Chapter 15:

Equilibrium State
Content:
 The Nature of the Equilibrium State
 The Equilibrium Constant Expression
 Relationship Involving the Equilibrium
Constant
 The magnitude of the Equilibrium
Constant
 Altering Equilibrium Conditions: Le
Chatelier’s Principle
 Equilibrium Calculations: Some
Illustrative Examples

1
 1- The Nature of the Equilibrium State

In a closed reaction vessel at constant temperature, a reaction

proceeds spontaneously toward equilibrium. At equilibrium, the
reaction quotient Q attains the same constant value, K irrespective of
the starting amounts of reactants and products.

2
The approach to equilibrium in the reaction

3
EX: At 25 C, K=9.14X10-6 for the reaction:
2Fe+3 (aq) + Hg2+2(aq)  2Fe+2 (aq) + 2Hg+2 (aq)
If equilibrium concentration of Fe+3, Fe+2 and Hg+2 are
0.015 M,0.0025 M and 0.0018 M, respectively, what is
the equilibrium concentration of Hg2+2?
Solution:
K = 9.14X10-6 = [Fe+2]2 * [ Hg+2]2 / [ Fe+3]2*[ Hg2+2]
9.14X10-6 = [0.0025]2 [0.0018]2 / [0.015]2 [x]
X = 0.0098 M

The Dynamic Nature of The Equilibrium Condition:

The equilibrium condition is dynamic, which the forward and
reverse reaction occurring not only indefinitely but also at exactly
the same rate.

4
2- The Equilibrium Constant Expression:

5
Equilibria Involving Gases:

Equilibria in Aqueous Solution:

6
Equilibria Involving Pure Liquids and solids:

Pure solids and liquids ae not included in equilibrium

constant expression

7
EX: Teeth are made principally from the mineral
hydroxyapatite, Ca5(po4)3OH, which can be dissolved in
acidic solution such as that produced by bacteria in the
mouth. The reaction that occurs is:
Ca5(po4)3OH(s)+4H+(aq) 5Ca+2(aq) + 3HPO4-2(aq)+H2O(l)
Write the equilibrium constant Kc expression for this
reaction.
Solution:
Kc = [Ca+2]5 * [HPO4-2]3 / [H+]4

3- Relation Involving the Equilibrium

Constant:
Relationship of K to the Balanced chemical
Equation:
--When we reverse an equation, we invert the value of K.
--When we multiply the coefficient in a balanced equation by a
common factor we raise the equilibrium constant to the
corresponding power.
--When we divide the coefficients in a balanced equation by a
common factor we take the corresponding root of the equilibrium
constant.

EX: Use data below:

N2(g) + 3H2(g) 2NH3
To determine the value of K at 298 K for the reaction
1/3 N2(g) + H2 (g)  2/3 NH3(g)
solution:

Kc (1/3) = 3√5.8X105 = 8.3X102

8
Combining equilibrium constant expression:
When individual equation are combined (that is, added) their
equilibrium constants are multiplied to obtain the equilibrium
constant for the overall reaction.

9
Relationship Between Kp and Kc for Reactions
Involving Gases:

The composition of a gas mixture can be specified

by giving the concentration, in moles per liter, of
the component gases.
Suppose we have nA moles of gas A, nB moles of
gas B, nc moles of gas C, and nD moles of gas D in
container of volume V at temperature T. Assuming
ideal gas behavior, the partial pressure of gas A is:

10
Relationship Between Kp and Kc for Reactions

Involving Gases:

EX: for the reaction:

2NH3(g) N2(g) + 3H2(g)
at 298 K, Kc = 2.8 X 10-9 what is the value of Kp
for this reaction?
Solution:
Kp = Kc (RT)Δvgas = 2.8X10-9 * (0.08314*298)2

= 1.7 X 10-6

11
 4- The Magnitude of the Equilibrium
Constant:

A very large value of K signifies that the reaction, as

written, exhibits a strong tendency to go to completion.
An equilibrium mixture constants about as much product
as can be formed from the given initial amounts of
reactants.

A very small value of K signifies that the reaction, as

written, exhibits very little tendency to occur. An
equilibrium mixture contains reactants, in essentially
their initial amounts, and very small amounts of product.

Table for Equilibrium constants of some common

Reaction:

12
EX: for the reaction: PCl3(g) PCl3(g) + Cl(g)
Kc=0.0454 at 261 ⁰C. if a vessel is filled with these
gases such that the initial partial pressure are
PPCl3(g)=2.19 atm, PCl(g)=0.88 atm, PPCl5(g)=19.7 atm,
in which direction will a net change occure?
Solution:
Kp = Kc (RT)Δv = 0.0454 (0.08314*534.15) = 2.02
Qp= (PPCl3(g) * PCl(g)) / PPCl5(g)
= (2.19 * 0.88) / 19.7 = 0.098
So the direction is to right  to increase the
product.

13
 5- Altering Equilibrium Conditions:
Le Chatelier’s Principle

At times, we want only to make qualitive

statements about a reversible reaction:
The direction of a net change, whether the amount
of a substance will have increased or decreased
when equilibrium is reached, and so on. Also we
may not have the data needed for a quantitative
calculation. In these cases, le Chatelier’s principle.
It essential
Meaning is stated here.

When an equilibrium system is subjected to a

change in temperature, pressure, or concentration
of a reacting species, the system responds by
attaining a new equilibrium that partially offsets the
impact of the change.

Effect of changing the amounts Reacting Species

on Equilibrium

14
EX: given the reaction:
2CO(g) + O2 (g)  2Co2(g)
what is the effect of adding O2(g) to a constant-volume
equilibrium mixture?
Solution:
Adding O2 means consuming O2 and producing more
CO2 to reach the equilibrium constant.

Effect of changes in pressure or volume on equilibrium:

1- Add or remove a gaseous reactant or product:

Changes Pgas. The effect of these actions on the
equilibrium condition is simply that caused by adding or
removing a reaction component.
2- Adding an inert gas to the constant volume
reaction mixture:

15
 This has the effect of increasing the total pressure,
but the partial pressure of the reacting species are all
unchanged. An inert gas added to a constant-volume
equilibrium mixture has no effect on the equilibrium
condition.
3- Change the pressure by changing the volume of
the system:
Describing the volume of the system increases the
pressure, and increasing the system volume decreases the
pressure. thus, the effect of this type of pressure change
is simply that of a volume change.

EX: The reaction N2O4 2NO2; is at equilibrium in a

3.00L cylinder. What would be the effect on the
concentrations of N2O4(g) and NO2(g) if the pressure were
doubled (that is, cylinder volume decreased to 1.50 L)?
Solution:
The increasing of pressure makes the reaction goes to the direction
of the lowest number of moles.

16
Effect of Temperature on Equilibrium

Raising the temperature of an equilibrium mixture shifts

the equilibrium condition in the direction of the
endothermic reaction Lowering the temperature causes a
shift in the direction of the exothermic reaction.

For endothermic reactions, K increases as temperature

increases. For exothermic reaction, K decreases as
temperature increases.

EX: The reaction N2O4 2NO2; has ΔrH⁰ = +57.2KJ.

Will the amount of NO2(g) formed from N2O4(g) be grater
at high or low temperature?
Solution:
K value increases as temperature increases
ΔrH⁰ + R  P
K=P/R

Effect of a Catalyst on Equilibrium:

Adding a catalyst to a reaction mixture speeds up both
the forward and reverse reactions. Equilibrium is
achieved more rapidly, but the equilibrium amounts are
unchanged by the catalyst.

A catalyst has no effect on the condition of equilibrium

in a reversible reaction.

EX: Equilibrium is established in a 3.00L flask at 1405K

for the reaction 2H2S(g) 2H2(g) + S2(g)
At equilibrium, there is 0.11 mol S2(g), 0.22 mol H2(g), and
2.78 mol H2S(g). what is the value of Kc for this reaction?

17
Solution:

KC = [H2]2 [S2] / [H2S]2 = (0.22/3)2 (0.11/3) / (2.78/3)

=2.3X10-4

EX: A 5.00 L evacuated flask is filled with 1.86 mol

NOBr. At equilibrium at 25⁰C, there is 0.082 mol of Br2
present. Determine Kc and Kp for the reaction
2NOBr(g) 2NO(g) + Br2(g).

Solution:
2NOBr(g) 2NO(g) + Br2(g).
i: 1.86 0 0
ch: -0.164 +0.164 +0.082
eq: 1.7 0.164 0.082

Kc = (0.164/5)2(0.082/5) / (1.7/5)2 = 1.5X10-4

Kp = Kc (RT)Δv = 1.5X10-4 *(0.08314*298)1
Kp = 3.7 X 10-3

EX: sodium hydrogen carbonate (baking soda)

decomposes at elevated temperatures and is one of
the sources of CO2(g) when this compound is used
in baking.
Determine Kc and Kp for the reaction:
2NaHCO(s)  Na2CO3(s) + H2O(g) + CO2(g)
Kp = 0.231 at 100 ⁰C.
What is the partial pressure of CO2(g) when this
equilibrium is established starting with NaHCO3(s)?

18
Solution:

Kp = PH2O PCO2 = 0.231

PH2O =PCO2 = √Kp =√0.231 = 0.481

EX: if 0.150 mol H2(g) and 0.2 mol I2(g) are

introduced into 15 L flask at 445 ⁰C and allowed to
come to equilibrium, how many moles of HI(g) will
be present?
H2(g) + I2(g)  2HI(g) Kc = 50.2 at 445 ⁰C
i 0.15 0.2 0
c -x -x +2x
eq 0.15-x 0.2-x +2x

Kc = (2x/15)2 / (0.15-x/15) (0.2-x/15) =

4x2=1.51-17.6x + 50.2x2
x1 = 0.25
x2 = 0.131
X = 0.131  [HI] = 2X =0.262 mol

EX: a solution is prepared with [V+3] = [Cr+2] =

0.100 M and [V+2] = [Cr+3] = 0.150 M
V+3(aq) + Cr+2(aq)  V+2(aq) + Cr+3(aq)

Kc = 7.2 X 102
What are the ion concentration when equilibrium is
established?
[Hint: the algebra can be greatly simplified by
extracting the square root of both sides of an
equation at the appropriate point]
19
Solution:

Qr = [V+2] [Cr+3] / [Cr+2] [V+3]

Qr = 0.15 * 0.15 / 0.01 * 0.01 = 225 < 7.2 X 102

V+3(aq) + Cr+2(aq)  V+2(aq) + Cr+3(aq)

I : 0.01 0.01 0.15 0.15

C: -x -x +x +x
Eq:0.01-x 0.01-x 0.15+x 0.15+x

Kc = (0.15+x)2 / (0.01-x)2 = 7.2 X 102

X = 0.0043 M

[Cr+2] = [V+3] = 0.0057 M

[Cr+3] = [V+2] = 0.154 M

Examples:

Ex: which of the following statements is

correct?
1- At equilibrium the reaction stops.
2- At equilibrium the rate constants for
the forward and reverse reactions are
equal
3-At equilibrium the rates of the
forward and reverse reactions are
equal.
4-At equilibrium the rates of the forward
and reverse reactions are zero.

20
EX: for the reaction given to the right, the diagram
represents an equilibrium mixture. the equilibrium
constant Kp at this temperature is, (do not use a
calculator)

1- 6 atm
2- 6 atm-1
3- 12 atm
4- 12 atm
5- 36 atm-1

EX: consider the following reaction:

CO(g) + 2H2 (g)  CH3OH(g)
1.0 mol of gaseous methanol was put into a 10.0L flask
at 483 K. The equilibrium mixture as depicted to the
right was found. The value of the equilibrium
constant, Kc is (you may use a calculator)

21
 1- 0.22 M-1
2- 0.15 M-2
3- 2.2 M-1
4- 1.5 M-2
5- 15 M-2

EX: the following reaction has the given equilibrium

constant at a high temperature:
C(s) + CO2(g)  2CO(g) Kp=0.5 atm
A mixture represented by the figure to the right was
prepared which of the following statements is correct?
) in the figure we have 4 molecules CO2 and 2 molecules CO
and each molecule equal 1 atm)
1- The mixture is at equilibrium
2- The reaction must proceed toward reactants to
reach equilibrium.
3- The reaction must proceed toward products to
reach equilibrium.

EX: consider the equilibrium:

CaCO3(s) CaO(s) + CO2 (g)
Assuming the system is at equilibrium, which of the
following statements is incorrect?

1- Adding solid CaCO3 will have the same result on

the pressure in the container as adding solid
CaO.
22
 2- Removing some CO2 from the container will result
in production of CaO.
3- Adding some CO2 to the container will result in
production of CaCO3.
4- Adding solid CaCO3 to the container will result in
the production of CaO.

EX: for the equilibrium reaction:

(NH4)2CO3(s) 2NH3(g) + CO2(g) +H2O(g)ΔH⁰=+33KJmol-1

The reaction is reactant favored at equilibrium if:

1-the reaction vessel volume decreased by a factor of 2 at
constant temperature.
2-additional reactant is added.
3-temperature is increased, reaction vessel volume
remains constant.
4-CO2 gas is removed from the reaction vessel.

EX: The reaction of coke and steam at high temperatures

produces H2 and CO gases.
At 1000K; Kp = 0.75.
H2O(g) + C(s)  H2(g) + CO(g)
If you have steam and carbon in an isolated container at
1000K at equilibrium. Which of the following would
increase the amount of H2 in the container?

1-increasing the volume of the container.

2-Decreasing the volume of the container.

23
3-Adding an atmosphere of N2(g).
4-Adding an atmosphere of CO(g).
5-None of the above.

EX: the reaction of coke, C(s), and steam at high

temperature produces H2 and CO gases,
H2O(g) + C(s)  H2(g) + CO(g) ΔrH=133KJmol-1

If you a mixture of reactants and products in an isolated

container at 1000L, which of the following would
increase the amount of H2 in the container?
1-Decrasing the temperature.
2-increasing the temperature.
3-adding carbon.
4-decreasing the volume of the container.
5-none of the above.

EX: giving the following equilibrium at 650K;

CO(g) + 2H2(g)  CH3OH(g) Keq = 6 X 10-5
H2O(g) + C(s) H2(g) + CO(g) Keq = 10 X 10-5
What is the equilibrium constant for the following
reaction?
C(s) + H2(g) + H2O(g)  CH3OH(g) Keq=?

1- 1.6 X 10-4
2- 6X 10-5
3- 1.6 X 10-9
4- 6 X 10-10
5- 6 X10-9

Examples From Second midterm 16/12/2016:

24
1* The solubilities of ammonium bromide NH4Br in
water at 0C, 20 C and 80 ⁰C are as follows:
T (⁰C) sol.gNH4Br/100g H2O
0 60.5
20 76.4
80 125
Which is the following fractional crystallization schemes
would produce the highest percent yield for the
recrystallization of ammonium bromide?
A- A solution containing 50.5g NH4Br in 100g
H2O at 20 ⁰C is cooled to 0 ⁰C.
B- A solution containing 115gNH4Br in 200g H2O at
80 ⁰C is cooled to 0 ⁰C.
C- A solution containing 120 g NH4Br in 100g
H2O at 80 ⁰C is cooled to 20 ⁰C.
D- A solution containing 100 g NH4Br in 100g
H2O at 80 ⁰C is cooled to 20 ⁰C.
E- A solution containing 95 g NH4Br in 175g H2O at
80 ⁰C is cooled to 0 ⁰C.

2* which definition is incorrect?

A- Mole fraction = moles of solute/moles of
solvent
B- Molality = moles of solute /kilogram of solvent
C- Weight percent = grams of solute / 100g of
solution
D- Molarity = moles of solute /liter of solution
E- Volume percent = volume of solute /volume of
solution

25
Examples From final exam 6/1/2017:

1* at certain temperature, Kc = 0.05 and ΔH= +39.6 KJ

for the reaction below:
2MgCl2(s) + O2  2MgO(s) + 2Cl2(g)
Calculate Kc for the reaction:
MgO(s) + Cl2(g)  MgCl2(s) + 1/2 O2(g)
And indicate whether the value will be larger or smaller
at a lower temperature.

A- 100, smaller
B- 0.224, larger
C- 0.004, smaller
D- 1, larger
E- 4.47, larger

End of Chapter 15
Abdulfettah Hazuri

26
27