2CH3CH2CH2OH+9O2 --> 6CO2+8H2O Based on the standard free energies of formation

Question

2CH3CH2CH2OH+9O2 --> 6CO2+8H2O

Based on the standard free energies of formation given in the table below, what is the standard free energy change for this reaction?


2CH3CH2CH2OH = 360.5
O2 =0
CO2 = 394.4
H2O = 228.6


Find Delta G

Explanation / Answer

For example, you could look up the heat of combustion of 1-propanol in a book. The question obviously wants something else. One approach is to use /H's (heats) of formation of products and reactants. Here they are: (1) 3C + 4H2 + 1/2 O2 ===>C3H8O(g) (1-propanol) /H = -61.17 kcal/mole (2) C+ O2 ===> CO2 /H = -94.05 kcal/mole (3) H2 + 1/2 O2 ===> H2O(g) /H = -59.56 Your reaction has 2C3H8O on the left. So multiply equation (1) by 2 and write it backwards. Don't forget to reverse the sign of /H and double that too. (4) 2C3H8O(g) ===> 6C + 8H2 + O2 /H = +122.34 kcal/mole Your reaction has 6CO2 on the right. So multiply equation (2) by 6: (5) 6C + 6O2 ===> 6CO2 /H = -564.30 kcal/mole Your reaction has 8H2O on the right, so multiply equation (3) by 8: (6) 8H2 + 4O2 ===> 8H2O /H = -476.48 kcal/mole Finally, add equations (4), (5), and (6) together, cancel the C's and H2's and the one O2 on each side, and add all the /H's. 122.34 - 564.30 - 476.48 = -918.44 kcal/mole

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