Question

Given the following bond enthalpy values, calculate the enthalpy of formation for gas phase isopropanol, C3H7OH at 25

Answer 1

The enthalpy of formation of isopropanol is given by the following equation:

3 C(graphite) ---> 3 C(g) ?H = 3(715.0 kJ/mole)
3 C(g) + 4 H2(g) + 1/2 O2(g) ---> CH3CH(OH)CH3

The bonds formed to make isopropanol are:

3 C-C = 3 X 338 = 1014
7 C-H = 7 X 411 = 2877
1 C-O =355
1 O-H =451
The bonds broken to make isopropanol are:
4 H-H = 4 X 432 = 1728
1/2 O=O = 1/2 X 498 = 249

deltaH of formation = B.E of reactants - B.E of products

delta H of formation = 1728 + 249 ) - ( 1014 +2877+355+451) = -2720

We have to consider the formation of Carbon as well = 2145

so enthalpy of formation = - 575