Draw the uncharged product of the following reaction: SO3 H2SO4
The Correct Answer and Explanation is :
In this reaction, sulfur trioxide reacts with sulfuric acid to form oleum (H₂S₂O₇), a compound often used in the manufacturing of sulfuric acid.
Here’s a breakdown of the reaction and its product:
Reaction:
[
SO₃ + H₂SO₄ \rightarrow H₂S₂O₇
]
Explanation:
- Reactants:
- SO₃ (Sulfur Trioxide): Sulfur trioxide is a highly reactive compound that is an important industrial chemical, typically in the production of sulfuric acid. It is a gas at room temperature and dissolves exothermically in water.
- H₂SO₄ (Sulfuric Acid): Sulfuric acid is a strong acid and is commonly used in industrial applications. It has a concentrated form and is also used in the formation of oleum when mixed with sulfur trioxide.
- Reaction Mechanism:
- Sulfur trioxide (SO₃) reacts with sulfuric acid (H₂SO₄) to produce oleum (H₂S₂O₇). Oleum is essentially a solution of sulfur trioxide in concentrated sulfuric acid. The product formed is a mixture of sulfuric acid and sulfur trioxide in varying concentrations.
- Product – Oleum (H₂S₂O₇):
- Oleum is a liquid and is often described as fuming sulfuric acid due to the presence of dissolved sulfur trioxide (SO₃).
- The general formula for oleum can be written as H₂S₂O₇, but in different cases, the formula might include x moles of SO₃ depending on the concentration.
- Oleum is an important intermediate for producing sulfuric acid, as it allows for the controlled addition of SO₃ to water, preventing the exothermic reaction that occurs when SO₃ is dissolved directly into water.
Visual Representation: The uncharged product, oleum (H₂S₂O₇), would be represented as a combination of sulfuric acid (H₂SO₄) and sulfur trioxide (SO₃), where one molecule of sulfuric acid reacts with one molecule of sulfur trioxide to form oleum. The product doesn’t carry any charge because both sulfuric acid and sulfur trioxide are neutral compounds.