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Hess's law

  1. May 17, 2006 #1
    hello i really struggle with this!

    i can never get the equations right or anything!

    i cant find any sites that help!

    i have a practical exam tomorow that could be about it

    please help thanks
     
  2. jcsd
  3. May 17, 2006 #2

    dav2008

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    Feel free to post a problem and show your approach. That would probably be the best way for us to help you because we can clearly see where you make your mistakes.

    All Hess's law is saying is that the standard enthalpy change for a reaction is equal to the sum of the heat of formation of the products minus the sum of the heat of formation of the ractants.

    A quick google search for "Hess's Law Tutorial" brings this website: http://www.wwnorton.com/chemistry/tutorials/ch11.htm Scroll down and click on the Hess's law tutorial (all the way at the bottom)
     
    Last edited: May 17, 2006
  4. May 18, 2006 #3
    if you've done them, it's the same thing as: delta S reaction= delta S formation of products - delta S formation of reactants. the same goes for delta G reaction.

    many things in chem deal with the answer being products minus reactants or products over reactants and such
     
  5. May 19, 2006 #4
    I always find it helpful to draw a diagram.
    Then you can follow the arrows to show you which way round the diagram is equal to the bit you want. Change the sign of any arrows you have to go backwards on. (I'll rustle up an example diagram and scan it in sometime this week.)
     
  6. May 20, 2006 #5

    Gokul43201

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    In its most general form, Hess' Law merely reiterates the fact that enthalpy is a state function, not a path function. But this is nothing new to us. What's new is the way we apply this to the determination of unknown enthalpies of certain reactions given enthalpies of certain other related reactions.

    The fact that enthalpy is a state variable allows us to add chamical equations (and their associated enthalpies) like we do algebraic equations (without any concern for the order in which terms are added). The problem in chemistry, thus gets reduced to one of solving simultaneous linear equations.
     
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