Oxidation of gaseous ClF by F2 yields liquid ClF3, an

Chapter 5, Problem 67P

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6.67 Oxidation of gaseous ClF by \(\mathrm{F}_{2}\) yields liquid \(\mathrm{ClF}_{3}\), an important fluorinating agent. Use the following thermochemical equations to calculate H°rxn for the production of \(\mathrm{ClF}_{3}\):

\(2 \mathrm{ClF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{OF}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}=167.5 \mathrm{~kJ}\)

\(2\mathrm{~F}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{OF}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}=-43.5 \mathrm{~kJ}\)

\(2 \mathrm{CIF}_{3}(\mathrm{I})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{OF}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}=394.1 \mathrm{~kJ}\)

Equation Transcription:

F2

ClF3

H°rxn

2CIF(g) + O2(g) CI2O(g) + OF2(g)        H° = 167.5 kJ

2F2(g) + O2(g) 2OF2(g)                        H° = -43.5 kJ

2CIF3(I) + 2O2(g) CI2O(g) + 3OF2(g)          H° = 394.1 kJ

Text Transcription:

F_2

ClF_3

Delta H°_rxn

2CIF(g) + O_2(g) rightarrow CI_2O(g) + OF_2(g)        Delta H° = 167.5 kJ

2F_2(g) + O_2(g) rightarrow 2OF_2(g)                        Delta H° = -43.5 kJ

2CIF_3(I) + 2O2(g) rightarrow CI_2O(g) + 3OF_2(g)          Delta H° = 394.1 kJ