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Two quick/dumb questions regarding thermodynamics

  1. Mar 3, 2015 #1
    First question is : why is ##\bigg(\frac{\partial \Delta H_{vap/fusion maybe something else} }{\partial T} \bigg)_p=\Delta Cp ## I can't wrap my head around that and nowhere other than my class can I find such an expression.

    The other question is regarding the calculation of stuff (say H, S, G, A) during chemical reactions as (for example, enthalpy of a reaction):

    ##H_{reaction}=\Bigg(\sum \nu_i \Delta Hº\Bigg)_{products}- \Bigg(\sum \nu_i \Delta Hº\Bigg)_{reactants}##

    Namely I don't get why adding Deltas ##\Delta## of formation (I'm excepting entropy here) will give the correct enthalpy of the reaction.

    I have a very naive argument that defends this expression which I will give just to see what happens:

    I imagine two mountains of different height. Here height is refers to the thermodynamic variable in question and its strictly a Delta with respect to the ground. If I have two states, my way to check the height between them is to substract one Delta from the other.

    This sounds dumb though.

    Any help would be nice. Thanks!
    Last edited: Mar 3, 2015
  2. jcsd
  3. Mar 4, 2015 #2


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    What makes you think this method wouldn't work?
    What do you think it should be?
  4. Mar 4, 2015 #3
    Well if I consider enthalpy for example, it's a bit mysterious. Why would the enthalpy of a reaction be equal to the difference of the enthalpies of formation? The only way I can make sense of it is through my dumb analogy of mountains and heights.

    Well, I don't know what it should be. I've finally seen a derivation on google after a lot of time, which makes sense to me at least. Namely they say:

    ##H_{reaction}=H_{products}-H_{reactants}## If I take the partial derivative of the LHS then its decomposed into two partial derivatives on the RHS which are just Cp's and its therefore a change in Cp.
  5. Mar 4, 2015 #4
    The answers to both your questions relate to the fact that the enthalpy of a system is a function of state and, in each case, you are determining the change in enthalpy between two equilibrium states. To do this, it doesn't matter what path you follow between the two equilibrium states. So you can choose any convenient route. These evaluations can be done by applying Hess's law. Are you familiar with the derivation and application of Hess's law?

    In the equation you wrote for the heat of reaction, those superscript 0's means that you are doing whatever is necessary to hold the final temperature the same as the initial temperature, and you are starting out with the pure reactant gases in stoichiometric proportions and going to the pure product gases in corresponding stoichiometric proportions (all at 1 atm.)

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