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Given ∆Hf = -607 kJ/mol and S = 417 J, find ∆H and ∆S

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Ni (s) + 4 CO(g) ↔ Ni(CO)4(g)

    For Ni(CO)4(g), ∆Hf = -607 kJ/mol and S = 417 J/mol⋅K at 298 K. Using these values and data in the Appendix of your text, calculate ∆H and ∆S for the above reaction.

    2. Relevant equations
    no idea

    3. The attempt at a solution
    I don't know how to solve this, because my lab class somehow ended up a week ahead of lecture, so i haven't even learned thermodynamics yet.

    I think i found the answers online but i need to show work.
    ∆S°= -409.5 J/K
    ∆H° =−160.8 kJ

    I'm trying to look up formulas right now and find something i can use.
     
  2. jcsd
  3. Nov 15, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Don't you have a textbook?

    We can't do your work for you, but have a look at http://chem.libretexts.org/Core/Phy...tions/Enthalpy/Standard_Enthalpy_Of_Formation
    https://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalpy.htm [Broken]
    https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm [Broken]
     
    Last edited by a moderator: May 8, 2017
  4. Nov 15, 2016 #3
    Hae can you define it clearly ∆H° is for and also∆s°
     
  5. Nov 15, 2016 #4
    Can you send the questions image
     
  6. Nov 15, 2016 #5
    heres the image if it helps

    boBmFEm.png
     
  7. Nov 15, 2016 #6
    Yeah i don't want to come across like i'm hoping someone will do the work for me, i really try to avoid posting things like this.

    I do have a textbook, but i didn't want to read the entire chapter in one night so i could answer one question. I was hoping someone could just tell me the equations and i could plug them in.

    I'll take a look at the links you posted tomorrow.
     
  8. Nov 15, 2016 #7
    Last edited by a moderator: May 8, 2017
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