Problem 56 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In order to get maximum calorific output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of the fuel. A burner which has been adjusted for methane as fuel (with \(x\) litre/hour of \(\mathrm{CH}_{4}\) and \(6 x\) litre/hour of \(\mathrm{O}_{2}\) ) is to be readjusted for butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\). In order to get the same calorific output, what should be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion, \(e t c\), are the same for both the fuels and the gases behave ideally. Heats of combustion : $$ \mathrm{CH}_{4}=809 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{C}_{4} \mathrm{H}_{10}=2878 \mathrm{~kJ} / \mathrm{mol} $$

Step-by-Step Solution

Answer

For equivalent calorific output, supply \( \frac{x \times 809}{2878} \) liters/hour of butane and \( 3 \times 6.5 \times \frac{x \times 809}{2878} \times 22.4 \) liters/hour of oxygen.

Step 1: Calculate Moles of Methane Consumed

First, calculate the moles of methane consumed per hour by using its volume and the fact that 1 mole of any ideal gas occupies 22.4 liters at standard conditions. Thus,\[ \text{Moles of CH}_4 = \frac{x}{22.4} \text{ moles/hour} \]

Step 2: Calculate Energy Released by Methane

Now, compute the energy released by the combustion of methane per hour using its heat of combustion:\[ \text{Energy from CH}_4 = \frac{x}{22.4} \times 809 \text{ kJ/hour} \]

Step 3: Calculate Required Moles of Butane for Equivalent Energy

To find the quantity of butane that releases the same amount of energy per hour, divide the energy requirement by the heat of combustion of butane:\[ \text{Moles of C}_4\text{H}_{10} = \frac{\frac{x}{22.4} \times 809}{2878} \text{ moles/hour} \]

Step 4: Convert Moles of Butane to Volume

Convert the moles of butane into liters, knowing that 1 mole of gas occupies 22.4 liters at standard conditions:\[ \text{Volume of C}_4\text{H}_{10} = \frac{x}{22.4} \times \frac{809}{2878} \times 22.4 \] \[ \text{Volume of C}_4\text{H}_{10} = \frac{x \times 809}{2878} \text{ liters/hour} \]

Step 5: Calculate Required Oxygen for Butane Combustion

The theoretical equation for the combustion of butane is:\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \to 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \]From this, 1 mole of \( \mathrm{C}_4\mathrm{H}_{10} \) requires 6.5 moles of \( \mathrm{O}_2 \). Given this, and the optimal condition requiring 3 times the theoretical oxygen amount,\[ \text{Liters of } \mathrm{O}_2 = 3 \times 6.5 \times \text{moles of } \mathrm{C}_4\mathrm{H}_{10} \times 22.4 \]Substitute the moles of \( \mathrm{C}_4\mathrm{H}_{10} \) calculated earlier:\[ \text{Volume of } O_2 = 3 \times 6.5 \times \frac{x \times 809}{2878} \times 22.4 \]

Step 6: Finalize Results

The rates of provision for butane and oxygen, ensuring equivalent calorific output to a methane system, are:\[ \text{Rate of supply of } \mathrm{C}_4\mathrm{H}_{10} = \frac{x \times 809}{2878} \text{ liters/hour} \]\[ \text{Rate of supply of } \mathrm{O}_2 = 3 \times 6.5 \times \frac{x \times 809}{2878} \times 22.4 \text{ liters/hour} \]

Key Concepts

Calorific Output

Calorific output refers to the amount of energy released when a fuel undergoes complete combustion. This energy is usually measured in kilojoules (kJ) per mole and is crucial in determining how much energy different fuels can produce. For example, methane and butane, common fuels in burners, have heats of combustion of 809 kJ/mol and 2878 kJ/mol respectively. This difference indicates how much energy each mole of fuel releases. In practical applications, knowing the calorific output helps in tuning burners and other combustion systems to achieve the desired energy efficiency. Thus, to match the energy output when switching from methane to butane, adjustments in oxygen supply are necessary, as shown in our example exercise.

Combustion Reaction

A combustion reaction involves the chemical reaction of a fuel with an oxidant, typically oxygen, leading to the production of heat and light. This process can be understood through a chemical equation which shows reactants transforming into products. For example, the combustion of butane can be described by the equation:

• \[\text{2 C}_4\text{H}_{10} + 13 \text{O}_2 \to 8 \text{CO}_2 + 10 \text{H}_2\text{O} \]

This reveals that 2 moles of butane require 13 moles of oxygen for complete combustion. Adjustments in fuel and oxygen supply ratios are crucial to meet the thermal demands and ensure efficient energy release. Moreover, understanding the stoichiometric coefficients in these equations is vital for calculating precise amounts of reactants needed for optimal energy output.

Ideal Gases

The concept of ideal gases is a simplifying assumption where gases are treated as if they respond predictably to changes in pressure, volume, and temperature according to the ideal gas law: \(PV = nRT\). This assumption simplifies calculations and is often a good approximation for real gases under many conditions. Ideal gas behavior assumes that the volume occupied by the gas particles themselves is negligible and that there are no intermolecular forces. In our exercise, we use the fact that at standard conditions, 1 mole of any ideal gas occupies 22.4 liters. This allows us to convert between volumes and moles effortlessly during calculations involving gases such as methane and butane.

Heat of Combustion

The heat of combustion is the total energy released when one mole of a substance is completely burned in oxygen. It's an essential measure of a fuel's energy efficiency and desirability. In the exercise, we compare the heats of combustion for methane (809 kJ/mol) and butane (2878 kJ/mol). These values guide how we adjust the amounts of fuel and oxygen to ensure consistent calorific outputs when changing fuels. Calculating the necessary supply rates of fuel and oxidant is directly linked to these measures, ensuring that the combustion system delivers the same energy output despite using different fuels.