Problem 64 - A Complete Resource Book in Chemistry for JEE Main

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 \(\mathrm{H}\), and \(100 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

The answer is (c) 300 mL of H₂ and 200 mL of NH₃.

Step 1: Analyze the Chemical Reaction

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

Step 2: Understand the Volume Ratios

According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. So, the stoichiometric coefficients in the reaction equation represent volume ratios. Thus, if \( 100 \mathrm{~mL} \) of \( \mathrm{N}_{2} \) is used, it will require \( 3 \times 100 \mathrm{~mL} = 300 \mathrm{~mL} \) of \( \mathrm{H}_{2} \).

Step 3: Calculate the Volume of NH3 Produced

From the balanced equation, 1 volume of \( \mathrm{N}_{2} \) produces 2 volumes of \( \mathrm{NH}_{3} \). Thus, the reaction of \( 100 \mathrm{~mL} \) of \( \mathrm{N}_{2} \) will produce \( 2 \times 100 \mathrm{~mL} = 200 \mathrm{~mL} \) of \( \mathrm{NH}_{3} \).

Step 4: Choose the Correct Option

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

Key Concepts

Avogadro's Law

Avogadro's Law is a fundamental principle in chemistry that relates the volume of a gas to the amount of substance. It states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. This means that if you have two different gases, say Gas A and Gas B, and they occupy the same volume under identical conditions of temperature and pressure, they will have the exact same number of molecules.

For example, if we take our reaction where nitrogen gas (\(\mathrm{N}_2\)) and hydrogen gas (\(\mathrm{H}_2\)) react to form ammonia (\(\mathrm{NH}_3\)), Avogadro’s Law allows us to use volume ratios directly from the balanced chemical equation. In simpler terms, if you know the volume of one gas involved in the reaction, you can use Avogadro’s Law to determine the volumes of the other gases, considering they are measured at the same conditions of temperature and pressure. This is crucial because the law makes calculations not only straightforward but also very intuitive. If you see that 1 mole of a gas reacts with 3 moles of another gas to produce 2 moles of a product, you can immediately equate these mole ratios to volume ratios.

Volume Ratios

Volume ratios are an extension of Avogadro's Law and are used to relate the volumes of gases involved in a chemical reaction. Since volumes of gases at the same temperature and pressure are proportional to their respective mole numbers, the volume ratios can be directly taken from the coefficients of the balanced chemical equation. By understanding volume ratios, we can determine how much of each reactant is needed to produce a certain volume of product or how much product will be formed from given reactants.

In our example reaction:

• 1 volume of \(\mathrm{N}_2\) (100 \mathrm{~mL}) reacts with 3 volumes of \(\mathrm{H}_2\) (300 \mathrm{~mL}).
• This reaction produces 2 volumes of \(\mathrm{NH}_3\) (200 \mathrm{~mL}).

The ratios are directly derived from the coefficients in the balanced chemical equation: 1:3:2 for \(\mathrm{N}_2\):\(\mathrm{H}_2\):\(\mathrm{NH}_3\) respectively. This makes understanding and applying volume ratios easier, as it aligns perfectly with the balanced chemical equation. So, you always start with the known quantity (volume, in this case) and multiply by the ratios from the equation to find other volumes.

Balanced Chemical Equation

A balanced chemical equation is essential in chemical stoichiometry because it provides the quantitative relationship between reactants and products. It ensures the law of conservation of mass is followed, meaning the amount of each element is the same on both sides of the reaction.

To translate this into our reaction:

• The balanced equation is \(\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g})\).
• The coefficients in front of each molecule (1 for \(\mathrm{N}_2\), 3 for \(\mathrm{H}_2\), and 2 for \(\mathrm{NH}_3\)) tell us the mole ratios for the reaction.

Each of these coefficients can also represent relative volumes for gases when dealing with problems set at the same conditions of temperature and pressure. This correlation between moles and volumes simplifies complex chemical equations into understandable, manageable calculations in a stoichiometric analysis. Just follow the steps:

• Identify the balanced equation.
• Use the stoichiometric coefficients to directly determine volume ratios.
• Apply Avogadro’s Law for calculating unknown volumes.

Understanding and using balanced chemical equations is your roadmap to navigating complex chemical reactions with confidence.