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Homework Help: Enthalpies of formation. burning methane in oxygen can produce three different carbo

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Enthalpies of Formation. burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO2 (g), and CO2 (g). A) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that H2O (l) is the only other product. B) Determine the standard enthalpies for the reactions in part A). C) Why, when the oxygen supply is adequate, is CO2 (g) the predominant carbon-containing product of the combustion of methane?

    2. Relevant equations

    Standard formations of enthalpies table was used for this question

    3. The attempt at a solution

    My attemp for A and B but have no idea how to figure out C I need help.

    A) and B) toegther is as follows

    1. 2CH4 (g) + 3(O2) (g) -----> 2CO (g) + 4H2O (l)
    ((2*-110.5)+(4*-285.8))-((3*0)+(2*-74.80)) = DH is -1214.6 kJ/mol

    2. CH4 (g) + 2(O2) (g) -------> CO2 (g) + 2H2O (l)
    ((1*-393.5)+(2*-285.8)) - ((1*-74.8)+(2*0)) = DH is -890 kJ/mol

    3. CH4 (g) + O2 (g) ----------> C (s) + 2H2O (l)
    ((1*1.88)+(2*-285.8)) - ((1*-74.80)+(1*0)) = DH is -494.92 kJ/mol

    I used the enthalpies of formation data tables and the formula
    DH= (sum of products) - (sum of reactants). Now how do I figure out C?

    I notice that B with CO2 as the product is -890.3 kJ/mol which is between A and C. or is it because when there are more number of moles of O2 and less only 1 mole of methane then CO2 forms as a product. I am still confused on how to answer C. Any help please???:confused::confused:
  2. jcsd
  3. Jul 23, 2010 #2


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    Science Advisor

    Re: Enthalpies of formation. burning methane in oxygen can produce three different c

    Looks fine. Just consider: Since you have all the oxygen you need, which reaction produces the most energy per mole of methane?
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