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Topic: Hess's Law  (Read 7682 times)

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777888

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Hess's Law
« on: January 11, 2005, 04:34:20 PM »
Q1.Write 3 balanced thermochemical eq'ns to represent the combustions of 1 mol each of octane, hydrogen, and carbon, given their molar enthalpies of combustion are -5.47MJ, -285.8kJ, and -393.5kJ/mol respectively. Then, use Hess's law to predict the enthalpy change for the formation of octane form its elements.
Target: 8C + 9H2 ->C8H18    delta H = ?

Work:
(1): C8H18 + 25/2 O2 -> 8CO2 + 9H2O delta H=-5.47x10^3kJ
(2): H2 + 1/2 O2 -> H2O  delta H=-285.8kJ
(3): C + O2 -> CO2   delta H=-393.5kJ

Using (1)times -1, (2) times 9, (3) times 8 and adding, I get:
8C +9 H2 -> C8H18
delta H=5.47x10^3 -2572.2 - 3148= -250.2kJ
But the answer in my book is -247.5kJ..I don't know who's wrong...my or the book!? Did I do something wrong?
Thanks for helping! :)
« Last Edit: January 11, 2005, 04:45:40 PM by 777888 »

777888

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Re:Hess's Law
« Reply #1 on: January 11, 2005, 04:40:45 PM »
Q2: Target: C2H5OH + O2 -> CH3COOH + H2O. The molar enthalpies of combustion of ethanol and acetic acid are respectively, -1367kJ/mol and -875kJ/mol. Write thermochemical eq'ns for the combustions, and use Hess's law to find the enthalpy change for the conversion of ethanol to acetic acid.
Work:
(1): C2H5OH + 11/4 O2 -> 2CO2 + 5/2 H2O  delta H=-1367kJ
(2): CH3COOH + 7/4 O2 -> 2CO2 + 3/2 H2O  delta H=-875kJ
Using (1)times 1, (2) times -1,and add, I get:
C2H5OH + O2 -> CH3COOH + H2O
delta H=-1367+875= -492kJ
But the answer in my book is 492kJ...oh no...where did I do wrong?

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