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...
(The entire section contains 171 words.)
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