Textbook question 4D.9

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Paige Lee 1A
Posts: 136
Joined: Sat Sep 07, 2019 12:16 am

Textbook question 4D.9

Postby Paige Lee 1A » Sat Feb 01, 2020 12:53 pm

Could someone please explain how to get the correct answer?

The enthalpy of formation of trinitrotoluene (TNT) is 267 kJ?mol21, and the density of TNT is 1.65 g?cm23. In princi- ple, 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 change per liter) for the reaction
4 C7H5N3O6(s) 1 21 O2(g) ¡ 28 CO2(g) 1 10 H2O(g) 16 N2(g)

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: Textbook question 4D.9

Postby Aman Sankineni 2L » Sat Feb 01, 2020 1:26 pm

That is correct the O2 and N2 are diatomic gases in their most stable form and thus their standard enthalpy of formation is zero. Using the enthalpies of formation we calculate that CO2 requires -393.51 kJ/mol while H20 requires -241.82. Given TNT's enthalpy of formation we calculate:

28(-393.51 kJ/mol)+10(-241.82)-4(-67)=-13168 kJ/mol.

TNT is negative while CO2 and H2O are positive because the enthalpy of the reaction is equal to the sum of the products minus the sum of the reactants.

-13168 kJ/mol is the value released for the total reaction so we divide this value by 4 getting 3292 kJ/mol before converting this value to kJ/L using the molar mass of TNT as well as its density.

Paige Lee 1A
Posts: 136
Joined: Sat Sep 07, 2019 12:16 am

Re: Textbook question 4D.9

Postby Paige Lee 1A » Sat Feb 01, 2020 8:22 pm

Aman Sankineni 2L wrote:That is correct the O2 and N2 are diatomic gases in their most stable form and thus their standard enthalpy of formation is zero. Using the enthalpies of formation we calculate that CO2 requires -393.51 kJ/mol while H20 requires -241.82. Given TNT's enthalpy of formation we calculate:

28(-393.51 kJ/mol)+10(-241.82)-4(-67)=-13168 kJ/mol.

TNT is negative while CO2 and H2O are positive because the enthalpy of the reaction is equal to the sum of the products minus the sum of the reactants.

-13168 kJ/mol is the value released for the total reaction so we divide this value by 4 getting 3292 kJ/mol before converting this value to kJ/L using the molar mass of TNT as well as its density.




Could you please explain how to go from 3292 kJ/mol to 23.9*10^3 kJ/L?

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

Re: Textbook question 4D.9

Postby Aman Sankineni 2L » Thu Feb 06, 2020 1:05 pm

To get energy density in kJ/L from kJ/L, you divide the amount of energy with the mass of one mole of TNT and multiply it by the density of TNT. The answer should be 23.9 x 10^3 kJ/L.

3292 / 227.13 x 1650=23.9 x 10^3 kJ/L


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