4D.9

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Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

4D.9

Postby Ellis Song 4I » Tue Feb 04, 2020 5:56 pm

The enthalpy of formation of trinitrotoluene (TNT) is -67 kJ/mol, and the density of TNT is 1.65 g/cm^3. In principle, it could be used as 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 fuel because it is sensitive to shock. Explore its potential as a rocket fuel by calculating its enthalpy density (enthalpy change per liter) for the reaction

4 C7H5N3O6 + 21 O2 -> 28 CO2 + 10 H2O + 6 N2

How would you approach this problem? The solutions manual says to use the enthalpies of formation but how/where are the enthalpies of formation found? Also why is the enthalpy of formation for TNT multiplied by 4?

BNgo_2L
Posts: 95
Joined: Wed Sep 11, 2019 12:17 am

Re: 4D.9

Postby BNgo_2L » Tue Feb 04, 2020 9:32 pm

All the enthalpies of formation are found in Appendix 2A of the textbook.

After finding the enthalpy for the reaction, you need to find how much energy is released per mole of TNT. The reaction requires 4 moles of TNT, but since you want the energy per mole of TNT, you divide the reaction enthalpy by 4.

Now, all you have to do if multiply this energy by the density of TNT while also trying to get the correct units (kJ/L). Therefore you would need to convert cm^3 to L and moles to grams of TNT. (kj/mol TNT divided by molar mass of TNT)(density of TNT)(conversion for cm^3 to L).

Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

Re: 4D.9

Postby Ellis Song 4I » Wed Feb 05, 2020 10:39 pm

BNgo_2L wrote:After finding the enthalpy for the reaction, you need to find how much energy is released per mole of TNT. The reaction requires 4 moles of TNT, but since you want the energy per mole of TNT, you divide the reaction enthalpy by 4.

Now, all you have to do if multiply this energy by the density of TNT while also trying to get the correct units (kJ/L). Therefore you would need to convert cm^3 to L and moles to grams of TNT. (kj/mol TNT divided by molar mass of TNT)(density of TNT)(conversion for cm^3 to L).


How do you know the reaction requires 4 moles of TNT?

chemboi
Posts: 101
Joined: Sat Jul 20, 2019 12:16 am

Re: 4D.9

Postby chemboi » Wed Feb 05, 2020 11:21 pm

The 4 in front of the TNT molecule in the given equation indicates this.


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