Question 6.2.3: Additional Answer Options

Marking criteria for option 2:

• ⁽¹⁾ Initial n(P) and n(Q2) and n(PQ) from table.
• ⁽²⁾ Change in n(P) = equilibrium n(P) – initial n(P).
• ⁽³⁾ USING ratio: P : Q2 : PQ = 2 : 1 : 2.
• ⁽⁴⁾ Equilibrium n(Q2) = initial n(Q2) + change in n(Q2) and Equilibrium n(PQ) = initial n(PQ) - change in n(PQ).
• ⁽⁵⁾ Divide equilibrium amounts of P and Q₂ and PQ by 2 dm³.
• ⁽⁶⁾ Correct Kc expression (formulae in square brackets).
• ⁽⁷⁾ Substitution of equilibrium concentrations into Kc expression.
• ⁽⁸⁾ Final answer: 10,889.

Marking criteria option 3:

• ⁽¹⁾ Initial n(P) = 4 mol and n(Q₂) = 2,4 mol and n(PQ) = 0.
• ⁽²⁾ Change in n(P) = equilibrium n(P) – initial n(P) = 2,8 mol.
• ⁽³⁾ USING ratio: P : Q2 : PQ = 2 : 1 : 2.
• ⁽⁴⁾ Equilibrium n(Q₂) = initial n(Q₂) + change in n(Q₂) and Equilibrium n(PQ) = initial n(PQ) - change in n(PQ).
• ⁽⁵⁾ Divide equilibrium amounts of P and Q₂ and PQ by 2 dm³.
• ⁽⁶⁾ Correct Kc expression (formulae in square brackets).
• ⁽⁷⁾ Substitution of equilibrium concentrations into Kc expression.
• ⁽⁸⁾ Final answer: 10,889.

Marking criteria option 4:

• ⁽¹⁾ Initial c(P) and c(Q2) and c(PQ) from table.
• ⁽²⁾ Change in c(P) = equilibrium c(P) – initial c(P).
• ⁽³⁾ USING ratio: P : Q2 : PQ = 2 : 1 : 2.
• ⁽⁴⁾ Equilibrium n(Q₂) = initial n(Q₂) + change in n(Q₂) and Equilibrium n(PQ) = initial n(PQ) - change in n(PQ).
• ⁽⁵⁾ Divide initial amounts of P and Q₂ and PQ by 2 dm3
• ⁽⁶⁾ Correct Kc expression (formulae in square brackets).
• ⁽⁷⁾ Substitution of equilibrium concentrations into Kc expression.
• ⁽⁸⁾ Final answer: 10,889.

Or

Option 2

Reverse reaction:

Forward reaction:

Or

Option 3

Or

Option 4