6.2.1: Additional Answer Options:

6.1

(The stage in a chemical reaction when the) rate of the forward reaction equals the rate of the reverse reaction.

Or

(The stage in a chemical reaction when the) concentrations of reactants and products remain constant.

Tips: 

• Remember to state that the rates of the reactions are equal not that the reactions are equal.
• Remember to state that the concentrations remain constant and not that the concentrations are equal. You also need to specify both reactants and products.

2 marks

(1 x 2 marks)

6.2.1

How to approach this question: 

• Calculate the number of moles in the 60,8 g of pure carbon dioxide.

Option 1

• One of the easiest ways to lay out your method is using a K_c table.

• Since there is no carbon monoxide in the reaction vessel, the original number of moles is 0.

• Only aqueous substances and gases are included in the K_c expression. Solids and pure liquids are left out of the expression since they do not technically have a concentration that changes.
• To get marks for the K_c expression, you need to ensure that you use square brackets since these indicate concentration.

There are more ways to answer this question. To view other options, click here.

7 marks

(7 x 1 mark)

Marking criteria:

• ⁽¹⁾ Substitute:

44 g∙mol^-1

• ⁽²⁾ Equilibrium concentration of CO₂ multiply by 3 dm³   and n(CO)_eq divide by 3 dm³
• ⁽³⁾ Use mole ratio:  1:2/n(CO) = 2n(CO₂)
• ⁽⁴⁾ n(CO₂)_change = n(CO₂)_initial - n(CO₂)_final n(CO)_eq = n(CO)_initial+ ∆n(CO)
• ⁽⁵⁾ Correct K_c expression (formulae in square brackets) 
• ⁽⁶⁾ Substitution of concentrations into K_c expression
• ⁽⁷⁾ Final answer: 12,24 (range: 11,85 – 12,66)

6.2.2

⁽¹⁾

⁽²⁾

⁽³⁾

Tips: 

• As K_C  does not include carbon, the only requirement is sufficient carbon available to react in the ratio with the other reactants.
• Since 1,22 mol of CO₂ reacted, according to the balanced equation C (s) + CO₂ (g) ⇌ 2CO₂ (g), there must be a minimum of 1,22 mol of carbon to react.
• The mass of the carbon is calculated using the molar mass of carbon.

3 marks

(3 x 1 mark)

Marking criteria:

• ⁽¹⁾ Use mol ratio: n(C) = n(CO₂)
• ⁽²⁾ Substitute: 12 g∙mol^-1
• ⁽³⁾ Final answer: 14,64 g

6.3.1

Remains the same.

Tip: adding carbon does not disturb the equilibrium since it is a solid.

1 mark

6.3.2

Decreases

• (When pressure is increased) the reaction that leads to the smaller amount/number of moles/volume of gas is favoured.
• The reverse reaction is favoured./More CO2 is formed.

Tips: 

• Changing the volume of a closed container containing gases disrupts its equilibrium.
• According to Le Chatelier's principle, the disturbance is an increase in pressure, therefore, the system will favour the reaction which decreases the pressure by producing fewer moles of gas.
• The balanced equation for the reaction is C (s) + CO₂ (g) ⇌ 2CO₂ (g).
• Since there is one mol of gaseous reactants and two moles of a gaseous product, the reverse reaction will be favoured as it produces fewer moles of gas and, therefore, will decrease the pressure.
• Remember to use the word favoured since an increase in pressure will increase both the forward and reverse reaction. This is because the particles collide more frequently when packed closer together however, the reverse reaction rate will increase more, therefore, it is favoured.
• A disturbance in an equilibrium favours a reaction, not a side. If you use the term 'side' instead of 'reaction', you are likely to lose the mark.

3 marks

(3 x 1 mark)

6.4.1

Endothermic

• When the temperature increases the mol/percentage CO(g) / product increases / forward reaction is favoured.
• An increase in temperature favours the endothermic reaction. 

Tips: 

• As the temperature increases, the percentage of CO₂ decreases (from 6,23 to 0,06) and the percentage of CO increases (from 93,77% to 99,94%). This indicates that the forward reaction is favoured because CO₂ is used and CO is made.
• An increase in temperature increases the rate of both the endothermic and exothermic reactions since the particles will have more kinetic energy. However, the rate of the endothermic reaction will increase more, therefore, the endothermic reaction is favoured.
• Since an increase in temperature favours the endothermic and the forward reaction, you can assume that the reactions are the same and, therefore, the forward reaction is endothermic.

3 marks

(3 x 1 mark)

6.4.2

How to approach this question:

• This was a particularly tricky question designed to differentiate the very top students from the top students.
• Try not to panic if you do not understand how to answer this question. It was difficult for a large number of teachers as well.

Option 1

Part 1

• Using the values for the equilibrium concentrations of both CO₂ and CO calculated in the K_c table for question 6.2.1, the amount of CO₂  at equilibrium is 0,162 mol and the amount of CO is 2,44 mol. Therefore, the total number of moles is 2,606 mol.

n_{total eq} = 0,162 + 2,44  ₍₁₎

= 2,606 moles

•  Now calculate the percentage that CO₂ and CO contribute towards the total number of moles in the container at equilibrium. You only need to calculate one of them to be able to determine the temperature.

Part 2

⁽²⁾

Or

⁽²⁾

Part 3

⁽³⁾

• The percentage of CO₂ is 6,23% and the percentage of CO is 93,63% at a temperature of 827°C.

Or

Option 2

• Using the concentrations at equilibrium.

Part 1

⁽¹⁾

Part 2

⁽²⁾

Or

⁽²⁾

Part 3

⁽³⁾

3 marks

(3 x 1 mark)

Marking criteria:

• ⁽¹⁾ Calculate the total volume/mol of gas at equilibrium: 0,162 + 2,44 = 2,606 mol∙dm^-3 or calculate the total concentration at equilibrium: 0,054 + 0,813 = 0,867 mol∙dm^-3
• ⁽²⁾ Calculate the percentage of any one gas (CO₂ or CO)
• ⁽³⁾ Final answer: T = 827 °C