Problem 68 - A Complete Resource Book in Chemistry for JEE Main

Question

In Haber's process, \(30 \mathrm{~L}\) of dihydrogen and \(30 \mathrm{~L}\) of dinitrogen were taken for reaction which yielded only \(50 \%\) of expected product. What is the composition of the gaseous mixture under afore-said conditions in the end? (a) \(20 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (b) \(20 \mathrm{~L} \mathrm{NH}_{3}, 20 \mathrm{~L} \mathrm{~N}_{2}, 20 \mathrm{~L} \mathrm{H}_{2}\) (c) \(10 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (d) \(20 \mathrm{~L} \mathrm{NH}_{3}, 10 \mathrm{~L} \mathrm{~N}_{2}, 30 \mathrm{~L} \mathrm{H}_{2}\)

Step-by-Step Solution

Answer

The composition of the mixture is 10 L \(\text{NH}_3\), 20 L \(\text{N}_2\), and 15 L \(\text{H}_2\). (Option c)

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the Haber process is: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]This equation tells us that 1 volume of \(\text{N}_2\) reacts with 3 volumes of \(\text{H}_2\) to yield 2 volumes of \(\text{NH}_3\).

Step 2: Determine the Proportions Required

Given 30 L of \(\text{N}_2\), you would need 90 L of \(\text{H}_2\) for a complete reaction according to the stoichiometry (1:3 ratio). However, only 30 L of \(\text{H}_2\) is available. This means \(\text{H}_2\) is the limiting reagent.

Step 3: Calculate the Volume of Ammonia Formed

From the limiting reagent \(\text{H}_2\), we calculate the ammonia produced. The available reaction is limited by \(\frac{30 \text{ L } \text{H}_2}{3} = 10 \text{ L}\) of \(\text{N}_2\). This reacts to form 20 L of \(\text{NH}_3\) in a complete reaction, but since the yield is 50%, it results in:\[ \frac{20}{2} = 10 \text{ L}\] of ammonia actually formed due to the 50% yield.

Step 4: Determine the Remaining Volumes of Reactants

Since \(10\text{ L}\) of \(\text{N}_2\) reacted, the remaining \(\text{N}_2\) is \(30 \text{ L} - 10 \text{ L} = 20 \text{ L}\). Similarly, \(30 \text{ L}\) of \(\text{H}_2\) was initially present; \(30\text{ L}\) was intended to react for 10L of \(\text{N}_2\), but the actual reaction due to 50% yield implies that 15 L of \(\text{H}_2\) remain unreacted.

Key Concepts

Limiting Reagent

When conducting a chemical reaction, it's crucial to identify the limiting reagent. This is the reactant that gets completely consumed first, limiting the amount of product formed. In Haber's process, where ammonia is produced from nitrogen and hydrogen gases, we must consider the stoichiometry of the reaction: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]This equation reveals that 1 volume of nitrogen (\(\text{N}_2\)) reacts with 3 volumes of hydrogen (\(\text{H}_2\)). If we begin with equal volumes of nitrogen and hydrogen (30 L each), and need 90 L of hydrogen for a full reaction with 30 L of nitrogen, hydrogen becomes the limiting reagent because there isn't enough to react completely according to the equation.

• The limiting reagent determines the maximum possible amount of product.
• When the limiting reagent is used up, the reaction stops even if other reactants are still available.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants transforming into products and includes important information like reactant ratios and reaction direction. In Haber's process, the balanced chemical equation:\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]demonstrates the exact quantities of reactants needed and the products generated. This balance is crucial for understanding how much of each reactant is required to produce a desired quantity of product.

• Balanced equations obey the law of conservation of mass.
• They provide the stoichiometric ratios needed for quantitative analysis in reactions.
• Using these equations allows us to determine limiting reagents and calculate yields effectively.

Here, one molecule of \(\text{N}_2\) combines with three molecules of \(\text{H}_2\) to produce two molecules of ammonia, indicating the precise molecular interaction in the reaction.

Reaction Yield

In real-world chemical reactions, the amount of product formed is often less than the theoretical maximum due to inefficiencies or other factors. The reaction yield quantifies this efficiency as a percentage, calculated by comparing the actual amount of product obtained to the amount expected from the stoichiometric calculations. In the context of the exercise, even though the stoichiometry indicated that 20 L of ammonia could be produced, the actual yield was only 50%, resulting in just 10 L of ammonia.

• Theoretical yield is what is calculated from the balanced chemical equation with complete reaction.
• Actual yield is what is practically obtained and may differ due to losses or side reactions.
• Reaction yield (%) is calculated as:\[ \text{Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \%\]

Understanding reaction yield helps in evaluating the efficiency of a process and in optimizing it for better product output.