CHEMISTRY A LEVEL(FORM SIX) NOTES – PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM(2)

Multiplying an equilibrium reaction by a number

When the stoichiometric coefficient of a balanced equation is multiplied by the same factor, the equilibrium constant for the new equation is the old equilibrium constant raised to the power of the multiplied factor.

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

K_c (i) =

If the equation (i) is multiplied by ½:

PCl₅ (g) ⇌ PCl₃ (g) + ½ Cl₂ (g)

Note: If an equilibrium reaction is multiplied by a factor n, then;

K_c (ii) = (K_c (i))ⁿ

Example:

The K_c for the reaction below is :

S (s) + O₂ (g) ⇌ SO₃ (g)

What is the K_c for 2SO₃ (g) ⇌ 2S (s) + 3O₂ (g)?

Equation (2) has been reversed and multiplied by 2.

When reversed:

K_c (i) = = 1.1 x 10⁶⁵

K_c (ii) =

K_c (ii) = = 9.09 x 10^-66

When multiplied

K_c (ii) =

3. Adding the equilibrium

The equilibrium constant for the reaction:

i) 2HCl (g) + H₂ (g) ⇌ Cl₂ (g) + H₂ (g)

K_c (i) = 4.17 x 10^-34 (at 25°C)

The equilibrium constant for reaction:

ii) I₂ (g) + Cl₂ (g) ⇌ 2ICl (g)

K_c (ii) = 2.1 x 10⁵ (at 25°C)

Calculate the equilibrium constant for the reaction:

iii) 2HCl (g) + I₂ (g) ⇌ 2ICl (g) + H₂ (g)

K_c (iii) = ?

When equation (i) + equation (ii) = equation (iii), hence:

K_c (iii) = K_c (ii) × K_c (i)

= (2.1 x 10⁵) × (4.17 x 10^-34)

K_c (iii) = 8.757 x 10^-29 (mol dm^-3)^1/2

Question 1:

The following are reactions which occur at 3500K:

i) 2H₂ (g) + O₂ (g) ⇌ 2H₂O (g)

K_p (i) = 26.4 atm^-1

ii) 2CO (g) + O₂ (g) ⇌ 2CO₂ (g)

K_p (ii) = 0.376 atm^-1

Calculate equilibrium constant for reaction:

CO₂ (g) + H₂ (g) ⇌ CO (g) + H₂O (g)

K_p (iii) = ?

Question 2:

Determine the equilibrium constant for reaction:

M₂ (g) + O₂ (g) + Br₂ (g) ⇌ NOBr (g)

Given that:

i) 2NO (g) + O₂ (g) ⇌ 2NO₂ (g), K_c = 2.4 x 10³⁰

ii) NO (g) + Br₂ (g) ⇌ NOBr (g), K_c = 1.4

Answers:

i) 2H₂ (g) + O₂ (g) ⇌ 2H₂O (g), K_p (i) = 26.4 atm^-1

ii) 2CO (g) + O₂ (g) ⇌ 2CO₂ (g), K_p (ii) = 0.376 atm^-1

iii) CO₂ (g) + H₂ (g) ⇌ CO (g) + H₂O (g), K_p (iii) = ?

Reverse (ii), multiply (i) and (ii) by ½:

i) (2H₂ (g) + O₂ (g) ⇌ 2H₂O (g))^½, K_p (i) = 5.138 atm^-1

Reverse (ii):

ii) 2CO₂ (g) ⇌ 2CO (g) + O₂ (g), K_p (ii) = 2.6595 atm^-1

Multiply (ii) by ½:

ii) CO₂ (g) ⇌ CO (g) + ½ O₂ (g), K_p (ii) = 1.63079

Add equation (i) + (ii), then:

K_p (iii) = K_p (i) × K_p (ii) = 5.138 × 1.63079 = 8.379 atm^-1

2.

i) 2NO (g) + O₂ (g), K_c (i) = 2.4 x 10³⁰

ii) NO (g) + Br₂ (g), K_c (ii) = 14

iii) N₂ (g) + Br₂ (g) + O₂ (g) ⇌ NOBr (g), K_c (iii) = ?

Reverse (i):

O₂ (g) + N₂ (g) ⇌ 2NO (g), K_c (i) = 4.167 x 10^-31

Multiply (i) by ½:

½ O₂ (g) + ½ N₂ (g) ⇌ NO (g), K_c (i) = 6.455 x 10^-16

Equation (i) + (ii) = (iii)

Therefore:

K_c (iii) = K_c (i) × K_c (ii) = 6.455 x 10^-16 × 1.4 = 9.0369 x 10^-16 (mol dm^-3)^1/2

3.

The equilibrium constants for the reactions which have been determined at 878K are as follows:

i) CO (s) + H₂ (g) ⇌ CO (s) + H₂O (g), K₁ = 67

ii) CO (s) + CO (g) ⇌ CO (s) + CO₂ (g), K₂ = 490

Using this information, calculate K’s (at the same temperature) for:

iii) CO₂ (g) + H₂ (g) ⇌ CO (g) + H₂O (g), K₃ = ?

iv) CO (g) + H₂O (g) ⇌ CO₂ (g) + H₂ (g), K₄ = ?

Reverse (ii):

CO₂ (g) + CO (s) ⇌ CO (g) + CO (s), K₂ = 2.0408 x 10^-3

Equation (i) + (ii) = (iii)

K₃ = K₁ × K₂ = 67 × 2.0408 x 10^-3 = 0.1367

To find K₄:

i) CO (s) + H₂ ⇌ CO (s) + H₂O (g), K₁ = 67

ii) CO (g) + CO (s) ⇌ CO₂ (g) + CO (s), K₂ = 490

iii) CO₂ (g) + H₂ (g) ⇌ CO (g) + H₂O (g), K₃ = 0.1367

iv) CO (g) + H₂O (g) ⇌ CO₂ (g) + H₂ (g), K₄ = ?

When equation (iii) is reversed, it is equal to equation (iv):

K₄ = 1 / K₃ = 7.315

4. The heterogeneous equilibrium

i) Fe (s) + H₂O ⇌ FeO (s) + H₂ (g)

ii) Fe (s) + CO₂ (g) ⇌ FeO (s) + CO (g)

Have been studied at 800°C and 1000°C. Also the rate (k) is constant = 1.81 at 800°C and 2.48 at 1000°C.

i) Why are the ratios constant?

ii) Calculate equilibrium constant at two temperatures of the reaction.

iii) H₂O (g) + CO (g) ⇌ H₂ (g) + CO₂ (g)

Answer:

i) The ratios are constant because for any system in equilibrium at a given temperature, the ratio of concentration of products to the concentration of reactants raised to the power of their mole ratios is constant.

ii) At 800°C:

K_p1 = 2

K_p2 = 1.81

K_p3 = ?

(i) Fe (s) + H₂O (l) ⇌ Fe (s) + H₂ (g), K_p (i) = 2

Reversed (ii):

CO (g) + FeO (s) ⇌ CO₂ (g) + Fe (s), K_p (ii) = 0.55

Equation (i) + (ii) = (iii)

Therefore:

K_p3 = K_p1 × K_p2 = 2 × 0.55 = 1.1

At 1000°C:

K_p1 = 1.49

K_p2 = 2.48

i) Fe (s) + H₂O (g) ⇌ FeO (s) + H₂ (g), K_p1 = 1.49

Reversed (ii):

CO (g) + FeO (s) ⇌ CO₂ (g) + Fe (s), K_p2 = 0.403

Equation (i) + (ii) = (iii)

K_p3 = K_p1 × K_p2 = 1.49 × 0.403 = 0.6

1:

When 1 mole of ethanoic acid is maintained at 25°C with 1 mole of ethanol, 1/3 of ethanoic acid remains when equilibrium is attained. How much would have remained if 3/4 of 1 mole of ethanol had been used instead of 1 mole at the same temperature?

CH₃CH₂OH + CH₃COOH ⇌ CH₃COOCH₂CH₃ + H₂O

At start: 1 mol, 1 mol, 0, 0

At time: 1-x, 1-x, x, x

At equilibrium: 1 – x = 1/3

K_c =

=

=

K_c = 4

4 =

4 =

4 = X²

3 – 7x + 4x² = x²

3x² – 7x + 3 = 0

a b c

x =

x =

=

=

X = or x =

X = 1.76 or 0.566

X cannot be 1.76

X = 0.566

0.566 = 0.184 moles or 23/125 moles

0.184 moles of ethanol would have remained.

2.

The equilibrium constant (K_c) for the reaction:

2HI (g) ⇌ H₂ (g) + I₂ (g) is 0.02 at 400°C. If 2 moles of H₂ and 1 mole of I₂ were mixed together in a 1.0 dm³ at 400°C, how many moles of HI, I₂ and H₂ would be present at equilibrium?

2HI (g) ⇌ H₂ (g) + I₂ (g), K_c = 0.02

Reverse the equation above:

H₂ (g) + I₂ (g) ⇌ 2HI (g), K_c = 1 / 0.02 = 50

At start: 2, 1, 0

At time: 2 – x, 1 – x, 2x = 50

At equilibrium:

K_c =

50 =

4x² = 50(2 – 3x + x²)

100 – 150x + 50x² = 4x²

100 – 150x + 46x² = 0

x =

x = or x =

x = 2.33 or 0.934

x cannot be 2.33

x = 0.934

At equilibrium:

Number of moles of H₂ = 2 – 0.934 = 1.066 moles

Number of moles of I₂ = 1 – 0.934 = 0.066 moles

Number of moles of HI = 2 × 0.934 = 1.868 moles

3.

The equilibrium constant for the reaction:

H₂ (g) + Br₂ (g) ⇌ 2HBr (g) at 1024K is 1.6 x 10⁵. Find the equilibrium pressure of all gases if 10 atm of HBr is introduced into a second container at 1024K.

PREDICTION OF DIRECTION AND EXTENT OF CHEMICAL EQUILIBRIUM

At each point in the progress of a reaction, it is possible to formulate the ratio of concentration having the same form as the equilibrium constant expression. This generalized ratio is called the reaction quotient (Q).

For the reaction:

Then:

Q_C =

Q_p =

Q_C differs from K_C in that the concentration in the expression is not necessarily the equilibrium concentration.

When the values of K_C and Q_C are compared, one can predict the direction of the chemical reaction.

• If Q_C < K_C, the system is not at equilibrium; the reactants must further be converted to products to achieve equilibrium. Therefore, the net reaction proceeds from left to right.
• If Q_C = K_C, the system is at equilibrium.
• If Q_C > K_C, the system is not at equilibrium; the products must be converted to reactants to achieve equilibrium. Therefore, the net reaction proceeds from right to left.

Example:

1. Consider the reaction:

At 250°C, K_c = 4.0 x 10^-2. If the concentration of Cl₂ and PCl₃ are both 0.30 M while that of PCl₅ is 3.0 M, is the system at equilibrium? If not, in which direction does the reaction proceed?

Q_c = 0.03

Q_c ≠ K_c, Q_c < K_c

The system is not at equilibrium and the reaction proceeds from left to right.

2.

At 200K, the K_p for the formation of NO is 4 x 10^-4.

N₂ (g) + O₂ (g) ⇌ 2NO (g)

If at 200K the partial pressure of N₂ is 0.5 atm and that of O₂ is 0.25 atm, that of NO is 4.2 x 10^-3 atm, decide whether the system is at equilibrium. If not, in which direction does the reaction proceed?

K_p = 4.0 x 10^-4

Q_p = 1.4112 x 10^-4

K_p ≠ Q_p, K_p > Q_p

The system is not at equilibrium; therefore, the reaction proceeds from left to right.

EQUILIBRIUM CONSTANT WITH DEGREE OF DISSOCIATION (α)

– It gives to what extent the reactants are converted to products by dissociation.

– This has to be treated similar to moles.

Example 1.

0.01 moles of PCl₅ was placed in 1L vessel at 210K. It was found to be 52.6% dissociated into PCl₃ and Cl₂. Calculate the K_c at that temperature.

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

At start: 0.01, 0, 0

At equilibrium: 0.01 – (5.26 x 10^-3), 5.26 x 10^-3, 5.26 x 10^-3

PCl₅ = 4.74 x 10^-3 moles

PCl₃ = 5.26 x 10^-3 moles

Cl₂ = 5.26 x 10^-3 moles

K_C =

K_C = 5.837 x 10^-3 moles

2.

At 1 atm and 85°C, N₂O₄ is 50% dissociated. Calculate the equilibrium constant in terms of pressure and calculate the degree of dissociation of the gas at 100°C and 55°C.

N₂O₄ ⇌ 2NO₂

Start: 1, 0

Equilibrium: 1 – α, 2α

V.D = ρ₂ / ρ₁ = 70.2 / 104.25 = 0.673

Degree of dissociation = 48.5%

DEGREE OF DISSOCIATION BY DENSITY MEASUREMENT

This method is used for the determination of degree of dissociation of gases in which 1 molecule produces 2 or more molecules.

i.e. PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

Thus at constant temperature and pressure, the volume increases. The density at constant pressure decreases.

The degree of dissociation can be calculated from the difference in density between the undissociated gas and that of partially dissociated gas at equilibrium.

If we start with 1 mole of the gas (PCl₅) and the degree of dissociation (α), then:

Moles at equilibrium: 1 – α, α, α

Total moles: 1 + α

Note:

The density of an ideal gas at constant temperature and pressure is inversely proportional to the number of moles for a given weight.

Hence the ratio of density:

=

Examples

1. When PCl₅ is heated, it gasifies and dissociates into PCl₃ and Cl₂. The density of the gas mixture at 200°C is 70.2. Find the degree of dissociation of PCl₅ at 200°C.

Observed density, ρ₂ = 70.2

Density before dissociation, ρ₁ = ?

V.D =

= 104.25

Degree of dissociation =

= 48.5%

DETERMINATION OF DEGREE OF DISSOCIATION BY MOLECULAR MASS

Molecular masses are proportional at constant temperature and pressure to the density of their gases; therefore we can substitute the molecular masses for the density in the degree of dissociation.

Where:

M₁ = Molecular mass of undissociated gas

M₂ = Average molecular mass of gases at equilibrium

Example 1:

1.588 g of N₂O₄ gives a total pressure of 1 atm when partially dissociated at equilibrium in a 500 cm³ glass vessel at 250°C. What is the degree of dissociation at this temperature?

N₂O₄ ⇌ 2NO₂

M₁ = (14 × 2) + (16 × 4) = 92 g mol^-1

M₂ = ?

From PV = nRT, where n =

M₂ =

= 77.7 g mol^-1

Degree of dissociation =

= 85.48%

FACTORS AFFECTING EQUILIBRIUM REACTION

These factors are as follows:

1. Temperature
2. Concentration
3. Pressure

∙ The first three affect both rates and position of chemical equilibrium (i, ii and iii).

∙ The other factors affect the rate of chemical equilibrium.

1) Temperature

a) Increasing the temperature increases the rate of reaction because usually at high temperature the collision factor increases; also the number of molecules having necessary activation energy is large.

b) Effect on the position of equilibrium is explained by using Le Chatelier’s principle which states that “when a system at equilibrium is subjected to a change, processes occur which tend to counteract the change” (If a system in equilibrium is disturbed (change in temperature and pressure), the system adjusts itself so as to oppose the disturbance).

Consider the reaction:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g) + Heat (negative)

If temperature is increased in the system, the equilibrium moves in a direction where there is absorption of heat; if temperature is decreased, the equilibrium moves in a direction where there is release of heat.

Effect of temperature on the position of equilibrium can be explored by Vant Hoff’s law of chemical equilibrium which states that:

“For any system in equilibrium, high temperature favours endothermic reactions and low temperature favours exothermic reactions.”

The way in which equilibrium constant changes with temperature is found both theoretically and experimentally governed by the following relationship:

ΔH_m = change in molar heat

K = Equilibrium constant

On integrating the equation above:

Where c = constant

If K₁ and K₂ are equilibrium constants corresponding to T₁ and T₂, the constant term can be eliminated from the equation above to give Vant Hoff’s equation:

ln K₁ =

ln K₂ =

Subtracting equation (1) from (2):

ln K₂ – ln K₁ =

But ln = 2.303 log

K_p2 = 0.113 at 298K, ΔH = 61.5 kJ mol^-1

i) What is the value of K_p at 0°C?

ii) At what temperature will K_p = 1?

Answers:

i) From:

K_p2 = 0.113

K_p1 = ?

T₁ = 273 K

T₂ = 298 K

ΔH = 61.5 kJ mol^-1

R = 8.314

ln K_p2 / K_p1 = 3211.9676 (3.0729896 x 10^-4)

= 0.987

To remove ‘log’ make 10^0.987 = 9.7

ii) From:

K_p2 = 0.113

K_p1 = ?

T₁ = 1

T₂ = 298 K

ΔH = 61.5 kJ mol^-1

R = 8.314

ln K_p2 / K_p1 = -0.9469

T₁ = 326.7 K

VARIATION OF EQUILIBRIUM CONSTANT WITH TEMPERATURE

According to Vant Hoff’s law of chemical equilibrium:

i) Exothermic reaction:

• Increasing temperature decreases the K_C value as the equilibrium shifts to the left to decrease the concentration of products and increase the concentration of reactants.
• Decreasing temperature increases the K_C value.

ii) Endothermic reaction:

• Increasing temperature increases the K_C value.
• Decreasing temperature decreases the K_C value.

2) Concentration

a) Effects on rate of reaction:

• Increase in concentration of reactants in the same volume increases the number of molecules, thus increasing the chances of collision between molecules and hence increases the rate of reaction.
• Decrease in concentration results in decrease in rate of reaction as it decreases the number of molecules per unit volume, hence less collision.

b) Effect on the position of equilibrium:

• This depends on whether the concentration of products or reactants has been increased or decreased.
• It also depends on Le Chatelier’s principle.

Example

1. Consider:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

State what happens when:

1. [SO₂] is increased at the same temperature
2. [O₂] is increased at the same temperature
3. [SO₃] is increased at the same temperature

According to Le Chatelier’s principle, the system will adjust itself so as to cancel out the effect by shifting the equilibrium to the right side, i.e., increase the number of products hence decreasing the amount of SO₂.

2. Using:

CH₃COOH (g) + CH₃CH₂OH (g) ⇌ CH₃COOCH₂CH₃ (g) + H₂O (l)

What happens when:

1. H₂O is added
2. CH₃CH₂OH is added
3. NaOH is added
4. Anhydrous copper (II) sulphate is added
5. CH₃COOCH₂CH₃ is added

Answers:

i) H₂O will react with CH₃COOH and concentration of CH₃COOH will decrease; hence forward reaction, equilibrium will lie on the products side.

ii) Forward reaction since it will get converted to products.

iv) Reverse reaction since H₂O will react with CuSO₄ to form CuSO₄.5H₂O.

3. The following reaction occurs in human body:

Hb (s) + O₂ (g) ⇌ HbO₂ (g)

i) What happens in the tissues?

ii) What happens in the lung?

Answers:

i) In the tissue, the amount of O₂ is less; hence according to Le Chatelier’s principle, the system will adjust itself so as to increase the amount of O₂ by favouring the decomposition of HbO₂ (backward reaction). Hence the equilibrium shifts to the left.

ii) In the lung, the amount of O₂ is a lot more; hence according to Le Chatelier’s principle, the system will adjust itself so as to decrease the amount of O₂ by favouring the forward reaction. Hence the equilibrium shifts to the right.

4. Consider the following equilibrium:

Orange yellow

What would you expect to see if:

1. Dilute NaOH is added to the equilibrium mixture
2. Dilute HCl is added to the equilibrium mixture

Answers:

i) When dilute NaOH is added to the mixture, it will react with H⁺ to form H₂O and Na⁺. This will decrease the concentration of H⁺ ions; hence equilibrium will shift to the left. The colour will change from yellow to orange.

ii) Dilute HCl dissociates to form H⁺ and Cl^–; therefore, when added to the mixture, the concentration of H⁺ ions increases. According to Le Chatelier’s principle, the equilibrium will adjust itself so that the concentration of H⁺ ions decreases; hence the backward reaction is favoured.

3) Pressure

a) Effect on the rate of reaction:

Increase in pressure increases the rate of reaction. This is because increase in pressure decreases the volume per unit volume, hence increases effective collision and thus increases the rate of reaction; vice versa is also true.

b) Effect on position of equilibrium:

Effect if number of moles of reactants and products differ.

Homogeneous gaseous equilibrium

Consider the reaction:

N₂O₄ ⇌ 2NO₂

What will be the effect on equilibrium when:

1. Pressure is increased?
2. Pressure is decreased?

Answers:

i) When the pressure is increased, according to Le Chatelier’s principle, the equilibrium will adjust itself so that the backward reaction is favoured.

ii) When pressure is decreased, the equilibrium will adjust itself so that the forward reaction is favoured.

Note: Pressure has no effect on position of equilibrium if the number of moles on reactants side equals the number of moles on products side.

Heterogeneous equilibrium

Position of equilibrium is affected by changing the partial pressure of the gases only.

Example:

If partial pressure of CO₂ is decreased, the equilibrium shifts to the right to increase the partial pressure of CO₂. Vice versa is true.

4) Catalyst

The catalyst speeds up the rate of both forward and backward reactions to some extent since it lowers the activation energy of the reaction.

Therefore, the equilibrium is not altered when catalyst is added. It only changes the rates at which the reaction approaches equilibrium.

5) Physical state and sunlight

They do not affect position of equilibrium, only the rate of reaction.

Assignment

Read and write on the important industrial applications of chemical equilibrium (specifically on Haber process and contact process).

MANUFACTURE OF AMMONIA (HABER PROCESS)

Haber’s process for the manufacture of ammonia involves direct combination (synthesis) of nitrogen and hydrogen:

1 mol N₂, 3 mol H₂, 2 mol NH₃

This reaction is:

1. Reversible
2. Exothermic
3. Proceeds with change in volume

According to Le Chatelier’s principle, the favourable conditions for formation of ammonia are:

1. Low temperature: The temperature should be kept as low as possible but at very low temperature, the rate of reaction becomes slow. It has been found that the yield of NH₃ is maximum at about 500°C which is the optimum temperature of the reaction.
2. High pressure: High pressure favours reaction which is accompanied by a decrease in volume. In actual practice, a pressure of 200-900 atm is employed in this process.
3. Catalyst: To increase the speed of the reaction, a catalyst should be finely divided iron containing molybdenum or alumina is used as a catalyst. Molybdenum or alumina (Al₂O₃) acts as a promoter and increases the efficiency of the catalyst. A mixture of iron oxide and potassium aluminate has been found to work more effectively. A catalyst iron oxide containing Al₂O₅ and K₂O is also used in the process.

A diagram to show the manufacture of ammonia by Born Haber process:

MANUFACTURE OF SULPHURIC ACID BY CONTACT PROCESS

In contact process, sulphur dioxide is oxidized by air in the presence of catalyst Vanadium pentaoxide. Sulphur trioxide produced is absorbed in concentrated H₂SO₄ to produce oleum (H₂S₂O₇). Oleum is then reacted with calculated amount of H₂O to form H₂SO₄ of the desired concentration.

The chemistry involved in the contact process is described as follows:

1. Production of SO₂: Sulphur dioxide is obtained by burning sulphur or iron pyrites.
2. Catalytic oxidation of SO₂ to SO₃: SO₂ is oxidized by air in the presence of a catalyst to give SO₃.
3. Conversion of SO₃ to oleum: SO₃ is dissolved in concentrated H₂SO₄ to produce oleum or fuming H₂SO₄.
4. Conversion of oleum to sulphuric acid: Oleum is diluted with a calculated amount of H₂O to get H₂SO₄ of desired concentration.

The great advantage of the contact process is that it produces a pure acid of any desired concentration and especially the fuming acid which is of great value in chemical industry.

THE CONTACT PROCESS