In the Contact process for the manufacture of sulfuric acid, the key stage is the reaction of sulfur dioxide and oxygen, in contact with a vanadium(V) oxide (V2O5) catalyst, to form sulfur trioxide - Leaving Cert Chemistry - Question 7 - 2016

1. Low temperature:

   • A low temperature favours the exothermic reaction in which sulfur trioxide is produced. Since the reaction releases heat, lowering the temperature will shift the equilibrium towards the products (2SO3), thereby increasing the yield.

   2. Explain:
   • High temperature would favour the endothermic reverse reaction which is not desirable in this case. Thus, to achieve maximum yield of SO3, lower temperatures are preferred.

1. High pressure:

   • Increasing pressure favours the side of the equilibrium with fewer moles of gas. In this case, the reactants have 3 moles (2SO2 + O2) and the products have 2 moles of SO3. Therefore, increasing the pressure will shift the equilibrium to the right, favouring the production of SO3.

   2. Explain:
   • Lower pressure would lead to a shift toward the side with more gas molecules, which would decrease the yield of sulfur trioxide.

A catalyst does not affect the position of equilibrium or the yield of products. Instead, it increases the rate at which equilibrium is reached by providing an alternative reaction pathway with a lower activation energy. Thus, while the presence of a catalyst will speed up the reaction process, it will not change the equilibrium position or the yield of SO3.

The equilibrium constant (Kc) expression for the reaction is given by: $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

1. Moles of Reactants:

   • Moles of SO2 = \frac{96 g}{32 g/mol} = 3 \text{ mol}
   • Moles of O2 = \frac{24 g}{32 g/mol} = 0.75 \text{ mol}

2. Initial Molar Concentrations:

   • SO2: ( [SO_2]_{initial} = \frac{3}{50} = 0.06 \text{ M} )
   • O2: ( [O_2]_{initial} = \frac{0.75}{50} = 0.015 \text{ M} )
   • SO3: At equilibrium, 112 g SO3 = \frac{112 g}{80 g/mol} = 1.4 \text{ mol} ; \Rightarrow ; [SO_3]_{equilibrium} = \frac{1.4}{50} = 0.028 \text{ M}

3. At Equilibrium Concentrations:

   • SO2: ( [SO_2]{equilibrium} = [SO_2]{initial} - x = 0.06 - 0.001 = 0.059 \text{ M} )
   • O2: ( [O_2]{equilibrium} = [O_2]{initial} - 0.001 = 0.014 \text{ M} )

4. Kc Calculation:

   • Substitute the equilibrium concentrations into the Kc expression: $K_c = \frac{(0.028)^2}{(0.059)^2(0.014)} = \frac{0.000784}{0.003481} = 224.4$

Thus, the value of Kc is approximately 224.4 under the given conditions.