The enthalpy of combustion of benzene can be determined using the heat produced during the combustion process. In this case, the heat produced by the combustion of 1 g of benzene in a bomb calorimeter at 298 K is 41.8 kJ.

Enthalpy of Combustion:
The enthalpy of combustion is defined as the heat released when one mole of a substance undergoes complete combustion with oxygen under standard conditions. It represents the energy content of the substance and is typically expressed in units of energy per mole (kJ/mol).

Heat Produced:
The heat produced during the combustion of benzene can be used to calculate the enthalpy of combustion. The heat produced in the bomb calorimeter is a measure of the energy released by the combustion reaction. In this case, 1 g of benzene produces 41.8 kJ of heat.

Conversion to Moles:
To calculate the enthalpy of combustion, it is necessary to convert the mass of benzene to moles. The molar mass of benzene is 78.11 g/mol. Therefore, 1 g of benzene is equal to 0.0128 moles (1 g / 78.11 g/mol).

Enthalpy Calculation:
The enthalpy of combustion can be calculated by dividing the heat produced by the number of moles of benzene consumed in the reaction. In this case, the enthalpy of combustion is given by:

Enthalpy of combustion (ΔH) = Heat produced / Moles of benzene

Substituting the values, we get:

ΔH = 41.8 kJ / 0.0128 moles ≈ 3259 kJ/mol

Therefore, the enthalpy of combustion of benzene is approximately 3259 kJ/mol.

Explanation:
The enthalpy of combustion represents the energy released when a substance undergoes complete combustion. In this case, benzene undergoes combustion and releases heat, which is measured in the bomb calorimeter. By knowing the mass of benzene and the heat produced, we can calculate the enthalpy of combustion.

The conversion of mass to moles allows us to relate the heat produced to the number of moles of benzene consumed. Dividing the heat produced by the number of moles gives us the enthalpy of combustion.

The calculated enthalpy of combustion for benzene is approximately 3259 kJ/mol. This value represents the energy content of benzene when it undergoes combustion with oxygen under standard conditions.

By understanding the enthalpy of combustion, we can gain insights into the energy content and reactivity of substances.

C6H6(l) +15/2O2(g) —>6CO2(g) + 3H2O(l)

∆ng= 6-7.5= -1. 5 (change in gaseous mole)
∆U or ∆E = -3263. 9KJ
∆H=∆U+∆ngRT
∆ng = -1. 5
R =8.314JK^-1 mol^-1
T =298k
so, ∆H= -3263. 9+(-1.5) 8.314×10^-3×298
=-3267.6 KJ


∆H= heat at constant pressure
∆U/∆E =heat at constant volume