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Enthalpy of Reactions - can someone help me real quick?

  1. Mar 17, 2006 #1
    I hate stoichiometry.

    Here's the problem:

    How do I figure out how many mols to multiply for each product and reactant? I've already been given the other part of the equation.

    Thanks
     
  2. jcsd
  3. Mar 17, 2006 #2
    The question should be worded "Complete combustion of" or its usually implied that the combustion goes to completion and that all the reactants are turned into products.

    Stoichiometrically you must find what ratios (how many moles of each reactant is needed) to acheive this complete combustion, that is, no reactants are left unburned/unreacted.

    Thats already provided in the starting reaction equation, but it is usually simple to work it out.

    Usually a question like this sais, Ethane is burned in "excess" oxygen, this implies that you have as much oxygen as you need. The total ammount of Carbon and Hydrogen after the reaction must be balenced with the amount before the reaction, you cant get matter from nowhere.

    C2H6(g)
    2x Carbon which when burned becomes Carbon Dioxide (CO2) requiring 2 O atoms ( or 1 O2 molecule, oxygen is naturally diatomic and found as paired Oxygen )
    6x hydrogen which when burned becomes Water (H2O) requiring another Oxygen atom for every 2 Hydrogen.

    this is written:
    C2H6 + ?O2 -> 2CO2 + 3H2O

    you can see that for it to balance you require 3.5 oxygen molecules to every 1 Ethane molecule, or 7 mol oxygen to 2 mol of Ethane.

    When considering the energy involved in the reaction, you must take into account these quantities of reactants. The more you burn, the more heat you get out.

    You have already done this:
    You multiplied it by the 4 moles you used in the reaction.

    The overall energy transfer in the reaction is given by the dH of Products - dH of Reactants.

    -3024.8 - (-169.4) = -3024.8 + 169.4 = -12855.4 kJ

    But remember that you used 2 moles of C2H6, and therefor need to divide by 2.

    -2855.4 kJ / 2 mol = -1427.9 kJ/mol

    You can then say that the Hreaction for the combustion of Ethane in Oxygen is -1427.9 kJ/mol.

    I hope i didnt confuse :s
     
  4. Mar 17, 2006 #3
    Whoops! This is an example question!! Sorry sorry - here's the real problem! This is what I'm trying to figure out how many mols to times by. :eek: :D

    Balance the equation (Already did this) and Calculate the enthalpy change for the following reaction:

    NH3(g) + O2(g) --> N2(g) + H2O(l)

    NH3(g)= -46.11 kj/mol

    O2(g) = 0 kJ/mol

    N2 = 0 kJ/mol

    H20= - 285.830 kJ/mol



    ETA- Thankyou soooo much for taking the time to respond. It means so much to me.
     
    Last edited: Mar 17, 2006
  5. Mar 17, 2006 #4
    Not quite, check again.

    [you wrote]
    Elements in Reactants:
    N x 1 (mono atomic in NH3)
    H x 3
    O x 2
    Elements in Products:
    N x 2 ( youve gained Nitrogen from nowhere ) (Diatomic in N2)
    H x 2 ( Youve lost Hydrogen somwhere )
    O x 1 ( youve lost oxygen also )

    This is like factoring numbers, you need to find a common factor between reactants/products that will allow you to make a complete reaction. To do this you mix the reactants in the correct ratios and *poof* you get the product.
     
  6. Mar 17, 2006 #5
    Ok! So here's the correct balanced equation:

    4NH3 + 3O2 ----> 2N2 + 6H2O


    Keeping in mind the problem:
    Now, I already know this (I didn't figure this out, it was provided to me):

    NH3(g)= -46.11 kj/mol

    O2(g) = 0 kJ/mol

    N2 = 0 kJ/mol

    H20= - 285.830 kJ/mol

    So now I need to figure out how many mols to times each amount by. This doesn't seem too tough so maybe I'm just missing something? How do I go about doing that?

    ------------

    Also, the example problem I was given, from the original post, this is the answer: [​IMG]
    Just thought it might help.

    Thanks SO much!
     
    Last edited: Mar 17, 2006
  7. Apr 28, 2008 #6
    i have to do this problem... (its in green)
    Assignment Problem
    Balance the equation and calculate the enthalpy change for the following reaction:

    NH3(g) + O2(g) --> N2(g) + H2O(l)

    NH3(g)= -46.11 kj/mol

    O2(g) = 0 kJ/mol

    N2 = 0 kJ/mol

    H20= - 285.830 kJ/mol


    and i dont get it at all. i think the balanced eqution is 4NH3 + 3O2 --> 2N2 +6H2O, but im not sure how to do the enthalpy change.

    can anyone help me please?????????!?!!!! :uhh:
     
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