Q22.49P - Chemistry: Molecular Nature Of Matter And Change

Question

Among the exothermic steps in the manufacture of sulfuric acid is the process of hydrating ${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}$. 

(a) Write two chemical reactions that show this process. 

(b) Why is the direct reaction of ${\mathbf{\text{SO}}}_{\mathbf{\text{3}}}$ with water not feasible?

Step-by-Step Solution

Answer

a) The two chemical reactions showing the process are, 

$$\begin{gathered} {\text{SO}}_{\text{3}}{\text{(g) + H}}_{\text{2}}\text{O(l)}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)} \\ {\text{SO}}_{\text{3}}{\text{(g) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}\to {\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}\text{(l).} \end{gathered}$$

b) The direct reaction of ${\text{SO}}_{\text{3}}$ with water is not feasible because the temperature at which the process occurs is too high.

Step 1: Concept Introduction

Sulfuric acid is created industrially by reacting water with sulphur trioxide which is made by combining sulphur dioxide and oxygen chemically using either the contact or chamber processes.

Step 2: Information Provided

a) The process of hydrating ${\text{SO}}_3$ can be described with the following equations:

$$\begin{gathered} {\mathrm{SO}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right) \\ {\mathrm{SO}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{H}}_2 {\mathrm{S}}_2{\mathrm{O}}_7\left(\mathrm{l}\right) \end{gathered}$$

Step 3: Information Provided

b)Because the temperature at which the process occurs is too high for a direct reaction of ${\text{SO}}_{\text{3}}$ with water, ${\text{SO}}_{\text{3}}$ would first meet water vapour, which catalyses its polymerization to  ${\text{(SO}}_{\text{3}}{\text{)}}_x$ and results in a smoke of solid particles that makes poor contact with water and yields little acid.

However, the previously formed ${\text{H}}_2{\text{SO}}_4$ absorbs ${\text{SO}}_{\text{3}}$ and forms ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}$ which is then hydrolyzed to give ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$:

${\mathrm{H}}_2 {\mathrm{S}}_2{\mathrm{O}}_7\left(\mathrm{l}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)$.