Notes on reaction rates, equilibrium and how to shift the equilibrium ( le Chatalier’s principle)
Reaction rates depend on many factors; you have investigated some of these:
Temperature
 Concentration and pressure
 Physical state (surface area) (or solid, liquid or gas (or plasma)
 Catalyst
What the chemical is and at what conditions are the most important

Reaction rate expressions take the form:
Rate = k [A]x [B]y [C]z
Where “k” is some numerical constant, [A] is the concentration of A; [B] is the concentration of B, etc… (x,y,
and z are some whole number that is referred to as the “order” and indicates the “power” of that particular chemical on
the overall rate).
For example: The following reaction occurs at 800 C:
2H2(g) + 2NO(g)  2H2O(g) + N2(g)
rate = k[NO(g)]2 [H2(g)]
This rate is 2nd order dependant on the concentration of nitrogen monoxide and 1st order dependant on the
concentration of hydrogen gas; we can say the rate expression is 3rd order overall (sum the exponents) which means that
if we double the concentration of everything the rate will go up by a factor of 8 (if we tripled the conc. Of everything
the rate would increase by a factor of 33 or 27 times)
Note: reaction rates must be experimentally
determined for non reversible reactions; fortunately
most reactions are reversible and we can calculate and
“equilibrium constant” based only on the balanced
chemical equation more on this later (see worksheet)
Note: reaction rates must be experimentally determined for non reversible reactions; fortunately most reactions are
reversible and we can calculate and “equilibrium constant” based only on the balanced chemical equation more on this
later (see worksheet)
Equilibrium is defined as when the rate of a forward reaction is equal
to the rate of a reverse reaction (and can be affected by temperature)
Equilibrium DOES NOT MEAN EQUAL QUANTITIES.
(it means that the rate of the forward reaction is the same as the rate of the reverse
reaction and that there is no desirable change in the amount of reactant and the amount of
product)
When a system is at equilibrium there does not appear to be any changing happening at
the macro level but there are changes happening at the micro level
(even when the un-employment rate is constant at 4.9% people are being hired and fired everyday)
(when people in highschool say that 25% of the students are “in a relationship”
some students are falling in when others are falling out)
For a reaction such as
A + 2B + C   3D + E; and all these chemicals can have their concentrations changed that
expression would be :
Keq = [D]3 [E] divided by [A] [B]2 [C] ; where the square brackets mean concentration(in
molarity, M) and the coefficients of the balanced equation are incorporated as exponents in the equilibrium constant
expression. Big value means products favored; small value more reactants at equilibrium.
This is more easily understood if written as
[D]3 [E]1
Keq =
[A]1 [B]2[C]1
( get the page of notes and do the worksheet on calculating equilibrium constants now; there is also a 2nd worksheet)
An Equilibrium constant is just a single number that lets you know if at the given conditions the reactants or the
products are favored. They vary widely in terms of their value; one (1) is the center.0008 is small 8000 is large
We don’t usually consider the units of this constant but you should know them: Molarity squared divided by Molarity to
the 5th power will result in a unit of per Molarity to the 3rd power or M-3
(the units completely cancel in the above equation(Molarity to the 4th power divided by Molarity to the 4th power)
le Chatalier’s principle
According to Le Chatelier’s Principle, when a system at equilibrium is subjected to a stress (a change in concentration,
temperature or pressure), the equilibrium will shift in the direction that tends to counteract the effect of the stress. We
can use this concept to maximize the particular chemical that we want in a chemical process. (more practice with a
worksheet later).
FOR EXAMPLE: methane reacting with oxygen to produce water and carbon-dioxide (combustion)
CH4(g) + 2 O2(g)   2 H2O(g) + CO2(g) + heat
This exothermic reaction is reversible but under normal conditions goes almost completely to products;
The reverse reaction is one of the ways we might try dealing with our surplus carbon dioxide problem.
If you flood the above system with water vapor you can reduce the amount of CO 2 at equilibrium by shifting
the equilibrium towards the reactants; you could also shift it to the left if you increased the temperature.