## Homework Problem 8.51

$\Delta U=q+w$

NinaSheridan
Posts: 30
Joined: Wed Sep 21, 2016 2:58 pm

### Homework Problem 8.51

For question 51 in chapter 8, the problem tells us the enthalpy of formation for TNT (-67kJ/mol) and the density (1.65g/cm^3). It then gives us the equation 4 C7H5N3O6 + 21 O2 --> 28 CO2 + 10 H20 + 6N2. It asks to find its enthalpy density.

I understand how to find the enthalpy of the equation, but it says only 1/4 of the energy will be released. Why is that? And after we divide the enthalpy by 4, we divide it by the molar mass and multiply by the density... I could use some clarification as to why we do all of this. Some individual steps make sense but all together it becomes hazy.

Yuchien Ma 2L
Posts: 46
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Homework Problem 8.51

I have the same problem. Since density is g*cm^-3, I end up with an extra "g" unit that I cannot cancel out.
I don't really understand the solutions manual's way of explaining it.

Emily_Lenh2A
Posts: 17
Joined: Sat Jul 09, 2016 3:00 am

### Re: Homework Problem 8.51

I believe the reason they say that only 1/4 of the energy will be released is because the enthalpy of the reaction is for 4 moles of TNT, and they want to find how much energy is released with only 1 mole of TNT. That way, you can use it more easily to find the enthalpy released per liter, which is what the question asks for.

As for the extra "g" unit, I had no problem with that when I converted 1.65 g/cm^3 to liters (using 10^3 cm^3/L) and then to mol TNT (mol TNT/227.1311 g TNT). Using this, I got mol TNT/L and used it to find the enthalpy released per liter.