# Thermochemical Analysis of Trinitroglycerin?

Jul 7, 2017

Any suggestions/mistakes, call it out!

#### Explanation:

A. Simply writing and balancing a chemical equation.
$2 {C}_{3} {H}_{5} {N}_{3} {O}_{9} \left(l\right) \to 3 {N}_{2} \left(g\right) + 6 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right) + {O}_{2} \left(g\right)$

B. Important note, reducing the equation to one mole of the compound of interest is helpful prior to calculation.
${C}_{3} {H}_{5} {N}_{3} {O}_{9} \left(l\right) \to \frac{3}{2} {N}_{2} \left(g\right) + 3 C {O}_{2} \left(g\right) + \frac{5}{2} {H}_{2} O \left(l\right) + {O}_{2} \left(g\right)$
DeltaH_(rxn)^° = sumnDeltaH_(P)^°-summDeltaH_(R)^°
$- 1541.4 \frac{k J}{m o l} = \left[\frac{3}{2} \cdot 0 + 3 \cdot \left(- 393.5\right) + \frac{5}{2} \left(- 285.83\right) + 0\right] - {C}_{3} {H}_{5} {N}_{3} {O}_{9}$
$- 1541.4 \frac{k J}{m o l} = - 1895 - {C}_{3} {H}_{5} {N}_{3} {O}_{9}$
DeltaH_(f)^°[C_3H_5N_3O_9] = 353.7(kJ)/(mol)

C. Unit conversion!
$M M \left[{C}_{3} {H}_{5} {N}_{3} {O}_{9}\right] = 227.11 \frac{g}{m o l}$
$0.60 m g \left(\frac{{10}^{-} 3 g}{m g}\right) \left(\frac{{C}_{3} {H}_{5} {N}_{3} {O}_{9}}{227.11 g}\right) \left(\frac{1541.4 k J}{m o l}\right) = 0.0041 k J = 4.07 J \left(\frac{c a l}{4.184 J}\right) = 0.9733 c a l$

D. Molecular; an ionic compound has a metal and nonmetal bonded, and a much higher boiling point due to the higher lattice energy.

E. Energy stored in trinitroglycerin is chemical potential energy. The decomposition is exothermic and produces a large amount of heat. This is transferred to the surroundings, whereby the gases do work on the solid rock and transfer kinetic energy to them. When the rocks are moving, potential and kinetic energy are balanced until the rocks hit the ground.