Benzene burns in oxygen according to the following reactions: (l) + 15/2 (g) - 3(l) + 6(g). If the standard enthalpies of formation of (l), (l) and (g) are 11.7, -68.1 and -94 kcal/mole, respectively

Question

Benzene burns in oxygen according to the following reactions: $C_6H_6$(l) + 15/2 $O_2$(g) -> 3$H_2O$(l) + 6$CO_2$(g). If the standard enthalpies of formation of $C_6H_6$(l), $H_2O$(l) and $CO_2$(g) are 11.7, -68.1 and -94 kcal/mole, respectively, the amount of heat that will liberate by burning 780 g of benzene is

Prashant Kotian · Accepted Answer

Correct answer: A. Enthalpy of combustion of 1 mole of benzene: δ H° = [6 * delta_f H°(, g) + 3 * delta_f H°(, l)] - [delta_f H°(, l)] = [6 * (-94) + 3 * (-68.1)] - [11.7] = -564 - 204.3 - 11.7 = -780 kcal/mol. Molar mass of benzene = 78 g/mol. Moles of benzene in 780 g: n = 780 g / 78 g/mol = 10 mol. Heat liberated by 10 moles of benzene = 10 * 780 kcal = 7800 kcal.

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