Question

Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: At a certain temperature, a chemist finds that a 7.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount

HNO 16.2 g 11.0 g 18.6 g H20 236.7 g 3 NO NO
Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.

Answer 1

K = 3.3

Step-by-step explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

K = 3.3