Question

he standard enthalpies of formation for S (g), F (g), SF4 (g), and SF6 (g) are +278.8, +79.0, -775, and -1209 kJ per mole, respectively. What is the energy of the S--F bond WITHOUT using the dissociation energy bond chart. (Step by step proof is needed)

Answer 1

Answer:

+60.54 kJ/mol

Explanation:

The enthalpy of formation of a compound is the sum of the enthalpies of the formation of its constituents and the bond of them. To form a compound, energy must be lost, so, they're negative.

For SF4, the enthalpy is formed by the energy of one S, two F, and 4 S-F bond (Hb):

H = - (278.8 + 4*79.0 + 4*Hb)

-775 = -(594.8 + 4Hb)

594.8 + 4Hb = 775

4Hb = 180.2

Hb = +45.05 kJ/mol

For SF6, the enthalpy is formed by the energy of one S, six F, and 6 S--F bonds (Hb):

H = -(278.8 + 6*79.0 + 6*Hb)

-1209 = -(752.8 + 6Hb)

752.8 + 6Hb = 1209

6Hb = 456.2

Hb = +76.03 kJ/mol

Thus, the energy of S--F bond must be the average of these two:

(45.05 + 76.03)/2 = +60.54 kJ/mol