1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

: chemistry qs: write the equillibrim constant for the reverse reaction

  1. URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    1. The problem statement, all variables and given/known data

    2. The equilibrium constant for the following reaction is 2.0 10^ 4

    2HBr(g)  H2(g)  Br2(g)


    2. Relevant equations

    b) What is the equilibrium constant for the reverse reaction?


    3. The attempt at a solution

    I don't think this is right but

    2.0 x 10^4 = (Br2)(H2)/(HBr)^2
    2.0 x 10^4 = (x)(x)/(2x)^2
    2.0 x 10^4 = 1/4

    so for the reverse reaction K = 4 * 2.0 x 10^4.
     
  2. jcsd
  3. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction


    You're right, it's not :) but this is right:


    If that is the forward reaction, what would be the equation for the reverse reaction? (it might be easier to to just think of it as k = (A)(B)/(C) for C --> A + B )
     
  4. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    In equilibrium the forward and reverse reaction rates are the same.

    Do you know what the equations for the reaction rates for each side are?
     
  5. Borek

    Staff: Mentor

    Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    While it can be done through reaction rates - it is unnecessary complication. It can be done much faster and much easier just by writing the equation for the reverse reaction (see zhermes post).
     
  6. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    ok it would be K = [HBr]^2/[H2][I2]

    but how does that give the the value of K?
     
  7. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    ok so the equation for the reverse reaction is 2HBr -> H2 + I2 and K = [HBr]^2/[H2][I2]

    ...how can i get the value of K from this
     
  8. Borek

    Staff: Mentor

    Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    [tex]K_1 = \frac {[H_2][Br_2]} {[HBr]^2} = 2 \times 10^4[/tex]

    [tex]K_2 = \frac {[HBr]^2} {[H_2][Br_2]} [/tex]

    If you still don't see the answer, try to calculate K1xK2.

    Edit: note, that most of your equations are incorrect as they have HBr on one side of the reaction, I2 on the second.
     
  9. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    so K1 * K2 =1

    that means K 2 must equal 1/2*10^ 4
     
    Last edited: Nov 29, 2010
  10. Borek

    Staff: Mentor

    Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    Wasn't that hard.
     
  11. Re: URGENT: chemistry qs: write the equillibrim constant for the reverse reaction

    Not quite, try multiplying K1 and K2 to check.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?