Problem 84
Question
A sulfuric acid plant produces a considerable amount of heat. This heat is used to generate electricity, which helps reduce operating costs. The synthesis of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) consists of three main chemical processes: (1) oxidation of \(\mathrm{S}\) to \(\mathrm{SO}_{2},(2)\) oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3},(3)\) the dissolving of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and its reaction with water to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If the third process produces \(130 \mathrm{~kJ} / \mathrm{mol}\), how much heat is produced in preparing a mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from a mole of \(\mathrm{S}\) ? How much heat is produced in preparing 5000 pounds of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)
Step-by-Step Solution
VerifiedKey Concepts
Thermochemistry
In any reaction, bonds are broken and new ones are formed, which involves changes in energy. In the context of sulfuric acid production:
- The oxidation of sulfur to sulfur dioxide (\(\text{SO}_2\)) and further to sulfur trioxide (\(\text{SO}_3\)) involves energy changes.
- When sulfur trioxide dissolves in sulfuric acid and reacts with water, it releases heat.
Chemical Processes
- Oxidation of sulfur (\(\text{S}\)) to sulfur dioxide (\(\text{SO}_2\)).
- Oxidation of sulfur dioxide to sulfur trioxide (\(\text{SO}_3\)).
- Dissolving of sulfur trioxide in existing sulfuric acid and then reacting it with water to yield more \(\text{H}_2\text{SO}_4\).
Molar Heat Calculation
- We know the third step in the \(\text{H}_2\text{SO}_4\) synthesis produces 130 kJ per mole.
- To find the total heat for large-scale production, such as 5000 pounds of sulfuric acid, we convert pounds to moles using molar mass.
- This involves converting pounds to grams, and dividing by the molar mass of sulfuric acid, which is approximately 98.086 g/mol.
- The final step is multiplying the number of moles by the energy per mole to find the total heat produced.