Given the following bond enthalpy values, calculate the enthalpy of formation for gas phase isopronanol, C3H7OH at 25 degrees BE (C-C)= 336 kJ mol-1 BE (C-H)= 4.20x10^2 kJ mol-1 BE(C-O)=342 kJ...
Given the following bond enthalpy values, calculate the enthalpy of formation for gas phase isopronanol, C3H7OH at 25 degrees
BE (C-C)= 336 kJ mol-1
BE (C-H)= 4.20x10^2 kJ mol-1
BE(C-O)=342 kJ mol-1
BE (O-H)= 468 kJ mol-1
BE(O=O) = 498 kJ mol-1
BE (H-H) =432 kJ mol-1
The enthalpy of formation of C(g) is 715.0 kJ mol-1
Isopropanol, also called isopropyl alcohol, has the following Lewis Dot structure:
(See image below)
There are eleven bonds. Tallying them up, we have the following;
- 2 C-C
- 7 C-H
- 1 C-O
- 1 O-H
To build this molecule, we need three gaseous carbon atoms (monatomic), 4 dihydrogen, and half of one dioxygen.
We need to add energy to shatter these molecules and vaporize them, and then energy will be released as the bonds form. So:
(Energy required to atomize 3 carbons, 1/2 O2, and 4 H2)
-(Energy released by formation of 2 C-C, 7 C-H, 1 C-O, 1 O-H)
You'll need data tables for enthalpy to complete these calculations. I suggest using only the values found in your textbook or provided by your instructor; I've found that online values can differ considerably.
3 (715) + .5(498) + 4(432) = 4122
2(336) + 7(420) + 1(342) + 1(468) = 4422
4122 - 4422 = -300Kj/mol
While the mathematical procedure remains the same, the exact answer largely depends on the bond values you use; for example, in the sources linked below, there is a calculation for isopropanol formation enthalpy that puts the value at -318, and the wikipedia source puts it at -261.