Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this: NO2(g) + H2O(l) →HNO3(aq) + NO(g)
At a certain temperature, a chemist finds that a reaction vessel containing a mixture of nitrogen dioxide, water, nitric acid, and nitrogen monoxide at equilibrium has the following composition
HNO3 15.5g
NO 16.6g
NO2 22.5g
H2O 189.0g
Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

0.29

Explanation:

Step 1: Write the balanced equation at equilibrium

3 NO₂(g) + H₂O(l) ⇄ 2 HNO₃(aq) +  NO(g)

Step 2: Calculate the molar concentration of the substances

We will not calculate the molarity of H₂O because pure liquids are not included in the equilibrium constant. We will use the following expression.

Molarity = mass of solute / molar mass of solute × liters of solution

Since the volume of the vessel is missing, we will assume it is 1 L to see the procedure.

[NO₂] = 22.5 / 46.01 g/mol × 1 L = 0.489 M

[HNO₃] = 15.5 g / 63.01 g/mol × 1 L = 0.246 M

[NO] = 16.6 g / 30.01 g/mol × 1 L = 0.553 M

Step 3: Calculate the value of the equilibrium constant Kc for this reaction

Kc = [HNO₃]² × [NO] / [NO₂]³

Kc = 0.246² × 0.553 / 0.489³

Kc = 0.29

12 is it do you think so