Problem 2 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\); identify dihydrogen \(\left(\mathrm{H}_{2}\right)\) as a limiting reagent in the following reaction mixtures. [Main April 9, 2019 (I)] (a) \(56 \mathrm{~g}\) of \(\mathrm{N}_{2}+10 \mathrm{~g}\) of \(\mathrm{H}_{2}(\mathrm{~b}) 35 \mathrm{~g}\) of \(\mathrm{N}_{2}+8 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (c) \(28 \mathrm{~g}\) of \(\mathrm{N}_{2}+6 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (d) \(14 \mathrm{~g}\) of \(\mathrm{N}_{2}+4 \mathrm{~g}\) of \(\mathrm{H}_{2}\)

Step-by-Step Solution

Answer

(a) Only in mixture (a), \(\mathrm{H}_2\) is the limiting reagent.

Step 1: Calculate Moles of Each Reactant

First, calculate the moles of each reactant, nitrogen (\(\mathrm{N}_2\)) and dihydrogen (\(\mathrm{H}_2\)), using their molar masses. The molar mass of \(\mathrm{N}_2\) is approximately 28 g/mol, and for \(\mathrm{H}_2\), it is about 2 g/mol. For each mixture: (a) Moles of \(\mathrm{N}_2 = \frac{56}{28} = 2\) moles, Moles of \(\mathrm{H}_2 = \frac{10}{2} = 5\) moles (b) Moles of \(\mathrm{N}_2 = \frac{35}{28} \approx 1.25\) moles, Moles of \(\mathrm{H}_2 = \frac{8}{2} = 4\) moles (c) Moles of \(\mathrm{N}_2 = \frac{28}{28} = 1\) mole, Moles of \(\mathrm{H}_2 = \frac{6}{2} = 3\) moles (d) Moles of \(\mathrm{N}_2 = \frac{14}{28} = 0.5\) moles, Moles of \(\mathrm{H}_2 = \frac{4}{2} = 2\) moles.

Step 2: Determine Stoichiometric Needs for Hydrogen

Next, use the reaction equation: \(\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g)\), where 1 mole of \(\mathrm{N}_2\) requires 3 moles of \(\mathrm{H}_2\). Calculate how much \(\mathrm{H}_2\) needed for each amount of \(\mathrm{N}_2\):(a) \(2\) moles of \(\mathrm{N}_2\) needs \(2 \times 3 = 6\) moles of \(\mathrm{H}_2\)(b) \(1.25\) moles of \(\mathrm{N}_2\) needs \(1.25 \times 3 = 3.75\) moles of \(\mathrm{H}_2\)(c) \(1\) mole of \(\mathrm{N}_2\) needs \(1 \times 3 = 3\) moles of \(\mathrm{H}_2\)(d) \(0.5\) moles of \(\mathrm{N}_2\) needs \(0.5 \times 3 = 1.5\) moles of \(\mathrm{H}_2\).

Step 3: Compare Available vs. Required Hydrogen

Compare the available moles of \(\mathrm{H}_2\) from Step 1 with the required moles calculated in Step 2 for each scenario:(a) Available: 5 moles; Required: 6 moles(b) Available: 4 moles; Required: 3.75 moles(c) Available: 3 moles; Required: 3 moles(d) Available: 2 moles; Required: 1.5 moles.

Step 4: Identify Limiting Reagent

Determine if \(\mathrm{H}_2\) is the limiting reactant (when available moles are less than required moles). In this case:(a) \(\mathrm{H}_2\) is the limiting reagent because only 5 moles are available and 6 moles are needed.(b) \(\mathrm{H}_2\) is not the limiting reagent because 4 moles are available and 3.75 moles are needed.(c) \(\mathrm{H}_2\) is not the limiting reagent because exactly 3 moles are available and 3 moles are needed.(d) \(\mathrm{H}_2\) is not the limiting reagent because 2 moles are available and only 1.5 moles are needed.

Key Concepts

Stoichiometry

Stoichiometry is like a recipe book for chemists. It provides a set of instructions on how much of each ingredient (reactant) you need to make a certain amount of product. This concept is rooted in the balanced chemical equation, which shows the relationship in the number of moles of reactants and products. For the reaction \( \mathrm{N}_{2} + 3\mathrm{H}_{2} \rightarrow 2\mathrm{NH}_{3} \), the coefficients in front of each molecule tell us how many moles of reactants are needed and how many moles of product will be formed.

Here’s how it works:

• According to the equation, 1 mole of \( \mathrm{N}_{2} \) reacts with 3 moles of \( \mathrm{H}_{2} \) to produce 2 moles of \( \mathrm{NH}_{3} \).
• The coefficients (1, 3, and 2) derive the mole ratio, which allows us to solve stoichiometry problems.

This mole ratio is critical for determining limiting reagents and excess reactants. By knowing the amounts of reactants present, we can calculate how much product will result, assuming complete reaction of the limiting reagent. The beauty of stoichiometry lies in its predictability, helping chemists design experiments and scale up reactions for practical applications.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of bonds, resulting in different substances. In the context of our example, the reaction between nitrogen and hydrogen to form ammonia \( \mathrm{NH}_{3} \) involves several key aspects:

• Reactants and Products:
  Reactants like \( \mathrm{N}_{2} \) and \( \mathrm{H}_{2} \) are transformed into products such as \( \mathrm{NH}_{3} \). The process is guided by the chemical equation that tells us exactly how these substances change.
• The Law of Conservation of Mass:
  This law states that mass cannot be created or destroyed in a chemical reaction. Thus, the mass of the products must equal the mass of the reactants. Balancing chemical equations ensures this principle is applied.
• Energy Transfer:
  Reactions often involve energy changes, either releasing energy (exothermic) or absorbing it (endothermic).

This specific reaction is widely known in industrial processes as the Haber process, which is crucial for synthesizing ammonia on a large scale, used primarily for fertilizers.

Mole Concept

The mole concept is the cornerstone of understanding chemical reactions on a tangible scale. A mole is a unit that helps chemists convert between atomic mass units and grams, allowing them to measure and manipulate quantities of substances.

Here’s how the mole concept was applied in the exercise:

• Calculating Moles:
  To determine how much hydrogen or nitrogen was present, we converted grams of each substance into moles using their respective molar masses:
  For nitrogen \( \mathrm{N}_2 \), \( \text{molar mass} = 28\, \text{g/mol} \)
  For hydrogen \( \mathrm{H}_2 \), \( \text{molar mass} = 2\, \text{g/mol} \)
• Understanding Limiting Reagents:
  The mole concept helps identify the limiting reagent by comparing the number of moles available against those needed from the reaction stoichiometry.

Using the mole concept ensures precise calculations in reactions, which is essential for obtaining desired outcomes whether in a lab setting or manufacturing plant. It also bridges the microscopic world of atoms with the macroscopic world we observe.