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Solid-Vapor Equilibrium

  1. Jan 17, 2008 #1
    Hey guys, I was just wondering if anybody could help me out with this problem?

    Calculate the heat (in KJ) required to transform 43.90 g of hydrazine from a solid at a temperature of 1.4 degreed C to a gas at 128 degrees C. report your answer to one decimal place.

    Data:
    molar mass of hydrazine: 32.045 g/mol
    melting point = 1.4 degrees C
    Boiling point = 114 degrees C
    Enthalpy of fusion = 12.6 KJ/mol
    Enthalpy of vaporization = 41.8 KJ/mol
    Molar heat capacity of the liquid phase = 98.9 J/mol x K
    Molar heat capacity of the gas phase = 49.6 J/mol x K

    q=ms(delta T)
    =(43.90 g)(98.9 J/mol x K)(387 K - 274.4 K)
    =488.9 KJ

    q=n(molar heat of vaporization)
    =(43.90 g)(1 mol/32.045 g/mol)(41.8 KJ/mol)
    =57.3 KJ

    q=ms(deltaT)
    =(43.90 g)(49.6 J/mol x K)(401 K - 387 K)
    =30.5 KJ

    488.9 KJ + 57.3 KJ + 30.5 KJ
    =576.7 KJ

    This seems right to me, however, I don't know why the enthalpy of fusion was listed here. Can anybody explain to me if an error has occurred here?
     
  2. jcsd
  3. Jan 21, 2008 #2
    At this step:
    you need to use the molar heat of fusion (melting) rather than vaporization to get the compound to the liquid state.

    then use this part:
    to get the substance to its boiling point.
    at this point you need to do q = n(molar heat of vaporization)
    to vaporize the substance.

    and then the last part:
    gets the substance from its boiling point to the desired temp of 128 C.

    also check your sig figs, especially when converting from degrees C --> K. They seem a bit off.
     
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