Chemical Equilibrium
Chapter 14
1
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Equilibrium
Equilibrium is a state in which there are no observable
changes as time goes by.
Chemical equilibrium is achieved when:
•
the rates of the forward and reverse reactions are equal and
•
the concentrations of the reactants and products remain
constant
Physical equilibrium
H2 O 𝑙 ⇄ H2 O 𝑔
Chemical equilibrium
N2 O4 𝑔 ⇄ 2NO2 𝑔
2
Equilibrium of N2 O4
N2 O4 𝑔 ⇄ 2NO2 𝑔
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equilibrium
equilibrium
Start with N2 O4
Start with NO2 & N2 O4
equilibrium
Start with NO2
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3
Equilibrium of N2 O4 (1)
N2 O4 𝑔 ⇄ 2NO2 𝑔
NO2 2
𝐾=
= 4.63 × 10−3
N2 O4
𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D
C
𝐾=
A
𝑐
D𝑑
𝑎 B 𝑏
Law of Mass Action
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4
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Equilibrium
of N2 O4 (2)
constant
5
Relationship of Equilibrium and K
C
𝐾=
A
𝑐
D𝑑
𝑎 B 𝑏
𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D
Equilibrium Will
𝐾≫1
Lie to the right
Favor products
𝐾≪1
Lie to the left
Favor reactants
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6
Example 14.1
Write expressions for 𝐾c , and 𝐾𝑝 if applicable, for the following
reversible reactions at equilibrium:
(a) HF 𝑎𝑞 + H2 O 𝑙 ⇌ H3 O+ 𝑎𝑞 + F − 𝑎𝑞
(b) 2NO 𝑔 + O2 𝑔 ⇌ 2NO2 𝑔
(c) CH3 COOH a𝑞 + C2 H5 OH a𝑞 ⇌ CH3 COOC2 H5 a𝑞 + H2 O 𝑙
9
Example 14.1 (1)
Strategy
Keep in mind the following facts: (1) the 𝐾𝑝 expression applies
only to gaseous reactions and (2) the concentration of solvent
(usually water) does not appear in the equilibrium constant
expression.
10
Example 14.1 (2)
Solution
(a) Because there are no gases present, 𝐾𝑝 does not apply and
we have only 𝐾c .
+
−
H
O
F
3
𝐾c′ =
HF H2 O
HF is a weak acid, so that the amount of water consumed in
acid ionizations is negligible compared with the total amount
of water present as solvent. Thus, we can rewrite the
equilibrium constant as
𝐇𝟑 𝐎+ 𝐅 −
𝑲c =
𝐇𝐅
11
Example 14.1 (3)
(b)
𝑲c =
𝐍𝐎2 2
𝐍𝐎 2 𝐎𝟐
𝑲𝑝 =
𝑷2No2
𝑷2NO 𝑷o2
(c) The equilibrium constant 𝐾c′ is given by
𝐾c′
CH3 COOC2 H5 H2 O
=
CH3 COOH C2 H5 OH
Because the water produced in the reaction is negligible
compared with the water solvent, the concentration of water
does not change. Thus, we can write the new equilibrium
constant as
𝐂𝐇3 𝐂𝐎𝐎𝐂2 𝐇5
𝑲c =
𝐂𝐇3 𝐂𝐎𝐎𝐇 𝐂2 𝐇5 𝐎𝐇
12
Example 14.2
The following equilibrium process has been studied at 230°C:
2NO 𝑔 + O2 𝑔 ⇌ 2NO2 𝑔
In one experiment, the concentrations of the reacting species
at equilibrium are found to be NO = 0.0542 M, O2 = 0.127 M,
and NO2 = 15.5 M. Calculate the equilibrium constant 𝐾c of
the reaction at this temperature.
13
Example 14.2 (1)
Strategy The concentrations given are equilibrium
concentrations. They have units of mol/L, so we can calculate
the equilibrium constant 𝐾c using the law of mass action
[Equation (14.2)].
Solution The equilibrium constant is given by
NO2 2
𝐾c =
NO 2 O2
Substituting the concentrations, we find that
15.5 2
𝟓
𝐾c =
=
𝟔.
𝟒𝟒
×
𝟏𝟎
0.0542 2 0.127
14
Heterogeneous Equilibrium
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 𝑠 ⇄ CaO 𝑠 + CO2 𝑔
𝐾c′
CaO CO2
=
CaCO3
𝐾𝑐 = CO2 =
𝐾c′ x
CaCO3
CaO
CaCO3 =constant
CaO =constant
𝐾𝑝 = 𝑃co2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
22
Heterogeneous Equilibrium (1)
CaCO3 𝑠 ⇄ CaO 𝑠 + CO2 𝑔
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𝑃CO2 = 𝐾𝑝
𝑃CO2 does not depend on the amount of CaCO3 or CaO
23
Example 14.6
Consider the following heterogeneous equilibrium:
CaCO3 𝑠 ⇌ CaO 𝑠 + CO2 𝑔
At 800° C, the pressure of CO2 is 0.236 atm. Calculate (a) 𝐾𝑝
and (b) 𝐾c for the reaction at this temperature.
28
Example 14.6 (1)
Strategy
Remember that pure solids do not appear in the equilibrium
constant expression. The relationship between 𝐾𝑝 and 𝐾c is
given by Equation (14.5).
Solution
(a) Using Equation (14.8) we write
𝐾𝑃 = 𝑃CO2
= 𝟎. 𝟐𝟑𝟔
29
Example 14.6 (2)
(b) From Equation (14.5), we know
𝐾𝑃 = 𝐾c 0.0821𝑇
∆𝑛
In this case, 𝑇 = 800 + 273 = 1073 K and ∆𝑛 = 1, so we
substitute these values in the equation and obtain
0.236 = 𝐾c 0.0821 × 1073
𝐾c = 𝟐. 𝟔𝟖 × 𝟏𝟎−𝟑
30
Equilibrium Constants
A+B⇄C+D
𝐾𝑐′
C+D⇄E+F
𝐾𝑐′′
A+B⇄E+F
𝐾𝑐
𝐾c′
C D
=
A B
𝐾𝑐′′
E F
=
C D
E F
𝐾c =
A B
𝐾𝑐 = 𝐾𝑐′ × 𝐾𝑐′′
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.
31
Equilibrium Constants (1)
N2 O4 𝑔 ⇄ 2NO2 𝑔
NO2 2
𝐾=
= 4.63 × 10−3
N2 O4
2NO2 𝑔 ⇄ N2 O4 𝑔
𝐾′
N2 O4
1
=
= = 216
2
NO2
𝐾
When the equation for a reversible reaction is written in
the opposite direction, the equilibrium constant becomes
the reciprocal of the original equilibrium constant.
32
Example 14.7
The reaction for the production of ammonia can be written in a
number of ways:
(a) N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔
(b)
1
N
2 2
(c)
1
N
3 2
𝑔
3
+ H2
2
𝑔 ⇌ NH3 𝑔
𝑔 + H2 𝑔 ⇌
2
NH3
3
𝑔
Write the equilibrium constant expression for each formulation.
(Express the concentrations of the reacting species in mol/L.)
(d) How are the equilibrium constants related to one another?
33
Example 14.7 (1)
Strategy
We are given three different expressions for the same reacting
system. Remember that the equilibrium constant expression
depends on how the equation is balanced, that is, on the
stoichiometric coefficients used in the equation.
34
Example 14.7 (2)
Solution
(a)
(b)
(c)
𝐾a =
NH3 2
N2 H2 3
𝐾b =
𝐾c =
NH3
1
N2 2
3
H2 2
2
NH3 3
1
N2 3 H2
35
Example 14.7 (3)
(d)
𝐾a = 𝐾b2
𝐾a = 𝐾c3
3
2
𝐾b2 = 𝐾c3 or 𝐾b = 𝐾c
36
Chemical Kinetics and Chemical Equilibrium
kf
ratef = 𝑘f A B
kr
rater = 𝑘r AB2
⎯⎯⎯
→ AB2
A+2B ⎯
2
Equilibrium
ratef = rater
𝑘f A B
2
= 𝑘r AB2
𝑘𝑓
AB2
= 𝐾𝑐 =
𝑘𝑟
A B2
38
Reaction Quotient
The reaction quotient 𝑸𝐜 is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant 𝐾𝑐 expression.
IF
•
𝑄𝑐 < 𝐾𝑐 system proceeds from left to right to reach equilibrium
•
𝑄𝑐 = 𝐾𝑐 the system is at equilibrium
•
𝑄𝑐 > 𝐾𝑐 system proceeds from right to left to reach equilibrium
39
Example 14.8
Strategy
We are given the initial amounts of the gases (in moles) in a
vessel of known volume (in liters), so we can calculate their
molar concentrations and hence the reaction quotient Q c . How
does a comparison of 𝑄c with 𝐾c enable us to determine if the
system is at equilibrium or, if not, in which direction will the net
reaction proceed to reach equilibrium?
40
Example 14.8 (1)
At the start of a reaction, there are 0.249 mol N2 , 3.21 × 10−2
mol H2 , and 6.42 × 10−4 mol NH3 in a 3.50 − L reaction vessel at
375°C. If the equilibrium constant 𝐾c for the reaction
N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔
is 1.2 at this temperature, decide whether the system is at
equilibrium. If it is not, predict which way the net reaction will
proceed.
41
Example 14.8 (2)
Solution
The initial concentrations of the reacting species are
N2
H2
0
0
NH3
0.249mol
=
= 0.0711 𝑀
3.50L
3.21 × 10−2 mol
=
= 9.17 × 10−3 𝑀
3.50L
0
6.42 × 10−4 mol
=
= 1.83 × 10−4 𝑀
3.50L
42
Example 14.8 (3)
Next we write
NH3 20
𝑄c =
N2 0 H2
3
0
1.83 × 10−4 2
=
0.0711 9.17 × 10−3
3
= 0.611
Because Qc is smaller than 𝐾c (1.2), the system is not at
equilibrium. The net result will be an increase in the
concentration of NH3 and a decrease in the concentrations
of N2 and H2 . That is, the net reaction will proceed from left
to right until equilibrium is reached.
43
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
44
Example 14.9
A mixture of 0.500 mol H2 and 0.500 mol l2 was placed in a
1.00 − L stainless-steel flask at 430° C. The equilibrium constant
𝐾c for the reaction H2 𝑔 + l2 𝑔 ⇌ 2Hl 𝑔 is 54.3 at this
temperature. Calculate the concentrations of H2 , l2 , and HI at
equilibrium
45
Example 14.9 (1)
Strategy
We are given the initial amounts of the gases (in moles) in a
vessel of known volume (in liters), so we can calculate their
molar concentrations. Because initially no HI was present, the
system could not be at equilibrium. Therefore, some H2 would
react with the same amount of l2 (why?) to form HI until
equilibrium was established.
46
Example 14.9 (2)
Solution We follow the preceding procedure to calculate the
equilibrium concentrations.
Step 1: The stoichiometry of the reaction is 1 mol H2 reacting
with 1 mol l2 to yield 2 mol HI. Let x be the depletion in
concentration (mol/L) of H2 and l2 at equilibrium. It
follows that the equilibrium concentration of HI must be
2𝑥. We summarize the changes in concentrations as
follows:
Initial (M):
Change (M):
Equilibrium (M):
H2 +
0.500
l2
0.500
⇌
2HI
0.000
−𝑥
−𝑥
+2𝑥
0.500 − 𝑥
0.500 − 𝑥
2𝑥
47
Example 14.9 (3)
Step 2: The equilibrium constant is given by
HI 2
𝐾c =
H2 I2
Substituting, we get
2𝑥 2
54.3 =
0.500 − 𝑥 0.500 − 𝑥
Taking the square root of both sides, we get
2𝑥
7.37 =
0.500 − 𝑥
𝑥 = 0.393 𝑀
48
Example 14.9 (4)
Step 3: At equilibrium, the concentrations are
H2 = 0.500 − 0.393 𝑀 = 𝟎. 𝟏𝟎𝟕𝑴
l2 = 0.500 − 0.393 𝑀 = 𝟎. 𝟏𝟎𝟕𝑴
Hl = 2 × 0.393𝑀 = 𝟎. 𝟕𝟖𝟔𝑴
Check You can check your answers by calculating 𝐾c using
the equilibrium concentrations. Remember that 𝐾c is a
constant for a particular reaction at a given temperature.
49
Example 14.10
For the same reaction and temperature as in Example 14.9,
H2 𝑔 + l2 𝑔 ⇌ 2HI 𝑔 , suppose that the initial concentrations
of H2 , l2 , and HI are 0.00623 𝑀, 0.00414 𝑀, and 0.0224 𝑀,
respectively. Calculate the concentrations of these species at
equilibrium.
50
Example 14.10 (1)
Strategy
From the initial concentrations we can calculate the reaction
quotient 𝑄c to see if the system is at equilibrium or, if not, in
which direction the net reaction will proceed to reach
equilibrium. A comparison of 𝑄c with 𝐾c also enables us to
determine if there will be a depletion in H2 and l2 or HI as
equilibrium is established.
51
Example 14.10 (2)
Solution
First we calculate 𝑄c as follows:
𝑄c =
HI
H2
2
0
0 I2 0
0.0224 2
=
= 19.5
0.00623 0.00414
Because 𝑄c (19.5) is smaller than 𝐾c (54.3), we conclude that the
net reaction will proceed from left to right until equilibrium is
reached (see Figure 14.4); that is, there will be a depletion of H2
and l2 and a gain in HI.
52
Example 14.10 (3)
Step 1: Let x be the depletion in concentration (mol/L) of H2
and I2 at equilibrium. From the stoichiometry of the
reaction it follows that the increase in concentration for
HI must be 2𝑥. Next we write
Initial (M):
Change (M):
Equilibrium (M):
H2 +
I2
⇌
2HI
0.00623
0.00414
0.0224
−𝑥
−𝑥
+2𝑥
0.00623 − 𝑥
0.00414 − 𝑥
0.0224 + 2𝑥
53
Example 14.10 (4)
Step 2: The equilibrium constant is
HI 2
𝐾c =
H2 l2
Substituting, we get
0.0224 + 2x 2
54.3 =
0.00623 − 𝑥 0.00414 − 𝑥
It is not possible to solve this equation by the square root
shortcut, as the starting concentrations [H2 ] and [I2 ] are
unequal. Instead, we must first carry out the multiplications
54.3 2.58 × 10−5 − 0.0104𝑥 + 𝑥 2 = 5.02 × 10−4 + 0.0896𝑥 + 4𝑥 2
54
Example 14.10 (5)
Collecting terms, we get
50. 3𝑥 2 − 0.654𝑥 + 8.98 × 10−4 = 0
This is a quadratic equation of the form a𝑥 2 + 𝑏𝑥 + 𝑐 = 0.
The solution for a quadratic equation (see Appendix 4) is
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
Here we have a = 50.3, 𝑏 = −0.654, and 𝑐 = 8.98 × 10−4 , so that
𝑥=
−0.654 2 − 4 50.3 8.98 × 10−4
2 × 50.3
𝑥 = 0.0114 𝑀
or
𝑥 = 0.00156 𝑀
0.654 ±
55
Example 14.10 (6)
The first solution is physically impossible because the amounts
of H2 and l2 reacted would be more than those originally
present. The second solution gives the correct answer. Note
that in solving quadratic equations of this type, one answer is
always physically impossible, so choosing a value for 𝑥 is easy.
Step 3: At equilibrium, the concentrations are
H2 = 0.00623 − 0.00156 𝑀 = 𝟎. 𝟎𝟎𝟒𝟔𝟕 𝑴
I2 = 0.00414 − 0.00156 𝑀 − 𝟎. 𝟎𝟎𝟐𝟓𝟖 𝑴
Hl = 0.0224 + 2 × 0.00156 𝑀 = 𝟎. 𝟎𝟐𝟓𝟓 𝑴
56
Le Châtelier’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
as the system reaches a new equilibrium position.
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• Changes in Concentration
N2 𝑔 + 3H2 𝑔 ⇄ 2NH3 𝑔
Equilibrium
shifts left to
offset stress
Add
NH3
58
Le Châtelier’s Principle (1)
• Changes in Concentration continued
𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D
Change
Increase concentration of product(s)
Decrease concentration of product(s)
Shifts the Equilibrium
left
right
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
right
left
59
Example 14.11
At 720° C, the equilibrium constant 𝐾c for the reaction
N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔
is 2.37 × 10−3 . In a certain experiment, the equilibrium
concentrations are N2 = 0.683 𝑀, H2 = 8.80 𝑀, and
NH3 = 1.05 𝑀. Suppose some NH3 is added to the mixture so
that its concentration is increased to 3.65 M. (a) Use Le
Châtelier’s principle to predict the shift in direction of the net
reaction to reach a new equilibrium. (b) Confirm your prediction
by calculating the reaction quotient 𝑄c and comparing its value
with 𝐾c .
60
Example 14.11 (1)
Strategy
(a) What is the stress applied to the system? How does the
system adjust to offset the stress?
(b) At the instant when some NH3 is added, the system is no
longer at equilibrium. How do we calculate the 𝑄c for the
reaction at this point? How does a comparison of 𝑄c with 𝐾c
tell us the direction of the net reaction to reach equilibrium.
61
Example 14.11 (2)
Solution
(a) The stress applied to the system is the addition of NH3 . To
offset this stress, some NH3 reacts to produce N2 and H2
until a new equilibrium is established. The net reaction
therefore shifts from right to left; that is,
N2 𝑔 + 3H2 𝑔 ⟵ 2NH3 𝑔
62
Example 14.11 (3)
(b) At the instant when some of the NH3 is added, the system
is no longer at equilibrium. The reaction quotient is given by
NH3 20
𝑄𝒄 =
N2 0 H2
3
0
2
3.65
=
0.683 8.80
3
= 2.86 × 10−2
Because this value is greater than 2.37 × 10−3 , the net
reaction shifts from right to left until 𝑄c equals 𝐾c .
63
Example 14.11 (4)
Figure 14.8 shows qualitatively the changes in concentrations
of the reacting species.
64
Le Châtelier’s Principle (2)
• Changes in Volume and Pressure
A 𝑔 +B 𝑔 ⇄C 𝑔
Change
Shifts the Equilibrium
Increase pressure Side with fewest moles of gas
Decrease pressure Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume Side with fewest moles of gas
65
Example 14.12
Consider the following equilibrium systems:
(a) 2PbS 𝑠 + 3O2 𝑔 ⇌ 2PbO 𝑠 + 2SO2 𝑔
(b) PCl5 𝑔 ⇌ PCl3 𝑔 + Cl2 𝑔
(c) H2 𝑔 + CO2 𝑔 ⇌ H2 O 𝑔 + CO 𝑔
Predict the direction of the net reaction in each case as a result
of increasing the pressure (decreasing the volume) on the
system at constant temperature.
66
Example 14.12 (1)
Strategy
A change in pressure can affect only the volume of a gas, but
not that of a solid because solids (and liquids) are much less
compressible. The stress applied is an increase in pressure.
According to Le Châtelier’s principle, the system will adjust to
partially offset this stress. In other words, the system will adjust
to decrease the pressure. This can be achieved by shifting to
the side of the equation that has fewer moles of gas. Recall that
pressure is directly proportional to moles of gas: 𝑃𝑉 = 𝑛𝑅𝑇
so 𝑃 ∝ 𝑛.
67
Example 14.12 (2)
Solution
(a) Consider only the gaseous molecules. In the balanced
equation, there are 3 moles of gaseous reactants and 2
moles of gaseous products. Therefore, the net reaction will
shift toward the products (to the right) when the pressure is
increased.
(b) The number of moles of products is 2 and that of reactants
is 1; therefore, the net reaction will shift to the left, toward
the reactant.
(c) The number of moles of products is equal to the number of
moles of reactants, so a change in pressure has no effect
on the equilibrium.
68
Le Châtelier’s Principle (3)
• Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
N2 O4 𝑔 ⇄ 2NO2 𝑔
∆𝐻 ° = 58.0 kJΤmol
69
Le Châtelier’s Principle (4)
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
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Copyright © McGraw-Hill Education. Permission required for reproduction or display
Catalyst lowers 𝐸a for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift
equilibrium.
70
Le Châtelier’s Principle - Summary
Change
Shift Equilibrium
Concentration
yes
Change Equilibrium
Constant
no
Pressure
yes*
no
Volume
yes*
no
yes
yes
no
no
Temperature
Catalyst
*Dependent on relative moles of gaseous reactants and products
71
Chapter 14 HW
On a piece(s) of paper, write your name and the
chapter number. For each slide, write the question
number and the letter corresponding to your answer
choice. On certain problems the required written work
must be shown for full credit.
* The homework assignment is worth a total of 16 pts.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Question 1
The reaction N2O4 (g) ↔ 2 NO2 (g) is endothermic.
Which of the following is true?
(Classify the reaction as homogeneous or heterogeneous)
A) Decreasing the pressure drives the reaction to the right.
B) Increasing the temperature drives the reaction to the right.
C) Decreasing the volume drives the reaction to the left.
D) All of the above.
Question 2
If an equilibrium for a reaction is said to lie to the
right then the equilibrium constant (K)…
(Write the equilibrium expression for Question 1)
A) is a negative value.
B) is a positive value less than one.
C) is a positive value greater than one.
D) is equal to one.
Question 3
The value of the equilibrium constant is not affected
by the concentrations of…
(Provide an example chemical equation and the expression
for K)
A) pure solids and liquids.
B) gases.
C) aqueous solutions.
D) all of the above.
Question 4
The value of an equilibrium constant (K) depends on…
(In your own words, write a definition for K)
A) the speed of the reaction.
B) the coefficients of the balanced chemical equation.
C) the starting concentrations of the reactants.
D) all of the above.
Question 5
For a particular set of concentrations, if the value of Q is
much greater than the value of the equilibrium constant, K,
for a reaction, then…
(In your own words, write a definition for Q)
A) the initial reaction rate will be fast, but slows down near
equilibrium.
B) the reaction will proceed to the right, but the rate cannot be
determined from the Q or K.
C) the reaction will proceed to the left, but the rate cannot be
determined from the Q or K.
D) the reaction will proceed to make more products if the value
of K is much greater than one.
Question 6
Consider the reaction of hydrogen gas and oxygen gas
to produce liquid water. Which of the following is true?
(Write the equilibrium expression of the reaction)
A) The position of the equilibrium is unaffected by the addition of
liquid water.
B) Removing some of the hydrogen gas has no effect on the
position of the equilibrium.
C) Reducing the volume of the reaction vessel has no effect on
the position of the equilibrium.
D) None of the above are true.
Question 7
Which of the following is (are) unaffected by
reversing the direction in which a chemical equation
is written?
A) The equilibrium concentrations.
B) The equilibrium constant value.
C) The mass action expression.
D) All of the above.
Question 8
The reaction of hydrogen gas with nitrogen gas to
produce ammonia gas is exothermic. Which of the
following is true?
(Write the equilibrium expression
A) Increasing the pressure of the reaction system will increase
production of ammonia.
B) Condensing the ammonia gas produced into liquid will
increase the production of ammonia.
C) Decreasing the temperature of the reaction will increase the
production of ammonia.
D) All of the above.
Question 9
Consider the chemical change AB. After the
reaction starts, but before the system reaches
equilibrium, the only process that takes places is A
being converted into B.
A) True
B) False
Question 10
Which of the following statements is true of a catalyst?
(Give an example of a catalyst and the corresponding reaction
it affects)
A) Its concentration stays constant throughout the reaction
B) it increases the rate of a reaction
C) it provides a new pathway for the reaction
D) all of the above
E) none of the above
Question 11
For reactions in gas phase, a relationship can be
established between Kc and Kp. For which of the following
processes will the two constants have the same numerical
value?
(Show a calculation that supports your answer)
A) formation of water vapor from hydrogen and oxygen.
B) formation of carbon dioxide from carbon monoxide and
oxygen.
C) synthesis of ammonia from hydrogen and nitrogen.
D) decomposition of hydrogen iodide into hydrogen and
iodine gas.
E) conversion of nitrogen dioxide into dinitrogen tetroxide.
Question 12
Calculate the equilibrium constant for the process AB, given
the following equilibrium concentrations:
[N2O4] = 0.0427 M ; [NO2] = 0.0141 M
(Show all calculation steps)
A) 6.60 x10-1
B) 215
C) 0.00466
D) 0.330
E) 3.03
Question 13
Consider the following reaction:
4 HCl (g) + O2 (g) ↔ 2 H2O (g) + 2 Cl2 (g) with DH = -114.4 kJ
One way to decrease the amount of chlorine present at
equilibrium is…
(Explain why your answer affects the equilibrium of the reaction)
A) Removing the water vapor while it is being formed.
B) Raising the temperature.
C) Adding more hydrogen chloride.
D) Decreasing the volume of the reaction vessel.
E) None of the above.
Question 14
The concentrations of pure solids are normally excluded in an
equilibrium constant expression because…
A) These substances do not participate in the processes that are at
equilibrium.
B) At the time the equilibrium is established, their absolute amounts
have not varied from the original amounts.
C) The concentration of a pure solid remains almost constant at a high
value throughout the reaction.
D) They do not play a vital role in the rate-determining step of the
reaction.
E) None of the above.
Question 15
At equilibrium…
(Using your answer, write a definition for equilibrium)
A) the rate constants for the forward and reverse
reactions are equal.
B) the reaction stops.
C) all reactants have been converted to products.
D) the forward and reverse rates are equal.
E) none of the above.
Question 16
Consider the following reaction:
CaCO3 (s) ↔ CO2 (g) + CaO (s)
What will happen to the system if more CaCO3 is added?
Explain how this condition affects the K value (concentration).
A) Nothing
B) The concentration of CO2 will increase.
C) less CaO will be produced.
D) the pressure will increase.
E) the amount of CO2 will decrease.
Question 17
Consider the following reaction:
2 H2 (g) + 2 Cl2 (g) ↔ 2 HCl (g) + ∆ (heat)
What will happen to the system if the volume of the
reaction vessel is decreased?
Explain how this condition affects the K values (concentration and
pressure) as well as if the value of one constant will change more.
A) more HCl will dissociate
B) amounts of H2 and Cl2 will decrease
C) less heat is produced
D) all of the above
E) none of the above
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