Problem 24 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In a constant volume calorimeter, \(3.5 \mathrm{~g}\) of a gas with molecular weight 28 was burnt in excess oxygen at \(298.0 \mathrm{~K}\). The temperature of the calorimeter was found to increase from \(298.0 \mathrm{~K}\) to \(298.45 \mathrm{~K}\) due to the combustion process. Given that the heat capacity of the calorimeter is \(2.5 \mathrm{~kJ} \mathrm{~K}^{-1}\), the numerical value for the enthalpy of combustion of the gas in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is

Step-by-Step Solution

Answer

The enthalpy of combustion is \(9 \text{ kJ/mol}\).

Step 1: Calculate the Heat Absorbed by the Calorimeter

First, determine the amount of heat absorbed by the calorimeter due to the temperature change. Use the formula \( q = C \times \Delta T \), where \( C \) is the heat capacity and \( \Delta T \) is the change in temperature. Given that \( C = 2.5 \text{ kJ K}^{-1} \) and \( \Delta T = 298.45 \text{ K} - 298.0 \text{ K} = 0.45 \text{ K} \), the heat absorbed is \( q = 2.5 \times 0.45 = 1.125 \text{ kJ} \).

Step 2: Calculate the Moles of Gas Burnt

Determine the number of moles of the gas that was burnt. Use the formula \( n = \frac{m}{M} \), where \( m \) is the mass of the gas and \( M \) is the molar mass. Here, \( m = 3.5 \text{ g} \) and \( M = 28 \text{ g/mol} \). Thus, \( n = \frac{3.5}{28} \approx 0.125 \text{ mol} \).

Step 3: Calculate the Enthalpy of Combustion per Mole

Finally, calculate the enthalpy of combustion per mole of gas. The enthalpy of combustion, \( \Delta H_c \), is the heat absorbed \( q \) divided by the number of moles \( n \). Thus, \( \Delta H_c = \frac{1.125 \text{ kJ}}{0.125 \text{ mol}} = 9 \text{ kJ/mol} \).

Key Concepts

Constant Volume Calorimeter

A constant volume calorimeter is a device used to measure the heat change during chemical reactions at a fixed volume. It is particularly useful for combustion reactions because it provides an enclosed environment where no gases can escape. This ensures that all the heat generated by the reaction can be accurately recorded.

In this specific experiment, a constant volume calorimeter is employed to analyze the combustion of a gas. Since the volume remains constant, the system is perfectly suited to measure the internal energy change, rather than changes in enthalpy directly, because enthalpy changes also incorporate pressure and volume changes. However, since the pressure change is generally negligible in gases with constant volume conditions, the heat measured gives a very good approximation of the enthalpy change.

Using constant volume calorimeters helps provide reliable and precise data, crucial for understanding thermodynamic properties like the enthalpy of combustion.

Heat Capacity

Heat capacity is the amount of heat needed to raise the temperature of a system by one degree Kelvin. It is a key value when measuring heat absorption in calorimetry.

In the context of this problem, the heat capacity of the calorimeter is given as 2.5 kJ K⁻¹. This means for every Kelvin the temperature increases, the calorimeter absorbs 2.5 kilojoules of energy. Understanding heat capacity allows us to calculate the total heat absorbed during the combustion process using the formula:

• \( q = C \times \Delta T \)

where \( C \) is the heat capacity and \( \Delta T \) is the temperature change.

The accurate measurement of heat capacity is essential, as it impacts the precision of the calculated heat change and, subsequently, any determination of enthalpy changes in the system.

Temperature Change

Temperature change is a core factor in calculating heat transfer within a reactive system, such as a calorimeter. It is determined by the difference between the final and initial temperatures.

For our experiment:

• Initial temperature (\( T_i \)) = 298.0 K
• Final temperature (\( T_f \)) = 298.45 K
• Temperature change (\( \Delta T \)) = 298.45 K - 298.0 K = 0.45 K

This change of 0.45 K allows us to determine how much energy was absorbed by the calorimeter over the course of the reaction through its heat capacity.

Temperature differences are critical for not only identifying the amount of energy exchanged but also in understanding the dynamics of the reaction itself.

Moles of Gas

Understanding the amount of substance involved in a reaction, in terms of moles, is crucial for accurate calorimetry measurements. Moles quantify the number of molecules in a given mass of a substance and are calculated using the formula:

• \( n = \frac{m}{M} \)

where \( n \) is the number of moles, \( m \) is the mass (in grams), and \( M \) is the molar mass (g/mol).

In this exercise:

• Mass \( m = 3.5 \text{ g} \)
• Molar Mass \( M = 28 \text{ g/mol} \)
• Moles \( n = \frac{3.5}{28} \approx 0.125 \text{ mol} \)

Knowing the moles of gas combusted allows you to link the heat absorbed by the calorimeter to the enthalpy change per mole, giving valuable insights into the energy efficiency of the reaction. Understanding this relationship aids in various calculations and analyses, especially when predicting the behavior of chemical systems in different conditions.