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Thermochemistry - please help, test tomorrow

  1. Oct 26, 2004 #1
    [tex]N_2_{(g)} + \frac{5}{2}O_2 \longrightarrow N_2O_5 + \mathfrm 533 kJ energy[/tex]
    533kJ energy is released as heat

    We are given that 101.7g N2 , 102.97g O2 reacted. And this happens at T= 25C and P = 1atm

    what is the delta H and delta U for this process per mol N2(g). And how much work is done by the surrounding?

    (H<- enthalpy , U<- internal energy of the system )
    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2


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    First of all, you need to write the coefficients correctly for the reaction.

    [tex]N_2_{(g)} + \frac{5}{2}O_2 \longrightarrow N_2O_5 + \mathfrm 533 kJ energy[/tex]

    Then find the molar amounts of nitrogen and oxygen gas, by using N=14 and O=16 gram/mol.

    If a thermal energy is given to the environment, then it is an exothermic reaction, thus [itex]\Delta H[/itex] must be negative, and must have the dimension [tex]\frac {mol}{L}[/tex].

    I have no idea what [itex]\displaystyle \Delta U[/itex] is, so another friend will help you in the following days, I think.
    Last edited: Oct 26, 2004
  4. Oct 26, 2004 #3
    i found the number of moles for each gas, but we are not given the volume;As you say [itex]\Delta H[/itex] = [tex]\frac {mol}{L}[/tex]. And how is -533kJ usefull in this problem?
  5. Oct 26, 2004 #4
    H = U + pv, so therefore [tex]\Delta H = \Delta U + \Delta (pV)[/tex]. First off, what is [tex]\Delta U[/tex]?

    You can't find [tex]\Delta (pV)[/tex] directly. But for an ideal gas, what is pV equal to ?
    Last edited: Oct 26, 2004
  6. Oct 26, 2004 #5
    For an ideal gas: PV=nRT
    but how do I get U, (H=U+PV ) if i also need H
  7. Oct 26, 2004 #6
    I just realized something, you say the reaction occurs at 1 atm and 25 C, so it is both a constant temperature and constant pressure reaction ?
  8. Oct 27, 2004 #7
    1 mole of a gas = 24 L (24dm^3)

    So you have your mass, so just change that into number of moles, then multiply by 24 L. That will find your Volume.
  9. Oct 27, 2004 #8
    OK I don't know how you did on your test but

    1) For an isothermal reaction for an ideal gas, [tex]\Delta U = C_v\Delta T[/tex] and since presumably [tex]\Delta T = 0[/tex], it follows that [tex]\Delta U = 0 [/tex]

    2) Under constant pressure, [tex]\Delta H = q[/tex]. If you don't know what [tex]\Delta H[/tex] is, go back and look at the thermodynamics section in your first-year chemistry book.

    3) Well now since you know q and [tex]\Delta U[/tex] , you can solve for w, right ?
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