Respuesta :
N2 & O2 <=> 2NO
0.2M & 0.2M <=> 0.4 M initially
it sounds like they add enough NO, to momentarily get [NO] to 0.7
but then that shifts in an attem,pt to re-equilibrate:
N2 & O2 <=> 2NO
[0.2+X] & [0.2+X] <=<= [0.7- 2X]
K = [NO]^2 / [N2] [O2]
4 = [0.7- 2X] ^2 / [0.2+X] [0.2+X]
4 = (0.49 - 2.8X + 4X^2) / (0.04 + 0.4X + X^2)
4 (0.04 + 0.4X + X^2) = (0.49 - 2.8X + 4X^2)
0.16 + 1.6X + 4X^2 = 0.49 - 2.8X + 4X^2
0.16 + 1.6X = 0.49 - 2.8X
1.6X & 2.8X = 0.49 - 0.16
4.4 X = 0.33
X = 0.075
so
final concentration of NO after equilibrium is re-established would be:
0.7- 2X
0.7- 2(0.075)
0.7- 0.15
0.55 Molar NO
================================
to check the answer:
N2 & O2 <=> 2NO
[0.2+X] & [0.2+X] <=<= [0.7- 2X]
[0.2+0.075] & [0.2+0.075] <=> [0.7- 2(0.075)]
[0.2+0.075] & [0.2+0.075] <=> [0.7- 0.15]
[0.275] & [0.275] <=> 0.55
K = [NO]^2 / [N2] [O2]
K = [0.55]^2 / [0.275] [0.275]
K = 0.3025 / 0.075625
K = 4
it checks out
your answer is
[NO] = 0.55 Molar
0.2M & 0.2M <=> 0.4 M initially
it sounds like they add enough NO, to momentarily get [NO] to 0.7
but then that shifts in an attem,pt to re-equilibrate:
N2 & O2 <=> 2NO
[0.2+X] & [0.2+X] <=<= [0.7- 2X]
K = [NO]^2 / [N2] [O2]
4 = [0.7- 2X] ^2 / [0.2+X] [0.2+X]
4 = (0.49 - 2.8X + 4X^2) / (0.04 + 0.4X + X^2)
4 (0.04 + 0.4X + X^2) = (0.49 - 2.8X + 4X^2)
0.16 + 1.6X + 4X^2 = 0.49 - 2.8X + 4X^2
0.16 + 1.6X = 0.49 - 2.8X
1.6X & 2.8X = 0.49 - 0.16
4.4 X = 0.33
X = 0.075
so
final concentration of NO after equilibrium is re-established would be:
0.7- 2X
0.7- 2(0.075)
0.7- 0.15
0.55 Molar NO
================================
to check the answer:
N2 & O2 <=> 2NO
[0.2+X] & [0.2+X] <=<= [0.7- 2X]
[0.2+0.075] & [0.2+0.075] <=> [0.7- 2(0.075)]
[0.2+0.075] & [0.2+0.075] <=> [0.7- 0.15]
[0.275] & [0.275] <=> 0.55
K = [NO]^2 / [N2] [O2]
K = [0.55]^2 / [0.275] [0.275]
K = 0.3025 / 0.075625
K = 4
it checks out
your answer is
[NO] = 0.55 Molar
