8.51 The enthalpy formation of trinitrotoluene

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Nathan Mariano 2G
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Joined: Fri Sep 28, 2018 12:29 am

8.51 The enthalpy formation of trinitrotoluene

Postby Nathan Mariano 2G » Sat Feb 09, 2019 4:19 pm

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 4 C7H5N3O6 (s) +  21 O2 (g) --> 28 CO2 (g) + 10 H2O (g) + 6 N2 (g).

After finding the total enthalpy for this reaction (-13168.48 kJ), why must you divide by four before calculating enthalpy density?

almaochoa2D
Posts: 67
Joined: Fri Sep 28, 2018 12:23 am

Re: 8.51 The enthalpy formation of trinitrotoluene

Postby almaochoa2D » Sun Feb 10, 2019 3:35 pm

Because there are 4 TNT and you want to find it for just 1 mole of TNT.


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