Activity 1

Write the name of a reaction in which all reactants and products are in the same phase (1) [1]

Solution

• Homogeneous ✓ [1]

Activity 2

Consider the following reversible reaction that is in equilibrium in a closed system (the symbols represent chemicals which are unnamed; if you see symbols like X or Y which aren’t on the Periodic Table, it often means you’ve got to work out what kinds of substances they are).

Step 4/5:
∆H < 0 

• the forward reaction is exothermic
• the reverse reaction is endothermic

In each of the following cases:

• Explain how the equilibrium will be affected (which reaction will be favoured)
• State how the concentration of the product A₂(g) will change.

CASE 1 The reaction system is heated. (3)   [3]

Solution
Case 1.
If the reaction system is heated, the temperature of the system increases.
According to Le Châtelier’s Principle, increasing the temperature of the reaction system (i.e. heating it) favours the reaction that will decrease the temperature of the system 

• ∴favours the endothermic ✓ reaction but ∆H < 0
• ∴the reverse reaction is endothermic
• ∴the reverse reaction is favoured✓∴the 3A₂(g) + BX₂Y₂(s) reacts faster than it is produced
• ∴the concentration of the A₂(g) decreases. ✓

(Remember: The opposite is true if the reaction vessel is cooled.)  [3]
hint
You must be able to use Le Châtelier’s Principle to identify the factor that will increase the yield of the products and what influence that factor will have on the reaction AT EQUILIBRIUM

CASE 2 Y₂(g) is added to the vessel at constant pressure and temperature  (3)  [3]
Solution
Case 2.
Adding Y₂(g) increases the concentration of the Y₂(g). ✓
According to Le Châtelier’s Principle, increasing the concentrationof Y₂(g) favours the reaction that decreases the concentration of Y₂(g)

• ∴the reaction that uses of Y2(g) as a reactant
• the forward reaction is favoured. ✓
• ∴the rate at which A₂(g) and BX₂Y₂(s) is produced, increases
• the concentration of the A2(g) increases. ✓  [3]

CASE 3 A₂(g) is removed from the vessel at constant pressure and temperature. (3)  [3]
Solution
Case 3.
Removing A₂(g) decreases the concentration of the A₂(g). ✓
According to Le Châtelier’s Principle, decreasing theconcentration of the A₂(g) favours the reaction that increases the concentration of the A₂(g);

• the reaction that makes A2(g) as a product
  ∴ the forward reaction is favoured. ✓
  the rate at which A₂(g) and BX₂Y₂(s) is produced, increases
  the concentration of the A₂(g) increases. ✓ [3]

CASE 4 Some of the BX₂Y₂(s) is removed from the system at  constant pressure and temperature. (3) [3]
Solution
Case 4.
Removing BX₂Y₂(s) has no effect on the equilibrium as it is a solid. ✓
Removing a solid from an equilibrium system does not disturb the concentration. ✓ (Solids in equilibrium reactions have very little change in mass, hence they are ignored)
The concentration of A₂(g) remain the same. ✓ [3]

CASE 5 The pressure on the system is increased (or the volume is decreased) (3)  [3]
Solution
Case 5.
According to Le Châtelier’s Principle, increasing the pressure (or decreasing the volume) of a gaseous system favours the reaction that decreases the pressure of the system by decreasing the total number of gaseous moles in the system. ✓

• the forward reaction is favoured. ✓
• the rate at which A₂(g) and BX₂Y₂(s) is produced, increases
• the concentration of the A₂(g) increases. ✓ [3]

hint
There are 4 gaseous moles on the left, and 3 gaseous moles on the right.

Activity 3

Say whether the statement is TRUE or FALSE: If the equilibrium constant for the reaction A₂(g) + B₂(g) ⇌ 2AB(g) is equal to K, then the equilibrium constant for the reverse reaction
2AB(g) ⇌ A₂(g) + B₂(g) is also equal to K. (3)   [3]

Solution
FALSE✓✓ . ... for the reverse reaction 2AB(g) ⇌ A₂(g) + B₂(g) is equal to ¹/_K  ✓✓  OR ... less (smaller) than K [3]

Activity 4

The industrial preparation of hydrogen gas is represented by the equation below:

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)           ∆H > 0

The reaction reaches equilibrium at 1 000 °C in a closed container.

1. State Le Châtelier’s principle. (2)
2. How will an increase in pressure at 1 000 °C (by decreasing the volume) affect the yield of hydrogen gas? Write down only INCREASES, DECREASES OR NO EFFECT. Explain the answer. (2)
3. Give TWO reasons why high temperatures are used for this reaction. (2)  [6]

hint
There are various ways to state Le Châtelier’s Principle correctly.

Solution

1. When the equilibrium in a closed system is disturbed the system will shift the equilibrium position (OR re-instate a new equilibrium)✓ as to favour the reaction that will oppose OR cancel OR counteract the change OR disturbance. ✓
   OR
   When a stress / change is placed on a system in equilibrium the system shifts the equilibrium ( position) OR re-instate a new equilibrium ✓ so as to remove OR cancel OR oppose the stress / change. ✓ (2)
   OR
   When the conditions affecting an equilibrium are changed, the equilibrium (position) shifts in such a way ✓ as to oppose the change OR cancel the change. ✓ (any two)
2. Decreases ✓ When the pressure is increased, the reverse reaction is favoured.✓
   OR
   The reaction that produced the smaller volume/amount of gas is favoured. ✓
   OR
   4 mol or volumes of gas produces ✓ 2 mol or volumes of gas. ✓ (2) (any two)
3. Products form at faster rate. ✓
   Higher yield of products. ✓ (2)  [6]

hint
In physical chemistry, saturation is the point at which a solution of a substance can dissolve no more of that substance. Additional amounts of the solute will appear as a separate phase (usually as a precipitate solid). This point of maximum concentration, the saturation point, depends on the temperature and pressure of the solution as well as the chemical nature of the substances involved.

Activity 5

1. The following hypothetical reaction reaches equilibrium in a closed container at a certain temperature:
X₂(g) + Y₂(g) ⇋ 2XY(g) ∆H < 0
Which ONE of the following changes will increase the AMOUNT of XY(g)?

1. Decrease in temperature
2. Increase in temperature
3. Increase in pressure
4. D. Decrease in pressure (2)

2. The equation below represents a chemical reaction at equilibrium in a closed container.
H₂(g) + I₂(g) ⇌ 2HI(g) ∆H < 0
Which ONE of the following changes will increase the yield of HI(g) in the above reaction?

1. Increase the temperature
2. Decrease the temperature
3. Increase the pressure by decreasing the volume
4. Decrease the pressure by increasing the volume (2)

3. A chemical reaction reaches equilibrium. Which ONE of the following statements regarding this equilibrium is TRUE?

1. The concentrations of the individual reactants and products are constant.
2. The concentrations of the individual reactants and products are equal.
3. The concentrations of the individual reactants are zero.
4. The concentrations of the individual products increase until the reaction stops. (2)

4. The reaction represented by the equation below reaches equilibrium.
2CrO₂⁴⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₂⁷⁻(aq) + H₂O(ℓ)
yellow                                    orange
Which ONE of the following changes to the reaction mixture will change its colour from yellow to orange?

1. Add a catalyst.
2. Add water to the reaction mixture.
3. Add a few drops of sodium hydroxide solution to the reaction mixture.
4. Add a few drops of concentrated hydrochloric acid to the reaction mixture. (2)

5. Give one word for the following phrase: 
The stage reached in a reversible chemical reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. (2)  [10]

Solution
1. A ✓✓ (2)
2. B✓✓ (2)
3. A ✓✓(2)
4. D ✓✓ (2)
5. (Dynamic/Chemical) equilibrium ✓✓ (2)  [10]

NB
Solids and pure liquids are omitted from the K_c expression as their concentration is [1], as multiplying by 1 has no effect.

Activity 6

Write K_c expressions for each of the following reactions:

1. N₂(g) + 3H₂(g) ⇋ 2NH₃(g)  (1)
2. CaCO₃(s) ⇋ CaO(s) + CO₂(g)   (1)
3. P₄(s) + 6Cℓ₂(g) ⇋ 4PCℓ₃(ℓ)  (1)    [3]

Solution

1. K_c =     [NH₃]³     
             [N₂]².[H₂]³
2. K_c =  [CO₂]
3. K_c =     1       
           [CL₂]⁶

Activity 7

The thermal decomposition of calcium carbonate (CaCO ₃) is an example of a heterogeneous equilibrium. The decomposition that takes place in a closed container can be represented by the following equation:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Initially 5 g of CaCO₃(s) is placed in a closed 500 cm³ container and then heated. Equilibrium is reached at 900°C.

1. Why is the above decomposition referred to as a heterogeneous equilibrium? (1)
2. Calculate the mass of unreacted CaCO₃(s) that remains in the container at equilibrium if Kc for the reaction is 0,0108 at 900°C. (19)
3. It is found that the value of Kc increases when the container is heated. Is the forward reaction exothermic or endothermic? Use Le Châtelier’s principle to explain your answer. (5)
4. The volume of the container is now decreased to 250 cm³ while the temperature is kept constant. How will each of the following be affected? Give a reason for your answer.
   4.1 The value of K_c.
   4.2 The number of mol CaCO₃(s) present in the equilibrium mixture. (10)
   4.3 The concentration of CO2(g) at the new equilibrium.
5. More CaCO₃(s) is now added to the equilibrium mixture in the 500 cm³ container. How will this change influence the number of moles of CO₂(g) in the equilibrium mixture? Give a reason for your answer. (2)  [37]

• hint; Refer back to the definition of a heterogeneous equilibrium

Solutions

1. Reactants and products are in different phases. 3 (1)
2. This answer takes six steps. Answers for questions 3.3 to 3.5 appear below these steps.

Step 1:
Convert the mass of the reactant to moles. ONLY moles may be used in the table.

• M(CaCO₃) = 40 + 12 + 3(16) = 100 g·mol^–1 
  n = m = 5   = 0,05 mol
        M  100

Step 2:
Write the expression for Kc and substitute all the known values – remember that the value of Kc was given.

• Kc = [CO₂]
  0,0108 = [CO₂]
  ∴ [CO₂] = 0,0108 mol·dm^–3 

Step 3:
Use the calculated CO₂(g) concentration and the formula for concentration (c = ⁿ/_V ) to calculate the number of moles of CO₂ in the equilibrium mixture. The volume of the container (500 cm³) was given. Convert the volume to dm³. Show all working.

• c = ​ n     ∴   0,0108  =  n   ∴  n = (0,5)(0,0108) = 0,0054 mol CO₂
         V                         0,5

Step 4:
Find the column with 2 values. Complete the mol used or produced line for this compound. (0 mol CO₂ initially + produced mol CO₂ = 0,0054 mol
∴ mol CO₂ produced = 0,0054 mol)

Step 5:
From the balanced equation ratio 1 : 1 : 1 determine the ratio of the moles used and produced. Therefore 1 : 1 : 1 becomes 0,0054 (from Step 3): 0,0054 : 0,0054 

Step 6
For the reactant CaCO₃ ... initial mol – mol used = final mol at equilibrium 
∴0,05 mol – 0,0054 = final mol at equilibrium = 0,0046 mol

∴0,0046 mol CaCO₃ remains unreacted at equilibrium
M(CaCO₃) = 40 + 12 + 3(16) = 100 g·mol^–1

• n = ​ m      ∴   0,0046  =  m   ∴  m = (0,0046)(10) = 0,46 g
         M                         100
  ∴ 0,46g CaCO₃  remains unreacted. (19)

Hint : Mass was asked, so convert the mol calculated to mass n = ​ m      
                                                                                                          M

3. Endothermic. ✓

• Kc increases ∴ the forward reaction or reaction to the right is favoured. ✓✓
• According to Le Châtelier’s principle, an increase in temperature ✓ the forward reaction or will favour the endothermic reaction reaction to the right must be endothermic. ✓ (5)

4.1 Remains the same. Only temperature affects the value of Kc.
4.2 Increases. ✓✓

• Decreasing the volume increases the pressure. According to Le Châtelier’s Principle, when the pressure is increased, the reaction which decreases the pressure by producing fewer moles of gas products, is favoured. ✓
  As 0 mol (g) ∴ 1 mol (g), the reverse reaction is favoured. ✓
• 4.3 Remains the same. ✓✓
  
  K_c = [CO₂]. The temperature remains constant ∴ K_c remains constant ∴ the concentration of CO₂ will remain constant. ✓ (10)

5. Remains the same. Adding a solid does not affect the equilibrium, it does not change the concentration of a reagent. ✓✓ (2)  [37]

Activity 8

 NB:

• When∆H > 0 it means the reaction to the right (the forward reaction) is endothermic
• When∆H < 0 the forward reaction is exothermic

hint

• These are the initial concentrations – before the new equilibrium is established. V = 1 dm₃ (given)
  ∴the values of the initial concentrations are equal to the initial number of moles.
  c= n  and c = n 
       V              1
  ∴ n = c(1)

The equation below represents an equilibrium reaction in a sealed 1dm³ container. 
NO₂(g) + NO(g) ⇋ N₂O(g) + O₂(g)                         ΔH > 0
Equilibrium was reached at a certain temperature and the value of K_c was 3,93. The concentration (in mol·dm–3) of each reactant and product in the container at equilibrium was:

• [NO₂] = 0,06 mol·dm^–3
• [NO] = 0,29 mol·dm^–3
• [N₂O] = 0,18 mol·dm^–3
• [O₂] = 0,38 mol·dm^–3

One of the conditions affecting the equilibrium is changed and a new equilibrium is established. At the new equilibrium, the concentration of the NO₂(g) is 0,12mol·dm^–3.

1. Calculate the Kc value at the new equilibrium. (11)
2. Which condition, concentration or temperature, was changed? (1)
3. Give an explanation for the answer to Question 4.2. (1)  [13]

Solutions

1. NO₂(g) + NO(g) + Energy ⇌ N₂O(g) + O₂(g)
   ∆ [NO₂] = [NO₂]_final − [NO₂]_initial = (0,12 – 0,06) = 0,06 mol NO₂ is formed. but NO₂ is the product of the reverse reaction
   ∴ the reverse reaction was favoured. 

   From the balanced reaction equation, 1 NO₂ : 1 NO : 1 N₂O : 1 O₂ 
   ∴0,06 mol N₂O + 0,06 mol O₂ → 0,06 mol NO₂ + 0,06 mol NO
    So at equilibrium:
   

2. Temperature.

3. K_c was initially 3,93 (given) now it is 0,91. If the K_c value changed, the temperature must have changed – only a change in temperature can change the value of K_c.  (1)

Activity 9

Consider the following reaction equation for the production of ammonia in the Haber process.
NB

• The Haber Process is the name of the chemical process to manufacture NH₃

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)     ∆H < 0
In each of the following cases:

1. State how Kc changes and
2. Briefly explain your answer.

1. The concentration of the nitrogen gas is increased. (1)
2. The pressure in the reaction system is decreased. (1)
3. FeO is added to the reaction system. (1)
4. The reaction system is heated. (5)   [8]

Solutions

1. Remains constant because... ✓ (1)
2. ... K_c is only affected by a change in temperature. ✓ (1)
3. K_c Decreases ✓ (1)
4. Heating the reaction system favours the endothermic, reverse reaction ✓ which decreases the temperature ✓ of the system ∴ the concentration of the forward product, [NH₃] ✓ decreases while the concentrations of the forward reactants, [N₂] and [H₂] increase ✓ ∴ K_c decreases. ✓ (5)  [8]

Activity 10

The reaction represented by the equation below reaches equilibrium.

Co(H₂O)₆²⁺(aq) + 4Cℓ–(aq)      ⇋    CoCℓ₄^2–(aq) + 6H₂O(ℓ)     ΔH>0
pink                                                   blue
Which ONE of the following changes to the reaction mixture will changes its colour from blue to pink?

1. Add a catalyst
2. Place the reaction mixture in a container with hot water.
3. Add a few drops of concentrated hydrochloric acid to the reaction mixture.
4. D. Add water to the reaction mixture. (2)  [2]

Solution
1. D [2]

Activity 11

0,25 mol of A and 0,15 mol of B are introduced into a 1 dm3 vessel. A and B react and reach an equilibrium which can be represented by the following equation:

hint: Make use of the graph to read off the values
The graph shows the number of moles of A(g), B(g) and AB(g) vs time, under varying conditions.

1. What are the concentrations of A(g), B(g) and AB(g) at t = 20 s? (3)
2. How can we be sure that the system has reached equilibrium at t = 20 s? (1)
3. 3. Give an explanation for the change that occurs at t = 25 s. (1)
4. Explain the change in concentrations of A(g) and B(g) between t = 25 s and t = 35 s. (3)
5. How does the equilibrium constant at t = 35 s compare with the equilibrium constant at t = 25 s? (No calculation is required.) (1)
6. Assuming that the volume was kept constant, what was done at t = 40 s that produced the change as shown? Explain your answer. (5)
7. Do any reactions occur during the interval t = 30 s and t = 40 s? Briefly explain your answer. (4)  [18]

Solutions

1. [A] = 0,15 mol·dm^–3 [B] = 0,05 mol·dm^–3 [AB] = 0,1 mol·dm^–3 (3)
2. The number of mol of A, B and AB remains constant from 16 s to 25 s. ✓(1)
3. The amount of B is increased by adding B to the system, thus increasing the [B]. ✓(1)
4. With the addition of B, [B] increases and the forward reaction is favoured ✓ according to Le Châtelier’s principle, so that some of B reacts with A to form more AB. Hence the [A] and [B] 3 decrease and [AB] increases. ✓ (3)
5. K_c remains constant provided the temperature remains constant. ✓ (1)
6. The temperature was lowered. ✓ It was not a concentration change, because there are no peaks at 25 s ✓. Therefore it must be a temperature change. From the graph, it can be seen that the forward reaction was favoured ✓ because there is an increase in the number of moles of AB, and a decrease in the number of moles of A and B when establishing the new equilibrium. ✓ According to the information given, the forward reaction is exothermic. ✓ According to Le Châtelier’s Principle, a decrease in temperature favours the exothermic reaction. (5)
7. Yes, but since equilibrium has been reached, ✓ the forward and reverse reactions take place at the same time ✓ and at the same ✓ rate, so that no changes occur in the concentrations of A, B or AB. ✓ (4)   [18]

Activity 12

Nitrogen and oxygen gases react in a sealed container according to the following equation:
O₂(g) + N₂(g)  ⇋ 2NO(g)
After the reaction reaches equilibrium, certain changes are made. The following graph of rate of reaction versus time represents the situation.

1. Write the equation for the reaction represented by the dashed line on this graph. (2)
2. What is represented by the section of the graph between the 10th and 15th minute? Explain your answer. (2)
3. A temperature change takes place at t = 15 minutes.
   3.1 Was the temperature increased or decreased at t = 15 minutes? (1)
   3.2 Explain whether the forward reaction is exothermic or endothermic. (3)
   3.3 What effect does this temperature change have on the equilibrium constant (K_c)? Explain your answer. (3)
4. A pressure change is introduced at t = 20 minutes.
   4.1 Was the pressure increased or decreased? (3)
   4.2 Explain how this change in pressure affects the amounts of each gas at equilibrium. (2)  [16]

Solutions

1. 2NO(g) → N₂(g) + O₂(g)   ✓✓ (2)
2. The system is in equilibrium ✓ because the rates of the forward and reverse reactions are equal. ✓ (2)
3. 3.1 Decreased. ✓ (1)
   3.2 The rates of both the forward and the reverse reactions decrease ✓, but the rate of the reverse reaction decreases more. ✓ The rate of the endothermic reaction will always decrease the most when a reaction mixture is cooled. Thus the reverse reaction is endothermic. That means that the forward reaction must be exothermic and was favoured by the decrease in temperature. ✓
   O₂(g) + N₂(g) ⇌ 2NO(g) + Energy (3)
   3.3 Increases. ✓
   Since Kc =      [NO]²   
                        [N₂][O₂]  (3)
   and the forward reaction is favoured  ✓ which causes an increase in [NO], the value of Kc increases. ✓  (3)
4. 4.1 Increased. ✓ The pressure is probably caused by a volume decrease. This results in the concentrations (c = ⁿ/_V) of the reactants, as well as the products, to increase ✓ , which in turn causes an equal increase in the forward and reverse reaction rates. ✓ (3)
   4.2 The amount (n) of reactants and products does not change, ✓ since 2 moles of gaseous reactant molecules ⇌ 2 moles of gaseous product molecules. ✓   (2)    [16]

Activity 13

A fertiliser company produces ammonia on a large scale at a temperature of 450°C. The balanced equation below represents the reaction that takes place in a sealed container.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)                ∆H < 0
To meet an increased demand for fertiliser, the management of the company instructs their engineer to make the necessary adjustments to increase the yield of ammonia. In a trial run on a small scale in the laboratory, the engineer makes adjustments to the TEMPERATURE, PRESSURE and CONCENTRATION of the equilibrium mixture. The graphs below represent the results obtained.

1. Identify the changes made to the equilibrium mixture at each of the following times:
   1.1 t₁ (2)
   1.2 t₂ (2)
   1.3 t₃ (2)
2. At which of the above time(s) did the change made to the reaction mixture lead to a higher yield of ammonia? Write down only t₁ and/or t₂ and/or t₃ (2)
3. The engineer now injects 5 mol N₂ and 5 mol H₂ into a 5 dm₃ sealed empty container. Equilibrium is reached at 450 °C. Upon analysis of the equilibrium mixture, he finds that the mass of NH₃ is 20,4 g. Calculate the value of the equilibrium constant (Kc) at 450 °C (9) [17]

Solutions

1. 1.1 The concentration of nitrogen ✓ is increased ✓ / More Nitrogen ✓ added ✓ (any two) (2)
   1.2 The pressure ✓ on the system is increased ✓ (2)
   1.3 The temperature ✓ is increased ✓ (2)
2.  t₁ 3and t₂ ✓ (2)
3. Pay special attention:
   This is how marks are allocated to the answer of this Kc calculation:
   • Use of n = ^m/_M ✓
   • n(NH₃) at equilibrium = 1,2 mol ✓
   • Using ratio n(N₂) : n(H₂) : n(NH₃) = 1 : 3 : 2 ✓
   • n(N₂) at equilibrium (initial - change) ✓
   • n(H₂) at equilibrium (initial - change) ✓
   • divide by volume (calculation of concentration) ✓
   • Kc expression ✓
   • Substitution into Kc expression ✓
   • Final answer ✓ (9)

  [17]

Activity 14

2 mol of NO₂(g) and an unknown amount of N₂O₄(g) are sealed in a 2 dm³ container, that is fitted with a plunger, at a certain temperature. The following reaction takes place:
2NO₂(g) ⇌ N₂O₄(g)
At equilibrium it is found that the NO₂ concentration is 0,4 mol·dm^–3. The equilibrium constant at this temperature is 2.

1. Calculate the initial amount (in mol) of N₂O₄(g) that was sealed in the container. (8)
   The plunger is now pushed into the container causing the pressure of the enclosed gas to increase by decreasing the volume
2. How will this change influence the amount of nitrogen dioxide at equilibrium? Only write down INCREASES, DECREASES or REMAINS THE SAME. (1)
3. Use Le Châtelier’s principle to explain your answer to QUESTION 2.1.2. (2)   [11]

Solutions
1. 

• Kc =   [N₂O₄]   
            [NO₂]²
  2 = (x + 0,6)2 
            (0,4)²
  x = 0,004 mol

2. Decreases  (1)
3. When the pressure is increased the system will try to decrease the pressure.   The forward reaction (2 mol to 1 mol) is favoured.  (2)    [11]

Activity 15

Name the industrial process for the production of ammonia (1)  [1]

Solution

• Haber (Bosch) process  [1]

Activity 16

Sulphuric acid is an important substance used in the manufacture of fertilisers. The equation below represents one of the steps in the industrial preparation of sulphuric acid.
2SO₂(g) + O₂(g) ⇋ 2SO₃(g)              ∆H < 0

1. Write down the name of the process used to prepare sulphuric acid in industry. (1)
2. Write down the NAME or FORMULA of the catalyst used in the process in question (1)
3. Is the forward reaction exothermic or endothermic? Give a reason for the answer. The reaction reaches equilibrium at a certain temperature in a 2 dm3 closed container. On analysis of the equilibrium mixture, it is found that 0,6 mol of SO₂(g), 0,5 mol of O₂(g) and 0,4 mol of SO₃(g) are present in the container. (1)
4. List THREE changes that can be made to this equilibrium to increase the yield of SO₃(g) (3)
5. The temperature is NOW increased and the reaction is allowed to reach equilibrium for the second time at the new temperature. On analysis of this new equilibrium mixture, it is found that 0,2 mol of for this reaction at the new temperature. (8)  [14]

Solutions

1. Contact process ✓ (1)
2. V₂O₅ (vanadium pentoxide) ✓ (1)
3. Exothermic∆H < 0 ✓ (1)
4. Any three: ✓✓ ✓ 
   Decrease temperature
   Increase pressure
   Increase concentration of both/any one of the reactants
   Remove SO₃ continuously (3)
5.                                 

   	SO2	O2	SO3
   Molar ratio	2	1	2
   Initial quantity (mol)	0,6	0,5	0,4
   Change (mol)	0,2	0,1	0,2
   Quantity at equilibrium	0,8	0,6	0,2
   Concentration (mol·dm–3)	0,4	0,3	0,1

   Kc =     [SO₃]²     =     (0,1)²    = 0,21
           [SO₂]²[O₂]     (0,4)²(0,3)

NB 
This how marks are allocated to the answer of this Kc calculation:

• Change in n(SO₃) = 0,2 mol ✓
• Ratio n(SO₂) : n(O₂) : n(SO₃) = 2 : 1 : 2 ✓
• n(SO₂) at equilibrium = initial + change ✓
• n(O₂) at equilibrium = initial + change ✓
• divide three equilibrium amounts by 2 (calculation of concentration) ✓
• K_c expression ✓
• Substitution into K_c expression ✓
• Final answer ✓ (8)   [14] 

Activity 17

The reaction below represents the catalysed step in the contact process:
2SO₂(g) + O₂(g) ⇋ 2SO₃(g)            ∆H <0

The reaction takes place in a closed container and reaches equilibrium at 427°C. How will a HIGHER temperature affect each of the following? Write down only INCREASES, DECREASES or REMAINS THE SAME.

1. The rate of production of SO₃(g) (2)
2. The yield of SO₃(g) (2)
3. The reaction is investigated on a small scale in the laboratory. Initially 4 mol of SO₂(g) and an unknown mass, x, of O₂(g) are sealed in a 2 dm^–3 flask and allowed to reach equilibrium at a certain temperature. At equilibrium it is found that the concentration of SO₃(g) present in the flask is 1,5 mol·dm^–3. Calculate the mass of O₂(g) initially present in the flask if the equilibrium constant (K_c) at this temperature is 4,5. (8) [12]

Solutions

1. Increases ✓✓ (2)
2. Decreases ✓✓ (2)
3. 

  (8)  [12]