Nitroglycerin, one of the most commonly used explosives, has the following struc

Question

Nitroglycerin, one of the most commonly used explosives, has the following structure The decomposition reaction is: 4 C_3H_5N_3O_9(I) rightarrow 12 CO_2(g) + 10 H_2O(g) + 6 N_2(g) + O_2(g) The explosive action is the result of the heat released and the large increase in gaseous volume. (a) Calculate the Delta H degree for the decomposition of one mole of nitroglycerin using standard enthalpy of formation values. Assume that the two O atoms in the NO_2 groups are attached to N with one single bond and one double bond. (b) Calculate the combined volume of the gases at STP. (c) Assuming an initial explosion temperature of 3000K, estimate the pressure exerted by the gases using the result from (b). (The standard enthalpy of formation of nitroglycerin is -371.1 k/mol). You didn't think the last one was going to be a freebie, did you?? To get you started you should probably divide the decomposition reaction by 4 to get the decomposition value for 1 mole of nitroglycerin. This should really not be that bad because all you are doing is calculating enthalpy of reaction values (as you did in chapter 6). For part b, look at the reaction given and decide how much volume of gases are being produced by one mol of nitroglycerin and then correlate that value to one mol of gas at STP. The final part (c) should be a gimme ellipsis it is a simple plug-n-chug. Consider it a gift! Good luck!

Explanation / Answer

Enthalpy of formation (KJ/mole) : CO2= -393.5, H2O(g)=-241.8, C3H5N3O9:-371.1 Kj/mole, N2=0 and O2=0 (elements in their standard states will have zero enthalpy)

For the decomposition reactin of nitroglycerine, 4C3H5N3O9---------à12CO2(g)+10H2O(g)+6N2(g)+ O2(g)

Enthalpy change of the reaction = sum of enthalpy of products- sum of enthalpy of reactants

=12* enthalpy change of CO2+10* enthalpy change of water+6* enthalpy change of N2+1* enthalpy change of O2- 4*Enthalpy change of glycerine, where, 12,10,6 and 1 are coefficients of CO2, H2O, N2 and O2 in the reaction

=12*(-393.5)+10*(-241.8)+0+0- (-371.1)=-6769 Kj

This is the heat liberted upon decomposition of 4 moles of nitroglycerine.

For decomposition of one mole, heat liberted= -6769/4= -1692.25 KJ/mole

At STP one mole of any gas occupies 22.4 L

1 mole of glycerine gives 12/4=3 moles of CO2, 10/4=2.5 moles of H2O, 6/4=1.5 moles of N2 and ¼=0.25 moles of O2.

Total moles of gases per mole of nitroglycerine= 3+2.5+1.5+0.25= 7.25

Volume of 7.25 moles of gases at STP= 7.24*22.4= 162.4 L

Since the reaction is carried out in a reactor, the volume remains the same

Hence from gas law PV= nRT, when volume and no of moles of gas are fixed,

P1/T1= P2/T2, where P1= 1 atm, T1= 27+273=300K,T2= 3000K

P2= P1T2/T1= 1*3000/300= 10 atm, pressure in the reactor.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.