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Kp of NH4Cl reaction

  1. Aug 31, 2011 #1
    A 9g sample of solid NH4Cl is heated in a 4L container to 924.4oC and decomposes according to the following equation.
    NH4Cl(s) -> HCl(g) + NH3(g)

    at equalibrium the partial pressure of NH3 (g) is 1.346 atm. Calculate the equilibrium constant for Kp for this reaction


    2. Relevant equations

    Kp = Pproductn/Preactantn

    or

    Kc equation and Pv = nRT

    3. The attempt at a solution

    First I used the Ideal gas equation the amount of NH3 in the system.
    So n = PV/RT
    n = (136383.45 Pa * (4 *10-3 L))/(8.3144*1197.4K)
    n = 5.48*10-5

    This means the concentration of the NH4Cl is.
    n = 9/54.45 = 0.1653
    0.1653 - 2(5.48*10-5) = 0.1652

    I assumed then that Cl would also have this concentration.
    So the Kc of the reaction = ([5.48*10-5*[5.48*10-5)/0.1652 = 1.817 * 10-8

    Then Kp = Kc(RT)(delta)n
    Kp= 1.817 * 10-8 * (8.3144*1197.4K)1
    Kp = 0.00018 = 1.8*10-4

    The more I look at this the more I feel there is a much easier solution, is my solution even right?
     
  2. jcsd
  3. Aug 31, 2011 #2

    Borek

    User Avatar

    Staff: Mentor

    You have to calculate Kp, you are given pressure, why do you calculate concentrations instead of using pressure directly?

    Besides, what is the activity of the solid (hint: it is always identical)? Does your calculation of NH4Cl make any sense in this context?
     
  4. Aug 31, 2011 #3
    Well, the thing that confused me is that NH4Cl is a solid, so it is not counted in the calculation for Kp. Doesn't that mean it will be [products]/0? Or do you just not divide the product by anything?

    What do you mean about the solid? it says its heated, and gas forms so presumably it gets smaller. My calculations would suggest that the reaction doesn't react very far.
    I'm not really clear on what Kp is. Pressure are equilibrium?
     
  5. Aug 31, 2011 #4

    Borek

    User Avatar

    Staff: Mentor

    Activity of solids is always 1, period. Whenever you see a solid in reaction equation, just put 1 in the reaction quotient.

    How come you are asked to calculate Kp, but you don't know what it is? Please check your book or lecture notes. Seems like you are missing many basic parts of the theory, and that impedes your ability to solve the question.
     
  6. Aug 31, 2011 #5
    Ah! So I am wrong.
    I assumed it was pressure at equilibrium. And after some searching I thought I had come up with the right answer.
    Thanks! I'll keep looking.
     
  7. Aug 31, 2011 #6
    If, the activity of a solid is always 1.
    Then would Kp =[Products]
    Kp = [ 5.48*10-5] * [ 5.48*10-5]?
     
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