Download Equilibrium (Sheet 1)

EQUILIBRIUM
SHEET 1
Section I
A system is in equilibrium when the two opposing reactions occur simultaneously at the same rate. Let us
examine the general reaction:
A+B
C+D
When this system is in equilibrium the rate at which A reacts with B to give the products C and D is exactly
equal to the rate at which the products C and D react to give the original reactants A and B. If this system is
undisturbed, then the concentration of the reactants and the concentration of the products will not change.
If R1 is the rate of the forward reaction, then R1 is proportional to the concentration of A times the concentration
of B, each expressed in moles per liter (m/l). The mathematical expression is given in Eq. 1.
Eq. 1:
R1  [A] [B]
When proportionality is changed to equality, a constant is added. Therefore, the rate of the forward
reaction, R1 is given in Eq. 2.
Eq. 2:
R1 = K1 [A] [B]
Where K1 is the rate constant for the reaction.
Using the same procedure for the reverse reaction and calling the rate for the reverse reaction R 2, then R2
given by Equation 3.
Eq. 3:
R2 = K2 [C] [D]
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, that is R1 = R2.
Substituting the values of R1 and R2 from Eq. 2 and 3 above, Eq. 4 is obtained.
Eq. 4:
K1 [A] [B] = K2[C] [D]
Dividing both sides of Eq. 4 by K2 [A] [B], Eq. 5 is obtained.
Eq. 5:
K1
K2

C D
A B
A constant divided by a constant is equal to a constant. That is, K1 divided by K1 is equal to Ke, known as
the equilibrium constant.
K1
 Ke
K2
The mathematical expression for the above reaction is given in Eq. 6.
Eq. 6:
Ke 
C D
A B
A more general equation for a chemical equilibrium is:
nA + mB
xC + yD
n, m, x, and y represent the number of moles of each substance appearing in the balanced reaction. The
equilibrium expression for the above reaction is given in Equation 7.
Ke 
Eq. 7:
 C x  D  y
A  n  B m
Keep in mind that the concentration of each of the reactants and the products is expressed in moles per liter.
Example 1
Suppose in a one liter reaction vessel at a given temperature an equilibrium mixture consisting of 0.1M of
carbon monoxide, 0.8M of H2O, 0.8M of carbon dioxide, and 0.5M of hydrogen is present. The following reaction
is given:
CO + H2O
CO2 + H2. Calculate Ke for the reaction.
Solution
The equilibrium expression for the above reaction is given in Equation 8.
Ke 
Eq. 8:
CO 2  H 2 
CO  H 2 O
By substituting the concentration of each of the reactants and each of the products in Eq. 8, Eq. 9 is
obtained.
.8.5
Eq. 9:
Ke 
5
.1.8
As long as the temperature of the above reaction does not change, the value of the equilibrium constant will
always be the same.
Example 2
Using the same reaction as in Example 1, if in a one liter vessel the equilibrium mixture consisted of 0.5
moles of steam, 0.5 moles CO2, and 0.2 moles of hydrogen. Calculate the concentration of carbon monoxide in
moles per liter.
Solution
Refer to Example 1 for the Ke value, then substitute the values in Equation 8. Solve the equation and arrive
at .04m/l, the concentration of CO.
.5.2
5
CO.5
[5] [CO] [.5] = [.5] [.2]
[CO] = .04 moles per liter
It is important to note that Ke, the equilibrium constant value for the reaction, did not change because the
temperature did not change.
EQUILIBRIUM
Sheet 1
Section II
1. In a one liter reaction vessel, the following equilibrium concentrations were found for the following reaction:
CO + Cl2
COCl2
The concentration of CO is .2 mol, Cl2 is .3 mol, and COCl2 is .8 mol. Calculate Ke.
2. In a one liter reaction vessel, an equilibrium mixture of N 2, H2, and NH3 which reacts according to N2 + H2
NH3. The concentration of N2 is .15, H2 is 1 and NH3 is .4. Calculate K
3. An equilibrium mixture CO + Cl2
in a 5 liter vessel. Calculate Ke.
4. The Ke for CO + H2O
COCl2 contained 2 moles of Cl2, 1 mole of CO, and 5 moles of COCl2
CO2 + H2 is 4 at a certain temperature.
The equilibrium concentrations were found to contain .3 moles of CO, .1 mole of H 2O, and .25 mole of CO2
in a liter.
(a) How many moles/liter of H2 were in the mixture?
(b) How many grams/liter of H2 were in the mixture?
(c) How many grams/200 ml of H2 were in the mixture?
5. An equilibrium mixture of SO2 + O2
reaction vessel. Calculate Ke.
Answers for Section II
1. 13.3
2. 1.07
3. 12.5
4. (a) .48
(b) .96
(c) .192
5. 16
SO3 contained 80 g SO3, 32g of SO2 , and 16g of O2 in a 2 liter
Le Chatelier Principle
Sheet 1
Section III
La Chatelier's principle states that if a stress such as a change in concentration, pressure or temperature is
applied to a system in equilibrium, the equilibrium will shift in a way that tends to undo the effect of the stress. For
example:
H2O + CO
H2 + CO2 + heat. If no stress is introduced into this system, then the concentration of H 2O,
CO, H2, and CO2 will not change. Now then, assume the concentration of H2O was increased, then effectively the
number of collisions between H2O molecules and CO molecules are increased, resulting in an increase in the number
of molecules possessing the activation energy, thus the rate of the forward reaction is increased and the equilibrium
is said to shift to the right, which means that the relative concentration of the reactant will decrease and the relative
concentration of the products will increase. It can be reasoned that if the concentration of CO is increased, then the
equilibrium will shift to the right, while increasing the concentration of hydrogen of carbon dioxide or both, i.e.,
increasing the concentration of the products and the equilibrium will shift to the left. On the other hand, if the
concentration of steam or the concentration of carbon monoxide or both are decreased, then the equilibrium will shift
to the left. Decreasing the concentration of either the hydrogen or the carbon dioxide or both will shift the
equilibrium to the right.
Chemical reactions are either exothermic or endothermic. In any chemical reaction in equilibrium the
endothermic and the exothermic reactions are taking place simultaneously. Refer to the reaction above; it is
exothermic reaction in the forward direction. For this reaction, an increase in temperature has the same effect on the
equilibrium as increasing the concentration of H2, i.e., if the temperature is increased, the equilibrium will shift to the
left. In an endothermic reaction increasing the temperature shifts the equilibrium to the right; on the other hand, for
the same reaction above, if the temperature of the reaction is decreased then the equilibrium will shift left; that is, the
relative concentration of the reactants will increase while the relative concentration of the products will decrease.
If a chemical reaction is in equilibrium and the concentration of the reactants or the concentration of the
products were increased or decreased, the Ke value does not change. But if the temperature of the reaction is
changed, then the value of the Ke will change; i.e., the change in temperature is the only factor that changes the Ke
value.
A catalyst is a substance that changes the rate of the chemical reaction. A positive catalyst increases the
rate of a chemical reaction while a negative catalyst decreases the rate of a chemical reaction. For example, in the
preparation of oxygen, manganese dioxide, MnO2, was used as a positive catalyst to increase the rate of
decomposing potassium chlorate to KCl and O2. Anti-oxidants are added to tires to reduce the rate at which the tires
oxidize. The addition of the catalyst has no effect on the final concentration of the reactants or the products; it
merely changes the time for establishing the equilibrium. Adding a catalyst has no effect on the Ke value.
EQUIILIBRIUM
Sheet 1
Section IV
Given the following equation in equilibrium:
CO + Cl2
COCl2 + heat
Discuss how the equilibrium shifts when the following stresses are introduced.
1. Doubling the concentration of CO.
2. Doubling the concentration of COCl2.
3. Decreasing the concentration of Cl2.
4. Adding a catalyst.
5. Increasing the temperature.
6. Decreasing the temperature.
7. Increasing the pressure.
Answers for Section IV
1. Eq. shifts to right
2. Eq. shifts to left
3. Eq. shifts to left
4. No shift
5. Eq. shifts to left
6. Eq. shifts to right
7. Eq. shifts to right