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Chapter 17
Equilibrium:
The Extent of Chemical Reactions
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Figure 17.1
Reaching equilibrium on the macroscopic and molecular levels.
N2O4(g) 2NO2(g)
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Equilibrium - the condition in which the concentrations of all the
reactants and products in a closed system cease to change with time
At equilibrium: rateforward = ratereverse
no further net change is observed because changes in one direction are
balanced by changes in the other but it doesn’t mean that the reaction
had stopped.
The amount of reactants and products are constant but they are not
necessarily equal.
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If rateforward = ratereverse then
kforward[reactants]m = kreverse[products]n
kforward [products]n
= = K the equilibrium constant
kreverse [reactants]m
The values of m and n are those of the coefficients in the balanced
chemical equation. The rates of the forward and reverse reactions are
equal, NOT the concentrations of reactants and products.
K is dependent only of the temperature.
Note: The terms for pure solids or pure liquids do not appear in
the equilibrium constant expression
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Equilibrium constant expression for
N2O4(g) 2NO2(g)
2
k fwd [NO ] 2 eq
K= =
k rev [N 2O 4 ]eq
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Practice Problem
Write the Equilibrium Constant for the combustion of Propane gas
C3H8(g) + O2(g)  CO2(g) + H2O(g)
1. Balance the Equation
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
[CO 2 ]3 [H 2O]4
Kc =
[C3 H 8 ]1[O 2 ]5
The subscript “c” in Kc indicates the equilibrium constant is based
on reactant and product concentrations
The value of “K” is usually shown as a unitless number,
BUT IT ACTUALLY DOES HAVE A UNIT EXPRESSION
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Equilibrium Constants in Terms of Partial Pressures
• For reactions involving gases, concentrations are
generally reported as partial pressures, and the
equilibrium expression is often written:
( PP ) ( PQ )
p q
Kp =
( PA ) ( PB )
a b
• where the partial pressure of each reactant and
product are given as Px in units of atmospheres (atm),
and Kp is the equilibrium constant when
concentration is given in partial pressures.
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Figure 17.2 The range of equilibrium constants
small K
large K
intermediate K
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N 2( g ) + O2( g ) ← 2 NO( g )
→ K c = 1.0 x10−30
Thus, a mixture of N2 and O2 will react to a very small extent to
produce NO at equilibrium.
N 2( g ) + 3H 2( g )
← 2 NH 3( g ) K c = 5.0 x108
→
Thus, a mixture of N2 and H2 will almost completely be
converted to NH3 at equilibrium.
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Relationsip between
Kc and Kp
Kp = Kc (RT)∆n(gas)
Δn = (no. of moles of gaseous products) – (no. of moles of gaseous reactants)
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Sample Problem 17.4 Converting Between Kc and Kp
PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot flue
gas of a coal-burning power plant for form lime (CaO), which
scrubs SO2 from the gas and forms gypsum. Find Kc for the
following reaction, if CO2 pressure is in atmospheres.
CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K)
PLAN: We know Kp and can calculate Kc after finding ∆ngas. R = 0.0821
L*atm/mol*K.
SOLUTION: ∆ngas = 1 - 0 since there is only a gaseous product and no
gaseous reactants.
Kp = Kc(RT)∆n Kc = Kp/(RT)∆n = (2.1x10-4) / (0.0821 x 1000)1 = 2.6x10-6
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Q - The Reaction Quotient
The reaction quotient, Q, is defined in the same way as the equilibrium
constant Kc except that the concentrations in the equilibrium constant
expression are not necessarily equilibrium
We use the molar concentrations of the substances in the reaction. This
is symbolized by 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]d
Q=
[A]a[B]b
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Figure 17.3 The change in Q during the N2O4-NO2 reaction.
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Figure 17.5 Reaction direction and the relative sizes of Q and K.
Reaction Reaction
Progress Progress
reactants products Equilibrium: reactants products
no net change
forward backward
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Sample Problem 17.1 Writing the Reaction Quotient from the Balanced
Equation
Write the reaction quotient expression, Qc, for each of the following
PROBLEM: 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: Be sure to balance the equations before writing the Qc expression.
SOLUTION:
[NO2]4[O2]
(a) 2 N2O5(g) 4NO2(g) + O2(g) Qc =
[N2O5]2
[CO2]3[H2O]4
(b) C3H8(g) + 5O2(g) 3CO2(g) + 4 2O(g)
H Qc =
[C3H8][O2]5
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Sample Problem 17.5 Comparing Q and K to Determine Reaction Direction
PROBLEM: For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 1000C. At
a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M.
Is the reaction at equilibrium. If not, in which direction is it
progressing?
PLAN: Write an expression for Qc, substitute with the values given, and
compare the Qc with the given Kc.
SOLUTION: [NO2]2 (0.55)2
Qc = = = 2.5
[N2O4] (0.12)
Qc is > Kc, therefore the reaction is not at equilibrium and will
proceed from right to left, from products to reactants, until Qc = Kc.
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Sample Problem 17.10 Predicting Reaction Direction and Calculating
Equilibrium Concentrations
PROBLEM: Given the reaction:
CH4(g) + 2H2S(g) CS2(g) + 4H2(g)
In one experiment, 4.0 M of CH4, 4.0 M of CS2, 8.0 M of H2S, and 8.0Mof H2 are
initially mixed in a vessel at 9600C. At this temperature, Kc = 0.036
In which direction will the reaction proceed to reach equilibrium?
SOLUTION:
[CH4]initial = 4.0M [CS2]initial = 4.0M
[H2S]initial = 8.0M [H2]initial = 8.0M
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Sample Problem 17.10 Predicting Reaction Direction and Calculating
Equilibrium Concentrations
continued
[CS2][H2]4 [4.0][8.0]4
Qc = = = 64
[CH4][H2S]2 [4.0][8.0]2
A Qc of 64 is >> than Kc = 0.036
The reaction will progress to the
left.
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LeChatelier’s Principle
When a chemical system at equilibrium
is subjected to a stress,
the system will return to equilibrium
by shifting to reduce the stress.
If the concentration increases, the system reacts to consume some of it.
If the concentration decreases, the system reacts to produce some of it.
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Factors Affecting Equilibrium
• Concentration Changes
• Pressure Changes
• Temperature Changes
• Addition of a Catalyst
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I. Concentration Changes
If the conc. of a substance is increased, the equilibrium will shift in a
way that will decrease the conc. of the substance that was added.
H 2( g ) + I 2( g ) ← 2 HI ( g )
→
Where will the reaction shift?
Decrease HI - forward
Decrease I2 - backward
H2 HI
Increase H2 - forward
I2
Increase HI - backward
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Note:
adding/removing a solid/liquid to an equilibrium system will not cause any
shift in the position of equilibrium.
addition of an inert gas such as He, Ar, Kr, etc. at constant volume,
pressure and temperature does not affect the equilibrium.
CaCO3( s ) ← CaO( s ) + CO2( g )
→
Increase CO2 : - backward
Increase CaO - No effect
Decrease CO2 - forward
- No effect
Adding Kr
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Practice Exercise: Predicting the Effect of a Change in Concentration
on the Equilibrium Position
PROBLEM: To improve air quality and obtain a useful product, chemists often
remove sulfur from coal and natural gas by treating the fuel
contaminant hydrogen sulfide with O2;
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
In what direction will the rection shift if
(a) O2 is added? (b) O2 is added?
(c) H2S is removed? (d) sulfur is added?
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I. Pressure Changes
+
lower P
(higher V)
more moles
of gas
higher P
(lower V)
fewer moles
of gas
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The effect of pressure (volume) on an equilibrium system.
An increase in pressure (decrease in volume) shifts the position of the
equilibrium in such a way as to decrease the number of moles of
gaseous component.
When the volume is increased (pressure decreased), a net reaction
occurs in the direction that produces more moles of gaseous
component
N 2( g ) + 3H 2( g ) ← 2 NH 3( g )
→
Decrease in pressure : backward
Decrease in volume: forward
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Sample Problem 17.12 Predicting the Effect of a Change in Volume
(Pressure) on the Equilibrium Position
PROBLEM: For the following reactions, predict the direction of the reaction
if the pressure is increased:
(a) CaCO3(s) CaO(s) + CO2(g)
(b) S(s) + 3F2(g) SF6(g)
(c) Cl2(g) + I2(g) 2ICl(g)
(a) CO2 is the only gas present. The equilibrium will shift to
SOLUTION:
the direction with less moles of gas. Answer: backward
(a) There are more moles of gaseous reactants than products.
Answer: forward
(c) There are an equal number of moles of gases on both sides of the
reaction. Answer: no effect
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The Effect of a Change in Temperature on an Equilibrium
Consider heat as a product or a reactant.
In an exothermic reaction, heat is a product, ∆H0rxn = negative
In an endothermic reaction, heat is a reactant, ∆H0rxn = positive
CaCO3( s ) ← CaO( s ) + CO2( g )
→ ∆H = −92.4kJ
CaCO3( s ) ← CaO( s ) + CO2( g ) + 92.4kJ
→
Increase temperature: backward
CO2
92.4 kJ
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Sample Problem 17.13 Predicting the Effect of a Change in Temperature on
the Equilibrium Position
PROBLEM: In what direction will the reaction shift if there is a decrease in
temperature
(a) CaO(s) + H2O(l) Ca(OH)2(aq) ∆H0 = -82kJ
(b) CaCO3(s) CaO(s) + CO2(g) ∆H0 = 178kJ
(c) SO2(g) S(s) + O2(g) ∆H0 = 297kJ
SOLUTION:
(a) CaO(s) + H2O(l) Ca(OH)2(aq) + heat
A decrease in temperature will shift the reaction to the right
(b) CaCO3(s) + heat CaO(s) + CO2(g)
The reaction will shift to the left
(c) SO2(g) + heat S(s) + O2(g)
The reaction will shift to the left
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Addition of a Catalyst
The presence of a catalyst has no effect on the position of
the chemical equilibrium, since a catalyst affects the rates
of the forward and reverse reactions equally.
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