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 \(\mathrm{H}_{2}\) and \(100 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

(c) 300 mL of \( \mathrm{H}_{2} \) and 200 mL of \( \mathrm{NH}_{3} \).

Step 1: Write the Balanced Chemical Equation

The reaction is given as \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \). This equation tells us 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: Determine Volume Ratios

According to the stoichiometry of the balanced equation, 1 volume of \( \mathrm{N}_{2} \) reacts with 3 volumes of \( \mathrm{H}_{2} \) to produce 2 volumes of \( \mathrm{NH}_{3} \).

Step 3: Calculate Volumes for \( \mathrm{H}_{2} \) and \( \mathrm{NH}_{3} \)

Given that \( 100 \mathrm{~mL} \) of \( \mathrm{N}_{2} \) reacts, the corresponding volume of \( \mathrm{H}_{2} \) required would be \( 3 \times 100 \mathrm{~mL} = 300 \mathrm{~mL} \). Consequently, the volume of \( \mathrm{NH}_{3} \) formed would be \( 2 \times 100 \mathrm{~mL} = 200 \mathrm{~mL} \).

Step 4: Matching the Results to Options

The calculated volumes are \( 300 \mathrm{~mL} \) of \( \mathrm{H}_{2} \) and \( 200 \mathrm{~mL} \) of \( \mathrm{NH}_{3} \). Option (c) matches these results.

Key Concepts

Chemical Reactions

When we talk about chemical reactions, we are discussing processes where substances called reactants are transformed into new substances known as products. This is represented by a chemical equation, which shows the reactants on the left and the products on the right. For instance, in the reaction \[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \],\(\mathrm{N}_2\) and \(\mathrm{H}_2\) are the reactants that produce the product, \(\mathrm{NH}_3\).
The arrow between these substances indicates the direction of the reaction. Sometimes reactions are reversible, which is shown with a double-headed arrow. Understanding the components and direction of a reaction helps us predict the quantities of products formed and reactants consumed.

• Reactants: Substances that start a chemical equation.
• Products: Substances formed from a chemical reaction.
• Reversible Reaction: Can proceed in both forward and backward directions.

Volume Ratios

Volume ratios in chemical reactions are vital for understanding how much of each substance is used or produced. In gaseous reactions, these ratios are often determined using the coefficients from the balanced chemical equation. These coefficients indicate the relative amounts of each substance involved.
For example, the equation \[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \] demonstrates this principle. It shows that one volume of \(\mathrm{N}_2\) reacts with three volumes of \(\mathrm{H}_2\) to form two volumes of \(\mathrm{NH}_3\).

• Volume of \(\mathrm{N}_2\): 1 part
• Volume of \(\mathrm{H}_2\): 3 parts
• Volume of \(\mathrm{NH}_3\): 2 parts

These ratios simplify calculations in reactions involving gases, allowing us to determine the volumes of products and reactants given any starting volume.

Mole Concept

The mole concept is a cornerstone in chemistry for relating the amount of substance to its chemical quantities. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\). This concept allows us to equate volumes of gases under standard conditions directly to moles because of Avogadro's law.
In our given chemical reaction, \[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \], one mole of \(\mathrm{N}_2\) reacts with three moles of \(\mathrm{H}_2\) to produce two moles of \(\mathrm{NH}_3\). When dealing with gases under the same temperature and pressure, the volume is directly proportional to the number of moles.
This means the concept not only assists in determining amounts in terms of particles but also in understanding volumetric relationships in gaseous reactions.

• 1 mole of any gas at STP occupies 22.4 liters.
• Direct relationship between moles and gas volumes.
• Avogadro's number helps bridge microscopic particle quantities to measurable macroscale amounts.