### 8.51

Posted:

**Thu Jan 18, 2018 2:28 pm**For homework problem 8.51, I understand how to come up with the calculation formatting, but I'm a bit confused with the signs of some of the values in the solutions manual. Can someone explain the reason the final answer is +23.9x10^3 kJ/L, a positive value? Energy is being released so I would originally expect to be working with a negative value in the middle of the calculations. Instead of explicitly saying that +3292 kJ/mol will be released, can't you technically say that the reaction enthalpy per mol of TNT is -3292 kJ/mol? Is that wrong?

Here is the question from the book?

8.51 The enthalpy of formation of trinitrotoluene (TNT) is 67 kJ·mol 1, and the density of TNT is 1.65 g·cm 3. In principle, it could be used as a rocket fuel, with the gases resulting from its decomposition streaming out of the rocket to give the required thrust. In practice, of course, it would be extremely dangerous as a fuel because it is sensitive to shock. Explore its potential as a rocket fuel by calculating its enthalpy density (enthalpy released per liter) for the reaction

Here is the question from the book?

8.51 The enthalpy of formation of trinitrotoluene (TNT) is 67 kJ·mol 1, and the density of TNT is 1.65 g·cm 3. In principle, it could be used as a rocket fuel, with the gases resulting from its decomposition streaming out of the rocket to give the required thrust. In practice, of course, it would be extremely dangerous as a fuel because it is sensitive to shock. Explore its potential as a rocket fuel by calculating its enthalpy density (enthalpy released per liter) for the reaction