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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

Sagot :

sebassandin

Answer:

[tex]K=1.12x10^9[/tex]

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

[tex]K=\frac{[NO_2]^2}{[NO]^2[O_2]}[/tex]

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

[tex]K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9[/tex]

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

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