Question

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.

Answer 1

Answer:

$K=1.12x10^9$

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:

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

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

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

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

Best regards!