Chemistry 112

2 NO(g) → ← N2(g) + O2(g)





Initial0.100/2.00 = 0.0500 M 0 0

Change–2x +x +x

Equilibrium0.0500 – 2x x x

The final conditions for the experiment give us [NO]e = 0.00044 mol/2.00 L = 0.00022 M.

This allows us to find x:

[NO]e = 0.0500 – 2x = 0.00022

or x = 0.0249 M = [N2]e = [O2]e

Now we have all the values we need to calculate Kc:

Recall that the relationship between Kp and Kc is

Kp = Kc(RT)Δn

Δn = 1 + 1 - 2 = 0

so

Kp = Kc(RT)0 = Kc = 1.3×104

 

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