Equilibrium

EQUILIBRIUM
REVERSIBLE REACTION &
DYNAMIC EQUILIBRIUM

Reversible Reactions and
Equilibrium
Some reaction can be reserved, for example reaction
formation of stalagmites and Stalactites
Consider the following reactions:
CaCO3(s) + CO2(aq) + H2O(l)  Ca2+(aq) + 2HCO3
-(aq) ..(1)
and
Ca2+(aq) + 2HCO3
-(aq)  CaCO3(s) + CO2(aq) + H2O(l) ..(2)
Reaction (2) is the reverse of reaction (1).
Chemical equilibrium occurs when a reaction
and its reverse reaction proceed at the same
rate.

Chemical Equilibrium
• Suppose we have the gaseous reactants I2
and H2. They undergo a synthesis reaction
to form HI:
H2 + I2 → 2HI
• As HI accumulates, some molecules have
enough energy to decompose to H2 and I2:
2HI → H2 + I2

• As this process
continues,
eventually the
rate of the
forward reaction
equals the rate of
the reverse
reaction.

• The final equilibrium
mixture will contain
both reactants and
products.

Characteristics of Equilibrium
An equilibrium reaction has four particular
features under constant conditions:
• it is dynamic
• the forward and reverse reactions occur at
the same rate
• the concentrations of reactants and
products remain constant at equilibrium
• it requires a closed system.

1. It is dynamic
The phrase dynamic equilibrium means that
the molecules or ions of reactants and
products are continuously reacting.
2. The forward and backward reactions
occur at the same rate
At equilibrium the rate of the forward
reaction equals the rate of the backward
reaction.

3. The concentrations of reactants and
products remain constant at equilibrium
The concentrations remain constant
because, at equilibrium, the rates of the
forward and backward reactions are equal.

• For example, in the reaction
H2(g) + I2(g) 2HI(g)

4. Equilibrium requires a closed system
A closed system is one in which none of the
reactants or products escapes from the
reaction mixture.

EQUILIBRIUM
Changing the Position of
Equilibrium

Le Châtelier’s Principle
• The Le Châtelier's principle states that:
when factors that influence an equilibrium are altered, the
equilibrium will shift to a new position that tends to
minimize those changes.
• Factors that influence equilibrium:
Change of Concentration, Change of pressure (for
gaseous), Change of temperature, and catalyst

Effect of Concentration Change on
Equilibrium
Changing the amount, or concentration, of
any reactant or product in a system at
equilibrium disturbs the equilibrium.
• The system will adjust to minimize the effects
of the change.

Effect of Concentration Change on
Equilibrium
Suppose carbon dioxide is added to the
system.
• This increase in the concentration of CO2 causes
the rate of the reverse reaction to increase.
• Adding a product to a reaction at equilibrium
pushes a reversible reaction in the direction of the
reactants.
H2CO3(aq) CO2(aq) + H2O(l)
Add CO2
Direction of shift

Suppose carbon dioxide is removed.
• This decrease in the concentration of CO2 causes
the rate of the reverse reaction to decrease.
• Removing a product always pulls a reversible
reaction in the direction of the products.
H2CO3(aq) CO2(aq) + H2O(l)
Add CO2
Direction of shift
Remove CO2
Direction of shift
Effect of Concentration Change
on Equilibrium

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or its affiliates. All Rights Reserved.
Effects of Pressure Change on Equilibrium
Initial equilibrium
Equilibrium is
disturbed by an
increase in
pressure.
When the plunger is pushed down, the volume
decreases and the pressure increases.
A new equilibrium
position is
established with
fewer molecules.
SMA SEMESTA BILINGUAL BOARDING SCHOOL CHEMISTRY

Effects of Pressure Change on Equilibrium
• If the volume of a gas mixture is compressed, the
overall gas pressure will increase. In which direction
the equilibrium will shift in either direction depends
on the reaction stoichiometry.
• However, there will be no effect to equilibrium if the
total gas pressure is increased by adding an inert gas
that is not part of the equilibrium system.

Reactions that shift right when pressure increases
and shift left when pressure decreases
Consider the reaction:
2SO2(g) + O2(g) ⇄ 2SO3(g),
1. The total moles of gas decreases as reaction
proceeds in the forward direction.
2. If pressure is increased by decreasing the volume
(compression), a forward reaction occurs to reduce
the stress.
3. Reactions that result in fewer moles of gas favor
high pressure conditions.

Reaction that shifts left when pressure increases,
but shifts right when pressure decreases
Consider the reaction: PCl5(g) ⇄ PCl3(g) + Cl2(g);
1. Forward reaction results in more gas molecules.
2. Pressure increases as reaction proceeds towards
equilibrium.
3. If mixture is compressed, pressure increases, and
reverse reaction occurs to reduce pressure;
4. If volume expands and pressure drops, forward
reaction occurs to compensate.
5. This type of reactions favors low pressure condition

Reactions not affected by pressure changes
Consider the following reactions:
1. CO(g) + H2O(g) ⇄ CO2(g) + H2(g);
2. H2(g) + Cl2(g) ⇄ 2HCl(g);
1. Reactions have same number of gas molecules in
reactants and products.
2. Reducing or increasing the volume will cause equal
effect on both sides – no net reaction will occur.
3. Equilibrium is not affected by change in pressure.

The Effect Temperature on Equilibrium
• Consider the following exothermic reaction:
N2(g) + 3H2(g) ⇄ 2NH3(g); DHo
= -92 kJ,
• The forward reaction produces heat => heat is a product.
• When heat is added to increase temperature, reverse reaction
will take place to absorb the heat;
• If heat is removed to reduce temperature, a net forward
reaction will occur to produce heat.
• Exothermic reactions favor low temperature conditions.

The Effect Temperature on Equilibrium
Consider the following endothermic reaction:
CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g), DHo
= 205 kJ
1. Endothermic reaction absorbs heat  heat is a reactant;
2. If heat is added to increasing the temperature, it will cause a
net forward reaction.
3. If heat is removed to reduce the temperature, it will cause a
net reverse reaction.
4. Endothermic reactions favor high temperature condition.

uncatalyzed catalyzed
14.5
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
The Effect Catalyst on Equilibrium

EQUILIBRIUM
Equilibrium Expression and
The Equilibrium Constant, Kc

the Equilibrium Constant, Kc
• The equilibrium constant (Kc) is the ratio of
product concentrations to reactant
concentrations at equilibrium.

Example
Write an expression for Kc:
N2(g) + 3H2(g) 2NH3(g)
2SO2(g) + O2(g) 2SO3(g)

• In equilibrium expressions involving a
solid, we ignore the solid. This is because
its concentration remains constant,
however much solid Is present. For
example:

What are the units of Kc?
• In the equilibrium expression each figure
within a square bracket represents the
concentration in mol dm–3.

The colorless gas dinitrogen tetroxide
(N2O4) and the brown gas nitrogen dioxide
(NO2) exist in equilibrium with each other.
Expressing and Calculating Kc
A liter of the gas mixture at equilibrium
contains 0.0045 mol of N2O4 and 0.030 mol
of NO2 at 10o
C. Write the expression for the
equilibrium constant (Kc) and calculate the
value of the constant for the reaction.
N2O4(g) 2NO2(g)

KNOWNS UNKNOWN
Kc (algebraic expression) = ?
Kc (numerical value) = ?
Analyze List the knowns and the unknowns.1
Modify the general expression for the
equilibrium constant and substitute the known
concentrations to calculate Kc.
[N2O4] = 0.0045 mol/L
[NO2] = 0.030 mol/L

• Start with the general expression for
the equilibrium constant.
Calculate Solve for the unknowns.2
Place the concentration of
the product in the
numerator and the
concentration of the
reactant in the
denominator. Raise each
concentration to the power
equal to its coefficient in
the chemical equation.
• Write the equilibrium constant
expression for this reaction.
Kc =
[C]c x [D]d
[A]a x [B]b
Kc =
[NO2]2
[N2O2]

Substitute the concentrations that are known
and calculate Kc.
Calculate Solve for the unknowns.3
You can ignore the unit
mol/L; chemists report
equilibrium constants
without a stated unit.
Kc =
(0.030 mol/L)2
(0.0045 mol/L) =
(0.030 mol/L x 0.030 mol/L)
(0.0045 mol/L)
Kc = 0.20 mol/L = 0.20

One mole of colorless hydrogen gas and one
mole of violet iodine vapor are sealed in a 1-L
flask and allowed to react at 450o
C. At
equilibrium, 1.56 mol of colorless hydrogen
iodide is present, together with some of the
reactant gases. Calculate Kc for the reaction.
Finding the Equilibrium Constant
H2(g) + I2(g) 2HI(g)

KNOWNS UNKNOWN
Kc = ?
Analyze List the knowns and the unknown.1
Find the concentrations of the reactants at
equilibrium. Then substitute the equilibrium
concentrations in the expression for the
equilibrium constant for this reaction.
[H2] (initial) = 1.00 mol/L
[I2] (initial) = 1.00 mol/L
[HI] (equilibrium) = 1.56 mol/L

First find out how much H2 and I2 are consumed
in the reaction.
Calculate Solve for the unknown.2
Let mol H2 used = mol I2 used = x.
The number of mol H2 and mol I2
used must equal the number of mol
HI formed (1.56 mol).
x + x = 1.56 mol
2x = 1.56 mol
x = 0.780 mol

• Calculate how much H2 and I2 remain in the
flask at equilibrium.
mol H2 = mol I2 = (1.00 mol – 0.780 mol) = 0.22 mol
• Write the expression
for Kc.
Kc =
[HI]2
[H2] x [I2]
Use the general
expression for Keq as a
guide:
Kc =
[C]c x [D]d
[A]a x [B]b

Substitute the equilibrium concentrations of
the reactants and products into the equation
and solve for Kc.
Kc =
(1.56 mol/L)2
0.22 mol/L x 0.22 mol/L
Kc =
1.56 mol/L x 1.56 mol/L
Kc = 5.0 x 101
0.22 mol/L x 0.22 mol/L

EQUILIBRIUM
Equilibrium in gas reactions:
The equilibrium constant, Kp

The Equilibrium Constant (Kp)
• When the reactants and products are
gases, we can determine the equilibrium
constant in terms of partial pressures.
• Dalton’s Law of Partial Pressures – the
total pressure in a system is equal to the
sum of the individual pressures.

Equilibrium Pressure
• Pressure is measured in atmospheres.
• If the values at equilibrium are given in atmospheres, then
solving for the constant is the same, but use Kp instead of
Kc.
• What makes them different:
Kc = equilibrium constant based on molarity and
concentration.
Kp = equilibrium constant based on partial pressures.

What are the units of Kp?
• The units of pressure are pascals, Pa. The
units of Kp depend on the form of the equilibrium
expression.

2 NO2 (g) 2 NO (g) + O2 (g)
• At equilibrium, the pressure of nitrogen
dioxide is 0.425 atm, nitrogen monoxide is
0.270 atm, and oxygen is 0.100 atm. Determine
the Kp for the above reaction.
How to Calculate Kp?

2 NO2 (g) 2 NO (g) + O2 (g)
• Determine the Kp for the above reaction
when at equilibrium, the pressure of nitrogen
dioxide is 0.425 atm, nitrogen monoxide is
0.270 atm, and oxygen is 0.100 atm.
• Kp = [NO]2 [O2]
[NO2]2
• Kp = [0.270]2 [0.100]
[0.425]2
• Kp = 0.0404

Relationship between Kc and Kp
If values at equilibrium are not in atmospheres, We
use the Ideal Gas Law, and derive the relationship
between Kc and Kp.
Kp = Kc (RT)Dn
Kp = Equilibrium constant in partial pressure
Kc = Equilibrium constant in molar concentrations
R = Ideal Gas constant (always 0.082)
T = Temperature (in Kelvin)
Dn = Change in moles of GASES ONLY
(moles of gas products) – (moles of gas reactants)

CO(g) + Cl2(g)  COCl2(g)
• At equilibrium the concentrations at 740C are [CO] =
0.012 M, [Cl2] = 0.054 M, and
[COCl2] = 0.14 M.
Calculate the equilibrium constants Kc and Kp.
• Start with the equilibrium expression for Kc:
Kc = [COCl2]
[CO][Cl2]
Kc = [0.14 M] = 216
[0.012][0.054]

CO(g) + Cl2(g)  COCl2(g)
• At equilibrium the concentrations at 740C are [CO] =
0.012 M, [Cl2] = 0.054 M, and
[COCl2] = 0.14 M.
Calculate the equilibrium constants Kc and Kp.
• To solve for Kp we use the Ideal gas law eq.
Kp = Kc (RT)Dn
Kp = 216 x (0.0821 x 347)(1-2)
Kp = 7.58

Equilibrium

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