Download Dynamic Equilibrium Powerpoint Note

Equilibrium
Dynamic Equilibrium in Chemical
Systems
What is Equilibrium?
Dynamic equilibrium is defined as a balance
between forward and reverse processes
occurring at the same rate.
Example
The forward “running” process is balancing
the reverse “treadmill” process with the
result being the person staying in one spot.
A change in any of these processes (ie. the
person slowing down) will result in a shift in
equililibrium (ie. The person getting closer
to the edge)
Equilibrium in Chemical
Systems
When observing changes in chemical
systems, it is important to have a closed
system (matter is contained, however
energy is free to enter/leave the
system).
An aqueous solution in a test tube is a
closed system only if no gases are used
or produced in the system.
Equilibrium in Chemical
Systems
Equilibrium is represented by a
sign.
For simplicity reasons, I will be using
“”
The left arrow represents the reverse
reaction and the right arrow represents
the forward reaction.
Types of Equilibrium
Solubility Equilibrium
A dynamic equilibrium between a solute
and a solvent in a saturated system.
Ie. NaCl(s)Na(aq) + Cl(aq)
The rate of dissolution is equal to the rate
of crystallization
Types of Equilibrium
Phase Equilibrium
a dynamic equilibrium between different
physical states of pure substance in a
closed system
Ie. H2O(l) H2O(s) at 0°C
The rate of freezing is equal to the rate of
melting (slush)
Types of Equilibrium
Chemical Reaction Equilibrium
a dynamic equilibrium between reactants
and products of a chemical reaction in a
closed system.
Some Reactions will always go to
completion (all reactants used up to form
products) and such are called quantitative
reactions.
Chemical Reaction Equilibrium
When a chemical reaction occurs, we often think
that the products formed are permanent, however
all reactions are reversible under the correct
conditions.
Some reasons why reactions go to completion are:
The reverse activation energy is too high (eg. Exothermic
reactions)
A gas is formed and released because of an open
system.
As product is formed it is removed thereby not allowing
an equilibrium to be reached
Example of Chemical Reaction
Equilibrium
Decomposition of Dinitrogen tetroxide
N2O4(g)2NO2(g)
Classes of Chemical Reactions
at Equilibrium
Calculating Concentrations at
Equilibrium
Example 1
Consider the following equation for the
formation of hydrogen fluoride from its
elements at SATP;
H2(g) + F2(g)  2 HF(g)
If the reaction begins with 1.00 mol/L
concentrations of H2(g) and F2(g) and no HF(g),
calculate the concentrations of H2(g) and HF(g)
at equilibrium if the equilibrium concentration of
F2(g) is measured to be 0.24 mol/L.
Calculating Concentrations at
Equilibrium
List given information
[H2(g)]initial = 1.00 mol/L
[F2(g)]initial = 1.00 mol/L
[HF(g)]initial = 0.00 mol/L
[F2(g)]equilibrium = 0.24 mol/L
Calculating Concentrations at
Equilibrium
Setup ICE tables (Initial Conc., Change
in Conc, and Equilibrium Conc.)
Calculating Concentrations at
Equilibrium
We know that the change from reactants to
products occurs at a ratio of 1:1:2 based on
the balanced equation.
We do not know however how much reactant
has been changed into product.
We will choose a variable to represent the
change in concentration (x)
Calculating Concentrations at
Equilibrium
With this in mind our ICE table looks
like:
Calculating Concentrations at
Equilibrium
[F2(g)]equilibrium = 0.24 mol/L
1.00 mol/L – x = 0.24 mol/L
x = 0.76 mol/L
[H2(g)] = 1.00 mol/L - x
1.00 mol/L - 0.76 mol/L
[H2(g)] = 0.24 mol/L
[HF(g)] = 2x
2(0.76 mol/L)
[HF(g)] = 1.52 mol/L
Example 2
When ammonia is heated, it decomposes into
nitrogen gas and hydrogen gas according to
the following equation.
2 NH3(g) N2(g) + 3 H2(g)
When 4.0 mol of NH3(g) is introduced into a 2.0 L rigid
container and heated to a particular temperature, the
amount of ammonia changes as show in the following
figure. Determine the equilibrium concentrations of the
other two entities.
Example 2
ICE table
Through Calculations:
X = 0.5 mol/L
[NH3(g)] = 1.0 mol/L
[N2(g)] = 0.5 mol/L
[H2(g)] = 1.5 mol/L
Example 3
In a gaseous reaction system, 0.200
mol of hydrogen gas, H2(g) is added to
0.200 mol of iodine vapour, I2(g), in a
rigid 2.00 L container at 448oC. At
equilibrium the system contains 0.040
mol of hydrogen gas, H2(g). Determine
the equilibrium concentrations of H2(g)
and HI(g).
Example 3
ICE table
Through Calculations
X = 0.080 mol/L
[I2(g)] = 0.020 mol/L
[HI(g)] = 0.160 mol/L
Homework
p. 428 # 1-5
p. 437 (top) # 6, 7 and
p. 437 (bottom) # 1-4, 7a, 7b, 8