14.2-14.3 The Concept of Dynamic Equilibrium –
The Equilibrium Constant (K)
How does this chapter differentiate form
the previous chapter of reaction rates?
• In the last chapter, we found that the speed of a
chemical reaction was determined by kinetics. We also
observed factors that affected reaction rates.
• In this chapter, we will determine how far a reaction
will go, based on the equilibrium constant (K).
– The equilibrium constant is experimentally determined.
• A large equilibrium constant = reaction nearly goes to completion
(pretty much all of the reactants react to form products). Here the
reaction lies far to the right at equilibrium (product concentrations
are favored).
• A small equilibrium constant = reaction barely proceeds at all (Pretty
much all reactant, remain unreacted). Here the reaction lies far to
the left at equilibrium (reactant concentration is favored).
The Concept of Dynamic Equilibrium
• When a reaction is shown with a yield sign pointing in
both directions, this means that the reaction is
reversible.
• In the case above, as nitrogen gas begins to react with
hydrogen gas to produce ammonia, the concentration
of the reactants begins to decrease, and the rate of the
forward reaction also slows down. At the same time
the product concentration begins to increase, and the
reverse reaction rate also increases. Eventually, the
two rate will be equal. This is the point when a
dynamic equilibrium has been reached
What is a dynamic equilibrium?
• A dynamic equilibrium for a chemical reaction is the condition
in which the rate of the forward reaction equals the rate of
the reverse reaction.
• The reason that rate is underlined above is to emphasize the
point that at equilibrium it’s the concentrations of the
reactants and products no longer change…but that does not
mean that the concentrations of the reactants and products
are the same!!!
• “Dynamic” means that the forward and reverse reactions are
still occurring (the reaction doesn’t stop), but they are
occurring at the same rate.
A Graph that Shows Equilibrium: Concentration vs. Time
• As you can see in the graph above, as the concentration of the
reactant begin to decrease, the concentration of the products
increase, until the rates at which the reactants and products are
formed, are constant. This is the point where the lines are horizontal
and a dynamic equilibrium has been established. The graphs vary
depending on the reaction, but in the end, an equilibrium of
reversible reactions will always be reached.
The Equilibrium Constant (K)
• The equilibrium constant (K) is a way to quantify the
concentrations of the reactants and products at equilibrium.
Let’s consider the reaction:
aA + bB
cC + dD
Here A and B are reactants, C and D are products, and a, b, c,
d are the stoichiometric coefficients.
The equilibrium constant (K), at equilibrium, is defined as the
ratio of the concentration of the products raised to their
stochiometric coefficients, divided by the concentration of the
reactants raised to their stochiometric coefficients.
Law of Mass Action
• Law of mass action is the relationship between the balanced chemical
equation and the expression of the equilibrium constant:
[C]c[D]d
K = -----------[A]a[B]b
Let’s try a practice problem:
Express the equilibrium constant for the following
reaction:
2N2O5(g)
[NO2]4[O2]
K = ---------------[N205]2
4NO2(g) + O2(g)
Let’s Try Another!!!
Express the equilibrium constant for the combustion
of propane as shown by the balanced equation:
C3H8(g) + 5O2(g)
[CO2]3[H20]4
K = ------------------[C3H8][O2]5
3CO2(g) + 4H2O(g)
The Significance of the Equilibrium Constant
• A large K ( K > 1): The numerator is greater than
the denominator. The forward reaction is favored
(products are favored).
• A small K ( K < 1): The reverse reaction is favored
(reactants are favored).
• If K = 1, neither direction is favored and the reaction
proceeds about half-way.
Let’s Try a Practice Problem!!!
The equilibrium constant for the reaction:
A(g)
B(g) is 10. A reaction mixture initially contains [A] = 1.1
M and [B] = 0.0 M. Which statement is true at equilibrium?
(a) The reaction mixture will contain [A] = 1.0 M and [B] = 0.1 M
(b) The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M
(c) The reaction mixture will contain equal concentrations of A
and B.
(b) The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M,
so that [B]/[A] = 10.
Relationships between the Equilibrium
Constant and the Chemical Equation
• 1.) If you reverse the equation, invert the
constant. Kreverse = 1/ Kforward = K’
• 2.) If you multiply the coefficients in the equation
by a factor, raise the equilibrium constant to
that same factor.
• 3.) If you add two or more individual chemical
equations to obtain an overall equation,
multiply the corresponding equilibrium
constants by each other to obtain an overall
equilibrium constant.
Let’s Try a Practice Problem!!!
The reaction A(g)
2B(g) has an equilibrium constant
of K = 0.010. What is the equilibrium constant for the
reaction B(g)
½A(g)?
(a) 1
(b) 10
(c) 100
(d) 0.0010
(b) 10, here’s why. First, the equilibrium constant for the
reverse reaction K’ = 1/K = 1/0.010 = 100. K’=100 for the
reaction 2B(g)
A(g). However the problem is asking
us to multiply the coefficients by ½ , so we must raise K’
to a factor of ½. 100½ = 10.
Let’s Try a Couple More Practice Problems!!!
Consider the following chemical equation and the
equilibrium constant at 25oC:
2COF2(g)
CO2(g) + CF4(g) K = 2.2x106
Calculate the equilibrium constant for the following
reaction at 25oC.
2CO2(g) + 2CF4(g)
4COF2(g) K’ = ?
K’ = 1/K = (1/(2.2X106))2 = 2.1X10-13
Let’s Try Another!!!
Predict the equilibrium constant for the first reaction
shown here given the equilibrium constants for the
second and third reactions:
CO2(g) +3H2(g)
CO(g) +H2O(g)
CO(g) +2H2(g)
CH3OH(g) + H2O(g) K1 = ?
CO2(g) + H2(g)
K2 = 1.0X105
CH3OH(g)
K3 = 1.4X107
Let’s rearrange the last two equations above, to
predict the equilibrium constant of the first  next
slide
CO2(g) +3H2(g)
CO(g) +H2O(g)
CO(g) +2H2(g)
CH3OH(g) + H2O(g) K1 = ?
CO2(g) + H2(g)
K2 = 1.0X105
CH3OH(g)
K3 = 1.4X107
By reversing the second reaction and keeping the third
reaction as it is, we can cancel substances that appear on
both sides of the reaction, and we will end up with the
first reaction:
CO2(g) + H2(g)
CO(g) +2H2(g)
CO2(g) +3H2(g)
CO(g) +H2O(g) K2’ = 1/(1.0X105)
CH3OH(g)
K3 = 1.0X107
CH3OH(g) + H2O(g) K1 = K2’ X K3 =
K1 = 1.4X102
14.2-14.4 pgs. 688-689 #’s 21(a & b), 24, 28, & 30
Read 14.4-14.7 pgs. 658-667