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Equilibrium of ammonia synthesis

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data
    In order to obtain commercial conversion levels in the production of ammonia, reactors operate between a pressure range from 150 to 300 atm, and a temperature range from 700 to 750 K. Calculate the mole fraction of N2 when the reaction reaches equilibrium at 300 atm and 723 K. The reactor's feedstock initially contained 25% in mole of N2 and 75% of H2. Under these conditions we have the following parameters:
    Ka = 0.0066
    γNH3 = 0.91
    γH2 = 1.09
    γN2 = 1.14

    2. Relevant equations
    [tex]\frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2 \rightarrow \textrm{NH}_3[/tex]
    [tex]K_a = K_p K_{\gamma}[/tex]

    3. The attempt at a solution
    First, I calculate the dummy constant Kγ:
    [tex]K_{\gamma} = \frac{\gamma_{NH_3}}{\gamma_{N_2}^{0.5} \gamma_{H_2}^{1.5}} = 0.7489[/tex]
    Now I can get Kp which is Ka/Kγ; 0.0088129. Now, I figured out it was easier working with the reversed reaction:
    [tex]\textrm{NH}_3 \rightarrow \frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2[/tex]
    The reversed reaction has Kp' = 113.47. Now, I applied the following method. Suppose we have the following initial conditions:
    NH3 - n moles
    N2 - 0 moles
    H2 - 0 moles
    When the reaction reaches equilibrium we will have:
    NH3 - n (1 + α) moles
    N2 - 0.5n α moles
    H2 - 1.5n α moles
    The total moles found in equilibrium are given by:
    [tex]n_T = n - n\alpha + 0.5n\alpha + 1.5n\alpha = n(1+\alpha)[/tex]
    Now, applying Pi = niP, the partial pressures of the components in equilibrium will be given by the following expressions:
    [tex]P_{NH_3} = \left( \frac{1-\alpha}{1+\alpha} \right) P[/tex]
    [tex]P_{N_2} = \left( \frac{0.5\alpha}{1+\alpha} \right) P[/tex]
    [tex]P_{H_2} = \left( \frac{1.5\alpha}{1+\alpha} \right) P[/tex]
    Kp' will be given by:
    [tex]K_p' = \frac{\left( \frac{0.5\alpha}{1+\alpha} \right)^{0.5} P^{0.5} \left( \frac{1.5\alpha}{1+\alpha} \right)^{1.5} P^{1.5}}{\left( \frac{1-\alpha}{1+\alpha} \right) P}[/tex]. Solving for α results in the following quadratic equation:
    [tex](K_p' + 1.299P)\alpha^2 - K_p' = 0[/tex]
    Plugging in Kp' = 113.47 and P = 300 atm and solving the equation results in α = 0.475. Now we can calculate the partial pressure of each of the components in equilibrium, which are 106.78 atm for ammonia, 48.31 atm for nitrogen and 144.91 atm for hydrogen. Finally, we calculate the mole fraction of N2:
    [tex]Y_{N_2} = \frac{48.31 \ atm}{300 \ atm} = 0.161[/tex]
    This result is apparently correct, although the procedure is a bit laborious; is there another method for solving this problem without involving the reversed reaction and α?

    Thanks in advance for any input!
     
    Last edited: Feb 25, 2015
  2. jcsd
  3. Feb 25, 2015 #2
    Sure. Start out with 0.25 moles of N2 and 0.75 moles of H2, and let them react until x moles of NH3 are formed. So, in the final equilibrium,

    moles of N2 = 0.25 - 0.5x

    moles of H2 = 0.75 - 1.5x

    total moles = 1 - x

    Then get the mole fractions and partial pressures.

    Chet
     
  4. Feb 26, 2015 #3
    Thanks, Chet! I got the same result using that method, and it sure is a bit shorter. I think I'll be working with this method for synthesis reactions and the other method for dissociation reactions. Is there a set of restrictions for using each method? Like knowing the initial proportion of reactants? In this case 25% nitrogen and 75% hydrogen.
     
  5. Feb 26, 2015 #4
    Yes, there's a restriction. You can't do it by the "reverse" method that you used unless you know in advance that the reactants were in stochiometric proportions.

    Chet
     
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