases and the reaction progresses
to the right to come back to K. Thus, [H2O] increases.
17-44
Sample Problem 17.11
(continued)
2H2S(g) + O2(g)
Q=
2S(s) + 2H2O(g)
[H2O]2
[H2S]2[O2]
(b) When O2 is added, Q decreases and the reaction progresses
to the right to come back to K. Thus, [H2S] decreases.
(c) When H2S is removed, Q increases and the reaction
progresses to the left to come back to K. Thus, [O2] increases.
(d) Sulfur is not part of the Q (or K) expression because it is a
solid. Therefore, as long as some sulfur is present, the reaction is
unaffected. [H2S] is unchanged.
17-45
Pressure effects on chemical equilibria
For reactions where ∆ngas is not zero:
If the reaction volume is decreased (pressure increased), the reaction
shifts so that the total number of gas molecules decreases.
If the reaction volume is increased (pressure decreased), the reaction
shifts so that the total number of gas molecules increases.
For reactions where ∆ngas is zero, there are no volume/pressure
effects on the equilibrium position.
Like [ ], pressure changes do not alter K.
17-46
Effect of pressure (volume) on chemical equilibria
+
lower P
(higher V)
more
moles of
gas
shift to left
Figure 17.8
17-47
higher P
(lower V)
fewer
moles of
gas
shift to right
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Sample Problem 17.12
PROBLEM:
Predicting the effect of a change in volume
(pressure) on the equilibrium position
How would you change the volume of each of the following
reactions to increase the yield of products.
(a) CaCO3(s)
(b) S(s) + 3F2(g)
(c) Cl2(g) + I2(g)
PLAN:
CaO(s) + CO2(g)
SF6(g)
2ICl(g)
When gases are present, a change in volume will affect the
concentration of the gas. If the volume decreases (pressure
increases), the reaction will shift to fewer moles of gas and vice versa.
SOLUTION: (a) CO2 is the only gas present. To increase its yield,
increase the volume (decrease the pressure).
(b) There are more moles of gaseous reactants than products, so decrease
the volume (increase the pressure) to shift the reaction to the right.
(c) There are an equal number of moles of gases on both sides of the
reaction. Therefore, a change in volume will have no effect.
17-48
Temperature effects on chemical equilibria
Unlike [ ] and pressure, temperature changes alter K.
General Correlations
A temperature rise will increase Kc for a system with
a positive ∆Horxn (endothermic).
A temperature rise will decrease Kc for a system with
a negative ∆Horxn (exothermic).
17-49
Sample Problem 17.13
Predicting the effect of a change in temperature
on the equilibrium position
PROBLEM: How does an increase in temperature affect the concentration of
the underlined substance and Kc for the following reactions?
(a) CaO(s) + H2O(l)
(b) CaCO3(s)
(c) SO2(g)
PLAN:
Ca(OH)2(aq) DHo = -82 kJ
CaO(s) + CO2(g) DHo = 178 kJ
S(s) + O2(g) DHo = 297 kJ
Express the heat of reaction as a reactant or a product. Then consider
the increase in temperature and its effect on Kc.
SOLUTION:
(a) CaO(s) + H2O(l)
Ca(OH)2(aq)  heat
An increase in temperature will shift the reaction to
the left, decrease [Ca(OH)2], and decrease Kc.
(b) CaCO3(s)  heat
CaO(s)  CO2(g)
The reaction will shift right, resulting in an increase in [CO2] and increase in Kc.
(c) SO2(g)  heat
S(s)  O2(g)
The reaction will shift right, resulting in an decrease in [SO2] and increase in Kc.
17-50
van’t Hoff Equation
ln (K2/K1) = -∆Horxn/R (1/T2 - 1/T1)
This equation allows calculation of K at one temperature if ∆Horxn and
K are known at another temperature.
17-51
Effect of a Catalyst
A catalyst shortens the time a reaction takes to reach equilibrium, but
has no effect on the equilibrium position.
17-52
17-53
End of Assigned Material
17-54
Liquid
ammonia
used as
fertilizer
Figure B17.1
17-55
Table B17.1 Effect of Temperature on Kc for Ammonia Synthesis
T (K)
Kc
200.
7.17 x 1015
300.
2.69 x 108
400.
3.94 x 104
500.
1.72 x 102
600.
4.53 x 100
700.
2.96 x 10-1
800.
3.96 x 10-2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
17-56
Percent yield of ammonia vs temperature (oC) at five
different operating pressures
Figure B17.2
17-57
Key stages in the Haber process for synthesizing ammonia
Figure B17.3
17-58
The metabolic pathway for the biosynthesis of
isoleucine from threonine
Figure B17.4
The effect of
inhibitor
binding on the
shape of an
active site
Figure B17.5
17-59