Nitrogen dioxide decomposes on heating according to the following equation

Question

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

Answer 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.

Answer 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
?

Answer 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

Answer 4

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

Answer 5

@bobpursley , Thanks alot i got it . Thanks Sir

Answer 6

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

Answer 7

how we got 2.4