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  1. #1

    Default Heat of formation question

    I learned from the chad video to calculate the heat of formation in easy way. For example,
    Q. given that the delta H of the following reaction is 92.2 kj/mol, what is the heat of formation of ammoni a gas?
    2NH3 -----> N2 (g) + 3H2 (g) - 92kj/mol
    N2 (g) + 3H2 ---------> 2NH3 92kj/mole, so heat of formation of ammonia gas is 92/2 = 46kj/mole

    Today, I have faced with similar problem and I solve it with same manner as above example, but I got it wrong. The question was

    Q. The combustion of ammonia is represented by this equation.
    4NH3 (g) + 5O2 (g) ----> 4NO (g) + 6H2O (g) Delta H = -904.8KJ
    (Enthalpy of formation data, NO(g) = 90 KJ/mol, H2O (g) = -241.8kj/mol)
    What is the enthalpy of formation of NH3(g)

    I solved it with exactly same nanner.
    4NO + 6H2O ---------- 4NH3 + 5O2, Delta H = (+) 904.8
    divided by 4 = 227, but the correct answer is -46 KJ per mole

    I figured out how to reach the correct answer, but I don't know why my calculation doesn't get the correct answer?

  2. #2
    Chad
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    The first question is an application of Hess' Law.
    You first need to figure out what you're being asked. The heat of formation of ammonia corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) --> NH3(g)
    We can easily see that this is similar the rxn provided and all we need to do is reverse the one provided and cut it in half to yield the above formation rxn.

    But the 2nd question is a little different based upon the data they supply you with. Again you first need to figure out what you're being asked. The heat of formation of ammonia still corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) --> NH3(g)
    Simply dividing the provided rxn by 4 doesn't get us this formation rxn for NH3. So we'll have to take a slightly different approach. Here we'll use the fact that the deltaH rxn = sum (delta H,products) - sum (delta H, reactants) and so:

    -904.8 = [4(x)+5(0)]-[4(90)+6(-241.8)]

    Solving for x gives -46.5kJ.

    Hope this helps.

  3. #3
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    Quote Originally Posted by chad View Post
    The first question is an application of Hess' Law.
    You first need to figure out what you're being asked. The heat of formation of ammonia corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) --> NH3(g)
    We can easily see that this is similar the rxn provided and all we need to do is reverse the one provided and cut it in half to yield the above formation rxn.

    But the 2nd question is a little different based upon the data they supply you with. Again you first need to figure out what you're being asked. The heat of formation of ammonia still corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) --> NH3(g)
    Simply dividing the provided rxn by 4 doesn't get us this formation rxn for NH3. So we'll have to take a slightly different approach. Here we'll use the fact that the deltaH rxn = sum (delta H,products) - sum (delta H, reactants) and so:

    -904.8 = [4(x)+5(0)]-[4(90)+6(-241.8)]

    Solving for x gives -46.5kJ.

    Hope this helps.
    Chad, I hope you don't mind if I follow up on your answer to this poster's question because I'm confused about it and am also doing heats of formation in my gen chem II course.

    I don't understand how the math works out on the last calculation.

    If I substitute in your value of -46.5 for every instance of x:

    -904.8 = [4(-46.5)+5(0)]-[360+(-1450.8)]

    -904.8 = [-186]-[-1090.8]

    -904.8 = [-186]+[1090.8]

    -904.8 ≠ 904.8

    Should the enthalpy of formation of NH3 be -498.9 instead of -46.5?

    I [think] my algebra is correct so I wonder if I'm misunderstanding the problem setup.

    Thanks.

  4. #4
    Chad
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    Hey Jake,
    Don't mind at all...especially when I've done something wrong (I lost sight of the original question). To set the record straight, the question is:
    Q. The combustion of ammonia is represented by this equation.
    4NH3 (g) + 5O2 (g) ----> 4NO (g) + 6H2O (g) Delta H = -904.8KJ
    (Enthalpy of formation data, NO(g) = 90 KJ/mol, H2O (g) = -241.8kj/mol)
    What is the enthalpy of formation of NH3(g)

    -904.8 = [4(90)+6(-241.8)]-[4(x)+5(0)]
    Solving for x does indeed yield -46.5kJ

    In the original thread I'd set up the calculation as reactants minus products whereas it should have been products minus reactants.
    Thanks for catching my error and hopefully this clears this one up.
    Last edited by chad; 08-06-2010 at 09:07 PM.

  5. #5
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    Thanks Chad! Just to clarify though, x should be -46.5 kJ right, because it's exothermic? You had that in the initial answer.

    Thanks for the VESPR video also, it is great. You're much better at explaining it than my prof is.

    Later,
    Jake

  6. #6
    Chad
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    Yes, you caught me again. I corrected it in the post above.

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