Answer parts (a)–(c) in a writing booklet - HSC - SSCE Chemistry - Question 32 - 2010 - Paper 1

When comparing the electrolysis of molten sodium chloride (NaCl) to that of aqueous sodium chloride, there are key differences in the products and reactions taking place.

1. Electrolysis of Molten Sodium Chloride:

   • In molten NaCl, the ions are free to move, resulting in the following reactions:

   • At the cathode (reduction):

     $Na^+ + e^- \rightarrow Na$

   • At the anode (oxidation):

     $2Cl^- \rightarrow Cl_2 + 2e^-$

   • Overall reaction:

     $2NaCl (l) \rightarrow 2Na (l) + Cl_2 (g)$

2. Electrolysis of Aqueous Sodium Chloride:

   • In aqueous NaCl, water also participates in the reactions, resulting in different products:

   • At the cathode (reduction):

     $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

   • At the anode (oxidation):

     $2Cl^- \rightarrow Cl_2 + 2e^-$

   • Overall reaction:

     $2NaCl (aq) + 2H_2O \rightarrow Cl_2 (g) + H_2 (g) + 2NaOH (aq)$

In summary, electrolysis of molten NaCl produces sodium and chlorine gas, while electrolysis of aqueous NaCl yields sodium hydroxide, chlorine gas, and hydrogen gas.

The equilibrium constant (K) for the reaction can be expressed in terms of the concentrations of the products and reactants.

If we consider the reaction:

$2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g)$

The equilibrium constant expression is given by:

$K = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

At time A, the concentrations can be calculated based on the moles and volume (10 L):

• For 0.80 moles of SO₂:

  $[SO_2] = \frac{0.80 \text{ moles}}{10 \text{ L}} = 0.08 ext{ M}$

• For 0.40 moles of O₂:

  $[O_2] = \frac{0.40 \text{ moles}}{10 \text{ L}} = 0.04 ext{ M}$

Assuming at time A, the equilibrium concentrations of SO₃ are at the maximum and calculated to be 0.20 M:

• Thus,

$K = \frac{(0.20)^2}{(0.08)^2(0.04)} = \frac{0.04}{0.00256} = 15.625$

A new equilibrium position at time B can be established due to several factors influencing the system.

1. Changes in Concentration: When the concentrations of the reactants or products change, the system responds by shifting the equilibrium position to restore balance, according to Le Chatelier's principle.

2. Temperature Changes: If the temperature of the system changes, the equilibrium will shift towards the endothermic or exothermic direction to minimize that change.

3. Volume Changes: Any alterations in the volume of the container affect the pressure, which in turn influences the direction of the equilibrium shift, particularly in gaseous reactions.

At time B, the concentrations of SO₂ and O₂ may have altered due to the formation of SO₃, leading to a new equilibrium being established as the system seeks to achieve stability again.