The standard molar enthalpies of formation of trinitrotoluene(l), (g) and (l) are 65, -395 and -285 — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard molar enthalpies of formation of trinitrotoluene(l), $CO_2$(g) and $H_2O$(l) are 65, -395 and -285 kJ/mol, respectively. The density of trinitrotoluene is 1.816 g/ml. Trinitrotoluene can be used as rocket fuel, with the gases resulting from its combustion streaming out of the rocket to give the required thrust. What is the enthalpy density for the combustion reaction of trinitrotoluene?
💡 Solution & Explanation
The formula of TNT is C7H5N3O6. Standard combustion reaction:<br>C7H5N3O6(l) + 21/4 $O_2$(g) → 7$CO_2$(g) + 5/2 $H_2O$(l) + 3/2 $N_2$(g).<br>Enthalpy of combustion ΔHc = [7 × ΔfH($CO_2$) + 2.5 × ΔfH($H_2O$)] - ΔfH(TNT) = [7 × (-395) + 2.5 × (-285)] - 65 = [-2765 - 712.5] - 65 = -3542.5 kJ/mol.<br>Molar mass of TNT = 227 g/mol. Volume of 1 mole of liquid TNT = M / d = 227 g / 1.816 g/ml = 125 ml = 0.125 l.<br>Enthalpy density (heat per unit volume) = ΔHc / V = -3542.5 kJ / 0.125 l = -28340 kJ/l = -28.34 MJ/l.