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Chemical Equilibrium
Mr. Matthew Totaro Legacy High School
AP Chemistry © 2012 Pearson Education, Inc.
Equilibrium
The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
© 2012 Pearson Education, Inc.
Equilibrium
The Concept of Equilibrium • As a system approaches
equilibrium, both the forward and reverse reactions are occurring.
• At equilibrium, the forward and reverse reactions are proceeding at the same rate.
© 2012 Pearson Education, Inc.
Equilibrium
A System at Equilibrium
Once equilibrium is achieved, the amount of each reactant and
product remains constant.
© 2012 Pearson Education, Inc.
Equilibrium
Dynamic Equilibrium • As the forward reaction slows and the reverse
reaction accelerates, eventually they reach the same rate
• Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal
• Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant – because the chemicals are being consumed and made at
the same rate
5
Equilibrium
Depicting Equilibrium
Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow:
N2O4(g) 2NO2(g)
© 2012 Pearson Education, Inc.
Equilibrium
When Country A citizens feel overcrowded, some will emigrate to Country B
An Analogy: Population Changes
7
However, after a time, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal
Equilibrium
The Equilibrium Constant
© 2012 Pearson Education, Inc.
Equilibrium
The Equilibrium Constant
• Forward reaction: N2O4(g) → 2NO2(g)
• Rate law: Rate = kf[N2O4]
© 2012 Pearson Education, Inc.
Equilibrium
The Equilibrium Constant
• Reverse reaction: 2 NO2(g) → N2O4(g)
• Rate law: Rate = kr[NO2]2
© 2012 Pearson Education, Inc.
Equilibrium
The Equilibrium Constant
• Therefore, at equilibrium Ratef = Rater
kf[N2O4] = kr[NO2]2 • Rewriting this, it becomes
kf kr
[NO2]2 [N2O4]
=
© 2012 Pearson Education, Inc.
Equilibrium
The Equilibrium Constant
The ratio of the rate constants is a constant at that temperature, and the expression becomes
Keq = kf kr
[NO2]2 [N2O4]
=
© 2012 Pearson Education, Inc.
Equilibrium
The Equilibrium Constant
• Consider the generalized reaction
• The equilibrium expression for this reaction would be
Kc = [C]c[D]d [A]a[B]b
aA + bB cC + dD
© 2012 Pearson Education, Inc.
Equilibrium
Equilibrium Constant • Even though the concentrations of reactants and
products are not equal at equilibrium, there is a relationship between them
• The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action
• For the general equation aA + bB → cC + dD, the Law of Mass Action gives the relationship below – the lowercase letters represent the coefficients of the balanced
chemical equation – always products over reactants
• K is called the equilibrium constant – unitless
14
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
Writing Equilibrium Constant Expressions
• For the reaction aA(aq) + bB(aq) ⇔ cC(aq) + dD(aq) the equilibrium constant expression is
• So for the reaction 2 N2O5(g) ⇔ 4 NO2(g) + O2(g) the equilibrium constant expression is
15
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
What Does the Value of Keq Imply? • When the value of Keq >> 1, we know that when the
reaction reaches equilibrium there will be many more product molecules present than reactant molecules – the position of equilibrium favors products
• When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules – the position of equilibrium favors reactants
16
Equilibrium
A Large Equilibrium Constant
17
Equilibrium
A Small Equilibrium Constant
18
Equilibrium
Practice – Write the equilibrium constant expressions, K, and predict the position of
equilibrium for the following
2 SO2(g) + O2(g) ⇔ 2 SO3(g) K = 8 x 1025
N2(g) + 2 O2(g) ⇔ 2 NO2(g) K = 3 x 10−17
favors products
favors reactants
19
Equilibrium
Relationships between K and Chemical Equations
• When the reaction is written backward, the equilibrium constant is inverted
For the reaction aA + bB ⇔ cC + dD the equilibrium constant expression is
For the reaction cC + dD ⇔ aA + bB the equilibrium constant expression is
20
Equilibrium
Relationships between K and Chemical Equations
• When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor
For the reaction aA + bB ⇔ cC the equilibrium constant expression is
For the reaction 2aA + 2bB ⇔ 2cC the equilibrium constant expression is
21
Equilibrium
Relationships between K and Chemical Equations
• When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations
For the reactions (1) aA ⇔ bB and (2) bB ⇔ cC the equilibrium constant expressions are
For the reaction aA ⇔ cC the equilibrium constant expression is
22
Equilibrium
Example: Compute the equilibrium constant at 25 °C for the reaction NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g) for
N2(g) + 3 H2(g) → 2 NH3(g), K = 3.7 x 108 at 25 °C
Equilibrium
Kbackward = 1/Kforward, Knew = Koldn
Example: Compute the equilibrium constant at 25 °C for the reaction NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g)
24
for N2(g) + 3 H2(g) ⇔ 2 NH3(g), K = 3.7 x 108 at 25 °C K for NH3(g) ⇔ 0.5N2(g) + 1.5H2(g), at 25 °C
Solution:
Conceptual Plan:
Relationships:
Given:
Find:
N2(g) + 3 H2(g) ⇔ 2 NH3(g) K1 = 3.7 x 108
K’ K
2 NH3(g) ⇔ N2(g) + 3 H2(g)
NH3(g) ⇔ 0.5 N2(g) + 1.5 H2(g)
Equilibrium
Calculate Kc for the reaction SnO2(s) + 2 CO(g) → Sn(s) + 2 CO2(g)
given the following information:
SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g) Kc = 8.12 H2(g) + CO2(g) → H2O(g) + CO(g) Kc = 0.771
Equilibrium
The Equilibrium Constant
Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp = (PCc) (PD
d) (PA
a) (PBb)
© 2012 Pearson Education, Inc.
Equilibrium
Relationship Between Kc and Kp
• From the ideal-gas law we know that
• Rearranging it, we get PV = nRT
P = RT n V
© 2012 Pearson Education, Inc.
Equilibrium
Relationship Between Kc and Kp
Plugging this into the expression for Kp,,for each substance, the relationship between Kc and Kp becomes
where
Kp = Kc (RT)∆n
∆n = (moles of gaseous product) − (moles of gaseous reactant)
© 2012 Pearson Education, Inc.
Equilibrium
Example: Find Kc for the reaction 2 NO(g) + O2(g) ⇔ 2 NO2(g), given Kp = 2.2 x 1012 @ 25 °C
29
K is a unitless number because there are more moles of reactant than product, Kc
should be larger than Kp, and it is
Kp = 2.2 x 1012
Kc
Check:
Solution:
Conceptual Plan:
Relationships:
Given:
Find:
Kp Kc
2 NO(g) + O2(g) ⇔ 2 NO2(g) ∆n = 2 − 3 = −1
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
Practice – Calculate the value of Kp or Kc for each of the following at 27 °C
2 SO2(g) + O2(g) ⇔ 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) ⇔ 2 NO2(g) Kp = 3 x 10−17
30
Equilibrium
Equilibrium Can Be Reached from Either Direction
As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial
concentrations of NO2 and N2O4 are.
© 2012 Pearson Education, Inc.
Equilibrium
Equilibrium Can Be Reached from Either Direction
These are the data from the last two trials from the table on the
previous slide.
© 2012 Pearson Education, Inc.
Equilibrium
Equilibrium Can Be Reached from Either Direction
It doesn’t matter whether we start with N2 and H2 or whether we start with NH3, we will have
the same proportions of all three substances at equilibrium.
© 2012 Pearson Education, Inc.
Equilibrium
Heterogeneous Equilibrium
© 2012 Pearson Education, Inc.
Equilibrium
The Concentrations of Solids and Liquids Are Essentially Constant
Both the concentrations of solids and liquids can be obtained by multiplying the
density of the substance by its molar mass—and both of these are constants at
constant temperature.
© 2012 Pearson Education, Inc.
Equilibrium
The Concentrations of Solids and Liquids Are Essentially Constant
Therefore, the concentrations of solids and liquids do not appear in the
equilibrium expression.
Kc = [Pb2+] [Cl−]2
PbCl2(s) Pb2+(aq) + 2Cl−(aq)
© 2012 Pearson Education, Inc.
Equilibrium
As long as some CaCO3 or CaO
remain in the system, the amount of CO2 above the solid will
remain the same.
CaCO3(s) CO2(g) +CaO(s)
© 2012 Pearson Education, Inc.
Equilibrium
HNO2(aq) + H2O(l) ⇔ H3O+(aq) + NO2
−(aq) K = 4.6 x 10−4
Ca(NO3)2(aq) + H2SO4(aq) ⇔ CaSO4(s) + 2 HNO3(aq) K = 1 x 104
Practice – Write the equilibrium constant expressions, K, and predict the position of
equilibrium for the following
38
Equilibrium
Equilibrium Calculations
© 2012 Pearson Education, Inc.
Equilibrium
An Equilibrium Problem A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M
I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at
448 °C for the reaction taking place, which is
H2(g) + I2(s) 2HI(g)
© 2012 Pearson Education, Inc.
Equilibrium
What Do We Know?
[H2], M [I2], M [HI], M
Initially 1.000 x 10−3 2.000 x 10−3 0
Change
At equilibrium 1.87 x 10−3
© 2012 Pearson Education, Inc.
Equilibrium
[HI] Increases by 1.87 x 10−3 M
[H2], M [I2], M [HI], M
Initially 1.000 x 10−3 2.000 x 10−3 0
Change +1.87 x 10−3
At equilibrium 1.87 x 10−3
© 2012 Pearson Education, Inc.
Equilibrium
Stoichiometry tells us [H2] and [I2] decrease by half as much.
[H2], M [I2], M [HI], M
Initially 1.000 x 10−3 2.000 x 10−3 0
Change −9.35 x 10−4 −9.35 x 10−4 +1.87 x 10−3
At equilibrium 1.87 x 10−3
© 2012 Pearson Education, Inc.
Equilibrium
We can now calculate the equilibrium concentrations of all three compounds
[H2], M [I2], M [HI], M
Initially 1.000 x 10−3 2.000 x 10−3 0
Change −9.35 x 10−4 −9.35 x 10−4 +1.87 x 10−3
At equilibrium 6.5 x 10−5 1.065 x 10−3 1.87 x 10−3
© 2012 Pearson Education, Inc.
Equilibrium
and, therefore, the equilibrium constant:
Kc = [HI]2 [H2] [I2]
= 51
= (1.87 x 10−3)2 (6.5 x 10−5)(1.065 x 10−3)
© 2012 Pearson Education, Inc.
Equilibrium
Example: Find the value of Kc for the reaction 2 CH4(g) ⇔ C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4]
= 0.115 M and the equilibrium [C2H2] = 0.035 M
46
+0.035
Equilibrium
Example: Find the value of Kc for the reaction 2 CH4(g) ⇔ C2H2(g) + 3 H2(g) at 1700 °C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M
47
+0.035 −2(0.035) +3(0.035)
0.045 0.105
Equilibrium
Practice –The following data were collected for the reaction 2 NO2(g) ⇔ N2O4(g) at 100 °C. Complete
the table and determine values of Kp and Kc.
48
Equilibrium
Practice – A second experiment was done and the following data were collected for the reaction 2 NO2(g) ⇔ N2O4(g) at 100 °C. Complete the table and determine values of Kp and
Kc. Compare them to the first experiment.
49
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
The Reaction Quotient (Q)
• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.
© 2012 Pearson Education, Inc.
Equilibrium
The Reaction Quotient • If a reaction mixture, containing
both reactants and products, is not at equilibrium; how can we determine in which direction it will proceed?
• The answer is to compare the current concentration ratios to the equilibrium constant
• The concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q
for the gas phase reaction aA + bB ⇔ cC
+ dD the reaction quotient is:
51
Equilibrium
The Reaction Quotient: Predicting the Direction of Change
• If Q > K, the reaction will proceed fastest in the reverse direction
– the [products] will decrease and [reactants] will increase • If Q < K, the reaction will proceed fastest in the forward
direction – the [products] will increase and [reactants] will decrease
• If Q = K, the reaction is at equilibrium – the [products] and [reactants] will not change
• If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction
• If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction
52
Equilibrium
If Q = K, the system is at equilibrium.
© 2012 Pearson Education, Inc.
Equilibrium
If Q > K, there is too much product, and the equilibrium shifts to the left.
© 2012 Pearson Education, Inc.
Equilibrium
If Q < K, there is too much reactant, and the equilibrium shifts to the right.
© 2012 Pearson Education, Inc.
Equilibrium
Q, K, and the Direction of Reaction
56
Equilibrium
If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Example: For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm. & PICl = 0.355 atm
57
for I2(g) + Cl2(g) ⇔ 2 ICl(g), Kp = 81.9 direction reaction will proceed
Solution:
Conceptual Plan:
Relationships:
Given:
Find:
I2(g) + Cl2(g) ⇔ 2 ICl(g) Kp = 81.9
Q PI2, PCl2, PICl
because Q (10.8) < K (81.9), the reaction will proceed to the right
Equilibrium
Practice – For the reaction CH4(g) + 2 H2S(g) ⇔ CS2(g) + 4 H2(g) Kc = 3.59 at 900 °C. For each of the following measured concentrations
determine whether the reaction is at equilibrium. If not at equilibrium, in which direction will the reaction proceed to get to equilibrium?
58
Equilibrium
Practice – A sample of PCl5(g) is placed in a 0.500 L container and heated to 160 °C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635
PCl5 ⇔ PCl3 + Cl2
59
0.203 mol0.500 L
0.203 mol0.500 L
Equilibrium
Le Châtelier’s Principle
© 2012 Pearson Education, Inc.
Equilibrium
Le Châtelier’s Principle “If a system at
equilibrium is disturbed by a change in
temperature, pressure, or the concentration of
one of the components, the
system will shift its equilibrium position so
as to counteract the effect of the disturbance.”
© 2012 Pearson Education, Inc.
Equilibrium
Le Châtelier’s Principle
• Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium
• It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance – disturbances all involve making the system open
62
Equilibrium
The result will be people moving from Country B into Country A faster than
people moving from Country A into Country B. This will continue until a
new equilibrium between the populations is established; the new populations will have different numbers of people than
the old ones.
An Analogy: Population Changes
63
When an influx of population enters Country B from somewhere outside Country A, it disturbs the
equilibrium established between Country A and Country B.
When the populations of Country A and Country B are in equilibrium, the emigration rates between the two countries are equal so the populations stay constant.
Equilibrium
Disturbing Equilibrium: Adding or Removing Reactants
• After equilibrium is established, a reactant is added – as long as the added reactant is included in the equilibrium
constant expression • i.e., not a solid or liquid
• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?
64
Equilibrium
Disturbing Equilibrium: Adding Reactants • Adding a reactant initially increases the rate of the forward
reaction, but has no initial effect on the rate of the reverse reaction
• The reaction proceeds to the right until equilibrium is restored
• At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant
– but not as little of the added reactant as you had before the addition • At the new equilibrium position, the concentrations of
reactants and products will be such that the value of the equilibrium constant is the same
65
Equilibrium
Disturbing Equilibrium: Adding or Removing Reactants
• After equilibrium is established, a reactant is removed – as long as the added reactant is included in the equilibrium
constant expression • i.e., not a solid or liquid
• How will this affect the rate of the forward reaction? • How will it affect the rate of the reverse reaction? • How will it affect the value of K?
66
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
Disturbing Equilibrium: Removing Reactants • Removing a reactant initially decreases the rate of the forward
reaction, but has no initial effect on the rate of the reverse reaction.
– so the reaction is going faster in reverse • The reaction proceeds to the left until equilibrium is restored • At the new equilibrium position, you will have less of the
products than before, more of the non-removed reactants than before, and more of the removed reactant
– but not as much of the removed reactant as you had before the removal • At the new equilibrium position, the concentrations of
reactants and products will be such that the value of the equilibrium constant is the same
67
Equilibrium
The Effect of Concentration Changes on Equilibrium
• Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found – that has the same K
• Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. – you can use this to drive a reaction to completion!
• Equilibrium shifts away from side with added chemicals or toward side with removed chemicals – Remember, adding more of a solid or liquid does not change its
concentration – and therefore has no effect on the equilibrium 68
Equilibrium
The Effect of Concentration Changes on Equilibrium
When NO2 is added, some of it combines to make more N2O4
69
Equilibrium
The Effect of Concentration Changes on Equilibrium
When N2O4 is added, some of it decomposes to make more NO2 70
Equilibrium
The Haber Process The transformation of nitrogen and hydrogen into
ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost
importance.
© 2012 Pearson Education, Inc.
Equilibrium
The Haber Process
If H2 is added to the system,
N2 will be consumed and
the two reagents will form more
NH3.
© 2012 Pearson Education, Inc.
Equilibrium
The Haber Process This apparatus
helps push the equilibrium to the right by
removing the ammonia
(NH3) from the system as a
liquid.
© 2012 Pearson Education, Inc.
Equilibrium
The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium
• Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right – increasing its partial pressure increases its
concentration – does not increase the partial pressure of the other
gases in the mixture • Adding an inert gas to the mixture has no effect
on the position of equilibrium – does not affect the partial pressures of the gases in
the reaction 74
Equilibrium
Effect of Volume Change on Equilibrium • Decreasing the volume of the container increases the
concentration of all the gases in the container – increases their partial pressures – It does not change the concentrations of solutions!!
• If their partial pressures increase, then the total pressure in the container will increase
• According to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure
• The way the system reduces the pressure is to reduce the number of gas molecules in the container
• When the volume decreases, the equilibrium shifts to the side with fewer gas molecules 75
Equilibrium
Disturbing Equilibrium: Changing the Volume
• After equilibrium is established, the container volume is decreased
• How will it affect the concentration of solids, liquid, solutions, and gases?
• How will this affect the total pressure of solids, liquid, and gases?
• How will it affect the value of K?
76
Equilibrium
Disturbing Equilibrium: Reducing the Volume • Decreasing the container volume increases the concentration of
all gases, but not solids, liquid, or solutions – same number of moles, but different number of liters, resulting in a
different molarity • Decreasing the container volume will increase the total pressure
– Boyle’s Law – if the total pressure increases, the partial pressures of all the gases will
increase – Dalton’s Law of Partial Pressures • Because the total pressure increases, the position of equilibrium
will shift to decrease the pressure by removing gas molecules – shift toward the side with fewer gas molecules
• At the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same
77
Tro: Chemistry: A Molecular Approach, 2/e
Equilibrium
The Effect of Volume Changes on Equilibrium
78
Because there are more gas molecules on the reactants
side of the reaction, when the pressure is increased the
position of equilibrium shifts toward the side with fewer molecules to decrease the
pressure
When the pressure is decreased by increasing the
volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant
side
Equilibrium
The Effect of Temperature Changes on Equilibrium Position
• Exothermic reactions release energy and endothermic reactions absorb energy
• If we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes – even though heat is not matter and not written
in a proper equation 79
Equilibrium
The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
• For an exothermic reaction, heat is a product • Increasing the temperature is like adding heat • According to Le Châtelier’s Principle, the
equilibrium will shift away from the added heat • Adding heat to an exothermic reaction will decrease
the concentrations of products and increase the concentrations of reactants
• Adding heat to an exothermic reaction will decrease the value of K
• How will decreasing the temperature affect the system?
aA + bB ⇔ cC + dD + Heat 80
Kc =C
c× D
d
A a× B
b
Equilibrium
The Effect of Temperature Changes on Equilibrium for Endothermic Reactions
• For an endothermic reaction, heat is a reactant • Increasing the temperature is like adding heat • According to Le Châtelier’s Principle, the
equilibrium will shift away from the added heat • Adding heat to an endothermic reaction will
decrease the concentrations of reactants and increase the concentrations of products
• Adding heat to an endothermic reaction will increase the value of K
• How will decreasing the temperature affect the system?
Heat + aA + bB ⇔ cC + dD 81
Kc =C
c× D
d
A a× B
b
Equilibrium
The Effect of Temperature Changes on Equilibrium
82
Equilibrium
Catalysts
© 2012 Pearson Education, Inc.
Equilibrium
Catalysts Catalysts increase the rate of both the forward and reverse reactions.
© 2012 Pearson Education, Inc.
Equilibrium
Catalysts When one uses a catalyst, equilibrium is achieved faster,
but the equilibrium composition remains unaltered.
© 2012 Pearson Education, Inc.
Equilibrium
Not Changing the Position of Equilibrium: The Effect of Catalysts
• Catalysts provide an alternative, more efficient mechanism
• Catalysts work for both forward and reverse reactions
• Catalysts affect the rate of the forward and reverse reactions by the same factor
• Therefore, catalysts do not affect the position of equilibrium
86
Equilibrium
Practice – Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) ⇔ 2 SO3(g) with
∆H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? – adding more O2 to the container – condensing and removing SO3 – compressing the gases – cooling the container – doubling the volume of the container – warming the mixture – adding the inert gas helium to the container – adding a catalyst to the mixture
87
Equilibrium
Practice – Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g) ⇔ 2 SO3(g) with
∆H° = −198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is restored? – adding more O2 to the container shift to SO3 – condensing and removing SO3 shift to SO3 – compressing the gases shift to SO3 – cooling the container shift to SO3 – doubling the volume of the container shift to SO2 – warming the mixture shift to SO2 – adding helium to the container no effect – adding a catalyst to the mixture no effect
88
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