Use the bond dissociation enthalpies from appendix 3 in Rayner Canham or some similar collection. (N.B. values from different sources may differ slightly.) Additional data are given at the bottom of the page.
1. Estimate the enthalpy of formation of the unknown compound BrCl3(l).
2. Estimate the enthalpy of vaporisation for chloroform, CHCl3.
3. Calculate the mean S-F bond energy in each of SF4 and SF6. Does the difference between them make sense?
Enthalpies of formation/ kJ.mol-1
CHCl3(l) -135 SF4(g) -775 SF6(g) -1209
Enthalpies of atomisation/ kJ.mol-1
S(g) 238 C(g) 715
Enthalpies of vaporisation/ kJ.mol-1
Br2(l) 15 BrCl3 50 (estimated)
1. Estimate the enthalpy of formation of BrCl3(l).
values in kJ.mol-1 Br(g) + 3Cl(g)
/ \
/|\ / \
| 1.5/\Hdiss(Cl2) = 1.5 x 240 \
| / \
| / \
| Br(g) + 1.5Cl2(g) \
| /|\ -3/\Hbond(Br-Cl) = -3 x 216
| 0.5/\Hdiss(Br2) = 0.5 x 190 \
| | \
H| 0.5Br2(g) + 1.5Cl2(g) \
| /|\ \
| /\Hvap(Br2) = 15 \|
| | BrCl3(g)
| 0.5Br2(l) + 1.5Cl2(g) |
| \ -/\Hvap(BrCl3) = -60
/\Hform(BrCl3) = x |
\ \|/
\| BrCl3(l)
x = (15 + 85 + 360 - 648 -60) kJ.mol-1
/\Hform = -248 kJ.mol-1
Even though BrCl3 seems likely to be exothermic it is not in
fact known. This is likely to be due to the fact that Cl2 is
a gas and the formation is not favored by entropy.
2. Estimate the enthalpy of vaporisation for chloroform, CHCl3.
Values in kJ.mol-1
Values in kJ.mol-1 C(g) + H(g) + 3Cl(g)
/|\ / \
| 1.5/\Hdiss(Cl2) = 1.5 x 240 \
| / \
| / -/\Hbond(C-H) = -411
| C(g) + H(g) + 1.5Cl2(g) \
| /|\ -3/\Hbond(C-Cl) = -3 x 327
| 0.5/\Hdiss(H2) = 0.5 x 432 \
| | \
H| C(g) + 0.5H2(g) + 1.5Cl2(g) \
| /|\ \
| /\Hatom(C) = 715 \|
| | CHCl3(g)
| C(s)+ 0.5H2(g) + 1.5Cl2(g) |
| \ -/\Hvap(CHCl3) = y
\ |
/\Hform(CHCl3) = -135 |
\ \|/
\| CHCl3(l)
-y = (981 + 411 - 360 - 216 - 715 - 135) kJ.mol-1
/\Hvap(CHCl3) = 34 kJ.mol-1 (This is good agreement with the
experimental value of 30 kJ.Mol-1.)
3. Calculate the mean S-F bond energy in each of SF4 and SF6.
Does the difference between them make sense?
Values in kJ.mol-1 S(g) + 4F(g)
/|\ /|\ |
| 2/\Hdiss(F2) = 2 x 155 |
| | |
H| S(g) + 2F2(g) -4/\Hbond(S-F) = -4z
| /|\ |
| /\Hatom(S) = 238 |
| | |
| S(s) + 2F2(g) |
\ |
\ |
/\Hform(SF4) = -775 |
\ \|/
\| SF4(g)
-4z = (-310 - 238 - 775) kJ.mol-1
/\Hbond(S-F) = z = 331 kJ.mol-1
Similarly for SF6
-6z' = (-465 - 238 - 1209) kJ.mol-1
/\Hbond(SF6) = z' = 317 kJ.mol-1
Since it is more difficult to raise an element to a higher
oxidation state it makes sense that the bond energy is
slightly lower. Looked at another way the "extra" two bonds
release only 296 kJ.mol-1 each.
Note on precision and significant figures.
The precision of the data used in these calculations is
variable, but very few of them are reliable to the last place
before the decimal point. While the arithmetic of the
calculation may throw up some figures beyond the decimal
point these are meaningless and must be rounded off. The
main source of the uncertainty comes from using mean values
for bond energies; these are mean values for a wide range of
compounds and may differ by as much as 10% in any particular
compound (see the values for S-F in problem 3.)
(C) H.J Clase 1999. Updated 2000.01.24