Chemistry

Nitrogen dioxide decomposes on heating according to the following equation

2NO2 (g) ↔ 2NO (g) + O2 (g)

When 4 mole of nitrogen dioxide were put into a 1dm3 container and heated to a constant temperature, the equilibrium mixture contained 0.8 mole of oxygen.
What is the numerical value of the equilibrium constant Kc, at the temperature of the experiment?

Answer is . 1.6^2 X 0.8 / 2.4^2

, I want to know how to solve it , i have MCAT exams please help me out , Thanks in advance

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  1. pretty easy.

    K=[NO]^2[O2]/ [NO2]^2

    note that each mole of NO2 creates one mole of NO, and a half mole of O2

    K=x^2*(.5x)^2/(4-x)^2 but .5x=.8, so x=1.6
    put that value for x in and you have it.

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    bobpursley
  2. @bobpursley , thanks for solution, but i still don't get it bro, what we did to 4 mole of N ? and 0.8 mol of O? help and how to write answer in this form
    1.6^2 X 0.8 / 2.4^2
    ?

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  3. Those values were given.
    4moles initially of NO2
    .8 moles of O2 in equalibirum

    the 2.4 is the final NO2 concentration, which is 4-1.6 or 4-2*.8

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    bobpursley
  4. In the equlibrium expression, the values of concentration are FINAL values.

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    bobpursley
  5. @bobpursley , Thanks alot i got it . Thanks Sir

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  6. can you please tell me how it has done? =x^2*(.5x)^2/(4-x)^2

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  7. how we got 2.4

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