Problem 38 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A gas mixture of \(3.67\) litres of ethylene and methane on complete combustion at \(25^{\circ} \mathrm{C}\) produces \(6.11\) litres of \(\mathrm{CO}_{2}\). Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are \(-1423\) and \(-891 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The heat evolved on burning one litre of the gas mixture is 56.92 kJ.

Step 1: Define the Combustion Reactions

The combustion reactions for ethylene (\(\text{C}_2\text{H}_4\)) and methane (\(\text{CH}_4\)) are given by: \[\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O},\quad\Delta H = -1423 \, \text{kJ/mol}\]\[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O},\quad\Delta H = -891 \, \text{kJ/mol}\].

Step 2: Calculate the Stoichiometric Combustion Output

From the reaction, combustion of one mole of \(\text{C}_2\text{H}_4\) yields \(2\) moles of \(\text{CO}_2\), and combustion of one mole of \(\text{CH}_4\) yields \(1\) mole of \(\text{CO}_2\). Using the ideal gas law, each mole of gas corresponds to \(22.414\) litres at \(25^{\circ} C\).

Step 3: Set Up the Equation System

Let \(x\) be the litres of \(\text{C}_2\text{H}_4\) and \(y\) be the litres of \(\text{CH}_4\). The total gas volume is: \[x + y = 3.67\text{ litres}\]The total \(\text{CO}_2\) volume produced is: \[2\frac{x}{22.414} + \frac{y}{22.414} = \frac{6.11}{22.414}\] This solves: \[0.08921x + 0.04461y = 0.2726\].

Step 4: Solve System of Equations for x and y

From equations: 1. \(x + y = 3.67\), 2. \(0.08921x + 0.04461y = 0.2726\).Let's solve equation (1) for \(y\): \[y = 3.67 - x\]. Substitute \(y\) in equation (2): \[0.08921x + 0.04461(3.67 - x) = 0.2726\].Solving gives \(x = 2.652\) litres and \(y = 1.018\) litres.

Step 5: Calculate Heat Evolved from Combustion

Using moles conversion, \(\text{litres of }\text{C}_2\text{H}_4 = \frac{2.652}{22.414} = 0.1183\) mol and \(\text{litres of }\text{CH}_4 = \frac{1.018}{22.414} = 0.04543\) mol.Calculate energy: \[\text{Energy from } \text{C}_2\text{H}_4 = 0.1183 \times 1423 = 168.4\text{kJ}\]\[\text{Energy from } \text{CH}_4 = 0.04543 \times 891 = 40.5\text{kJ}\] Total energy: \(208.9 \text{kJ for } 3.67 \text{ litres}\).

Step 6: Calculate Heat Evolved per Litre

The heat evolved per litre of gas mixture is \[\frac{208.9}{3.67} = 56.92\text{kJ/L}\].

Key Concepts

Heat of Combustion

The heat of combustion is a crucial concept in chemistry, representing the amount of heat energy released when a substance combusts completely with oxygen. It is typically expressed in kilojoules per mole (kJ/mol).
In the original exercise, the heat of combustion values for ethylene (\(-1423 \, \text{kJ/mol}\) ) and methane (\(-891 \, \text{kJ/mol}\) ), are given. These values indicate that when 1 mole of these substances is burned fully, 1423 and 891 kilojoules of heat are released, respectively.

• Higher absolute values indicate a higher energy release.
• The negative sign denotes that the process is exothermic (releases heat).

Understanding the heat of combustion helps in calculating the total energy released in a combustion reaction, like the one in the original problem. Knowing this helps in assessing the energy efficiency of fuels, determining how much energy can be harnessed from a given amount of fuel.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the calculation of reactants and products in chemical reactions, ensuring the balanced reactions align with the laws of conservation of mass.
In the exercise, to find out how much carbon dioxide (\(\text{CO}_2\) ) is produced, stoichiometry is used. It is crucial to establish the relationship between the amounts (in moles or volumes) of different compounds involved in reactions.

Balancing Equations

Chemical reactions are written as balanced equations, like those for ethylene and methane combustion:

• Ethylene: \(\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}\)
• Methane: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)

Each component appears in integers, reflecting balanced atomic and molecular quantities. From these equations, we derive conversion factors:
- 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{CO}_2\).- 1 mole of \(\text{CH}_4\) produces 1 mole of \(\text{CO}_2\).
It is from these relations the exercise computes the total \(\text{CO}_2\) produced and solves the equation system to find the mixture proportions.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates pressure, volume, temperature, and amount of gas molecules through the equation \(PV = nRT\) . Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.
In the problem, the ideal gas law is employed to establish the relationship between the gas moles and their volumes, assuming all gases behave ideally at the given temperature of \(25^{\circ} \text{C}\).

• Under ideal conditions, 1 mole of gas occupies 22.414 liters at \(25^{\circ} \text{C}\).
• Helps correlate volume to the mole count, crucial in evaluating the combustion reactions.
• Assumes gas particles have elastic collisions and occupy negligible space.

Through the ideal gas law, connections are made between the measured volume of gases (like 3.67 liters of gas mixture) and the stoichiometric calculations (in moles), facilitating the energy calculations leveraged in determining the combustion heat.