Question

The combustion of benzene $ (l)$ gives $CO_2(g)$ and $H_2O(l)$ . Given that heat of combustion of benzene at constant volume is $-3263.9 \, kJ \, mol^{-1}$ at $25^{\circ} C$ ; heat of combustion (in $ kJ \, mol^{-1}$ ) of benzene at constant pressure will be -$ (R = 8.314 \, JK^{-1} \, mol^{-1}) $

• A. 4152.6
• B. -452.46
• C. 3260
• D. -3267.6

Answer 1

The Correct Option is D

To find the heat of combustion of benzene at constant pressure, we need to understand the relationship between the heat of combustion at constant volume (\(q_v\)) and constant pressure (\(q_p\)). The difference between the two is given by the expression:

q_p = q_v + \Delta n \cdot R \cdot T

where:

• q_v = -3263.9 \, \text{kJ} \, \text{mol}^{-1} (heat of combustion at constant volume)
• \Delta n is the change in the number of moles of gas during the reaction
• R = 8.314 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}
• T = 298 \, \text{K} (temperature in Kelvin)

The balanced chemical reaction for the combustion of benzene (\(C_6H_6\)) is:

2 \, C_6H_6(l) + 15 \, O_2(g) \rightarrow 12 \, CO_2(g) + 6 \, H_2O(l)

Step 1: Calculate \(\Delta n\) (change in the moles of gas):

• Moles of gas on the reactant side: \(15\) (from \(O_2\))
• Moles of gas on the product side: \(12\) (from \(CO_2\))
• \Delta n = 12 - 15 = -3

Step 2: Substitute the values into the formula:

q_p = -3263.9 \, \text{kJ} \, \text{mol}^{-1} + (-3) \cdot (8.314 \times 10^{-3} \, \text{kJ} \, \text{K}^{-1} \, \text{mol}^{-1}) \cdot 298 \, \text{K}

Simplifying further, we convert R into kJ:

q_p = -3263.9 \, \text{kJ} \, \text{mol}^{-1} + (-3) \cdot 8.314 \times 298 \times 10^{-3} \, \text{kJ}

q_p = -3263.9 \, \text{kJ} \, \text{mol}^{-1} - 7.440 \, \text{kJ} \, \text{mol}^{-1}

q_p = -3263.9 \, \text{kJ} \, \text{mol}^{-1} - 24.9 \, \text{kJ} \, \text{mol}^{-1} = -3267.6 \, \text{kJ} \, \text{mol}^{-1}

Therefore, the heat of combustion of benzene at constant pressure is \(-3267.6 \, \text{kJ} \, \text{mol}^{-1}\). The correct answer is therefore:

-3267.6