Science Questions and Answers

Start Your Free Trial

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

Expert Answers info

Blake Douglas eNotes educator | Certified Educator

calendarEducator since 2013

write1,065 answers

starTop subjects are Literature, Science, and Math

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.)

Unlock This Answer Now

check Approved by eNotes Editorial