Question

The combustion of benzene (L) gives CO\(_2\) (g) and H\(_2\)O (L). Given that heat of combustion of benzene at constant volume is -3263.9 kJ/mol at 25°C, heat of combustion (in kJ/mol) of benzene at constant pressure will be: (R = 8.314J\(K^-^1\) \(mol^-^1\))

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

Hint:

For reactions involving gases, the heat of combustion at constant pressure is adjusted by the work done due to changes in gas volume, which is linked to the change in the number of moles of gas.

Answer 1

The Correct Option is D

Step 1: {Understanding the Problem}
The given reaction is the combustion of benzene. We are given the heat of combustion at constant volume, and we need to find the heat of combustion at constant pressure. For the combustion of benzene: \[ {C}_6{H}_6(l) + \frac{15}{2} {O}_2(g) \rightarrow 6{CO}_2(g) + 3{H}_2{O}(l) \] Step 2: {Using the Relation between Heat at Constant Volume and Pressure}
The change in enthalpy is related to the change in internal energy by the equation: \[ \Delta H = \Delta U + \Delta n_{{g}} RT \] where \( \Delta n_{{g}} \) is the change in the number of moles of gas between products and reactants. In the given reaction, we calculate the change in the number of moles of gas: \[ \Delta n_{{g}} = (6 { mol CO}_2) - \left( \frac{15}{2} { mol O}_2 \right) = 6 - 7.5 = -1.5 \] Thus, we have: \[ \Delta H = \Delta U + (-1.5) \times (8.314 \times 10^{-3} \times 298) \] \[ \Delta H = -3263.9 + (-1.5) \times 2.478 \approx -3267.6 \, {kJ/mol} \] Thus, the correct answer is (D).