Question

In the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\), when 100 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) has reacted, the volumes of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are: (a) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(300 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (b) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (c) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (d) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(100 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\)

Short answer

(c) 300 mL of H2 and 200 mL of NH3.

Step-by-step solution

Analyze Reaction Stoichiometry

The balanced chemical equation is \(\mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). This shows that 1 mole of \(\mathrm{N}_{2}\) reacts with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\).

Calculate Volume of H2

According to the stoichiometry, for every 1 volume of \(\mathrm{N}_{2}\), 3 volumes of \(\mathrm{H}_{2}\) are required. Therefore, if 100 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) reacts, \(3 \times 100 \mathrm{~mL} = 300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) is needed.

Calculate Volume of NH3 Produced

The reaction also shows that for every 1 volume of \(\mathrm{N}_{2}\) used, 2 volumes of \(\mathrm{NH}_{3}\) are produced. So, if 100 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) has reacted, \(2 \times 100 \mathrm{~mL} = 200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) is produced.

Choose the Correct Option

Based on the calculations, \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) is used and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) is produced. The correct answer is option (c).

Key concepts

Balanced Chemical Equation

To understand chemical reactions, we always start with a balanced chemical equation. This is vital because it ensures that the same number of atoms for each element are present on both sides of the equation. In our exercise, the balanced equation is \[ \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \].
This equation tells us a few important aspects of the reaction:

• The reactants are nitrogen (\( \mathrm{N}_2 \)) and hydrogen (\( \mathrm{H}_2 \)).
• The product is ammonia (\( \mathrm{NH}_3 \)).
• The stoichiometric coefficients (1, 3, 2) tell us the molar ratios: 1 mole of \( \mathrm{N}_2 \) reacts with 3 moles of \( \mathrm{H}_2 \) to produce 2 moles of \( \mathrm{NH}_3 \).

Balancing equations is essential in stoichiometry as it forms the base for any quantitative analysis of a chemical reaction. It helps us know exactly how much of each reactant is needed and what quantities of products will be formed.

Volume of Gases in Chemical Reactions

In gas reactions like ours, the concept of volume ratios follows Avogadro's Law, which states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. So, when we talk about gases, the coefficients in the balanced equation can also represent volume ratios.
For example, our reaction shows:

• 1 volume of \( \mathrm{N}_2 \) gas requires 3 volumes of \( \mathrm{H}_2 \) gas.
• It produces 2 volumes of \( \mathrm{NH}_3 \) gas.

When 100 mL of \( \mathrm{N}_2 \) reacts, we can directly use these ratios:

• To determine the volume of \( \mathrm{H}_2 \) required: multiply the volume of \( \mathrm{N}_2 \) by 3, leading to 300 mL.
• To determine the volume of \( \mathrm{NH}_3 \) formed: multiply the volume of \( \mathrm{N}_2 \) by 2, resulting in 200 mL.

This straightforward correlation due to Avogadro’s hypothesis makes calculations with gases much simpler compared to reactions involving liquids or solids.

Reaction Yield Calculations

Reaction yield calculations help us determine the efficiency of a reaction. These calculations involve comparing the actual amount of product obtained with the theoretical amount, which is derived from stoichiometry.
In this exercise, if the reaction went perfectly according to the balanced chemical equation and stoichiometry, using 100 mL of \( \mathrm{N}_2 \) should:

• Consume 300 mL of \( \mathrm{H}_2 \).
• Produce 200 mL of \( \mathrm{NH}_3 \).

The theoretical yield is based on these stoichiometric ratios. However, actual reactions may not always achieve these yields due to factors like side reactions or incomplete conversions.
To calculate the percent yield, you can use the formula:\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \%\]Knowing the percent yield can indicate how successful a reaction is and may guide improvements in efficiency for practical applications.