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Equilibrium Pressure

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data
    H2 (g) + Cl2(g) [tex]\Updownarrow[/tex] 2HCl (g) Kp = 2.5 × 1033
    NH3(g) + HCl(g) [tex]\Updownarrow[/tex] NH4Cl(s) Kp = 2.1 × 1015
    N2(g) + 4H2(g) +Cl2 [tex]\Updownarrow[/tex] 2NH4Cl(s) Kp = 3.9 × 1070

    Determine the Kp for N2(g) + 3H2 (g) [tex]\Updownarrow[/tex] 2NH3(g).

    2. Relevant equations
    Don't know/none


    3. The attempt at a solution
    In order to get to N2(g) + 3H2 (g) [tex]\Updownarrow[/tex] 2NH3(g), I would have to multiple/flip equations so that they result in the desired reaction set.
    -(H2 (g) + Cl2(g) [tex]\Updownarrow[/tex] 2HCl (g) Kp = 2.5 × 1033)
    -2(NH3(g) + HCl(g) [tex]\Updownarrow[/tex] NH4Cl(s) Kp = 2.1 × 1015)
    N2(g) + 4H2(g) +Cl2 [tex]\Updownarrow[/tex] 2NH4Cl(s) Kp = 3.9 × 1070

    These would cancel out to the desired reaction.

    In Hess's Law, I understand that multiplying a step would mean its enthalpy gets multiplied by that number. If I flip a step, its enthalpy would inverse its sign.

    In voltage calculation from standard reduction potentials, reversing the sign would inverse the potential for the step but multiplying the step does not affect the potential.

    Originally I would just follow Hess's Law to calculate but I never did this for Kp and so I'm not sure the answer would be correct. How do I approach this problem and solving for Kp.
     
  2. jcsd
  3. Apr 16, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Just combine these reactions so that everything cancels out - you will be left with Kp=f(Kp1,Kp2,Kp3) (indices just to signal these are constants for each reaction given). That's all.
     
  4. Apr 16, 2009 #3
    Ok thanks.

    On a side note, I think the latex reference for some of the arrows are wrong.
     
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