1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equilibrium constant

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the equilibrium constant for the reaction
    H2(g) + I2(g) <---> 2HI(g)
    at 80oC. The vapor pressure of solid iodine is 0.0216 bar at that temperature. If 52.7g of solid iodine are placed in a 10.0 L vessel at 80.0oC, what is the minimum amount of hydrogen gas that must be introduced in order to remove all of the solid iodine?

    2. Relevant equations
    ΔrG = ΔfG(products) - ΔfG(reactants)
    ln K = -ΔrG / RT

    3. The attempt at a solution
    ΔfG for HI(g) = 1.70, for H2(g) = 0 and for I2(g) = 19.33 KJ mol-1
    ΔrG = 2(1.7) - 0 - 19.33
    ΔrG = -15.93 KJ mol-1
    ln K = -ΔrG / RT
    ln k = -(-15.93E3) / (8/314)(353) = 5.42
    k = 225.9

    Now I need to find the minimum amount of hydrogen gas and I have no idea where to start. Thanks for any help.
     
  2. jcsd
  3. Nov 1, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    You need enough hydrogen to shift equilibrium right till the pressure of iodine is below 0.0216 bar.

    --
    methods
     
  4. Nov 1, 2009 #3
    I'm not sure how to do that.. do I use the equilibrium constant? How would I incorporate the amount of iodine given?
     
  5. Nov 1, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    General approach is to write three equations - one will be the equilibrium constant, one iodine mass balance (mass of iodine present in HI and present as unreacted I2) and mass balance for hydrogen (again - in HI and unreacted). Fourth equation is condition mentioned above. Solve for total number of moles of hydrogen.

    --
    methods
     
  6. Nov 1, 2009 #5
    EDIT
    I'm not sure what a mass balance is..
    mass of iodine is 126.9 g/mol mass of H is 1.008g/mol
    mass of iodine in HI is 126.9 g/mol
    mass of iodine in I2 is 253.8 g/mol and there are 2 moles so 507.6 g
    mass of H in HI is 1.008 g
    mass of H in H2 is 2.016
    k=[C]c/[A]ab
    Sorry I'm still really confused.. where I need to include the 10.0L as the volume of the vessel? And the 52.7g of iodine..
     
    Last edited: Nov 1, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook