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

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caledon | High School Teacher | (Level 3) Senior Educator

Posted on

Isopropanol, also called isopropyl alchohol, has the following Lewis Dot structure;

   H  H  H

    |  |  |

H-C-C-C-H

    |  |  |

   HO-H H

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.

Sources:

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caledon | High School Teacher | (Level 3) Senior Educator

Posted on

That dot structure didn't turn out well at all. Just use this one.

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