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 = 8.314 J\(K^{-1}\) \(mol^{-1}\))

• 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: {Problem Identification}
The problem concerns the combustion of benzene. We are provided with the heat of combustion at constant volume and are tasked with determining the heat of combustion at constant pressure.The balanced chemical equation for the combustion of benzene is:\[{C}_6{H}_6(l) + \frac{15}{2} {O}_2(g) \rightarrow 6{CO}_2(g) + 3{H}_2{O}(l)\]Step 2: {Application of Enthalpy-Internal Energy Relationship}
The relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) is given by:\[\Delta H = \Delta U + \Delta n_{{g}} RT\]where \( \Delta n_{{g}} \) represents the change in the number of moles of gaseous reactants and products.For the given reaction, the change in the number of moles of gas is calculated as follows:\[\Delta n_{{g}} = (6 \text{ mol CO}_2) - \left( \frac{15}{2} \text{ mol O}_2 \right) = 6 - 7.5 = -1.5\]Substituting the values into the equation:\[\Delta H = \Delta U + (-1.5) \times (8.314 \times 10^{-3} \times 298)\]\[\Delta H = -3263.9 + (-1.5) \times 2.478 \approx -3267.6 \, {kJ/mol}\]Therefore, the correct option is (D).