Question 14.9: Chemical reactions can be used to produce flames for heating......

Chemical reactions can be used to produce flames for heating. The nonnuclear reaction with the highest attainable flame temperature (approximately 6000°C, about the surface temperature of the sun) is that between hydrogen and fluorine:

H_{2}(g) + F_{2}(g) → 2\ HF(g)

Use the molar bond enthalpies in Table 14.5 to estimate the value of \Delta H^{\circ}_{rxn} for this equation. Compare this to the value calculated using the molar enthalpies of formation in Table 14.3.

TABLE 14.3 standard enthalpies of formation, ΔH^{\circ}_{f}, for various substances a  25°c
Substance Formula ΔH^{\circ}_{f}/kJ\cdot mol^{-1} Substance Formula ΔH^{\circ}_{f}/kJ\cdot mol^{-1}
aluminum oxide Al_{2}O_{3}(s) -1675.7 hydrogen fluoride HF(g) -273.3
ammonia NH_{3}(g) -45.9 hydrogen iodide HI(g) +26.5
benzene C_{6}H_{6}(l) +49.1 hydrogen peroxide H_{2}O_{2}(l) -187.8
benzoic acid C_{6}H_{5}COOH(s) -385.2 iodine vapor I_{2}(g) +62.4
bromine vapor Br_{2}(g) +30.9 magnesium carbonate MgCO_{3}(s) -1095.8
butane C_{4}H_{10}(g) -125.7 magnesium oxide MgO(s) -601.6
calcium carbonate CaCO_{3}(s) -1207.6 magnesium sulfide MgS(s) -346.0
carbon (diamond) C(s) +1.897 methane CH_{4}(g) -74.6
carbon (graphite) C(s) 0 methanol (methyl alcohol) CH_{3}OH(l)
CH_{3}OH(g)
-239.2
-201.0
carbon (buckminster fullerene) C_{60}(s) +2327.0 methyl chloride CH_{3}Cl(g) -81.9
carbon dioxide CO_{2}(g) -393.5 nitrogen dioxide NO_{2}(g) +33.2
carbon monoxide CO(g) -110.5 nitrogen oxide NO(g) +91.3
carbon tetrachloride CCl_{4}(l)
CCl_{4}(g)
-128.2
-95.7
dinitrogen tetroxide N_{2}O_{4}(g)
N_{2}O_{4}(l)
+11.1
-19.5
chromium (III) oxide Cr_{2}O_{3}(s) -1139.7 octane C_{8}H_{18}(l) -250.1
cyclohexane C_{6}H_{12}(l) -156.4 pentane C_{5}H_{12}(l) -173.5
ethane C_{2}H_{6}(g) -84.0 propane C_{3}H_{8}(g) -103.8
ethanol (ethyl alcohol) CH_{3}CH_{2}OH(l) -277.6 sodium carbonate Na_{2}CO_{3}(s) -1130.7
ethene (ethylene) C_{2}H_{4}(g) +52.4 sodium oxide Na_{2}O(s) -414.2
ethyne (acetylene) C_{2}H_{2}(g) +227.4 sucrose C_{12}H_{22}O_{11}(s) -2226.1
freon-12 (dichloro difluoromethane) CF_{2}Cl_{2}(g) -477.4 sulfur dioxide SO_{2}(g) -296.8
glucose C_{6}H_{12}O_{6}(s) -1273.3 sulfur trioxide SO_{3}(g) -395.7
hexane C_{6}H_{14}(l) -198.7 tin(IV) oxide SnO_{2}(s) -577.6
hydrazine N_{2}H_{4}(l)
N_{2}H_{4}(g)
+50.6
+95.4
water H_{2}O(l)
H_{2}O(g)
-285.8
-241.8
hydrogen bromide HBr(g) -36.3
hydrogen chloride HCl(g) -92.3
Data from CRC Handbook of Chemistry and Physics, 86th Ed., Ed. David R. Lide, CRC Press, 2005–2006. (More thermodynamic data are given in Appendix D.)

 

TABLE 14.5 average molar bond enthalpies
Bond Molar bond enthalpy, H_{bond}/kJ\cdot mol^{-1} Bond Molar bond enthalpy, H_{bond}/kJ\cdot mol^{-1}
OH 464 C≡N 890
OO 142 NH 390
CO 351 NN 159
O=O 502 N=N 418
C=O 730 N≡N 945
CC 347 FF 155
C=C 615 ClCl 243
C≡C 811 BrBr 192
C–H 414 HH 435
CF 439 HF 565
CCl 331 HCl 431
CBr 276 HBr 368
CN 293 HS 364
C=N 615
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

The reaction involves the breaking of one hydrogen–hydrogen bond and one fluorine–fluorine bond and the formation of two hydrogen– fluorine bonds. Thus,

ΔH^{\circ}_{rxn} ≈ H_{bond}(H–H) + H_{bond} (F–F) – 2\ H_{bond} (H–F)\\= (435\ kJ·mol^{–1}) + (155\ kJ·mol^{–1}) – (2)(565\ kJ·mol^{–1}) \\ = – 540\ kJ·mol^{–1}

Because this reaction is the formation reaction for two moles of HF(g) from its constituent elements, the experimentally determined value of \Delta H^{\circ}_{rxn} is twice the molar enthalpy of formation of HF(g) listed in Table 14.3, or – 546\ kJ·mol^{–1}, a value with less than a 2% difference from that which we found using average molar bond enthalpies.

Related Answered Questions