In the Classroom
The Relative Explosive Power of Some Explosives Marten J. ten Hoor J.W. Frisolaan 40, 9602 GJ Hoogezand, The Netherlands;
[email protected] Explosions may be characterized as being of a physical, chemical, or nuclear nature. For example, a physical explosion occurs when steam in a boiler is overheated to the point of rupture. Nuclear explosions have been well-known since the end of the Second World War. In this article, only chemical explosions will be considered. Such an explosion may happen if a chemical reaction fulfils three conditions; it should be fast, produce gas, and release energy. Efficiency of an Explosive During the short time it takes for an explosion to occur, the gas produced attains a high pressure. Initially, the gas occupies the small volume of the explosive substance (1). Using the ideal gas law, the increase in pressure, ∆p, can be calculated as,
∆p =
R ρVg ∆T nR ∆T nR ρ ∆T = = Vi m mVm
(1)
where R is the gas constant, n is the total amount of the gas produced, Vi is the initial volume of the explosive, m its mass, ρ its density, and ∆T the increase in temperature. The total volume of the produced gas, denoted by Vg, equals Vg = nVm, where Vm = 22.414 L mol᎑1 is the molar volume of an ideal gas at STP. The effect of the sudden increase in pressure, ∆p, depends on the design of the container in which the explosion takes place. If the container walls are not strong enough, the container may be blown apart. The device is classified as a bomb. However, if the container has strong walls and resembles a champagne bottle, its “cork” may be forced out. The device is now classified as a gun. It seems reasonable to relate the efficiency of an explosive to ∆p. However, choosing ∆p as a measure of the “explosive power” of an explosive creates a problem that is difficult to solve. This problem is the determination of ∆T or, more specifically, of the final temperature, Tf, of the mixture of the reaction products. The direct measurement of Tf would be a hazardous undertaking, and its calculation (see Appendix 1) is not easy. To obtain a more user-friendly explosive power, ∆T is replaced by Q, the energy released in an explosion. A larger value of Q will lead to a larger value of ∆T. Now, the explosive power, EP, of an explosive can be defined as: EP =
Q Vg m
(2)
Apart from Q, this definition contains some of the key quantities of eq 1, but not the density of the explosive.1 It should be noted that this explosive power has nothing to do with the physical concept of power. According to eq 2, the EP is an extensive property. But, if we want to compare the efficiencies of different explosives with each other, we need an intensive property, like ∆p.
Expressing the explosive power per unit mass, we define the “relative explosive power”, REP, as: REP =
Q Vg EP = m m2
(3)
This expression has been in use for a long time. Debus (2), who attributed it to Berthelot, used it to compare different mixtures of gunpowder with each other, and it is still used today (3) to compare modern explosives with each other. The values of Vg and m are determined by the chemical equation of the explosion. If the enthalpies of formation of all reactants and products are known, then Q may be found as Q = ᎑∆H(reaction). The REP can now be calculated for any explosive and is conveniently expressed in units of kJ L g᎑2. Types of Explosives Explosives can be classified either as primary or secondary explosives. A primary explosive has a low REP but can be easily detonated. Secondary explosives cannot be easily detonated and are generally more powerful than primary explosives. The explosion of a secondary explosive is usually initiated by detonation of a primary explosive. Details of this have been given by Harris (4). Here, only secondary explosives are considered. The chemical reaction that culminates in an explosion is usually a redox reaction (eq 14 is an exception). Thus the explosive must consist of (at least) one oxidizing agent and one reducing agent. These reagents must be thoroughly mixed so the reaction can proceed rapidly. If both reagents are part of the same molecule, the degree of mixing is maximized. Hence, suitable pure substances can be explosives. A mixture of pure explosives may have a larger REP than each of its components. In spite of the large number of existing explosives, the search for new explosives continues. Of interest are several unknown compounds for which the molecular structures suggest high REP values. Their synthesis is a challenge to chemists in general and to those associated with the military in particular. REP values of some explosives from each of the groups mentioned above shall be determined. Details about these and other explosives may be found in Akhavan’s monograph (1). Pure Explosives The following pure explosives shall be considered: ammonium perchlorate (NH 4 ClO 4 ); ammonium nitrate (NH 4 NO 3 ); ethylene glycol dinitrate or EGDN (C 2H4N 2O6); glyceryl trinitrate or GTN (C3H 5N3O 9); pentaerythrityl tetranitrate or PETN (C5H8N4O12); and 2,4,6-trinitrotoluene or TNT (C7H5N3O6). The REP values can be determined from the explosion equations. For the first
JChemEd.chem.wisc.edu • Vol. 80 No. 12 December 2003 • Journal of Chemical Education
1397
In the Classroom
four substances the reaction equations are: 2 NH4ClO4(s)
Cl2(g) + 4 H2O(g)
Table 1. Molar Masses and Standard Enthalpies of Formation of Some Substances
(4)
C2H4N2O6(l)
+ N2(g) + 2 O2(g) 2 NH4 NO3(s)
4 H2O(g) + 2 N2(g) + O2(g)
C2H4N2O6(l)
2 CO2(g) + 2 H2O(g) + N2(g) (6)
4 C3H5N3O9(l)
12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)
M/(g mol᎑1)
Substance
(5)
152.06
259a
C3H5N3O9(l)
227.09
370.9
C5H8N4O12(s)
316.14
538.6
C7H5N3O6(s)
227.13
26a
H2O2(l)
(7)
᎑∆fH⬚/(kJ mol᎑1)
34.02
187.8
NH4ClO4(s)
117.49
295.3
NH4NO3(s)
80.04
365.6
CO(g)
28.01
110.5
CO2(g)
44.01
393.5
H2O(g)
18.02
241.8
a
Data from ref 7. All other values are from ref 6.
Both PETN and TNT contain too little oxygen to permit complete combustion. A number of equations could describe the explosive reaction. Akhavan (5) gives sets of rules by which “fairly good approximations” may be obtained. These sets of rules are reproduced in Appendix 2. In general, application of the rules yield explosion equations with different REP values. The equations for PETN and TNT with the largest REP values are: 3 C5H8N4O12(s)
C(s) + 4 CO(g) + 10 CO2(g) + 12 H2O(g) + 6 N2(g)
2C7H5N3O6(s)
6 C(s) + 6 CO(g) + 2 CO2(g)
(8)
Table 2. Relative Explosive Power, REP, and Oxygen Balance, OB, of Some Explosives Explosive
REP/(kJ L g᎑2)
OB
NH4ClO4
1.223
2
NH4NO3
1.445
1
C3H5N3O9 (GTN)
4.456
0.5
C2H4N2O6 (EGDN)
4.903
0
C5H8N4O12 (PETN)
4.515
᎑2
C7H5N3O6 (TNT)
3.270
᎑10.5
(9)
+ 3 H2(g) + 2 H2O(g) + 3N2(g)
Equations 4–9 form the basis for the calculation of the REP values of the corresponding explosives. The value of Vg can be deduced from each of these equations by counting the number of molecules of the gaseous products. Using the molar masses and enthalpies of formation (Table 1; ref 6 ), m and Q can be calculated. The REP values found are presented in Table 2. A sample calculation is given in Appendix 2. An interesting property of an explosive is its oxygen balance, OB. It is defined as the number of oxygen atoms in the explosive that are not needed for complete combustion. It is measured per molecule (or formula unit) of explosive. A negative OB implies that the explosive is oxygen deficient. The OB of an explosive can be deduced from its explosion equation. For example, the products C, CO, and H2 in eq 9 would have been oxidized to CO2 and H2O if 21 more oxygen atoms had been present. Hence, the OB of TNT is ᎑10.5: 21 atoms of O (two atoms of oxygen per C, one atom of oxygen per CO, and one atom oxygen per H2) per 2 molecules of TNT. Alternatively, OB can be calculated from the formula of the explosive. If the molecular formula is written as CaHbNcOd, then OB = d − 2a − 1 2 b
(10)
There seems to be a relationship between REP and OB. As seen in Table 2, REP is larger if the absolute value of OB is smaller. If OB is positive, the redundant oxygen is not used 1398
for combustion, and Q turns out to be smaller than it could have been. If OB is negative, parts of the products have not been burned completely, and again Q is smaller than it could have been. Only if OB is equal to zero, the maximum quantity of energy is released. Mixtures of Pure Explosives An explosive with negative OB can be burned completely if it is mixed with the proper quantity of an explosive with positive OB. Such a mixture might have a larger REP than each of its components. To illustrate this, we consider a mixture of explosives with a small positive OB (e.g., GTN) and a small negative OB (e.g., PETN). The optimum mixture must contain four moles of GTN for every mole of PETN. The explosion equation is:
4C3H5N3O9(l) + C5H8N4O12(s) 17CO2(g) + 14H2O(g) + 8N2(g)
(11)
The REP of this mixture is found to be 4.694 kJ L g᎑2, which is slightly larger than the REP of either GTN or PETN. As the REP is not much larger, one may wonder whether there is any use for this mixture in practice. A more successful explosive mixture can be made from a pure explosive with a large negative OB, such as TNT. Op-
Journal of Chemical Education • Vol. 80 No. 12 December 2003 • JChemEd.chem.wisc.edu
In the Classroom Table 3. Properties of Some Mixtures of Pure Explosives Mass % TNT
REP/(kJ L g᎑2)
TNT and NH4ClO4
26.91
3.638
TNT and NH4NO3
21.28
3.821
4.55
4.686
Mixture
TNT and GTN
The value of BE (single NN-bond) used here pertains to a hypothetical N2 molecule in which the atoms are linked together by a single, unstrained bond. But in N8, the single NNbonds would be strained, hence a smaller value should have been used. Thus octa-azacubane’s REP value would be even larger. In any case, N8 would turn out to be an extremely powerful chemical explosive. Summary
timum mixtures of TNT and NH4ClO4, NH4NO3, or GTN, have appreciably larger REP values than each of their components. REP values of such mixtures are given in Table 3. Hydrogen peroxide is used as an oxidant in rocket fuels (8). Although not known as an explosive, H2O2 can furnish the oxygen that TNT lacks. The explosion equation of the optimum mixture is 2 C7H5N3O6(l) + 21 H2O2(l)
(12)
14 CO2(g) + 26 H2O(g) + 3 N2(g)
and its REP value turns out to be 5.505 kJ L g᎑2, which is exceptionally large. Possible Future Explosives Brown (9) gives the structural formulas of two interesting future compounds, octanitrocubane, C8(NO2)8, and the nitrogen allotrope octa-azacubane, N8 . The explosion equations of these potential explosives are very simple: C8(NO2)8(?)
N8(?)
8CO2(g) + 4N2(g)
4N2(g)
(13) (14)
The REP values cannot be calculated because the enthalpies of formation of the reactants are not known. But estimates of REPs can be made. As a result of their cubic form, both molecules would be subject to small-angle strain (10), hence we may expect that both octanitrocubane and octa-azacubane would have positive enthalpies of formation. If we assume that the enthalpy of formation of C8(NO2)8 is equal to zero, then its REP value, calculated from eq 13, would be 3.931 kJ L g᎑2. However since it is likely that C8(NO2)8 would have a positive enthalpy of formation, its REP value could be appreciably larger, and C8(NO2)8 would be a powerful explosive. If N8 happens to be a gas, then the Q of eq 14 may be calculated from the appropriate bond enthalpies, BE. The bond enthalpy is defined as the enthalpy change of the process in which the bond is broken: Q ⫽ 4 BE (triple NN bond) ⫺ 12 BE (single NN bond) (15)
In the literature several values of these bond enthalpies are given: BE (triple NN-bond) values between 941.4 (11) and 945.6 (12) kJ mol᎑1; and BE (single NN-bond) values between 160 (13) and 180 (14) kJ mol᎑1. Taking mean values of 943.5 kJ mol᎑1 for BE (triple NN-bond) and 170 kJ mol᎑1 for BE (single NN-bond), eq 15 yields Q = 1734 kJ. Octaazacubane’s REP value can now be found to be 12.4 kJ L g᎑2!
Factory-made explosives have a REP value of about 4.5 kJ L g᎑2, but those explosives that would cause a really big bang have not yet been made. Note 1. One of the referees of the original version of this paper suggested that ρ should be incorporated in the definition of EP. This sympathetic suggestion has not been followed for two reasons: (i) EP of eq 2 leads naturally to REP of eq 3, which is used by “explosive chemists” and (ii) the density of a future explosive is not known and difficult to estimate. A comparison of its REP with those of existing explosives would not be possible.
Literature Cited 1. Akhavan, J. The Chemistry of Explosives; RSC: Cambridge, United Kingdom, 1998; p 19. 2. Debus, H. Ann. Chem. 1891, 265, 257–315. 3. Akhavan, J. The Chemistry of Explosives; RSC: Cambridge, United Kingdom, 1998; p 85. 4. Harris, B. W. J. Chem. Educ. 1987, 64, 541–544. 5. Akhavan, J. The Chemistry of Explosives; RSC: Cambridge, United Kingdom, 1998; pp 73–76. 6. Handbook of Chemistry and Physics, 80th ed.; Lide, D. R., Ed.; CRC Press: Boca Raton, FL, 1999. 7. Akhavan, J. The Chemistry of Explosives; RSC: Cambridge, United Kingdom, 1998; p 79. 8. Shaw, B. L. Inorganic Hydrides; Pergamon Press: Oxford, United Kingdom, 1967; p 74. 9. Brown, G. I. The Big Bang. A History of Explosives; Sutton Publishing: Stroud, United Kingdom, 1998; Appendix I. 10. March, J. Advanced Organic Chemistry. Reactions, Mechanisms, and Structure, 2nd ed.; McGraw-Hill International Book Company: London, United Kingdom, 1977; p 141. 11. Toon, E. R.; Ellis, G. L. Foundations of Chemistry, 2nd ed.; Holt, Rinehart and Winston: New York, 1973; p 368. 12. Mackay, K. M.; Mackay, R. A. Introduction to Modern Inorganic Chemistry, 3rd ed.; International Textbook Company: London, United Kingdom, 1981; p 270. 13. Cotton, F. A.; Wilkinson, G. Advanced Inorganic Chemistry. A Comprehensive Text, 3rd ed.; Interscience: New York, 1972; p 113. 14. Hägg, G. Allmän och oorganisk kemi, 5th ed.; Almqvist & Wiksell: Stockholm, Sweden, 1969; p 155. 15. Bromberg, J. P. Physical Chemistry; Allyn and Bacon: Boston, MA, 1980; p 90. 16. Atkins, P. W. Physical Chemistry; OUP: Oxford, United Kingdom, 1978; p 108. 17. Castellan, G. W. Physical Chemistry, 2nd ed.; Addison-Wesley: Reading, MA, 1973; p 127.
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Appendix 1 Assuming that Q, the energy released in the explosion, only heats up the reaction products, ∆T can be calculated from Q = CV ∆T
(A1)
where CV denotes the heat capacity at constant volume (i.e., the initial volume of the explosive) of the reaction products. That is, CV =
∑nj CV, j j
(A2)
where nj is the amount of the jth reaction product, and CV,j denotes its molar heat capacity at constant volume. Unfortunately, we cannot use the easily accessible standard values of heat capacities in eq A2. With increasing temperature the numerical values of the various CV,j increase as well. Hence, eqs A2 and A1 must be replaced by CV (T ) =
∑nj CV, j (T ) j
Kistiakowsky–Wilson Rules 1. Carbon atoms are converted to carbon monoxide. 2. If any oxygen remains, then hydrogen is oxidized to water. 3. If any oxygen still remains, then carbon monoxide is oxidized to carbon dioxide. 4. All the nitrogen is converted to nitrogen gas.
Using these rules the explosion equation of TNT becomes 2 C(s) + 12 CO(g) + 5 H2(g) + 3 N2(g) (A7)
2 C7H5N3O6(s)
and its REP value is found to be 2.768 kJ L g᎑2.
Modified Kistiakowsky–Wilson Rules 1. Hydrogen atoms are converted to water. 2. If any oxygen remains, then carbon is converted to carbon monoxide. 3. Rules 3–4 are identical to the Kistiakowsky–Wilson rules.
(A3)
and
The explosion equation of TNT now becomes 7C(s) + 7CO(g) + 5H2O(g) + 3N2(g) (A8)
2C7H5N3O6(s)
Q =
Tf
CV (T ) dT
(A4)
Ti
respectively. The limits of integration in eq A4, Ti and Tf, are the initial and final temperatures. Substituting the standard temperature of 298.15 K for Ti, Tf is varied until the calculated value of Q agrees with the one found from Q = ᎑∆H(reaction). In this way ∆T = Tf − Ti can be found, if values of the various CV,j(T ) are available. Most physical chemistry texts give power series in T of Cp,j(T ), the molar heat capacity at constant pressure. For example (15)
C p, j (T ) = a j + bj T + cj T 2
(A5)
and the REP value is 3.145 kJ L g᎑2.
Springall Roberts Rules Rules 1–4 are identical to the Kistiakowsky–Wilson rules. 5. One third of the carbon monoxide formed is converted to carbon and carbon dioxide. 6. One sixth of the original amount of carbon monoxide is converted to carbon and water (using some of the initially formed hydrogen).
Using rules 5 and 6, eq A7 is changed to 6C(s) + 6CO(g) + 2CO2(g)
2C7H5N3O6(s)
and (16) C p, j (T ) = A j + Bj T + Cj T −2
+ 3H2(g) + 2H2O(g) + 3N2(g)
(A6)
Using the appropriate relationships between Cp,j and CV,j (17) these power series can be converted into expansions for CV,j(T ). However, eqs A5 and A6 yield reliable results only if T is smaller than 1500 K and 2000 K, respectively. Hence, the method outlined above can be used for primary explosives, but usually not for secondary explosives. For organic secondary explosives Tf is estimated to lie between 4000 and 5000 K, and the method would yield results that may be quite wrong. As expansions that are valid for high temperatures seem to be unattainable, reliable values of Tf cannot be determined. Thus choosing ∆p as a measure of the explosive power of a secondary explosive is not a user-friendly option. Appendix 2 Akhavan (5) gives three sets of rules that can be used to find the explosion equation of an oxygen-deficient explosive: the Kistiakowsky–Wilson rules, the Modified Kistiakowsky– Wilson rules, and the Springall Roberts rules, which are really extended Kistiakowsky–Wilson rules. The results obtained using these sets of rules are illustrated by the explosion equations of TNT that follow. 1400
(A9)
which gives a REP value of 3.270 kJ L g᎑2. Since the Springall Roberts rules seem to be an improvement of the other two sets, eq A9 is assumed to represent the explosion equation of TNT best. In the text eq A9 appears as eq 9. The associated REP value is calculated as follows. m = n ( TNT ) M ( TNT ) = 2 mol × 227.13 g mol −1 = 454.266 g Vg = nVm = 16 mol × 22.414 L mol −1 = 358.62 L Q = −∆H ( reaction) = − 6 mol × ∆ f H (CO) + 2 mol × ∆ f H ( CO2 ) + 2 mol × ∆ f H (H 2O) − 2 mol × ∆ f H
=
[(6 × 110.5) + (2 × 393.5) + (2 × 241.8) − (2 × 26) ] kJ
REP =
QVg m
2
=
= 1881.6 kJ
1881.6 kJ × 358.62 L
( 454.26 g )2
( TNT)
= 3.270 kJ L g −2
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