Equilibria

Equilibria

7.1 Chemical equilibria: reversible reactions, dynamic equilibrium

Chemical Equilibria

• Reversible reaction : a reaction which can go forwards or backwards depending on the conditions.

• Dynamic equilibrium : for a reversible reaction in a closed system, dynamic equilibrium occurs when the rate of the forward and backward reactions is equal. The concentrations of products and reactants remain constant despite the fact particles are continually reacting.

• Le Chatelier’s principle : if a dynamic equilibrium is subject to changing conditions, the position of equilibrium will shift to counteract this change.

What is dynamic equilibrium?

The rate of the forward reaction equals the rate of the reverse reaction, hence the concentrations of the reactants and products remains constant.

CONCENTRATION CHANGE IN A REACTION

• As the rate of reaction is dependant on the concentration of reactants...

• the forward reaction starts off fast but slows as the reactants get less concentrated

EQUILIBRIUM REACTIONS

• Initially, there is no backward reaction but, as products form, it speeds up and provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium.

• In an equilibrium reaction, not all the reactants end up as products; there is not a 100% conversion.

Multiple Choice

At what time the reaction reached equilibrium?

• t₁

• t₂

• t₃

• t₄

What is Le Chatelier’s principle?

If a system in dynamic equilibrium is

subjected to a change, then the position

of equilibrium will shift to minimise this

change.

Multiple Choice

Le Chatelier's Principle states that....

If a (dynamic) equilibrium is disturbed by changing the conditions, the position of equilibrium................

• moves to increase the change

• moves to counteract the change

• does not change

How does the concentration of reactants

affect the position of equilibrium?

• Increasing the concentration of reactants causes the position of equilibrium to shift right in order to reduce the concentration of reactants and form more products.

• The reverse occurs if the concentration of reactants is decreased.

  
  

Multiple Choice

For the reaction...

SO₂ + O₂ <=> SO₃

If the concentration of SO₂ is increased, the equilibrium position of the reaction will shift ___________.

• to the left

• to the right

Multiple Choice

For the reaction...

SO₂ + O₂ <=> SO₃

If the equilibrium position shifts to the right, the concentration of O₂ will ___________.

• increase

• decrease

• remain the same

Multiple Choice

For the reaction...

N₂ + O₂ <=> 2NO

If O₂ is removed, the concentration of N₂ will _______.

• increase

• decrease

• remain the same

How does increasing the temperature affect the position of equilibrium?

• For an equilibrium where the forward reaction is exothermic, increasing the temperature will shift the equilibrium left (so more endothermic reactions occur) to take in more heat energy and reduce the temperature.

• The reverse is true when the forward reaction is endothermic.

How does decreasing the temperature affect the position of equilibrium?

• For an equilibrium where the forward reaction is exothermic, decreasing the temperature will shift the position of equilibrium to the right (so more exothermic reactions occur) to release more heat energy and increase the temperature.

• The reverse is true when the forward reaction is endothermic.

Multiple Choice

For the reaction...

N₂ + O₂ <=> 2NO: Δ H = +182 kJ mol^-1.

If the temperature is increased the equilibrium position will shift _______.

• to the left

• to the right

How does pressure affect the position of equilibrium?

• We consider the number of gaseous molecules only.

  Increasing the pressure will cause the position of equilibrium to shift to the side with the fewest gaseous molecules in order to increase the pressure.

• The opposite occurs if pressure is decreased.

• If there is an equal number of gaseous molecules on both sides of the equation, changing the pressure will have no effect on the position of equilibrium.

Multiple Choice

Changes in pressure will only affect substances that are in the __________ state.

• solid

• liquid

• gaseous

Multiple Choice

For the reaction...

N_{2 (g)} + 3 H_{2 (g)} <=> 2 NH_{3 (g)}

If the pressure in the system is increased, which substance(s) will increase in concentration?

• N₂ (as 2 mol gas → 4 mol gas)

• H₂ (as 2 mol gas → 4 mol gas)

• N₂ and H₂ (as 2 mol gas → 4 mol gas)

• NH₃ (as 4 mol gas → 2 mol gas)

How does the presence of a catalyst affect the position of equilibrium and the magnitude of the equilibrium constant?

• The presence of a catalyst doesn’t affect the position of equilibrium.

• The magnitude of the equilibrium constant therefore is unaffected.

• It does however increase the rate of the forward and reverse reactions so equilibrium is established sooner.

Multiple Choice

What is the purpose of a catalyst on an equilibrium reaction?

• to increase the rate of forward reaction

• to increase rate of backward reaction

• in increase both forward and backward reaction rates

• no effect

The Equilibrium Law

“If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant”

THE EQUILIBRIUM CONSTANT Kc

• VALUE OF Kc

  AFFECTED by a change of temperature

• NOT AFFECTED by a change in concentration of reactants or products a change of pressure adding a catalyst

Kc

• Homogeneous equilibria have all substances in the same phase. In heterogeneous equilibria, substances are in different phases.

• For this general equilibrium equation, all substances are (g), (l) or (aq)

• aA + bB ⇋ cC + dD

  This equation is used below to show how to calculate Kc

• In heterogeneous equilibria, don’t include solids in the Kc equation.

Kp

• To work out Kp , more calculations are required:

• Mole fractions - in a mixture of gases, the mole fraction of gas A is:  $X_A=\frac{number\ of\ moles\ of\ gas\ A}{total\ number\ of\ moles\ of\ gas}$ 

• Partial pressures - in a mixture of gases, the partial pressure is the pressure that one gas would exert if it occupied the whole container:  $P_A=mole\ fraction\ of\ gas\ A\times total\ pressure$  

How does concentration affect the magnitude of the equilibrium constant?

• Changing the concentration of a reactant or product means that the system is no longer in equilibrium.

• The concentrations of the reactants and products now must change so the ratio and hence Kc is restored.

• Kc is therefore unaffected by concentration changes

How does temperature affect the magnitude of the equilibrium constant?

• If the forward reaction is exothermic, increasing the temperature shifts the position of equilibrium to the left so Kc decreases.

• If the forward reaction is endothermic, increasing the temperature shifts the equilibrium to the right so Kc increases.

• The reverse is true if temperature is decreased.

Multiple Choice

Given: 2A(g) <=> 2B(g) + C(g). At a particular temperature, K_c = 16000.

Raising the pressure, by decreasing the volume of the container, will...

• cause the value of K_c to increase

• cause the value of K_c to decrease

• have no effect on the value of K_c as temperature does not change

• favour the forward reaction

How does pressure affect the magnitude of the equilibrium constant?

• Kc remains the same:

• Doubling the pressure will double both the partial pressures and concentrations of the species on both sides of the equation.

• The system is no longer in equilibrium so partial pressures of reactants and products must change to keep Kc the same.

• New equilibrium position will be reached whereby Kc is restored (the ratio of the Kc expression is the same as before).

Multiple Choice

For the reaction...

H_{2 (g)} + Cl_{2 (g)} <=> 2HCl _(g) + heat

If the pressure in the system is increased, the equilibrium position will _______.

• shift to the left

• shift to the right

• not shift position at all as 2 mol gas <=> 2 mol gas

What does the equilibrium constant tell you?

• The position of equilibrium of a reaction.

• The magnitude indicates whether there are more reactants or products in an equilibrium system.

Multiple Choice

Consider the following reaction:

2SO_2(g) + O_2(g) <=> 2SO_3(g): ΔΗ = - 197 kJ mol ^-1

Which of the following will NOT shift the equilibrium position to the right?

• Adding more O₂

• Adding a catalyst

• increasing the pressure

• Lowering the temperature

Kc

Multiply all the concentrations of the products to the power of their balancing number.

Divided this by the product of all the concentrations of the reactants to the power of their balancing number.

This is Kc, the equilibrium constant.

Multiple Choice

What are [A] and [B]?

• concentration of reactants

• concentration of products

Multiple Choice

What is the K_c expression for this reaction?

2 NO_(g) + O_2(g) ⇌ 2 NO_2(g)

• $Kc=\frac{[NO_2]^2}{[NO]^2+[O_2]}$

• 
  

  $Kc=\frac{[NO_2]^2}{[NO]^2[O_2]}$

• $Kc=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Multiple Choice

An equilibrium constant with a large value, e.g. K_c = 1000, indicates…

• A very fast reaction

• Mostly products at equilibrium

• Mostly reactants at equilibrium

• Nothing useful at all

What is Kp?

• The equilibrium constant for reactions in the gaseous phase.

• Similar to Kc but it uses partial pressures instead of concentrations.

How do you calculate the partial pressure of a gas?

• For gas A:

• Partial pressure of A, p(A)= Mole fraction, X_A x Total Pressure

Multiple Select

Write the Kp expression for the following reaction:

H₂(g) + I₂(g) → 2HI(g)

• $Kp\ =\ \frac{p\left[HI\right]^2}{p\left[H_2\right]\times p\left[I_2\right]}$

• $Kp\ =\ \frac{p\left[HI\right]^2}{p\left[H\right]\times p\left[I\right]}$

Multiple Choice

Use the K_p expression to work out the units for the equilibrium constant for the equation below:

H₂(g) + I₂(g) → 2HI(g)

• kPa²

• kPa

• no unit

Multiple Choice

The following reaction :

SO_{2 (g)} + NO_{2 (g)} <=> SO_{3 (g)} + NO (g)

had reached a state of equilibrium, was found to contain

0.40 mol L^-1 SO₃ , and 0.30 mol L^-1 NO,

0.15 mol L^-1NO₂ , and 0.20 mol L^-1 SO₂.

Calculate the equilibrium constant for this reaction.

• 4

• 0.42

• 0.25

• 1

Multiple Choice

The Haber process is the industrial process for making...

• Margarine

• Ammonia

• Sulphuric acid

What conditions are used for the Haber process?

• 400 - 450℃

• 200 atm

• The iron catalyst used in the Haber process has no effect on the position of equilibrium. Instead, it increases the rate at which the equilibrium is established.

Multiple Choice

How is nitrogen obtained for use in the Haber process?

• By the filtration of liquid air

• By the fractional distillation of liquid air

Haber Process

• The Haber process produces ammonia from nitrogen and hydrogen: N₂ (g) + 3H₂ (g) ⇋ 2NH₃ (g).

  Nitrogen is obtained from the air and hydrogen from natural gas .

• Once the equilibrium is established, gases leaving the reactor are cooled in order to liquify the ammonia and separate it from the unreacted nitrogen and hydrogen.

• The unreacted nitrogen and hydrogen is recycled back into the reactor.

What temperature would you expect to

be used for the Haber process and why it is compromised when used in industry?

• The forward reaction in the Haber process is exothermic (ΔH = -92 kJ mol^-1).

• Using Le Chatelier’s principle, a low temperature would be favoured in order to shift the position of equilibrium to the right.

• However, a relatively higher temperature (400 - 450°C) is used to increase the rate of reaction . This temperature is a compromise.

What pressure would you expect to be used for the Haber process and why it is compromised in industry?

• There are more molecules on the left side of the equation, suggesting a high pressure should be used (according to Le Chatelier’s principle) in order to shift the position of equilibrium to the right.

• However, high pressures are expensive to maintain and they have safety risks, so a lower pressure of 200 atm is used.

Multiple Select

Ammonia is used for:

• cleaning agent

• fertiliser

• explosive production

• soap production

Multiple Select

What would happen if the temperature of a Haber process is reduced to below 450\degree C

450°C

? (Choose 2)

• Less ammonia would be produced

• More ammonia would be produced

• The reaction would be slower

• The reaction would be faster

Multiple Choice

Why isn't surface area a factor of the Haber Process?

• Only liquids have surface area

• N and H are gases

• Nitrogen is solidified into Ammonia

• Hydrogen is melted into Ammonia

Multiple Choice

The Haber process involves an exothermic reaction. What is the effect on the yield when the temperature is increased?

• More ammonia is produced

• Less ammonia is produced

• No change in ammonia produced

Multiple Choice

Explain why only a moderately high temperature is used in industry for the production of Ammonia.

• Produces a higher % yield but at a slower rate

• Produces a higher % yield but at a faster rate

• Produces a lower % yield but at a faster rate

• Produces a lower % yield but at a slower rate

What is the Contact process?

• Sulfur dioxide - heat sulfur in air

• Oxygen - air

What conditions are used for the Contact process?

• 400 - 450℃

• 1 - 2 atm

• V₂O₅ catalyst used in the Contact process has no effect on the position of equilibrium. Instead, it increases the rate at which the equilibrium is established.

Several stages to produce sulfuric acid.

• How sulfur dioxide is made: sulfur or sulfur ores (e.g. FeS₂) are heated in excess air: S(s) + O₂ (g) → SO₂ (g)

• Sulfur dioxide to sulfur trioxide: 2SO₂ (g) + O₂ (g) ⇋ 2SO₃ (g)

• Sulfur trioxide to concentrated sulfuric acid: sulfur trioxide is dissolved in concentrated sulfuric acid (as adding it to water would create a fog of sulfuric acid).

  H₂SO₄ (l) + SO₃ (g) → H₂S₂O₇ (l)

• The product (oleum) is then dissolved in water: H₂S₂O₇ (l) + H₂O(l) → 2H₂SO₄ (l)

What temperature would you expect to be used for the Contact process and why it is compromised in industry?

• The forward reaction is exothermic so a low temperature would give the greatest yield.

• However, a low temperature results in a slow rate of reaction and so a higher temperature may be used to strike a balance between yield and rate.

What pressure would you expect to be used for the Contact process and why it is compromised in industry?

• According to Le Chatelier’s principle, a high pressure would give the greatest yield of sulfur trioxide.

• However, even at pressures close to atmospheric pressure, 99.5% of SO₂ is converted into SO₃ so increasing the pressure would only see a minute improvement in yield that wouldn’t be economically worthwhile.

Multiple Choice

In the Contact process, what happens to the equilibrium when the pressure is increased?

• Shifts to the left

• Shifts to the right

• No change in equilibrium

Multiple Select

What are the uses of sulfuric acid? Choose the correct answers.

• In the manufacture of detergents and fertilisers

• As a bleach

• As a battery acid

• As food preservative

Multiple Select

2SO_{2 (g)} + O_{2 (g)} ⇌ 2SO_{3 (g)}

The equation above refers to the 2nd stage of Contact Process.

What are the optimum conditions to favour the yield of sulfur trioxide?

• Low Pressure

• High Pressure

• Low Temperature

• High Temperature

• Iron(II) oxide as catalyst

Poll

Can you complete this activity within 2 hours?