### 8.51

Posted:

**Sat Jan 13, 2018 9:25 pm**Can someone please explain this problem to me? I'm super confused! Are we just supposed to use enthalpies of formation and if so what do we do next?

Thanks!

Thanks!

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=25448

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Posted: **Sat Jan 13, 2018 9:25 pm**

Can someone please explain this problem to me? I'm super confused! Are we just supposed to use enthalpies of formation and if so what do we do next?

Thanks!

Thanks!

Posted: **Sat Jan 13, 2018 11:14 pm**

Yes, you use enthalpies of formation to find the enthalpy of reaction for the reaction.

28(-393.51kJ/mol)+10(-241.82kJ/mol)-4(067kJ/mol)=-13168kJ/mol (the energy released per mole of reaction)

Since it is asking "per mole," 1/4 of the energy (3292kJ/mol) will be released per mole of TNT consumed.

In order to get the energy density in "kJ/L," you need to take 3292kJ/mol and divide it by the mass of one mole of TNT (227.14 g/mol) and then multiply it by the density of TNT (1.65 g/cm^3) and then convert cm^3 to L (10^3cm^3/L).

The answer should be 23.9x10^3kJ/L

28(-393.51kJ/mol)+10(-241.82kJ/mol)-4(067kJ/mol)=-13168kJ/mol (the energy released per mole of reaction)

Since it is asking "per mole," 1/4 of the energy (3292kJ/mol) will be released per mole of TNT consumed.

In order to get the energy density in "kJ/L," you need to take 3292kJ/mol and divide it by the mass of one mole of TNT (227.14 g/mol) and then multiply it by the density of TNT (1.65 g/cm^3) and then convert cm^3 to L (10^3cm^3/L).

The answer should be 23.9x10^3kJ/L