Consider the reaction represented by the equation: 2SO\(_{2(g)}\) + O\(_{2(g)}\) 2SO\(_{3(g)}\). \(\Delta\)H = 188KJ. (a) Write an expression for the equili...

Important correction: the equation gives \(\Delta H=+188\text{ kJ}\), which means that the forward reaction is endothermic. Therefore, the energy diagram and temperature effect below follow the equation as written. In the Contact Process, this reaction is more commonly written with \(\Delta H=-188\text{ kJ}\), meaning it is exothermic. If the intended sign was negative, the temperature effect would be the opposite.

(a) Equilibrium constant expression

For

\[2SO_{2(g)}+O_{2(g)}\rightleftharpoons2SO_{3(g)}\]

the concentration equilibrium constant is:

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

The powers are taken from the balanced equation. The formula must contain \(SO_2\), not \(SO\).

(b) Energy profile for the forward reaction

Because \(\Delta H\) is positive as stated, the products are at a higher energy level than the reactants. A catalyst, \(V_2O_5\), provides an alternative pathway with a lower activation energy. It does not change \(\Delta H\), the reactant energy, or the product energy.

(c) Effects on the position of equilibrium

1. Increase in temperature: equilibrium shifts to the right, producing more \(SO_3\). The forward reaction is endothermic as written, so it absorbs heat; increasing temperature favours the heat-absorbing direction.
   
   If the intended value was \(\Delta H=-188\text{ kJ}\), the forward reaction would be exothermic and increasing temperature would shift equilibrium to the left, decreasing the yield of \(SO_3\).
2. Increase in pressure: equilibrium shifts to the right. There are three moles of gaseous reactants, \(2SO_2+O_2\), but only two moles of gaseous product, \(2SO_3\). Higher pressure favours the side with fewer moles of gas.
3. Removal of some \(SO_3\): equilibrium shifts to the right. Removing a product lowers its concentration, so the forward reaction occurs to replace some of the removed \(SO_3\).
4. Presence of \(V_2O_5\): there is no change in the position of equilibrium. \(V_2O_5\) is a catalyst: it increases the rates of both forward and backward reactions and allows equilibrium to be reached faster, but it does not change the equilibrium composition.

(d)(i) Conversion of sulphur(VI) oxide to tetraoxosulphate(VI) acid in the Contact Process

Sulphur(VI) oxide, \(SO_3\), is absorbed in concentrated tetraoxosulphate(VI) acid to form oleum. Oleum is then diluted with water:

\[SO_{3(g)}+H_2SO_{4(l)}\rightarrow H_2S_2O_{7(l)}\]

\[H_2S_2O_{7(l)}+H_2O_{(l)}\rightarrow2H_2SO_{4(aq)}\]

\(H_2S_2O_7\) is oleum. Direct addition of \(SO_3\) to water is avoided because it forms a fine mist of acid.

(d)(ii) Two uses of tetraoxosulphate(VI) acid, \(H_2SO_4\)

• Manufacture of fertilisers, for example superphosphate and ammonium sulfate.
• Manufacture of detergents, paints, and pigments.

Examination reminder: a catalyst changes the activation energy and reaction rate, but it does not change \(\Delta H\), \(K_c\), or the position of equilibrium.