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Homework Help: Determine the enthalpy of formation to find the enthalpy of reaction

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data

    NH3 + O2 --> NO + H2O

    Enthalpies of formation

    NO: 91.3 kJ
    H2O: -241.8 kJ
    NH3: -45.9 kJ

    2. Relevant equations

    delta h = sum*moles enthalpy of formation of products - sum*moles enthalpy of formation of reactants

    3. The attempt at a solution

    = [(1)(91.3) + (1)(-241.8)]-[(1)(-45.9)]
    = -104.kJ

    The answer in the back says -902 kJ. What am I missing?
  2. jcsd
  3. Nov 2, 2013 #2
    Your (chemical) equation is not balanced.
  4. Nov 2, 2013 #3
    But I thought it didnt matter if I multiplied by the number of moles right in the formula because its sum*moles enthalpy of formation?
  5. Nov 2, 2013 #4
    Actually that brings up an interesting issue that I didn't catch at first. Check the units of the Enthalpy of Formation, are you sure its kJ, or is it KJ/mol? Multiplying by the stoichiometric coefficient is appropriate only in one of those cases.

    What I originally referred to was your chemical equation. Hydrogens are not balanced. Unbalanced chemical equation leads to the wrong coefficients in calculations. Always check that your equation is balanced, it is a routine 'trick' used by pedagogues to make sure you are awake and paying attention rather than plugging and chugging.
  6. Nov 2, 2013 #5
    It is kJ/mol
  7. Nov 2, 2013 #6
    So even though moles are being multiplied I still have to balance?
  8. Nov 2, 2013 #7
    Of course! The stoichiometric coefficients are determined from a BALANCED equation. Otherwise you are violating energy/mass conservation.
  9. Nov 2, 2013 #8
    Well, yes...., It does matter. If you balance the equation first, and then calculate the heat of reaction, you get the right answer.
  10. Nov 2, 2013 #9
    Ok the reason I ask this is because in class we did this example question:
    Small amounts of oxygen gas can be produced in a laboratory by heating potassium chlorate. Calculate the enthalpy change of this reaction at SATP using standard enthalpies of formation.

    So for this question my teacher did KClO3 --> 3/2 O2 + KCl
    and used the formula to solve for -38.8 kJ.

    My question is that if everything is in proportions then if I multiply all the coefficients of this equation by 2, to get whole numbers, the law of conservation of mass will be followed, so theoretically shouldnt the value be the same, but it isnt obviously, it yields a larger value. I ask this because I do not know which balanced reaction equation to use, the one with fractions or the one with whole numbers, because it makes a difference in the final answer?
  11. Nov 2, 2013 #10
    Ah, I think I see your confusion now. The simple answer is that a question like this implies that the reaction happened 6.02 x 10^23 (1 mol) times. In reality, you would weigh or otherwise measure some amount of reactants and calculate how many moles of each reactant were used and how many moles of each product were produced. Those would then become your coefficients and will not be nice looking integers, but they will have the same ratios.

    Also, you are correct that multiples of the coefficients retain the correct proportions but the convention is to use the most reduced integer coefficients in balanced equations.
  12. Nov 2, 2013 #11
    So when you say "implies that the reaction happened 6.02 x 10^23 (1 mol) times" essentially where should the coefficient be 1 in the equation? How do we know?
  13. Nov 2, 2013 #12
    Yes. You are certainly correct. That has to be precisely specified. In your problem, the form of the balanced reaction that gives the answer in your book is:

    4 NH3 + 5 O2 --> 4NO + 6 H2O

    If we divided all the coefficients by 4, the heat of the reaction (expressed that way) would be 1/4 as much.
  14. Nov 2, 2013 #13
    Last edited: Nov 2, 2013
  15. Nov 2, 2013 #14
    It wouldn't necessarily show up anywhere. Take this example:

    CH3 + O2 -> CO2 + H2O

    This is unbalanced...As written we have somehow done nuclear chemistry and made an H atom an O atom and violated energy/mass conservation. Balancing we have:

    2 CH3 + 7/2 O2 -> 2 CO2 + 3 H2O

    Using convention of integers we have:

    4 CH3 + 7 O2 -> 4 CO2 + 6 H2O

    Now in an idealized situation, you just use the coefficients as listed because you assume that the reaction happened 1 mol times which implies that you measured out exactly 2 moles of methane, 3.5 moles of oxygen and reacted them to yield 2 moles of carbon dioxide and 3 moles of water (or 4, 7, 4, and 6 as written in the last equation). Once again though, that is an idealized situation which forces some assumptions on you.

    In reality you would, maybe, use 2.85 L of oxygen (use PV=nRT to figure out how many moles at that temp/pressure) and 3.26 L of methane (figure out how many moles). Then you determine limiting reagent etc, and figure out how many moles of each reactant disappeared and how many moles of each product appeared. You will NOT get numbers as nice as 2, 3 or even 3.5. You will get numbers, which are limited by your measurement errors, like 0.0432 moles (I didn't do the calculations, just making up the numbers) for oxygen but from there you will know that oxygen:methane mole ratio is 7/4 = (7/2)/2 which implies that 0.0247 moles of methane were reacted. So in real life, it won't matter whether you use the most reduced integer coefficients or multiples of them, it only matters that you are using a balance equation as its not the absolute values of each coefficient which matter, it is the ratios of the coefficients which give the correct information.

    What I meant that a reaction happens 1 mol times is that you imagine mixing exactly 4 moles of methane and 7 moles of oxygen and "watch" 4 molecules of methane react with 7 molecules of oxygen 6.02 x 10^23 (1 mole) times (meaning 4 moles of methane reacted with 7 moles of oxygen). In this case it really matters how you choose to write the balanced equation, but understand that this is simply an idealization and in real life you will have to measure some amount of each which will not result in exactly 2 mols of methane but however many moles of methane reacted, 7/4 (=7/2/2) times that amount of oxygen needs to have reacted. You can then figure out how much carbon dioxide and water you made by using the ratios of the stoichiometric coefficients in the same way.

    I hope this helps you understand what you are doing.
    Last edited: Nov 2, 2013
  16. Nov 2, 2013 #15
    Yes. That's right. The problem in the book is not specified precisely enough. Big surprise. Not every problem in a text book is posed well enough to remove all ambiguity. Get used to it.

    The thing you should be happy about is that you knew enough to ask the right questions, and you learned something. Well done.

  17. Nov 2, 2013 #16
    All right then I think I understand essentially the sort of questions I have in store. Thank you all for you help and encouragement!
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