Question

IIeat of combustion of \(\mathrm{CII}_{4}, \mathrm{C}_{2} \mathrm{II}_{4}\) and \(\mathrm{C}_{2} \mathrm{II}_{6}\) are 890 , 1411 and \(1560 \mathrm{~kJ} / \mathrm{mol}\), respectively. Which has the lowest calorific fucl value in \(\mathrm{kJ} / \mathrm{g}\) ? (1) \(\mathrm{CH}_{4}\) (2) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (3) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (4) All

Short answer

\(\text{C}_2\text{H}_4\) has the lowest calorific fuel value at 50.393 kJ/g.

Step-by-step solution

- Determine the molar mass of each compound

Calculate the molar mass of each compound: \(\text{CH}_4\): Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol Molar mass of \(\text{CH}_4\) = 12 + (4 * 1) = 16 g/mol. \(\text{C}_2\text{H}_4\): Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol Molar mass of \(\text{C}_2\text{H}_4\) = (2 * 12) + (4 * 1) = 24 + 4 = 28 g/mol. \(\text{C}_2\text{H}_6\): Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol Molar mass of \(\text{C}_2\text{H}_6\) = (2 * 12) + (6 * 1) = 24 + 6 = 30 g/mol.

- Compute the calorific fuel value for each compound

Use the given heat of combustion and the molar mass to calculate the calorific fuel value for each compound in kJ/g. For \(\text{CH}_4\): \( \frac{890\text{ kJ/mol}}{16\text{ g/mol}} = 55.625\text{ kJ/g} \) For \(\text{C}_2\text{H}_4\): \( \frac{1411\text{ kJ/mol}}{28\text{ g/mol}} = 50.393\text{ kJ/g} \) For \(\text{C}_2\text{H}_6\): \( \frac{1560\text{ kJ/mol}}{30\text{ g/mol}} = 52\text{ kJ/g} \)

- Identify the compound with the lowest calorific fuel value

Compare the calorific fuel values calculated in Step 2. \(\text{CH}_4\) = 55.625 kJ/g \(\text{C}_2\text{H}_4\) = 50.393 kJ/g \(\text{C}_2\text{H}_6\) = 52 kJ/g \(\text{C}_2\text{H}_4\) has the lowest calorific fuel value at 50.393 kJ/g.

Key concepts

heat of combustion

The heat of combustion is a crucial concept in thermodynamics and energy calculations in chemistry. It represents the amount of heat released when one mole of a substance burns completely in oxygen. This value is essential for determining how much energy can be obtained from a fuel.
For example, in the problem above, we used the heats of combustion for \(\text{CH}_4\), \(\text{C}_2\text{H}_4\), and \(\text{C}_2\text{H}_6\) which are 890 kJ/mol, 1411 kJ/mol, and 1560 kJ/mol respectively.
These values help us understand the potential energy output and efficiency of various fuels, making them critical for industries and everyday applications.

molar mass

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is the sum of the atomic masses of all the atoms in a molecule.
To calculate the molar mass, you need to add up the respective atomic masses of the elements in the compound. For instance:

• For \(\text{CH}_4\), the molar mass is calculated as 12 g/mol (carbon) + 4 * 1 g/mol (hydrogen) = 16 g/mol.
• For \(\text{C}_2\text{H}_4\), it is calculated as 2 * 12 g/mol (carbon) + 4 * 1 g/mol (hydrogen) = 28 g/mol.
• For \(\text{C}_2\text{H}_6\), the calculation is 2 * 12 g/mol (carbon) + 6 * 1 g/mol (hydrogen) = 30 g/mol.

Molar mass is important for various calculations, including determining the amount of a substance needed in a reaction and converting between mass and moles.

energy calculations in chemistry

Energy calculations are central to understanding how much energy is involved in chemical reactions. These calculations often involve the heat of combustion and the molar mass of the substances.
For instance, to find the calorific fuel value (energy per gram) of different fuels, you divide the heat of combustion by the molar mass. Here's how it's done for the given compounds:

• For \(\text{CH}_4\), the calorific fuel value is \(\frac{890 \text{ kJ/mol}}{16 \text{ g/mol}} = 55.625 \text{ kJ/g}\).
• For \(\text{C}_2\text{H}_4\), it is \(\frac{1411 \text{ kJ/mol}}{28 \text{ g/mol}} = 50.393 \text{ kJ/g}\).
• For \(\text{C}_2\text{H}_6\), the value is \(\frac{1560 \text{ kJ/mol}}{30 \text{ g/mol}} = 52 \text{ kJ/g}\).

These calculations help compare the energy outputs of different fuels, leading to better energy management and utilization choices.

thermodynamics

Thermodynamics is the study of energy changes, especially heat and work, in physical processes and chemical reactions. It provides the framework for understanding how energy is transferred and transformed.
Key principles in thermodynamics include:

• First Law: Energy cannot be created or destroyed; it can only change forms. This is the law of energy conservation.
• Second Law: The total entropy of a system and its surroundings always increases over time. Entropy is a measure of disorder or randomness.
• Third Law: As temperature approaches absolute zero, the entropy of a perfect crystal approaches a constant minimum.

In our example, the heat of combustion of various fuels is a thermodynamic quantity. By calculating and comparing these values, we can make informed decisions on fuel efficiency and energy utilization. Thermodynamics helps us understand these energy changes in greater detail.