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Physical Chemistry

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data
    For the reaction CH3OH(l) → CH4 (g)+ 1/2 O2 (g)

    (a) Calculate ∆Ho298

    (b) Calculate ∆Eo298

    (c) Write an equation that would allow you to determine ∆H at 500°C and 1atm

    2. Relevant equations
    ∆H=∆E+∆(PV)
    ∆H=Σ∆Hproducts-Σ∆Hreactants

    3. The attempt at a solution

    a) [∆(CH4)+.5(∆O2)]-∆(CH3OH)=[-74.87 kJ/mol + .5(0 kJ/mol)]-(-238.4 kJ/mol)=∆o298=163.53 kJ/mol

    b) I'm lost on how to find the change in internal energy of the system here... Is there a PV integral for work, maybe? If I can't solve it, I may be a smart ass and just write "energy is conserved, so ΔE is 0 for the universe"

    c) ΔGo=ΔHo-TΔSo would be my approach here, any feedback on whether this is correct and why would be appreciated
     
    Last edited: Oct 8, 2015
  2. jcsd
  3. Oct 8, 2015 #2
    Part b):

    To get ΔE, you have to subtract Δ(PV). What is the molar volume of methanol liquid? You can use the ideal gas law to get the molar volume of methane and oxygen.

    In part c), are you asking about ΔG or ΔH. There's nothing after the Δ.

    Chet
     
  4. Oct 8, 2015 #3
    Thank you! I will try the enthalpy=energy+pv equation.
    in (c) it was meant to be delta H, I've edited it so it should appear now.
     
  5. Oct 8, 2015 #4
    For part c), you just start with ΔH at 298K and 1 atm, and use molar heat capacities in conjunction with Hess's law to get ΔH at 500 K. Remember Hess's law?

    Chet
     
  6. Oct 8, 2015 #5
    Yup! The sum of the state functions of the products minus the sum of the state functions of the reactants is equal to the sum of the state function of the reaction; change in enthalpy is a state function, so it applies here.
     
  7. Oct 8, 2015 #6
    So, you're able to do part (c) now, right?

    Chet
     
  8. Oct 8, 2015 #7
    Would it look something like:
    ΔH500=ΔHo+(500-298)*___kJ/mol*___mol
    for each component of the reaction?
    Where ___kJ/mol is the molar heat capacity at constant pressure, and ___mol is the quantity of substance
     
  9. Oct 8, 2015 #8
    yes, but you would have to cool the reactant down to 298 and heat the products back up to 500.
     
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