Download Chemical Equilibrium is reached when

Chapter 15
Chemical Equilibrium
The Concept of Equilibrium:
Equilibrium: is a state in which there are no observable changes as time goes by.
Chemical Equilibrium is reached when:
1- The rates of forward and reverse reactions are equal.
2- The concentrations of the reactants and products no longer change with time.
Note that a chemical equilibrium reaction involves different substances as reactants and products.
Physical Equilibrium: equilibrium between two phases of the same substance.
H2O(l)  H2O(g)
Chemists are particularly interested in Chemical Equilibrium as
N2O4(g)  2NO2(g)
This reaction can be monitored easily because N2O4 is colorless gas, whereas NO2 has a dark brown
color.
Start with NO2
The Equilibrium Constant:
Start with N2O4
Start with N2O4 and NO2.
Analysis of the data at equilibrium shows that although the ratio of [NO2]/[N2O4] gives scattered
values, the ratio [NO2]2/[N2O4] gives nearly constant value (Table 15.1). This value is called
equilibrium constant, K, for the reaction at 25oC.
K = [NO2]2/[N2O4]
For the following general reversible reaction:
aA + bB  cC + dD
K = [C]c[D]d/[A]a[B]b
This equation is the mathematical form of the law of mass action. It relates the concentrations of
reactants and products at equilibrium in terms of a quantity called equilibrium constant.
 Equilibrium will lie to the right and favor the products, if K >> 1.
 Equilibrium will lie to the left and favor the reactants, if K << 1.
Ways of Expressing Equilibrium Constants:
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4(g)  2NO2(g)
Kc = [NO2]2/[N2O4]
c denotes that the concentrations of the reacting species are expressed in
moles per liters. Kc is constant for a specific reaction at specific temperature.
1
In gaseous reactions, concentrations of reactants and products can be expressed in terms of their
partial pressures (in atm.), since P = (n/V)RT (P is directly proportional to the molar
concentrations).
Kp = P2NO2/PN2O4
In general, Kc is not equal to Kp, because the partial pressures of the reactants and products are not
equal to their concentrations expressed in moles per liter.
Consider the following equilibrium in the gas phase:
aA(g)  bB(g)
Kc 
[ B]b
[ A] a
and
Kp 
PBb
PAa
PA V  n A RT
PB V  n B RT
 n RT 
or , PA   A

 V 
 n RT 
PB   B

 V 
 n B RT

 V
KP 
 n A RT

 V



b



a
K P  K c ( RT ) n

ba
[ B]b
(
RT
)
[ A] a
Where, n  b  a
n  moles of gaseous products  moles of gaseous reac tan ts
K p  K c (0.0821T ) n
In general, Kc is not equal to Kp except in the special case when n = 0.
As another example of homogenous equilibrium, consider the ionization of acetic acid in water:
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
Kc` = [CH3COO-][H3O+]/[CH3COOH][H2O]
(We use prime for Kc here to distinguish it from the final form of equilibrium constant).
However, in 1 L of water we have 55.5 M of water which is very large compared with the
concentrations of other species in solution, and we assume that it doesn’t change during the course
of a reaction.
Kc = [CH3COO-][H3O+]/[CH3COOH]
Kc = Kc`[H2O]
Note that it is general practice not to include units for the equilibrium constant.
Examples 15.1, 15.2, 15.3.
Heterogenous equilibrium applies to reactions in which reactants and products are in different
phases.
For example:
CaCO3(s)  CaO(s) + CO2(g).
Kc` = [CaO][CO2]/[CaCO3]
The concentration of a solid is usually constant (because it is an intensive property) and can be
combined to the equilibrium constant. Therefore, the equilibrium constant of the reaction can be
written as:
Kc = [CO2(g)] = Kc`[CaCO3]/[CaO]
We can express the equilibrium constant as: Kp = PCO2
The concentration of solids and pure liquids are not included in the expression for the
equilibrium constant.
2
Example 15.4.
The Form of K and the Equilibrium Equation:
When the equation for a reversible reaction is written in the opposite direction, the equilibrium
constant becomes the reciprocal of the original equilibrium constant.
N2O4(g)  2NO2(g)
K1 = [NO2]2/[N2O4] = 4.63 * 10-3.
2NO2(g)  N2O4(g)
K2 = [N2O4]/[NO2]2 = 1/Kc = 216.
K1 = 1/K2 or K1K2 = 1
Summary of the Rules for Writing Equilibrium Constant Expressions:
•The concentrations of the reacting species in the condensed phase are expressed in M. In the
gaseous phase, the concentrations can be expressed in M or in atm.
•The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium
constant expressions.
•The equilibrium constant is a dimensionless quantity.
•In quoting a value for the equilibrium constant, you must specify the balanced equation and the
temperature.
•If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the
overall reaction is given by the product of the equilibrium constants of the individual reactions.
What does the equilibrium constant tell us?
1- It helps us to predict the direction in which a reaction mixture will proceed to achieve
equilibrium.
2- It helps us to calculate the concentrations of reactants and products in equilibrium system.
Predicting the Direction of a Reaction:
The reaction quotient (Qc) is obtained by substituting the initial concentrations of the reactants and
products into the equilibrium constant (Kc) expression.
If
 Qc < Kc The system proceeds from left to right to reach equilibrium.
 Qc = Kc The system is at equilibrium.

Qc > Kc The system proceeds from right to left to reach equilibrium.
Example 15.5
Calculating Equilibrium Concentrations:
3
Let us consider the following system, which involves a pair of geometric isomers:
cis-stilbene

trans-stilbene
Initial (M):
0.850
0
Change (M):
-x
+x
Equilibrium (M):
(0.850 – x))
Kc = [trans-stilbene]/[cis-stilbene] = 24
24.0 = x/(0.850 – x)
x = 0.816
[cis-stilbene] = (0.850 – 0.816) M = 0.034 M.
[trans-stilbene] = 0.816 M.
x
so:
We can summarize our approach to solve equilibrium constant problems as:
1- Express the equilibrium concentrations of all species in terms of the initial concentrations and
a single unknown x, which represents the change in concentration.
2- Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing
the value of the equilibrium constant, solve for x.
3- Having solved for x, calculate the equilibrium concentrations of all species.
Examples 15.6, 15.7..
Factors That Affect Chemical Equilibrium:
Le Chatelier's Principle: If an external stress is applied to a system at equilibrium, the system
adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium
position.
The word “stress” means a change in concentration, pressure, volume, or temperature that removes
a system from the equilibrium state.
Changes in concentration:
 Increasing the concentration of the products shifts the equilibrium to the left, and
decreasing the concentration of the products shifts the equilibrium to the right.
 Increasing the concentration of the reactants shifts the equilibrium to the right, and
decreasing the concentration of the reactants shifts the equilibrium to the left.
Consider the reaction:
FeSCN2+(aq)  Fe3+(aq) + SCN-(aq)
1- Adding NaSCN to the solution will increase the concentration of SCN-, the
equilibrium shifts from right to left.
2- Adding Fe(NO3)3 to the solution will increase the concentration of Fe3+, the
equilibrium shifts from right to left.
3- Adding H2C2O4 to the solution will increase the concentration of C2O42-, which
binds strongly with Fe3+ to form the stable ion Fe(C2O4)33-, this results in decreasing
the concentration of Fe3+ and the equilibrium shifts from left to right.
Deep red color  Yellow color
In general, consider this general reaction:
aA + bB  cC + dD
Change
1- Increase the concentration of the products
2- decrease the concentration of the products
3- Increase the concentration of the reactants
4- decrease the concentration of the reactants
4
Shifts the Equilibrium
To the left
To the right
To the right
To the left
Example 15.8.
Changes in Pressure and Volume:
Concentrations of liquids and solids are not affected by changes in pressure because liquids and
solids are incompressible, but gases are greatly affected by changes in pressure.
PV = nRT
P = (n/V)RT
(n/V) is the concentration of the gas in M, and it varies directly with P.
For the reaction:
N2O4(g)  2NO2(g)
Increasing the pressure will increase the concentration of both N2O4 and NO2, and because the
concentration of NO2 is squared, the increase in the numerator more than the denominator. The
system is no longer at equilibrium, so we write Qc = [NO2]o2/[N2O4]o
Thus Qc > Kc and the net reaction will shift to the left until Qc = Kc.
1. In general, an increase in pressure (decrease in volume) favors the net reaction that
decreases the total number of moles of gases, and a decrease in pressure (increase
in volume) favors the net reaction that increases the total number of moles of gases.
2. Apressure has no effect in the position of the position of equilibrium, when there is
no change in the number of moles.
3. Adding an inert gas (helium, for example) to the equilibrium system at constant
volume, increases the total gas pressure and decreases the mole fraction of both
reactant and product: but the partial pressure of each gas does not change. Thus, the
pressure of an inert gas in such a case does not affect the equilibrium.
Example 15.9
Change in Temperature:
Only temperature changes will alter the equilibrium constant, and that is why we always specify the
temperature when giving the value of Kc.
1. Increasing the temperature of an exothermic reaction will shift the reaction to left
(more reactants and less products).
2. Increasing the temperature of endothermic reaction will shift the reaction to right
(more products and less reactants).
In summary, a temperature increase favors the endothermic direction and a temperature decrease
favors the exothermic direction.
The Effect of a Catalyst:
1. Catalyst lowers Ea for both forward and reverse reactions.
2. Catalyst does not change equilibrium constant or shift equilibrium.Example 15.10
Summary of Factors That May Affect the Equilibrium Position
Le Chatelier's Principle
Change
Concentration
Pressure
Temperature
Volume
Catalyst
Shift Equilibrium
Yes
Yes
Yes
Yes
No
Change Equilibrium Constant
No
No
Yes
No
No
Selected Problems: 8 a,b, 9, 11, 12, 13, 16, 17, 18, 20 – 24, 28, 29, 31, 33 – 35, 41 – 50.
5