Calculation of Heat of Reaction at Constant Pressure and Constant Volume for the Combustion of 2 Moles of Liquid Benzene at 298 K

1. Introduction:
The heat of reaction, also known as enthalpy change (∆H), is the amount of heat energy released or absorbed during a chemical reaction. It can be expressed at constant pressure (∆H_p) or constant volume (∆H_v). In this case, we will calculate the difference between the two for the combustion of 2 moles of liquid benzene at 298 K.

2. Combustion Reaction of Benzene:
The combustion reaction for benzene (C₆H₆) can be represented as follows:
C₆H₆ (l) + 15/2 O₂ (g) → 6 CO₂ (g) + 3 H₂O (g)

3. Calculation of Heat of Reaction at Constant Pressure (∆H_p):
To calculate ∆H_p, we need to refer to the standard enthalpies of formation (∆H_f) for the reactants and products involved in the combustion reaction. Given that benzene is in liquid form, its standard enthalpy of formation is considered zero. The standard enthalpies of formation for CO₂ and H₂O are -393.5 kJ/mol and -285.8 kJ/mol, respectively. Using Hess's law, we can calculate ∆H_p as follows:

∆H_p = ∑∆H_f(products) - ∑∆H_f(reactants)
∆H_p = [6 mol(-393.5 kJ/mol) + 3 mol(-285.8 kJ/mol)] - [0 + 15/2 mol(0)]
∆H_p = -3993.3 kJ

4. Calculation of Heat of Reaction at Constant Volume (∆H_v):
To calculate ∆H_v, we need to consider the change in internal energy (∆U) and the work done (∆W) during the reaction. At constant volume, ∆U = q_v (heat exchanged) since no work is done. Therefore, ∆H_v = ∆U.

5. Calculation of ∆U:
The change in internal energy (∆U) can be calculated using the equation:
∆U = q_v = ∆H - P∆V

Since the combustion reaction is carried out at constant volume, ∆V = 0. Therefore, P∆V = 0, and the equation simplifies to:
∆U = ∆H

6. Conclusion:
In this case, the difference