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Energy changes and rates of reaction

  1. Apr 4, 2017 #1
    1. The problem statement, all variables and given/known data
    The complete combustion of methane is represented in the following equation:
    CH4(g) + 2O2 -> CO2(g) +2H2O(g) +802.5 kJ
    Assume that natural gas is essentially methane. The emission level of Carbon monoxide, CO(g) from the burning of natural gas is 19.14kg/MJ of energy produced.
    2. Relevant equations
    What mass of CO(g) will be emitted when 100g of natural gas is burned?

    3. The attempt at a solution
    I attempted to convert the 100g of methane into mol (100g/16.04g per mol) and then i multiplied the mol by it's standard molar enthalpy of formation on wikipedia (-74.9KJ/mol) to find the KJ then converted it into MJ and multiplied the methane mol with 19.14Kg/MJ. My answer was wrong. The textbook answer is 95.7 kg.
     
  2. jcsd
  3. Apr 4, 2017 #2

    DrClaude

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    Staff: Mentor

    Don't you also need an equation for the partial burning leading to the formation of CO?
     
  4. Apr 4, 2017 #3

    DrClaude

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    The units seem to be off. 19.14kg/MJ doesn't make sense, as well as producing 95.7 kg of CO when burning 100 g of methane :confused:
     
  5. Apr 4, 2017 #4
    You're complicating it. 802.5 kJ (I think you've got the sign wrong) is the energy produced from the burning of natural gas. Just calculate the total energy produced for 100 g of methane and use the total energy to find amount of CO emission.
     
  6. Apr 4, 2017 #5

    DrClaude

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    I agree that this can be done instead of my suggestion of using a second reaction, if the production of CO is negligible. But not if it is 19.14kg/MJ :smile:
     
  7. Apr 4, 2017 #6
    Can you please explain?
     
  8. Apr 4, 2017 #7

    DrClaude

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    If the production of CO is important, then you can't simply take that combustion of 1 mole of CH4 will release 802.5 kJ. You have to weigh that value with the energy produced by the reaction leading to CO, which releases less energy than the complete combustion.
     
  9. Apr 4, 2017 #8
    Aah, I see your point. Thanks :smile:

    Anyway, my method does get the OP's textbook answer even though it's technically incorrect.
     
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