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Help with calorimetry

  1. Feb 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.070 g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.464 ∘C to 32.473 ∘C.

    determine delta H and delta U given the heat capacity of the calorimeter is 5.112 kJ/mol

    2. Relevant equations
    [tex] \Delta H = \Delta U + \Delta n_{gas} RT [/tex]

    3. The attempt at a solution
    i calculated delta U just fine and got -4904 kJ/mol

    I know the answer for delta H is - 4909 kJ/mol

    how do i calculate the the second term in the equation up there? from the combustion reaction i foudnm that delta n is -2. R is 8.314. But since T is not constant, how do i accoutnm for that? Do i average it?

    thanks for your help
     
  2. jcsd
  3. Feb 3, 2017 #2

    Charles Link

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    I think you need to compute the initial ## PV=n R T ## for the ## O_2 ## molecules at the initial temperature, and the final ## PV=n R T ## for the resulting components at the final temperature, and subtract the first from the second. When the ## O_2 ## combines with the carbon to make ## CO_2 ## there is no change in the pressure-volume, while when it combines with the ## H ## to make ## H_2 O ##, (presumably in the vapor form) the pressure-volume gets doubled, as you get two ## H_2 O ## molecules for each ## O_2 ##. Perhaps @Chestermiller could help answer this to see if he agrees. ## \\ ## Editing... Since the ## \Delta PV ## to get the ## \Delta H ## from the ## \Delta U ## appears to be ## -5 \,kJ/mole ##, does that mean that the water that forms is in the liquid phase? 1 mole of ## C_{10} H_8 ## corresponds to 2 moles of ## O_2 ## needed to make 4 moles of ## H_2 O ##.(Initial PV of ## O_2 ## that gets converted to liquid ## H_2 O ## ): 2(8.314)(297)=5000 joules that apparently goes into liquid form. That would explain the -5 kJ/mole in the difference between ## \Delta U ## and ## \Delta H ##. ## \\ ## Meanwhile, the change that occurs in the PV as ## O_2 ## goes from ## O_2 ## at the initial temperature to the same number of moles of ## CO_2 ## at the final temperature is most likely insignificant: 10 moles of ## O_2 ## and/or ## CO_2 ## and ## \Delta T=8 ## degrees. ## \Delta (PV)=10(8.314)8=665 \, joules ##. (Less than 1 kJ). It could make the -5 kJ into a -4kJ . Otherwise insignificant.
     
    Last edited: Feb 3, 2017
  4. Feb 3, 2017 #3
    I get -4904 kJ/mole for ##\Delta U##

    For ##\Delta (PV)##, I get ((14)(273.2+32.473)-(12)(273.2+24.464))(8.314)=+5882 J/mole = 5.8 kJ/mole

    So, I get ##\Delta H = -4904 + 6 = -4898\ kJ/mole##

    This neglects the sensible heat increase of the products.
     
  5. Feb 3, 2017 #4

    Charles Link

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    This seems to be more logical (with the ## H_2 O ## going to the vapor phase upon combustion) than going to the liquid phase, which is apparently what the answer that the OP gives as the "correct" answer must be assuming. Thank you @Chestermiller .
     
  6. Feb 3, 2017 #5

    Borek

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    If the final temp is 33°C most of the water will be probably liquid.
     
  7. Feb 3, 2017 #6

    Charles Link

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    In this sense, I think the problem statement really needs to be more specific. It does leave us guessing a little, and says nothing about the volume of gas that the calorimeter holds=if the volume is small, then certainly, a good deal of condensation would result. It also doesn't go into detail about how the combustion is initialized, but that is most likely beyond the scope of the exercise.
     
  8. Feb 3, 2017 #7

    Borek

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    In a typical bomb calorimeter sample is ignited electrically and the amount of heat from ignition is known and subtracted (no data for that here). To make things easier we add few drops of water to the sample before ignition to make sure head space was already saturated with vapor. Then we can assume all produced water was liquid (and ignore the change in amount of vapor water, head space is rather small).
     
  9. Feb 3, 2017 #8
    Oh. I meant to do it for the water being liquid in the end!!!! Sorry. In that case, the number of moles decreases by 2, rather than increases by two. So it would be about 5 kJ less, rather than 5 kJ more. Thanks guys.
     
  10. Feb 3, 2017 #9
    If we really did want to get the molar changes in enthalpy and internal energy for the reaction at the initial temperature and at constant volume, we would have to take into account the additional heat that would have to be removed to cool the 4 moles of liquid water and 10 moles of carbon dioxide back down to the initial temperature (at constant volume). Anyone want to take a shot at it? Just roughing it out in my head, I get a number of about 3-4 kJ.
     
  11. Feb 3, 2017 #10

    Charles Link

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    ## \Delta U=n C_v \Delta T ## for the ## CO_2 ##, where ## C_v=(3/2)R ##. This gives ## \Delta U=(10)(3/2)(8.314)(-8) \, joules ## for the ## CO_2 ## portion, and an additional ## \Delta U=4(18)(4.184)(-8) joules ## for the liquid water. (## C_v ## for water is 1 cal/gram which is 4.184 joules/gram which is 4.184(18) joules/mole). Some quick arithmetic gives me -996-2400=-3400 joules, very nearly what @Chestermiller estimated. :)
     
  12. Feb 3, 2017 #11
    I think 1.5 R is pretty low for CO2. But we're certainly in the right ballpark.
     
  13. Feb 3, 2017 #12

    Charles Link

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    My mistake. For a diatomic gas, ## C_v= (5/2)R ## and for a more complex molecule, it would be even higher. (I had originally thought of this, but mistakenly thought it didn't apply for ## C_v ##. )
     
  14. Feb 3, 2017 #13
    I looked up co2, and it was about 3.6R
     
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