Question

nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this:

2hno₃(aq)+no(g)→3no₂(g)+h₂o(l)

at a certain temperature, a chemist finds that a 8.8 l reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition:

calculate the value of the equilibrium constant k_c for this reaction. round your answer to 2 significant digits.

k_c=□

Explanation:

Step1: Calculate the molarity of each compound

• Molar mass of \(HNO_3\): \(M_{HNO_3}=1 + 14+3\times16 = 63\space g/mol\). Molarity \(c_{HNO_3}=\frac{n}{V}=\frac{\frac{6.2\space g}{63\space g/mol}}{8.8\space L}\approx0.0112\space M\)
• Molar mass of \(NO\): \(M_{NO}=14 + 16=30\space g/mol\). Molarity \(c_{NO}=\frac{n}{V}=\frac{\frac{14.3\space g}{30\space g/mol}}{8.8\space L}\approx0.0541\space M\)
• Molar mass of \(NO_2\): \(M_{NO_2}=14+2\times16 = 46\space g/mol\). Molarity \(c_{NO_2}=\frac{n}{V}=\frac{\frac{5.3\space g}{46\space g/mol}}{8.8\space L}\approx0.0130\space M\)
• Molar mass of \(H_2O\): \(M_{H_2O}=2\times1 + 16=18\space g/mol\). Molarity \(c_{H_2O}=\frac{n}{V}=\frac{\frac{167.6\space g}{18\space g/mol}}{8.8\space L}\approx1.06\space M\). Since \(H_2O(l)\) is a liquid, its concentration is considered constant and not included in the \(K_c\) expression for this reaction (as it is in a dilute aqueous solution and its activity is close to 1).

Step2: Write the equilibrium - constant expression

For the reaction \(2HNO_3(aq)+NO(g)
ightleftharpoons3NO_2(g)+H_2O(l)\), the equilibrium - constant expression \(K_c=\frac{[NO_2]^3}{[HNO_3]^2[NO]}\)

Step3: Substitute the molarity values into the \(K_c\) expression

\(K_c=\frac{(0.0130)^3}{(0.0112)^2\times0.0541}\)

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Answer:

\(0.32\)