Chapter 14 Worksheet 1 (ws14.1)
Chemical Equilibrium, the Equilibrium Constant (Keq), and the Reaction Quotient (Q)
As a reaction proceeds in a closed system, the concentrations of reactants and products change until, at
equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. At that point there is
no net change in the concentrations of reactants or products. For example, for the simple reaction: A = B:
Regardless of the initial concentrations, the concentrations of reactants and products present at equilbrium
are related through the equilibrium constant (Keq) as follows:
For the reaction: aA + bB ⇌ cC + dD
If all reactants and products are in solution then:
Keq = Kc =
[C]ceq [D] deq
b
[A] aeq [B] eq
(The subscript “eq” indicates “equilibrium” concentration. The subscript “c” stands for
“concentration”. The exponents are the coefficients from the balanced equation.)
If all reactants and products are ideal gases then either partial pressures or concentrations can be used in the
equilibrium constant expression. In terms of partial pressures:
Keq = Kp =
(PC ) ceq (PD ) deq
b
(PA ) aeq (PB ) eq
(The subscript “p” stands for “pressure”.)
The numerical value of the two equilibrium constants above will be different. Remember the ideal gas law
shows that the partial pressure of a gas is directly proportional to its molarity:
P=
nRT
= MRT (M is molarity and R = 0.0821 L-atm/mol-K).
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Using this relationship, it is easy to show that (see pages 605-606 in your textbook) :
Kp = Kc(RT)n where n = moles of gaseous products – moles of gaseous reactants
For reactions involving both gases and aqueous solutions, molar concentrations and partial pressures can
appear in the equilibrium constant expression.
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Important points about equilibrium constants:
1. Every reaction has a characteristic equilibrium constant that depends only on temperature. The
value of the equilibrium constant is independent of the initial concentrations of reactants and
products.
2. Equilibrium constants have no units. This is because each concentration or pressure is actually a
ratio of the concentration or pressure to their standard values. Standard concentration for a substance
in solution is 1 M and standard pressure for a gas is 1 atm.
3. The concentrations of pure solids or pure liquids do not appear explicitly in the equilibrium
constant expression. The concentrations of pure solids and liquids are constant (density divided by
molar mass). These constant concentrations are simply incorporated into the equilibrium constant.
Alternative explanation: The standard concentration for any pure substance is simply that of the pure
substance itself (which doesn’t change). Thus, the ratio that would appear in the equilibrium constant
expression is always one (see number 2 above).
4. All reactants and products must be present at equilibrium. Even though pure solids and liquids do
not appear in an equilibrium constant expression, if they participate in a reaction they must be present
to establish equilibrium.
5. The numerical value of the equilibrium constant expresses the tendency for reactants to be
converted to products. If Keq >> 1 we say that “products are favored”. If Keq << 1 we say that
“reactants are favored”. A very large equilibrium constant ( >1000) means that the reaction will
nearly go to completion. A very small equilibrium constant (<0.001) means very little product will be
present at equilibrium (or that the reverse reaction goes to completion).
6. Changing the initial concentrations of reactants or products does NOT change the equilibrium
constant. The equilibrium concentrations of reactants and products will change but Keq will not
change. The value of Keq depends only on the identity of the reactants and products and on
temperature. Every reaction has a unique Keq at a fixed temperature.
7. .A catalyst does NOT change the equilibrium constant. A catalyst increases the rates of the
forward and reverse reactions by the same factor. (See optional document on my website for
details.)
8. Changing the temperature DOES change the equilibrium constant. Increasing the temperature
increases the equilibrium constant for an endothermic reaction and decreases the equilibrium constant
for an exothermic reaction. Increasing the temperature shifts equilibrium in the endothermic
direction. (See optional document on my website for details.)
9. The equilibrium constant for a reaction in the reverse direction is the reciprocal of the
equilibrium constant in the forward direction.
10. The equilibrium constant for a reaction that has been multiplied by a number is the original
equilibrium constant raised to a power equal to that number.
11. The equilibrium constant for a net reaction produced by adding two or more steps is the product
of the equilibrium constants for the individual steps.
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Why is Keq a “constant” and why does it have its particular form?
Remember that at equilibrium:
rate of forward reaction = rate of reverse reaction
Consider the reaction: aA + bB ⇌ cC + dD
If this reaction occurred in a single elementary step then at equilibrium (fill in the derivation below):
What if the reaction occurred in multiple elementary steps? At equilibrium every step is at equilibrium.
Using this fact, you can derive exactly the same expression for the equilibrium constant. (See optional
document on my website. The derivation also appears on page 627 of your textbook.)
Equilibrium constants are independent of mechanism! (Unlike rate laws!)
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Writing Equilibrium Constant Expressions
1.
Write the equilibrium constant expression for each of the following reactions.
a. 2 H2O2 (g) ⇌ 2 H2O (g) + O2 (g)
b. 6 H2O2 (g) ⇌ 6 H2O (g) + 3 O2 (g)
c. The reverse of the reaction in part a
d. 2 PbS (s) + 3 O2 (g)
⇌ 2 PbO (s) + 2 SO2 (g)
e. MgCl2(s) ⇌ Mg2+ (aq) + 2 Cl- (aq)
f. The reverse of the reaction in part e
2. Consider the following mechanism
N2 (g) + O2 (g) ⇌ 2 NO (g)
Keq =K1
2 NO (g) + O2 (g) ⇌ 2 NO2(g)
Keq =K2
Overall: N2 (g) + O2 (g) ⇌ 2 NO2(g)
Keq = K1K2
(SHOW THAT THIS IS TRUE.)
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Reaction quotient (Q)
At any point during a reaction, if we know the concentrations of reactants and products, we can calculate the
reaction quotient (Q).
Q=
[C ] c [ D ] d
[ A] a [ B ]b
(notice that the concentrations are NOT necessarily equilibrium concentrations)
Make sure you understand the difference between Q and Keq:
Q tells you how far a reaction is from equilibrium.
Keq tells you where the reaction is trying to go.
By comparing the values of Q and Keq, you can determine whether or not the reaction is at
equilibrium. If the reaction is not at equilibrium, you can decide whether the (net) reaction is going
in the forward or reverse direction.
If Q = Keq, then ratef = rater, the reaction is at equilibrium and no NET reaction occurs
If Q < Keq then ratef > rater, reaction will proceed from left to right ( → ; more products)
If Q > Keq, then ratef < rater, reaction will proceed from right to left ( ← ; more reactants)
Note: ratef is the rate of the forward reaction and rater is the rate of the reverse reaction.
1. At 1000 K, the value of Kp for the reaction 2SO3(g) = 2SO2(g) + O2(g) is 0.338. Calculate the value
for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial
pressures of reactants are PSO3 = 2 x 10-3 atm; PSO2 = 5 x 10-3 atm; PO2 = 3 x 10-2 atm.
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2. Chemists always want to find ways to increase the yield of a reaction. It would appear that
equilibrium is the chemist’s worst enemy! (At least when a reaction has a small equilibrium constant.
Read important point 5.) Two common ways for a chemist to increase the yield of a reaction are
described below. Explain why they increase the yield.
a. Method 1: Add more of one of the reactants.
b. Method 2: Remove one of the products as it is being made.
3. At 573 K, the equilibrium constant for: N2(g) + 3 H2(g) ⇌ 2 NH3(g) is 4.34 x 10-3. At 700 K,
Kc = 1.04 x 10-4.
a. Is this an endothermic or exothermic reaction (in the forward direction)? (See important
points 1 and 8.) Explain your answer.
b. The formation of ammonia from nitrogen gas and hydrogen gas is an important industrial
process (called the “Haber process”). As you can see from the information above, the
equilibrium constant for this reaction decreases as the temperature increases. Why, then, is the
reaction performed at very high temperature? How do you think the yield of ammonia is
maximized?
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