CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

PHYSICAL CHEMISTRYPHYSICAL CHEMISTRY

Chemical Equilibrium1. Neuralization produces salt and water. However, salt and

water does not react to produce acid and alkali.

2. NaOH(aq) + HCl (aq) NaCl (aq) + H2O (l)3. The above reaction proceeds in one direction: ~

Irreversible Reaction

4. Hydrogen gas and iodine vapor reacts to form hydrogen iodide, at the same time, HI can dissociate to form hydrogen gas and iodine vapor.

5. Both forward and reverse (backward) reaction are called Reversible Reaction

6. When the rates of the forward and reverse reaction are equal, the concentration of reactants and products remain constant.

[Chemical Equilibrium]

H2 (g) + I2 (g) H-I(g)

Concept of Dynamic Equilibrium

1. Consider a non reversible reaction: A B

2. Consider a reversible reaction: X Y

[B]

[A]

concentration

T1 time

concentration [Y]

[X]

T1 time

Concept of Dynamic Equilibrium

1. When we plot the rate of reaction, both the forward reaction and reverse reaction:

2. At time T1, the rate of the forward reaction becomes equal to the rate of the reverse reaction.

3. Thereafter, the concentration of X and Y does not change.

4. We say the system has achieved dynamic equilibrium.

5. The concentration of X and Y at dynamic equilibrium is called equilibrium concentration.

rate Forward reaction

Reverse reaction

T1 time

1. Consider the decomposition of N2O4 to NO2… N2O4 2NO2

2. For a general reaction: aA + bB cC + dD

3. At equlibrium: [C]c[D]d = Equilibrium Constant Equilibrium Constant [Kc]

[A]a[B]b

Equilibrium Constant

Initial [ ] Initial [ ] Equal. [ ] Equal. [ ] EquilibriumConstant

EquilibriumConstant

[N2O4]o [NO2]o [N2O4] [NO2] [NO2]/ [N2O4]

[NO2]^2/ [N2O4]

0.670 0.000 0.643 0.055 0.085 0.00465

0.000 0.200 0.089 0.021 0.227 0.00463

1. Write the general formula of equilibrium constant for the following reaction and state its unit.

a) CH3COOH + C2H5OH CH3COOC2H5 + H2O

b) PCl5 PCl3 + Cl2c) N2 + 3H2 2NH3

2. Practice: Consider the following reaction: 2A + B 2C An equilibrium mixture was found to contain 0.25

moldm-3 of A, 0.56 moldm-3 of B and 1.24 moldm-3 of C at 450 K. Calculate the equilibrium constant, Kc.

[43.93 mol-

1dm3]

Equilibrium Constant For Homogeneous System

Dependence of Kc on Chemical Equation

1. Consider the following reaction: 2SO3 (g) 2SO2 (g) + O2(g)

2. Kc = [SO2]2[O2]/ [SO3]2 = 400

3. If we were interested in the formation of SO3:

2SO2 (g) + O2(g) 2SO3 (g)

K1 = [SO3]2 / [SO2]2[O2] @ 1/([SO2]2[O2]/ [SO3]2 ) = 1/Kc = 1/400

4. In short, 2SO3 (g) 2SO2 (g) + O2(g)

5.5. KK1 1 x Kx K-1-1 = 1 = 1

6. If the equation is written in the following way: SO3 (g) SO2 (g) + ½ O2(g)

K = [SO2][O2]½ / [SO3] = [[SO2]2[O2]/ [SO3]2 ]½ = [Kc]½ = (400)1/2

K1

K-1

Equilibrium Constant For Homogeneous System

1. For a gas, the concentration is directly proportional to its pressure (partial pressure).

2. N2 (g) + 3H2(g) 2NH3 (g) Kp = p2

NH3/(pN2 x p3H2)

3. Practice Phosgene (COCl2) dissociate according to the equation:

COCl2 (g) CO (g) + Cl2 (g)

The equilibrium pressure of COCl2, CO and Cl2 at 360 K are 0.67, 0.45 and 1.60 atm. Calculate the equilibrium constant Kpat the same temperature.

[1.07 atm]

Adding Chemical Equation

1. If a reaction can be expressed as the sum of two or more reaction, K for the overall reaction is the product of the equilibrium constants of the individual reaction.

K (overall reaction) = K1 x K2

For example:SO2 (g) + ½ O2 (g) SO3 (g) K1 = 2.2

NO2 (g) NO(g) + ½ O2 (g) K2= 4.0

Adding the equations eliminates ½ O2, the result is:

SO2 (g) + NO2 (g) SO3 (g) + NO (g)

K = 2.2 x 4.0 = 8.8

Adding Chemical Equation: PRACTICE

1. Consider the reaction by which the air pollutant, NO is made from the elements in a automobile engine:

N2 (g) + O2(g) 2NO (g)

a) Write the equilibrium constant expression for the reaction.

b) At 25oC, K for this reaction is 4.2 x 10-31. Calculate the K for the reaction

½ N2 (g) + ½ O2(g) NO (g) (6.5x 10-16)

c) At 25oC, K for this reaction N2 (g) + 2O2(g) 2NO2 (g)

is 1.0 x 10-18. Calculate K at 25oC for the reaction: 2NO (g) + O2 (g) 2NO2 (g)

(2.4x1012)

Heterogeneous System

1. Consider the following equilibrium in a closed system: CaCO3 (s) CaO(s) + CO2 (g)

2. The concentration of a solid like its density, is a constant. In heterogeneous system, solids, pure liquids and solvents can be omitted from the equilibrium constant expression.

3. Thus, the equilibrium constant: Kp = PCO2

Let’s try to write the equilibrium constant: (NH4)2S (s) 2NH3 (g) + H2S (g)

The Relationship between Kc and Kp

1. Kp = Kc (RT)∆ n

2. Consider the following general equation: aA + bB cC + dD

∆n = (c+d) – (a+b)

Practice:The equilibrium constant, Kp for the following reaction is 3.00

x 104 Pa at 450 K. PCl5(g) PCl3 (g) + Cl2 (g)

Calculate Kc for the reaction at the same temperature. 0.125

PRACTICE

1. Kc of the following reaction is 4.0 at 298 K. CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O

(l) A mixture containing 2 mol of CH3COOH and 2 mol of

C2H5OH were allowed to come to equilibrium at 298 K.

a) Calculate the number of moles of ester produced.b) Sketch a concentration/time graph for the system.

2. N2(g) + 3H2(g) 2NH3(g) A mixture containing 1 mol of N2 and 3 mol of

hydrogen are left to attain equilibrium at 650oC and a total pressure of 200 atm. The equilibrium mixture was found to contain 20% ammonia. Calculate the equilibrium constant Kp.

Reaction Quotient and Direction of Reaction

1. The reaction quotient (Q) is obtained by substituting the initial concentrations into the expression for Kc.

2. H2 (g) + I2(g) 2HI (g) The Kc for the above reaction at 200oC is 25. A mixture containing 1 M of H2 and I2 and 1.2 M of HI

was left standing at 200oC. Determine the net direaction of reaction.

Q > Kc

There is too much product. A net reverse reaction ( from right to left) will happen to decrease Q.

Q < Kc

There is too little product. A net forward reaction ( from left to right) will happen to increase Q.

Q = Kc

The system is in equilibrium. No net reaction occur

Le Chatelier’s Principle

When a system at equilibrium is subjected to a change in temperature, pressure or concentration of a reactant or product, the system will, if possible, shift its equilibrium position so as to partially counteract the effect of the change.

Factors that affect equilibrium system

1. The equilibrium constant of reversible reaction is only dependent on temperature. It does not depend on changes in concentration of the reactants or products, pressure or the use of catalyst.

Change Rate Rate constant

Equilibrium constant

Equilibrium composition

Conc. Change Unchanged Unchanged Change

Pressure Change Unchanged Unchanged Changed

Temperature Change Change Change Change

Catalyst Change Change Unchanged Unchanged

Effect of Concentration and concentration

[B]

[A]

[B]

[A]

Addition of A

New equilibrium

A B

Effect of Concentration and concentration

Addition of B

New equilibrium

[B][B]

[A]

[A]

time

Conc.

Effect of Temperature

1. Low temperature would favor exothermic reaction, while high temperature would favor endothermic reaction. Decreasing Temperature

∆H = -ve A + B C + D ∆ H = + ve

Increasing Temperature

Kc

temperature

Kc

temperature

Exothermic Endothermic

Relationship between Kc and Temperature

1. Van’t Hoff Equation: ln K = -∆H/RT + C where K = equilibrium constant H = enthalpy change of reaction R = gas constant T = temperature in (K) C = constant

2. For endothermic reaction, where ∆H = positive, a plot of ln k against 1/T would give a straight line with a negative slope:

3. For exothermic reaction, where ∆H = negative, a plot of ln k against 1/T would give a straight line with a positive slope:

Relationship between Kc and Temperature

Practice

• Ammonium hydrogen sulphide dissociates according to the equation:

NH4HS (s) H2S (g) + NH3 (g)

At 9.5 oC, the total pressure at equilibrium is 23.4 kPa. When the temperature is raised to 25.5 oC, the total pressure is 67.8 kPa. Calculate the value of ∆H for this reaction.

+92.6 kJ

Effect of Catalyst

1. Catalyst does not affect systems that are ready in equilibrium.

2. For system that have not reached equilibrium, it merely speeds up the rate of the forward and reverse reaction by the same factor, so that equilibrium is reached in a shorter time.

3. A catalysts does not affect the yield of a reaction. The same amount is obtained regardless of whether a catalyst is present or not.

The effect of a Noble Gas on an Equilibrium

At constant Pressure

• If a noble gas such as argon or neon is added to a gaseous equilibrium mixture at constant pressure, the partial pressure for the gases in the system is lowered.

• Thus, according to Le Chatelier’s Principle, the presence of a noble gas favors the direction that increases the number of moles of gas.

PCl3 (g) + Cl2 (g) PCl5 (g)

2H2O (g) 2H2 (g) + O2 (g)

H2 (g) + I2(g) 2HI (g)

The effect of a Noble Gas on an Equilibrium

At constant Volumes

• If a noble gas such as argon or neon is added to a gaseous equilibrium mixture at constant volume, no effect on the position and composition of the mixture

Equilibrium and Industrial Process

Haber Process• N2 (g) + 3H2 (g) 2NH3 (g) ∆H = -92 kJ/mol• According to the Le Chatelier’s Principle, the forward

reaction would be favored by high pressure as it is accompanied by a decrease in the number of moles of gas. The pressure normally used is 200 – 1000 atm.

• The forward reaction is exothermic. Hence, it is favored by low temperature. However, a low temperature will slow down the rate of reaction. A compromise temperature would be 450 -500o C.

• Fine divided iron is added as catalyst to increase the rate of reaction and shorten the time required to reach equilibrium.

Equilibrium and Industrial Process

Contact Process• 2SO2 (g) + O2 (g) 2SO3 (g) ∆H = -197 kJ/mol • Vanadium (V) oxide

• The rate of production of SO3 is greatly improved by: a) increasing the concentration (Pressure) of SO2 and O2

b) increasing the temperature c) Using catalyst 3. Because of exothermic, low temperature and high pressure

is needed.4. At low temperature, rate of reaction become slow. (450 C)5. In theory, application of higher pressures will give a higher

yield of sulphur trioxide. However, in practice, it only has small effect on the yield of SO3. (1atm)

Equilibrium and Industrial Process

PracticeThe oxidation of sulphur dioxide to sulphur trioxide is

reversible reaction.The table below shows the equilibrium partial pressure of

SO2, O2, SO3 at 700o C.

a) Calculate the Kp.b) What are the initial partial pressure of SO2 and O2?

3.51, 0.59, 0.56

Gas SO2 O2 SO3

Partial Pressure atm

0.27

0.40

0.32