Thermodynamics equilibrium constant problem

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  • #1
Samuel1321
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Homework Statement



N2 + 3H2 --> 2NH3 (all gases)

Why is such a high pressure needed? Calculate the equilibrium constant at 500k then estimate the percentage conversion at equilibrium at 1 bar total pressure, assuming the stoichiometric ratio of N2:H2 is 1:3.

Repeat the process at 50 bar.

Homework Equations



Van Hoff equation : ln(keq) at final temp = -((delta G)/RTi) - ((delta H)/R)(1/Tf-1/Ti)

Keq = ((PNH3)^2)/((PH2)^3)(PN2)

PNH3 + PH2 + PN2 = 1 bar


The Attempt at a Solution


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I solved for keq at 500k. I have no idea how to estimate the percentage conversion at equilibrium at 1 bar total pressure since we aren't given any of the partial pressures.

I tried using molar ratios to solve but it doesn't really make sense. Here's what I did 1x + 3x + 2x = 1, 6x=1 x=0.33 = 33%. For 50 bar 1x+3x+2x=50, then 6x=50, x would be over 100%.
 

Answers and Replies

  • #2
22,321
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What are the initial partial pressures at 1 bar? What are the initial partial pressures at 50 bars?
 
  • #3
Samuel1321
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What are the initial partial pressures at 1 bar? What are the initial partial pressures at 50 bars?

They did not say what the initial pressures are.
 
  • #4
22,321
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They did not say what the initial pressures are.
You know the total pressure and the mole fractions of H2 and N2. So, from Dalton's law of partial pressures, what are the initial partial pressures of H2 and N2?
 
  • #5
Samuel1321
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You know the total pressure and the mole fractions of H2 and N2. So, from Dalton's law of partial pressures, what are the initial partial pressures of H2 and N2?

So assuming that there is no product in at the beginning of the reaction, partial pressure of N2 would be 1/3 bar and the partial pressure of H2 would be 2/3 bar.
 
  • #6
22,321
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So assuming that there is no product in at the beginning of the reaction, partial pressure of N2 would be 1/3 bar and the partial pressure of H2 would be 2/3 bar.
No, 1/4 and 3/4. These are the mole fractions.
 
  • #7
Samuel1321
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No, 1/4 and 3/4. These are the mole fractions.

Oh right, just had a brain fart, don't know how I missed that. So how should I proceed? Should I set up an equation for Keq?
 
  • #9
22,321
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Let x be the fraction of the reactants converted to ammonia. What are the new mole fractions of N2 and H2, and what is the mole fraction ammonia.
 
  • #10
Samuel1321
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Let x be the fraction of the reactants converted to ammonia. What are the new mole fractions of N2 and H2, and what is the mole fraction ammonia.
Thanks for the reply! I appreciate the help! I asked my professor and I figured it out.
 

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