Thermodynamics equilibrium constant problem

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Discussion Overview

The discussion revolves around a thermodynamics problem involving the equilibrium constant for the reaction of nitrogen and hydrogen to form ammonia. Participants explore the calculation of the equilibrium constant at a specified temperature and the estimation of percentage conversion at different total pressures, addressing the implications of initial partial pressures and stoichiometric ratios.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the reaction N2 + 3H2 --> 2NH3 and questions the necessity of high pressure, seeking to calculate the equilibrium constant at 500K and estimate percentage conversion at 1 bar and 50 bar total pressure.
  • Several participants inquire about the initial partial pressures at both 1 bar and 50 bar, noting the absence of this information in the problem statement.
  • Another participant suggests using Dalton's law of partial pressures to derive initial partial pressures based on total pressure and mole fractions, proposing that if no product is present initially, the partial pressures of N2 and H2 can be calculated.
  • There is a correction regarding the mole fractions of N2 and H2, with a participant initially miscalculating them but later acknowledging the correct values.
  • One participant suggests setting up an ICE table to calculate final partial pressures at equilibrium after determining initial conditions.
  • Another participant introduces a variable x to represent the fraction of reactants converted to ammonia, prompting further calculations of new mole fractions.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate initial partial pressures and the use of mole fractions, but there is no consensus on the specific values or methods to proceed with the calculations. The discussion remains unresolved regarding the exact approach to estimating percentage conversion.

Contextual Notes

The discussion highlights limitations in the provided problem statement, particularly the lack of initial partial pressures and the assumptions made regarding the absence of products at the start of the reaction. There are also unresolved mathematical steps related to the calculations of equilibrium conditions.

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%.
 
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What are the initial partial pressures at 1 bar? What are the initial partial pressures at 50 bars?
 
Chestermiller said:
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.
 
Samuel1321 said:
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?
 
Chestermiller said:
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.
 
Samuel1321 said:
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.
 
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Chestermiller said:
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?
 
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
Chestermiller said:
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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