Problem 65 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

1 litre of mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) is taken. The mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volumes are measured under the same conditions. Find the composition of mixture by volume.

Step-by-Step Solution

Answer

60% \(\mathrm{CO}_{2}\) and 40% \(\mathrm{CO}\) by volume.

Step 1: Understanding the Reaction

The mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) is passed over red hot charcoal. The \(\mathrm{CO}_{2}\) reacts with carbon (charcoal) to form \(\mathrm{CO}\), as per the reaction: \(\mathrm{CO}_{2} + \mathrm{C} \rightarrow 2 \mathrm{CO}\). Therefore, only \(\mathrm{CO}_{2}\) volume will change, turning into more \(\mathrm{CO}\).

Step 2: Let Variables Represent the Composition

Let \(x\) be the initial volume of \(\mathrm{CO}_{2}\) and \(y\) be the initial volume of \(\mathrm{CO}\). Since the total initial volume is 1 liter, we have the equation \(x + y = 1\).

Step 3: Analyze the Volume Change

As \(\mathrm{CO}_{2}\) converts into \(\mathrm{CO}\), remember that 1 liter of \(\mathrm{CO}_{2}\) produces 2 liters of \(\mathrm{CO}\). The final volume of \(\mathrm{CO}\) is 1.6 liters, which includes both original and converted volumes.

Step 4: Set Up Equations

From the behavior of the volumes, we have two equations: \(x + y = 1\) (initial condition) and \(y + 2x = 1.6\) (final volume condition, since converted \(\mathrm{CO}_{2}\) forms additional \(\mathrm{CO}\)).

Step 5: Solve the Equations

From the first equation \(x + y = 1\), we can express \(y = 1 - x\). Substitute this into the second equation: \(1 - x + 2x = 1.6\). Solving this gives \(x = 0.6\). Then \(y = 1 - 0.6 = 0.4\).

Step 6: Interpret the Results

The initial volumes are \(x = 0.6\) liters of \(\mathrm{CO}_{2}\) and \(y = 0.4\) liters of \(\mathrm{CO}\). Therefore, the composition of the mixture by volume is 60% \(\mathrm{CO}_{2}\) and 40% \(\mathrm{CO}\).

Key Concepts

Chemical Reactions

When we talk about chemical reactions, we're discussing a process where substances called reactants transform into new substances called products. In this exercise, a reaction occurs between carbon dioxide \(\mathrm{CO}_2\) and carbon (charcoal) to produce carbon monoxide \(\mathrm{CO}\). This specific reaction is represented as: \(\mathrm{CO}_2 + \mathrm{C} \rightarrow 2\mathrm{CO}\). So, what happens here is that one molecule of \(\mathrm{CO}_2\) interacts with carbon and transforms into two molecules of \(\mathrm{CO}\).

• Reactants: \(\mathrm{CO}_2\) and \(\mathrm{C}\)
• Products: \(\mathrm{CO}\)

What's particularly important to note in this reaction is the change in volume. Because one molecule of \(\mathrm{CO}_2\) transforms into two molecules of \(\mathrm{CO}\), the volume of the gases changes significantly. By understanding this reaction, we can predict how the volume will increase as the reaction proceeds.

Gas Laws

Gas laws are essential for understanding how gases behave under different conditions. In this problem, the volume of the gas changes as a result of a chemical reaction. Since the conditions such as temperature and pressure remain constant, we can apply the Ideal Gas Law properties to understand these changes.
The Ideal Gas Law is typically stated as \(PV = nRT\), where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature.

In this example, the temperature and pressure are constant, so the volume change is directly related to the change in the amount of gas, or number of moles. This aligns with our understanding that one mole of \(\mathrm{CO}_2\) becomes two moles of \(\mathrm{CO}\), thus increasing the volume under unchanged temperature and pressure conditions.

Volume Calculations

Volume calculation plays a critical role in solving exercises related to gaseous mixtures. In our example, the transformation of \(\mathrm{CO}_2\) to \(\mathrm{CO}\) results in a change in the total gas volume. Initially, we have a mixture of 1 liter, but once the reaction is complete, it's increased to 1.6 liters. Here's a simple breakdown of the calculations that helped achieve this understanding:

• Let \(x\) represent the initial volume of \(\mathrm{CO}_2\).
• Let \(y\) represent the initial volume of \(\mathrm{CO}\).
• The sum of these, \(x + y\), equals the initial total volume of 1 liter.
• The new equation, after the reaction \(y + 2x = 1.6\), represents the final volume.

By solving these equations, we find \(x = 0.6\) liters for \(\mathrm{CO}_{2}\) and \(y = 0.4\) liters for \(\mathrm{CO}\). This yields a ratio of 60% \(\mathrm{CO}_2\) and 40% \(\mathrm{CO}\) in the mixture before any reaction took place. Accurate volume calculations help determine reactant composition, crucial for many chemical process simulations and industrial applications.