AIIMS 2003 Chemistry Calorimetry MCQ Question

To find the molar heat of decomposition for ammonium nitrate (NH₄NO₃) based on the provided data, we will follow these steps:

Step 1: Calculate the Total Heat Absorbed by the Calorimeter

The heat absorbed by the calorimeter can be calculated using the formula:

$$q = C \cdot \Delta T$$

where:

• $q$ is the heat absorbed (in kJ),
• $C$ is the heat capacity of the system (in kJ/g/°C or kJ/g/K),
• $\Delta T$ is the change in temperature (in °C or K).

Given Data:

• The heat capacity $C = 1.23 \, \text{kJ/g/K}$
• The temperature change $\Delta T = 6.12 \, \text{K}$
• The mass of the sample $m = 1 \, \text{g}$

Substituting the values into the formula:

$$q = (1.23 \, \text{kJ/g/K}) \times (6.12 \, \text{K}) = 7.52 \, \text{kJ}$$

Step 2: Determine the Molar Heat of Decomposition

Now, we need to convert the heat absorbed by the calorimeter to the molar heat of decomposition. First, we must find the number of moles of NH₄NO₃ in the 1 gram sample.

The molar mass of NH₄NO₃ is calculated as follows:

• N: 14.01 g/mol (2 Nitrogens)
• H: 1.01 g/mol (4 Hydrogens)
• O: 16.00 g/mol (3 Oxygens)

Calculating the molar mass:

$$\text{Molar mass of } NH_{4}NO_{3} = 2(14.01) + 4(1.01) + 3(16.00) = 14.01 \times 2 + 4.04 + 48.00 = 80.05 \, \text{g/mol}$$

Now, we can calculate the number of moles in 1 gram:

$$\text{Moles of } NH_{4}NO_{3} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \, \text{g}}{80.05 \, \text{g/mol}} \approx 0.0125 \, \text{mol}$$

Next, we can find the molar heat of decomposition by dividing the total heat absorbed by the number of moles:

$$\text{Molar heat of decomposition} = \frac{q}{\text{moles}} = \frac{7.52 \, \text{kJ}}{0.0125 \, \text{mol}} \approx 601.6 \, \text{kJ/mol}$$

Since decomposition is an exothermic process, we express this value as a negative:

$$\text{Molar heat of decomposition} \approx -602 \, \text{kJ/mol}$$

Final Answer

Thus, the molar heat of decomposition for NH₄NO₃ is approximately -602 kJ/mol, which corresponds to option D.

Explanation of Other Options

• Option A (-7.53 kJ/mol): This value seems to be incorrectly derived, likely from a miscalculation of heat or moles.
• Option B (-398.1 kJ/mol): This is also incorrect and appears to be a miscalculation, potentially based on an incorrect interpretation of the heat or conversion factors.
• Option C (-16.1 kJ/mol): Again, this value does not reflect the calculations based on the given data and is significantly lower than expected based on the heat absorbed.

Conclusion

After careful calculations, we confirm that the correct answer is D) -602 kJ/mol, indicating a significant amount of energy released during the decomposition of ammonium nitrate.