Nitrogen and oxygen react to produce nitric oxide according to the following equ

Question

Nitrogen and oxygen react to produce nitric oxide according to the following equation: N2 (g) + O2 (g) 2 NO (g) The equilibrium constant for this reaction is 1.70 x 10-3. Suppose that 0.110 mol N2 and 0.330 mol O2 are mixed in a 2.00-L reaction vessel. What will be the concentrations of N2, O2, and NO when equilibrium is established? (Hint: assume that the amounts of N2 and O2 that react are small—less than 10% of the starting amounts—and check your assumption when you have solved the equation.).Show all your steps.

Explanation / Answer

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

Kc = [NO[^2/[N2][O2] = 1.70 x 10^-3

initial [N2] = 0.110 mol/2 L = 0.055 M

initial [O2] = 0.330 mol/2 L = 0.165 M

ICE chart,

           N2(g) + O2(g) --> 2NO(g)

I         0.055     0.165         -       

C          -x          -x            +2x

E     0.055-x   0.165-x        2x

So,

1.70 x 10^-3 = (2x)^2/(0.055-x)(0.165-x)

4x^2 = 1.70 x 10^-3x^2 - 3.74 x 10^-4x + 1.53 x 10^-5

4x^2 + 3.74 x 10^-4x - 1.53 x 10^-5 = 0

x = 1.91 x 10^-3 M

so at equilibrium,

[N2] = 0.055 - 1.91 x 10^-3 = 0.05309 M

[O2] = 0.165 - 1.91 x 10^-3 = 0.16309 M

[NO] = 2 x 1.91 x 10^-3 = 3.82 x 10^-3 M

with 10% approximation,

1.70 x 10^-3 = (2x)^2/(0.055)(0.165)

4x^2 = 1.53 x 10^-5

x = 1.96 x 10^-3 M

so at equilibrium,

[N2] = 0.055 - 1.96 x 10^-3 = 0.05331 M

[O2] = 0.165 - 1.96 x 10^-3 = 0.16304 M

[NO] = 2 x 1.96 x 10^-3 = 3.92 x 10^-3 M

So the value at equilibrium does not change much with 10% with or without approximation.

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