Problem 54 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A mixture of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) occupies 40 litres at \(1.00\) atm and at \(400 \mathrm{~K}\). The mixture reacts completely with \(130 \mathrm{~g}\) of \(\mathrm{O}_{2}\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assuming ideal gas behaviour, calculate the mole fractions of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) in the mixture.

Step-by-Step Solution

Answer

Mole fractions are: C2H4: 0.34, C2H6: 0.66.

Step 1: Write the balanced chemical equations

For ethane (C2H6) reacting with oxygen, the equation is:\[ 2\,\text{C}_2\text{H}_6 + 7\,\text{O}_2 \rightarrow 4\,\text{CO}_2 + 6\,\text{H}_2\text{O} \]For ethene (C2H4) reacting with oxygen, the equation is:\[ \text{C}_2\text{H}_4 + 3\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 2\,\text{H}_2\text{O} \]

Step 2: Calculate moles of oxygen

Given, oxygen mass = 130 g. Molar mass of \(\text{O}_2\) = 32 g/mol.\[ n_{O_2} = \frac{130}{32} = 4.0625\,\text{mol} \]

Step 3: Use Ideal Gas Law to calculate moles of the gas mixture

Use the ideal gas law, \(PV = nRT\), to find total moles \(n\).\[ P = 1.00\,\text{atm},\ V = 40\,\text{L},\ R = 0.0821\,\text{L atm/mol K},\ T = 400\,\text{K} \]\[ n = \frac{PV}{RT} = \frac{1.00 \times 40}{0.0821 \times 400} = 1.22\,\text{mol} \]

Step 4: Set up system of equations for the mixture

Let \(x\) be the moles of \(\text{C}_2\text{H}_6\) and \(y\) be the moles of \(\text{C}_2\text{H}_4\). Total moles equation:\[ x + y = 1.22 \]And the oxygen consumption equation:\[ \frac{7}{2}x + 3y = 4.0625 \]

Step 5: Solve the system of equations

Using the two equations:1. \(x + y = 1.22\)2. \(3.5x + 3y = 4.0625\)From equation 1: \(x = 1.22 - y\)Substitute in equation 2:\[ 3.5(1.22 - y) + 3y = 4.0625 \]\[ 4.27 - 3.5y + 3y = 4.0625 \]\[ 0.5y = 4.27 - 4.0625 \]\[ 0.5y = 0.2075 \]\[ y = 0.415 \]Substitute \(y = 0.415\) in equation 1:\[ x = 1.22 - 0.415 = 0.805 \]

Step 6: Calculate mole fractions

To find the mole fraction, divide the moles of each component by the total moles.\[ \text{Mole fraction of } \text{C}_2\text{H}_4 = \frac{y}{x + y} = \frac{0.415}{1.22} \approx 0.34 \]\[ \text{Mole fraction of } \text{C}_2\text{H}_6 = \frac{x}{x + y} = \frac{0.805}{1.22} \approx 0.66 \]

Key Concepts

Mole Fraction in Gas Mixtures

The mole fraction is an expression of concentration, and it is particularly useful when dealing with gas mixtures. It is represented as the ratio of the moles of a particular component to the total moles in the mixture. The mole fraction does not have units and is expressed as a number between 0 and 1. This value helps to understand how each component contributes to the overall mixture.
For example, when calculating the mole fraction of ethene \(\left( \text{C}_2\text{H}_4 \right)\) and ethane \(\left( \text{C}_2\text{H}_6 \right)\) in a mixture, we consider the calculated number of moles of each component and the total number of moles of all substances present.
In our example problem, we determined that ethene had a mole fraction of approximately 0.34, while ethane had a mole fraction of approximately 0.66. Thus, ethane constitutes a larger portion of the gas mixture than ethene.

Stoichiometry and Chemical Reactions

Stoichiometry is crucial to solve chemical equations and reactions, ensuring the law of conservation of mass is followed. Balancing chemical equations forms the foundation of stoichiometry. It describes the quantitative relationships between reactants and products in a chemical reaction.
In the problem where ethane and ethene react with oxygen, we use stoichiometry to determine how much oxygen is consumed. The balanced equations for these reactions are specified as:

• For ethane: \ 2\,\text{C}_2\text{H}_6 + 7\,\text{O}_2 \rightarrow 4\,\text{CO}_2 + 6\,\text{H}_2\text{O} \
• For ethene: \ \text{C}_2\text{H}_4 + 3\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 2\,\text{H}_2\text{O}

These equations tell us how many moles of \(\text{O}_2\) are needed for complete combustion of each gas. By applying stoichiometry, we can assess the total moles of \(\text{O}_2\) used in the example and thereby solve for the moles of ethane and ethene.

Understanding Gas Mixtures

A gas mixture consists of two or more gases that are mixed but not chemically combined, often with each gas behaving independently in the mixture. Understanding gas mixtures is central to many scientific fields, including chemistry and physics. The Ideal Gas Law comes in handy when dealing with gas mixtures. It helps predict the behavior of gas mixtures under varying conditions of temperature and pressure.
The Ideal Gas Law is given by \(PV = nRT\), where each symbol represents:

• \P\: Pressure of the gas,
• \V\: Volume of the gas,
• \: Number of moles of the gas,
• \R\: Ideal gas constant, and
• \T\: Temperature in Kelvin.

By using this equation, we can calculate the total moles, \(n\), present in the mixture. In our exercise, the moles were computed to provide insight into the composition and behavior of the gas mixture. This law assumes the gas behaves ideally, meaning intermolecular forces are negligible, and the gas particles themselves take up no volume.