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Chem 2: Molecular Equilibria

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data
    The standard Gibb's free energy for the reaction N2+3H2<--->2NH3 is -33.0 KJ at 298 K. Everything is a gas. Calculate Kp for this reaction.


    2. Relevant equations

    I have in my notes that Keq=e^(-Delta G/RT) where R is the constant 0.008314 KJ/K. I also have a formula that converts between Kp and Keq: Kp= Kc(.0821Xtemperature)^(Moles of gasseous products-moles of gasseous reactants).


    3. The attempt at a solution
    My attempt at a solution was to use the first formula (which I think gives Kc) and then convert it into Kp using the second formula.
    My answer ended up differing from the one given in my homework though. Here is my work:

    Keq= e^(33.0/(0.008314x298)= 6.09x10^5

    Kp= (6.09x10^5)(0.0821x298)^(-2) = not the correct answer

    The answer is: Kp=5.97x10^5
     
    Last edited: Apr 5, 2007
  2. jcsd
  3. Apr 7, 2007 #2
    I think here, you have to use R=8.314 joules/mol . Keq can be calculated in the same way as you showed.

    That is Kc. Kp=Kc(RT)^(delta Ng)
     
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