Problem 8 - General Chemistry: Principles and Modern Applications

Question

Write balanced equations to represent the complete combustion of each of the following in excess oxygen: (a) propylene, \(\mathrm{C}_{3} \mathrm{H}_{6} ;\) (b) thiobenzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COSH}\) (c) glycerol, \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

The balanced equations for the combustion of propylene, thiobenzoic acid, and glycerol are: \(\mathrm{C}_{3} \mathrm{H}_{6} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COSH} + 8\mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O} + \mathrm{SO}_{2}\), and \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\) respectively.

Step 1: Combustion of Propylene

Propylene \(\mathrm{C}_{3} \mathrm{H}_{6}\) reacts with oxygen \(\mathrm{O}_{2}\) to produce carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\). The unbalanced equation is: \(\mathrm{C}_{3} \mathrm{H}_{6} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\). Balancing, we obtain: \(\mathrm{C}_{3} \mathrm{H}_{6} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O}\)

Step 2: Combustion of Thiobenzoic Acid

Thiobenzoic Acid \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COSH}\) will also react with \(\mathrm{O}_{2}\) to produce carbon dioxide \(\mathrm{CO}_{2}\), water \(\mathrm{H}_{2}\mathrm{O}\) and sulfur \(\mathrm{SO}_{2}\). The unbalanced equation is: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COSH} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} + \mathrm{SO}_{2}\). After balancing, the equation becomes: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COSH} + 8\mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O} + \mathrm{SO}_{2}\)

Step 3: Combustion of Glycerol

Glycerol \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\) reacts with \(\mathrm{O}_{2}\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\). The unbalanced equation is: \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\). After balancing, this becomes: \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides of the equation. This is crucial in chemical reactions to satisfy the law of conservation of mass. During the combustion reaction, it is important to add coefficients in front of the reactants and products instead of changing the subscripts of the compounds to balance the equation.
In the example of propylene combustion, first, identify the number of carbon, hydrogen, and oxygen atoms on both sides of the unbalanced equation. Propylene \( \mathrm{C}_{3} \mathrm{H}_{6} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)
To balance it:

• Count the carbon atoms: there are 3 carbon atoms in propylene, so you need 3 \( \mathrm{CO}_{2}\)
• Count the hydrogen atoms: there are 6 hydrogen atoms in propylene, so you need 3 \( \mathrm{H}_{2}\mathrm{O}\)
• Balance the oxygen: Finally, 3 \( \mathrm{CO}_{2}\) and 3 \( \mathrm{H}_{2}\mathrm{O}\) need a total of 9 oxygen atoms, which comes from 5 \( \mathrm{O}_{2}\)

Balancing requires patience and practice to ensure each atom type balances across the reaction equation.

Combustion of Hydrocarbons

The combustion of hydrocarbons is a key chemical reaction where hydrocarbons react with oxygen to produce carbon dioxide and water. This process is important for energy production because it releases energy in the form of heat.
Propylene is a simple hydrocarbon following the general formula for alkenes \( \mathrm{C}_n \mathrm{H}_{2n}\). When combusted in excess oxygen, as with any hydrocarbon, the reaction produces \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2}\mathrm{O}\). For example:

• The combustion equation for propylene:
  \( \mathrm{C}_{3} \mathrm{H}_{6} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O} \)
• Note there must always be sufficient oxygen to ensure complete combustion.

Not every combustion reaction is as straightforward as for hydrocarbons. The complete combustion of substances like thiobenzoic acid also involves sulfur, which results in sulfur dioxide \( \mathrm{SO}_{2} \) production. Understanding these reactions is fundamental for applications related to fuel energy and environmental considerations.

Chemical Stoichiometry

Chemical stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In combustion reactions, stoichiometry is used to understand and calculate the amounts needed for complete reactions.
When working through stoichiometry, the balanced chemical equation is critical as it provides the proportions of reactants and products involved:

• For instance, in the combustion of glycerol:
  \( \mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O} \)
• This tells us for every molecule of glycerol, five molecules of oxygen are required.
• In the same way, the equation indicates the production of three molecules of carbon dioxide and four molecules of water.

Thus, stoichiometry helps to determine how much oxygen is needed or how much product will be formed, allowing precise control over the chemical processes.