8.51 Sig Figs, Sign  [ENDORSED]

Moderators: Chem_Mod, Chem_Admin

Michael_Johanis_2K
Posts: 26
Joined: Thu Jul 28, 2016 3:00 am
Been upvoted: 1 time

8.51 Sig Figs, Sign

Postby Michael_Johanis_2K » Thu Jan 19, 2017 11:06 pm

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).

[6 mol(0 kJ)/mol + 10 mol(-241.82 kJ/mol) + 29 mol*-393.5 kJ/mol] - [21 mol* 0kJ/mol + 4mol(-67 kJ/mol)]
=-13168.2 kJ
= -13168.2 kJ per REACTION

-13168.2 kJ / 4 mol (1 mol/ molar mass) (1.65 g / 1 cm^3) (1000 cm^3 / 1L)
= -23915.09122 kJ/L
= -24000 kJ/L (2 sig figs, negative sign)

Why does the book say that the answer is +23.9 *10^3 kJ/L?

Isn't the answer limited by 2 sig figs (-67 kJ/mol)? Also, shouldn't the answer be negative because the enthalpy is negative?

EllisJang2O
Posts: 18
Joined: Wed Sep 21, 2016 2:55 pm

Re: 8.51 Sig Figs, Sign

Postby EllisJang2O » Fri Jan 20, 2017 5:58 pm

The "-67 kJ/mol" value is used in addition/subtraction, in which the answer is rounded to the same precision as the least precise number. Here, the least precise number is to the left of the decimal point (a whole number). That's why the answer is -13168 kJ/mol.
Then in multiplication/division, the least number of significant figures determines the number of significant figures in the answer. Here, the least number of sigfigs is 3 (from 1.65). Therefore the answer is 23.9 x 10^3 kJ/L.

EllisJang2O
Posts: 18
Joined: Wed Sep 21, 2016 2:55 pm

Re: 8.51 Sig Figs, Sign  [ENDORSED]

Postby EllisJang2O » Fri Jan 20, 2017 6:11 pm

When enthalpy is negative it just means heat is released from the reaction. Since the question asks for the enthalpy density (enthalpy released per liter), you would simply put the positive number. It would be unconventional to say negative 23.9 x 10^3 kJ/L released (aka negative heat released ???).

Tim Foster 2A
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am

Re: 8.51 Sig Figs, Sign

Postby Tim Foster 2A » Thu Jan 18, 2018 5:35 pm

Why is the answer to this problem written as "23.9 KJ/L x 10^3" and not "2.39 KJ/L x 10^4"?


Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 1 guest