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Estimate the standard enthalpy of formation

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data

    The standard enthalpy of formation of the metallocene bis(benzene) chromium was measured in a calorimeter. It was found for the reaction [itex] Cr(C_6H_6)_2(s) --> Cr(s) + 2(C_6H_6)(g) [/itex] that [itex] \Delta_r U~at~583 K = +8.0 kJ/mol [/itex]. Find the corresponding reaction enthalpy and estimate the standard enthalpy of formation of the compound at 583 K. The constant-pressure molar heat capacity of benzene is 136.1 J/K mol in its liquid range and 81.67 J/K mol as a gas.

    2. Relevant equations

    [itex] \Delta H = delta U +\Delta nRT [/itex]

    3. The attempt at a solution

    The first part is easy, [itex] \Delta H = +17.7 kJ/mol [/itex].

    I'm at a complete loss for estimating the standard enthalpy of formation. I've tried using Kirchoff's rule including phase change enthalpy and my result is +110.8 kJ/mol. I've tried other less sensible ways but those solutions are even further from the textbook answer of +116.0 kJ/mol.

    I used Kirchoffs rule for C6H6(l) --> C6H6(g) with [itex] \Delta_r H = 49.0 kJ/mol [/itex] for C6H6(l). The enthalpy of vaporization is 30.8 kJ/mol.

    I'm tired right now and don't want to type out all of my work. I'll do it tomorrow when I wake up. If anyone wants to help before then I thank you.
    Last edited: Sep 13, 2014
  2. jcsd
  3. Sep 13, 2014 #2
    OK, so here is how I approached the problem.

    Using Kirchoffs law and a temperature difference of [itex] \Delta T = 583 - 298 = 285 K[/itex], I calculated ## \Delta C_{p} = \sum(products)C_{p,m}-\sum(reactants)C_{p,m}## for the reaction [itex]C_6H_6(l) \rightarrow C_6H_6(g)[/itex], where the product molar heat capacity is that given for the gas, the reactant is that for the liquid.

    I then solved [itex]\Delta_f H (T_2) = \Delta_f H (T_1) + \int \Delta C_p[/itex] using [itex]\Delta_f H(T_1) = 49.0~kJ/mol[/itex].

    Since there is a phase change, I wasn't sure how to include that information so I solved it the same as above, using [itex]\Delta_{vap}H = 30.8~kJ/mol[/itex] for [itex]T_1 = 298~K[/itex], then added it to the above amount.

    Using the reaction enthalpy of [itex]17.7~kJ/mol[/itex] and Hess's Law, I solved for the enthalpy of formation at 583 K for [tex]Cr(C_6H_6)_2[/tex].

    Now, I understand some problems with this approach by I'm not sure how to solve it otherwise with the given information. The temperature range is massive, so Kirchoff's Law is expected to give a poor answer. I'm also not sure how to incorporate a phase change into Kirchoff's Law. Any help or guidance is appreciated.
    Last edited: Sep 13, 2014
  4. Sep 13, 2014 #3
    This is kind of a meaty question, and it tests many aspects of your understanding. What is your definition of the "standard enthalpy of formation" of a compound?

  5. Sep 13, 2014 #4
    My definition for the standard enthalpy of formation is the standard reaction enthalpy for the formation of that compound from its elements in their reference states, so [itex] \Delta_f H = 49.0~kJ/mol~for~6C(graphite)+3H_2(g) \rightarrow C_6H_6(l)[/itex].
  6. Sep 13, 2014 #5
    This is not quite correct. The standard heat of formation of a compound at a specified temperature is equal to the change in enthalpy in going from the elemental constituents at that temperature to the compound at that temperature (at a pressure of 1 atm). Therefore, in getting the standard heat of formation of your compound, you also have to consider the change in sensible heat of the C and H2 in going from 298 to 583, since these affect the heat of formation of benzene(g) at 583.

  7. Sep 13, 2014 #6

    [itex] 6C(gr)+3H_2(g) \rightarrow C_6H_6(l) [/itex]
    [itex] 6C(gr)+3H_2(g) \rightarrow C_6H_6(g) [/itex]

    Using the values [itex] C_{p,m} [/itex] for hydrogen, carbon, and benzene in liquid form at 298 K, solve for [itex] \Delta_f H [/itex] at 583 K. Then add in the molar enthalpy of vaporization to find [itex] \Delta_f H [/itex] for gaseous benzene?
    Last edited: Sep 13, 2014
  8. Sep 13, 2014 #7
    If I solve it as above, I get an enthalpy of formation of the metallocene equal to 141.0 kJ/mol.

    If I use the molar heat capacities to calculate the enthalpy of formation for gaseous benzene and ignore the liquid molar heat capacity and enthalpy of vaporization altogether, I get an enthalpy of formation of the metallocene equal to 116.3 kJ/mol, which is close to the books answer of 116.0.
  9. Sep 13, 2014 #8
    Actually, first you do the vaporization at 298, and then use the vapor phase heat capacity of benzene to go from 298 to 583. If you want to be a little more precise, you also include the enthalpy change in lowering the pressure of the liquid benzene at 298 from 1 atm to the equilibrium vapor pressure. Since, at 298, benzene is essentially an ideal gas between the equilibrium vapor pressure and 1 atm., you do not need to add a correction for the enthalpy change due to pressure on the gas.

    Alternately, to get even greater accuracy, you can get the change in enthalpy of the liquid benzene from 298 to 378.1 (which is the temperature at which the vapor pressure of benzene is 1 atm.), then you include the heat of vaporization at 378.1, then you add the change in enthalpy from 378.1 to 583 for the benzene vapor.

    Hope this makes sense.

  10. Sep 13, 2014 #9
    Yes, I do see.

    I had tried your alternative method but didn't account for the vapor pressure of benzene, instead I used the boiling temperature of benzene which was 353.2 K.

    I really appreciate the help.
  11. Sep 13, 2014 #10
    Oh boy. Senior Moment!!! I added the 80C to 298K instead of 273K. Thanks for picking up on that.

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