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Homework Help: Chemical equilibrium help

  1. Nov 28, 2008 #1
    Chemical equilibrium help!!!!

    For a reaction 2SO2(g) + O2(g) ↔ 2 SO3(g), 0.1mol of each SO2 and SO3 are mixed in a 2.0L flask at 27 degrees Celsius. After Equilibrium total pressure is 2.78atm.
    Calculate a) The mole fraction of O2 at equilibrium
    b) The value of Kp

    I dont know how to find out the mole fraction of O2...
    I did the problem finding Kp.. but I got a large number of 19.31....
    And my Nt at equilibrium= n(4-alpha)

    Any ideas??
  2. jcsd
  3. Nov 28, 2008 #2
    Re: Chemical equilibrium help!!!!

    Here is my work:

    2SO2(g) + O2(g) ↔ 2 SO3(g)
    2n 0 ↔ 2n
    -2n@ -n@ ↔ +2n@

    2n-2n@ -n@ ↔ 2n +2n@ Where Nt= 2n-2n@-n@+2n+2n@
    2n(1-@) -n@ ↔ 2n(1+@) Nt=4n-n@ Nt= n(4-@)

    2n(1-@)/n(4-@) -n@/(n4-@) ↔ 2n(1+@)/ (n (4-@))

    2(1-@)/(4-@) Pt 1/4 Pt ↔ 2(1+@)/ (4-@))

    Where Kp= (P SO3(g))^2 / ((P SO2)^2 * (P O2))

    Kp= ((4(1+@)^2) / ((1-@)^2 ) ) 1/Pt

    Where Nt at equilibrium= ( 2.78atm*2L) /( 0.08206 atm dm3 mol-1 k-1)(301.15)
    Nt= .2249 moles

    solving for @
    .2449= .1(4-@)
    @= 1.751

    Plugging in @ to find Kp.... I got 19.31....... but i'ts wrong it's supposed to be 0.356
  4. Nov 29, 2008 #3
    Re: Chemical equilibrium help!!!!

    try to convert the Kp to Kc.
  5. Nov 30, 2008 #4


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    Homework Helper
    Gold Member

    Re: Chemical equilibrium help!!!!

    You probably haven't received help because your answer is too hard to read.
    I don't know what @ or Nt mean.
    And just above the input box there are symbols X2 and X2 which allow you to easily write things like [SO3]2 which help legibility.

    The way I would find it easiest:

    Total S (sulphur) is 0.1 M if I am not mistaken.
    From the pressure work out the total molarity of the gas.
    The difference is the molarity of O2.
    Last edited: Nov 30, 2008
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