Question

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M....

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.

N2(g) + O2(g) <-----> 2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

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Answer #1

[N2]=[O2]=0.200 M and [NO]=0.500 M.

K = (NO^2)/(N2)(O2)

K = (0.5^2)/(0.2*0.2) =6.25

If

[NO] = 0.80

[N2] = [O2] = 0.2

then

in equilibrium

[NO] = 0.80 +2x

[N2] = [O2] = 0.2 -x

x here will be negative since the reaciton goes backwards (the shift goes to left)

then

K = (0.80 +2x)^2/(0.2 -x)^2

sqrt(6.25) = (0.80 +2x)/(0.2-x)

2.5(0.2-x) = (0.80 +2x)

0.5 - 2.5x = 0.80 + 2x

-4.5x = 0.8-0.5

x = -(0.8-0.5 )/4.5) = -0.0666

as expected, this is negative

so

[NO] = 0.80 +2*(-0.0666) =0.6668

[N2] = [O2] = 0.2 --0.0666 = 0.2666

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