Final answer:
Burning 1 ml of tetraethyl lead releases 31,878.5 joules of energy, calculated using its density (1.65 g/ml) and the enthalpy change per gram (-19.29 kJ/g or -19290 J/g).
Step-by-step explanation:
If the ∆h° rxn for the combustion of tetraethyl lead is -5900 kJ/mol, and the energy released by burning 1 ml of tetraethyl lead needs to be determined, we can start by finding the density of tetraethyl lead to convert volume to mass. Considering that the standard enthalpy (∆h° comb) of tetraethyl lead is given as -19.29 kJ/g, converting this value to joules gives us -19290 J/g. To find out the energy released by 1 ml of tetraethyl lead, we need to know its density to convert 1 ml to grams since the given enthalpy change is per gram.
Tetraethyl lead’s density is typically about 1.65 g/ml. Hence, for 1 ml of tetraethyl lead which is approximately 1.65 grams, the energy released can be calculated as follows: 1.65 g x -19290 J/g = -31878.5 J. Therefore, burning 1 ml of tetraethyl lead releases 31878.5 joules of energy.