At equilibrium the following mathematical relationship exists.
Substances that are excluded from this equation are SOLIDS AND LIQUIDS. They
CANNOT change concentrations (they are pure), so they can not be included in
this equilibrium expression.
Example of equilibrium constants
Reaction (all substances in the gas phase)
Equilibrium Expression
3 O2 <===> 2 O3
N2 + 3 H2 <===> 2 NH3
H2 + I2 <===> 2 HI
PCl5 <===> PCl3
+ Cl2
SO2 + (1/2) O2
<===> SO3
Why don't we include solids or liquids?
ANIMATION-Notice the change when more solid is
added to the system.
Hint there is no change.
What do equilibrium constants with
solids look like?
The equilibrium produced on heating carbon with steam
Everything is exactly the same as before in the equilibrium constant
expression, except that you leave out the solid carbon.
The equilibrium produced between copper and silver ions
Both the copper on the left-hand side and the silver on the right are
solids. Both are left out of the equilibrium constant expression.
The equilibrium produced on heating calcium carbonate
This equilibrium is only established if the calcium carbonate is heated
in a closed system, preventing the carbon dioxide from escaping.
The only thing in this equilibrium which isn't a solid is the carbon
dioxide. That is all that is left in the equilibrium constant
expression.
How do we calculate K when given concentrations?
Calculate the equilibrium constant (Keq) for the following
reaction:
H2 + I2 <==> 2 HI
[H2] = 0.0505 M
[I2] = 0.0498 M
[HI] = 0.389 M
The first thing to do is write the equilibrium expression for the reaction as
written in the problem. This is what to write:
Now, all you have to do is substitute numbers into the problem. Keq
is what we want to find, so that's our "x."
Here is what we get:
Keq
=
(0.389)2
_______________
(0.0505) (0.0498)
Keq= 60.2
Solving this and rounding to the correct number of sig figs (remember
those??), we get 60.2
Given Moles and Liters
What is the equilibrium constant for the gaseous reaction
2 N2 + O2 <--> 2 N2O
if, at equilibrium, 0.00335 mole N2, 0.000464 mol O2
and 0.545 mol N2O are in a 1.00-L container.
Solution:
The mass action expression for this reaction is
KC
=
[N2O]2
[N2]2[O2]
Using the given values in the equation,
KC
=
(0.545)2
(0.00335)2(0.000464)
KC
=
0.297025
(5.20724) x 10–9
KC = 5.7 x 107
What does K imply?
Large K > 1 products are "favored"
K = 1 neither reactants nor products are favored
Small K < 1 reactants are "favored"
Practice Problems-Highlight to
reveal answer
ANSWERS
2SO2(g) + O2(g) <-----> 2SO3(g)
K = [SO3]2
/[SO2]2 [O2]
SO2(g) + 1/2O2(g) <-----> SO3(g)
K= [SO3] / [SO2]
[O2]1/2
4NH3(g) + 5O2(g) <-----> 4NO(g)
+ 6H2O(g)
K= [NO]4 [H2O]6
/ [NH3]4
[O2]5
Pb(NO3)2(s) <-----> PbO(g)
+ 2NO2(g)
+ 1/2O2(g)
K = [PbO] [NO2]2
[O2]1/2
Calculations
At a certain temperature, the equilibrium mixture of PCl5
, PCl3 , and Cl2
has the following concentrations:
[PCl3] = 0.035 M,
[PCl5]
= 0.017 M, [Cl2]
= 0.074 M
Calculate Kc for the reaction PCl3(g)
+ Cl2(g)
<--> PCl5(g)
