Chemical Explosive - Example of Thermochemical Calculations

Example of Thermochemical Calculations

The PETN reaction will be examined as an example of thermo-chemical calculations.

PETN: C(CH₂ONO₂)₄

Molecular weight = 316.15 g/mol

Heat of formation = 119.4 kcal/mol

(1) Balance the chemical reaction equation. Using table 1, priority 4 gives the first reaction products:

5C + 12O → 5CO + 7O

Next, the hydrogen combines with remaining oxygen:

8H + 7O → 4H₂O + 3O

Then the remaining oxygen will combine with the CO to form CO and CO₂.

5CO + 3O → 2CO + 3CO₂

Finally the remaining nitrogen forms in its natural state (N₂).

4N → 2N₂

The balanced reaction equation is:

C(CH₂ONO₂)₄ → 2CO + 4H₂O + 3CO₂ + 2N₂

(2) Determine the number of molar volumes of gas per mole. Since the molar volume of one gas is equal to the molar volume of any other gas, and since all the products of the PETN reaction are gaseous, the resulting number of molar volumes of gas (N_m) is:

N_m = 2 + 4 + 3 + 2 = 11 V_molar/mol

(3) Determine the potential (capacity for doing work). If the total heat liberated by an explosive under constant volume conditions (Q_m) is converted to the equivalent work units, the result is the potential of that explosive.

The heat liberated at constant volume (Q_mv) is equivalent to the heat liberated at constant pressure (Q_mp) plus that heat converted to work in expanding the surrounding medium. Hence, Q_mv = Q_mp + work (converted).

a. Q_mp = Q_fi (products) − Q_fk (reactants)

where: Q_f = heat of formation (see table 1)

For the PETN reaction:

Q_mp = 2(26.343) + 4(57.81) + 3(94.39) − (119.4) = 447.87 kcal/mol

(If the compound produced a metallic oxide, that heat of formation would be included in Q_mp.)

b. Work = 0.572N_m = 0.572(11) = 6.292 kcal/mol

As previously stated, Q_mv converted to equivalent work units is taken as the potential of the explosive.

c. Potential J = Q_mv (4.185 × 106 kg)(MW) = 454.16 (4.185 × 106) 316.15 = 6.01 × 106 J kg

This product may then be used to find the relative strength (RS) of PETN, which is

d. RS = Pot (PETN) = 6.01 × 106 = 2.21 Pot (TNT) 2.72 × 106