caculate atm?

Question

At 2200°C, K = 0.000050 for the following reaction.
N2(g) + O2(g) 2 NO(g)

The gases N2 and O2 were placed in a flask at initial pressures of 0.91 and 0.25 atm, respectively. What is the value of 'x' (half of the partial pressure of NO at equilibrium)?

Answer 1

N2 O2 NO
initial .91 .25 0
change -x -x +2x
eq. (.91-x) (.25-x) +2x

K= (products)/(reactants)
K = [NO]^2 / [N2] [O2]

4x^2 / (.91 - x) (.25-x) = .000050

Solve for x.

.000050x^2 - .000058x + .00001375 = 4x^2

3.99995x^2 + .000058x - .00001375 = 0

x = .00185atm





And for giggles:

Equilibrium Partial Pressures :
N2 => 0.90815atm
O2 => 0.24815atm
NO => .0036atm

(Found by plugging x into the "equilibrium section" at the beginning)

Answer 2

N2 + O2 = 2NO

The number of moles of gaz is not changed after the reaction. YOu have to moles of gaz in the left of equqtion and two moles of gas after the reaction. I would say 0.91+0.25 = 1.16 after