Problem #1
6 mol of SO2 and 4 mol of O2 are placed into a 1 L flask at
temperature, T. The equilibrium concentration of SO3 is found to be
4 mol/L. Determine Kc.
2SO2(g) + O2(g) 2 SO3(g)
Kc = [SO3]2 / [SO2]2[O2]
Answer: Easiest way to determine equilibrium concentrations is to set up
a table as follows:
2 SO2
+
O2
2 SO3
initial conc
6
4
0
change
-4
-2
+4
final (eq) conc
2
2
4
note: change of +4 for [SO3] is found by difference, changes for [SO2]
and [O2 ], -4, -2, are found from the stoichiometry of the reaction
Now Kc = [SO3]2 / [SO2]2 [O2] = 42 / 22 x 2 = 2
Problem #2
Here we start with 1.00 atm of N2O4 (g) and Kp = 11 at 100oC
Calculate the partial pressures of each gas at equilibrium
Answer:
Initial
change
equilibrium
N2O4 (g)
1.0 atm
-x
(1.0-x)
2 NO2 (g)
0.0 atm
+2x
2x
set up Equilibrium constant expression :
Kp = 11 = PNO2 2
= (2x)2
PN2O4
(1.0 -x)
4x2 +11x -11 = O
using quadratic formula
we get x = 0.78
therefore P N2O4 = 1.0 0.78 = 0.22 atm
P NO2 = 2x= 2 (0.78) = 1.56 atm at equilibrium
1
Problem #3
2.0 moles of Br2 is placed in a 2.0 L flask and it quickly dissociates into
Br · radicals under UV irradiation. Only 10 % of the Br2 dissociates.
Calculate K c .
Answer is Kc = 0.044 (ask me or Joe Antle for solution)
13.6 Factors that Alter the Composition of an
Equilibrium Mixture.
Once a system is at equilibrium, we can change the ratio of
reagents to products (i.e. disturb the equilbrium) by;
1)
2)
3)
adding or removing one or more reagents or products
(changing the conc n or partial pressure of those
substances)
compressing or expanding the system (when dealing with
gaseous reactions)
changing the temperature
We would like to know how to maximize a product yield for a
reaction with a minimum of energy (and money) input. If a reaction
goes nearly to completion, then this isn t much of a problem.
However, most reactions don t go to near completion (what does
this say about their equilibrium constants?) and so we must adjust
experimental conditions so the reaction proceeds as favourably as
possible.
2
Once the equilibrium has been disturbed, the reaction will
then increase in rate either in a forward direction or in the
reverse direction in order to re-establish equilibrium again.
This fact was first stated by Le Chatelier in 1884 and is
known as Le Chatelier s Principle
If a system at equilibrium is disturbed by a change in
concentration, pressure or temperature, the system will, if
possible, shift so as to counteract the change (restore
equilibrium conditions)
13.7 Altering an Equilibrium Mixture: Changes in
Concentration
N2 (g) + 3 H2 (g)
2 NH3 (g)
Kc = 0.291 at 700 K.
At first, the system is in
equilibrium, with [N2] =
0.50 mol/L, [H2] = 3.00
mol/L, and [NH3] = 1.98
mol/L.
At a point in time, the
concentration of N2
is increased to 1.50
mol/L by adding N2 to
the reaction system.
3
Le Chatelier s Principle states the system will re-establish
equilibrium by reacting in such a way as to decrease the
stress to the system. Since we have added a reactant, the
reaction should proceed towards products to minimize the
amount of extra N2 in the system.
N2 (g) + 3 H2 (g)
2 NH3 (g)
This is reflected in the
Figure where the reactant
concentrations decrease
and the product
Concentration increases
until a new equilibrium
mixture is reached.
Using the concentrations of the ammonia example, we saw before the
introduction of more nitrogen that the reaction quotient would be:
2
Qc
NH 3
N2 H 2
3
(1.98) 2
(0.50)(3.00)3
0.29
Kc
The system is at equilibrium. Obviously, if we add nitrogen to the
system, the reaction quotient will change to:
2
Qc
NH3
3
N2 H2
(1.98)2
0.0968 Kc
(1.50)(3.00)3
we see the reaction quotient is now less than the equilibrium constant,
meaning the reaction must move from left to right to reach the (new!)
equilibrium mixture. At the new equilibrium, the concentrations are
found to be [N2] = 1.31 mol/L, [H2] = 2.43mol/L, and [NH3] = 2.36 mol/L.
The reaction quotient equals the equilibrium constant!
2
Qc
NH3
N2 H 2
3
(2.36)2
(1.31)(2.43)3
0.296 Kc
4
Problem 13.16
Consider the equilibrium for the water-gas shift reaction:
CO (g) + H2O (g) CO2 (g) + H2 (g)
Use Le Chatelier s Principle to predict how the concentration of
H2 will change when the equilibrium is disturbed by:
a) Adding CO:
Answer: More CO (a reactant) means the reaction will shift to reduce the
amount of reactants by creating products. Therefore the concentration of H2
will increase.
b) Adding CO2:
Answer: More CO2 (a product) means the reaction will shift to reduce the
amount of products by creating reactants. Therefore the concentration of H2
will decrease.
Problem 13.16 (continued)
CO (g) + H2O (g) CO2 (g) + H2 (g)
c) Removing H2O:
Answer: Less H2O (a reactant) means the reaction will shift to increase
the amount of reactants by consuming products. Therefore the
concentration of H2 will decrease.
d) Removing CO2; also account for the change using
the reaction quotient Qc:
Answer: Less CO2 (a product) means the reaction will shift to
increase the amount of products by consuming reactants. Therefore
the concentration of H2 will increase.
Reaction quotient: Imagine the initial equilibrium concentrations
of the chemicals as written from left to right in the balanced equation
are w, x, y, and z. The equilibrium constant will be Kc = (yz) / (wx) If
we remove, let s say, half of the CO2 (concentration of y/2), the reaction
quotient will be Qc = (yz)/(2wx) which will be less than Kc. This means
the reaction proceeds from left to right, meaning more hydrogen is
formed.
5
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