13.1: The Concept of Equilibrium
N2O4 (g)
•
2 NO2 (g)
Chemical equilibrium occurs when a reaction and its reverse reaction
proceed at the same rate.
13.1 The Concept of Equilibrium
N2O4 (g)
2 NO2 (g)
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
Equilibrium Can Be Reached from Either
Direction
Equilibrium Can Be Reached from Either
Direction
It does not 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.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
Equilibrium Can Be Reached from Either
Direction
It does not 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.
13.2 The Equilibrium Constant
For a general equilibrium reaction:
dD + eE
aA + bB
Equilibrium constant Kc = [products]
[reactants]
=
[D]d [E]e
[A]a [B]b
The constant expression depends only on the stoichiometry
of the reaction, not on its mechanism.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
13.2 The Equilibrium Constant
• Gases
(PC)c (PD)d
Kp = (P )a (P )b
A
B
From the ideal gas law we know that
PV = nRT
Therefore
n
P = V RT
Kp = Kc (RT)∆n
∆n = (moles of gaseous product) − (moles of gaseous reactant)
13.3: Interpreting and Working with Equilibrium
constants
• If K >> 1, the reaction
favours the product, i.e.
the product predominates
at equilibrium.
• If K << 1, the reaction
favours the reactant, i.e.
the reactant
predominates at
equilibrium.
Figure 13.6
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
Manipulating Equilibrium Constants
• The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to the power equal to that
number.
N2O4 (g)
[NO ]2
2 NO2 (g) Kc = [N O2 ] = 0.212 at 100°C
2 4
2 N2O4 (g)
[NO ]4
4 NO2 (g) Kc = [N O2 ]2 = (0.212)2 at 100°C
2 4
13.5/13.6 The Reaction Quotient (Q)
• To calculate Q, one substitutes the initial
concentrations on reactants and products into
the equilibrium expression.
• Q gives the same ratio the equilibrium
expression gives, but for a system that is not
at equilibrium.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
The Reaction Quotient (Q)
If Q = K
The system is at equilibrium.
Figure 13.8
If Q > K
There is too much product and the
equilibrium shifts to the left.
Figure 13.8
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
If Q < K
There is too much reactant and the
equilibrium shifts to the right.
Figure 13.8
“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.”
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
Change in Reactant or Product
Concentrations
•
E.g. If H2 is added to the
system, N2 will be
consumed and the two
reagents will form more
NH3.
•
Adding a substance will
shift the reaction so as to
re-establish the equilibrium
by consuming part of that
substance.
Figure 13.4
Figure 13.10
The Effect of Changes in Temperature
Co(H2O)62+(aq) + 4 Cl(aq)
CoCl4 (aq) + 6 H2O(l)
Figure 13.13
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia
The Effect of Catalysts
Figure 13.14
Catalysts increase
the rate of both
the forward and
reverse reactions
but do not change
the composition
of the equilibrium
mixture.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Education Australia