(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow.
(ii) Write a balanced chemical equation for each of the processes I, II, III and IV
(iv) Using Le Chatelier's principle, explain briefly why increasing the temperature would not favour the reaction in II
Calculate the volume of unused oxygen gas when \(40\ \text{cm}^3\) of hydrogen gas is sparked with \(30\text{cm}^3\) of oxygen gas
(c) Calcium carbonate of mass 1.0 g was heated until there was no further change.
(a) The Contact process. The scheme reads: X(gas) + Y(solid) \(\xrightarrow{I}\) Sulphur(IV) oxide; Oxygen + Sulphur(IV) oxide \(\xrightarrow{II}\) Sulphur(VI) oxide \(\xrightarrow{III}\) Oleum \(\xrightarrow{IV}\) Concentrated H2SO4.
(i) X and Y. The first step burns a solid in a gas to give SO2, so X is oxygen (air) and Y is sulphur.
(ii) Balanced equations.
Process I: \[ S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)} \]
Process II: \[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \]
Process III (SO3 absorbed in concentrated H2SO4 to form oleum): \[ SO_{3(g)} + H_2SO_{4(l)} \rightarrow H_2S_2O_{7(l)} \]
Process IV (oleum diluted with water): \[ H_2S_2O_{7(l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4(l)} \]
(iii) Catalyst in process II: vanadium(V) oxide, V2O5 (platinum can also be used).
(iv) Le Chatelier's principle. Process II is exothermic in the forward direction. By Le Chatelier's principle, raising the temperature shifts the equilibrium position in the endothermic (backward) direction so as to absorb the added heat. This reduces the yield of SO3, so a high temperature does not favour the forward reaction; a moderate temperature (about 450 C) is used instead.
(v) Two uses of SO2:
- As a bleaching agent (for wood pulp, straw, silk).
- As a food preservative and fumigant/disinfectant (it also serves as the starting gas for making H2SO4).
(b) Volume of unused oxygen. From \(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\), H2 reacts with O2 in the ratio 2 : 1 by volume.
\[ \text{O}_2\ \text{needed} = \frac{1}{2} \times 40 = 20\ \text{cm}^3 \]
\[ \text{Unused O}_2 = 30 - 20 = \mathbf{10\ cm^3} \]
(c) Heating calcium trioxocarbonate(IV).
\[ CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} \]
Moles of CaCO3: molar mass \(= 40.0 + 12.0 + (3\times16.0) = 100\ g\,mol^{-1}\).
\[ n = \frac{1.0}{100} = 0.01\ \text{mol} \]
Mass of residue (CaO): molar mass of CaO \(= 40.0 + 16.0 = 56\ g\,mol^{-1}\).
\[ m_{CaO} = 0.01 \times 56 = \mathbf{0.56\ g} \]
Volume of CO2 at s.t.p.: \(n_{CO_2} = 0.01\ mol\).
\[ V = 0.01 \times 22.4 = 0.224\ dm^3 = \mathbf{224\ cm^3} \]
Volume at 15 C and 760 mmHg: pressure is unchanged (760 mmHg = s.t.p. pressure), so use \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\) with \(T_1 = 273\ K,\ T_2 = 273 + 15 = 288\ K\).
\[ V_2 = 0.224 \times \frac{288}{273} = \mathbf{0.236\ dm^3\ (\approx 236\ cm^3)} \]
