Questions: Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) At a certain temperature, a chemist finds that a 2.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount HNO3 23.5 g NO 10.2 g NO2 6.8 g H2O 238.1 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits. Kc=

Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

At a certain temperature, a chemist finds that a 2.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition:
compound amount
HNO3 23.5 g
NO 10.2 g
NO2 6.8 g
H2O 238.1 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.
Kc=

Transcript text: Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: \[ 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \rightarrow 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \] At a certain temperature, a chemist finds that a 2.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: \begin{tabular}{|c|c|} \hline compound & amount \\ \hline $\mathrm{HNO}_{3}$ & 23.5 g \\ \hline NO & 10.2 g \\ \hline $\mathrm{NO}_{2}$ & 6.8 g \\ \hline $\mathrm{H}_{2} \mathrm{O}$ & 238.1 g \\ \hline \end{tabular} Calculate the value of the equilibrium constant $K_{c}$ for this reaction. Round your answer to 2 significant digits. \[ K_{c}= \] $\square$

Solution

Solution Steps

Step 1: Calculate Moles of Each Compound

First, we need to convert the mass of each compound to moles using their molar masses:

Molar mass of $\mathrm{HNO}_3$ = 63.01 g/mol
Molar mass of $\mathrm{NO}$ = 30.01 g/mol
Molar mass of $\mathrm{NO}_2$ = 46.01 g/mol
Molar mass of $\mathrm{H}_2\mathrm{O}$ = 18.02 g/mol

Calculate moles:

\[ \text{Moles of } \mathrm{HNO}_3 = \frac{23.5 \, \text{g}}{63.01 \, \text{g/mol}} = 0.3730 \, \text{mol} \]

\[ \text{Moles of } \mathrm{NO} = \frac{10.2 \, \text{g}}{30.01 \, \text{g/mol}} = 0.3399 \, \text{mol} \]

\[ \text{Moles of } \mathrm{NO}_2 = \frac{6.8 \, \text{g}}{46.01 \, \text{g/mol}} = 0.1478 \, \text{mol} \]

\[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{238.1 \, \text{g}}{18.02 \, \text{g/mol}} = 13.21 \, \text{mol} \]

Step 2: Calculate Concentrations

Next, calculate the concentration of each compound by dividing the moles by the volume of the reaction vessel (2.7 L):

\[ [\mathrm{HNO}_3] = \frac{0.3730 \, \text{mol}}{2.7 \, \text{L}} = 0.1381 \, \text{M} \]

\[ [\mathrm{NO}] = \frac{0.3399 \, \text{mol}}{2.7 \, \text{L}} = 0.1259 \, \text{M} \]

\[ [\mathrm{NO}_2] = \frac{0.1478 \, \text{mol}}{2.7 \, \text{L}} = 0.05474 \, \text{M} \]

\[ [\mathrm{H}_2\mathrm{O}] = \frac{13.21 \, \text{mol}}{2.7 \, \text{L}} = 4.889 \, \text{M} \]

Step 3: Write the Expression for $K_c$

The equilibrium constant expression for the reaction is:

\[ K_c = \frac{[\mathrm{NO}_2]^3 [\mathrm{H}_2\mathrm{O}]}{[\mathrm{HNO}_3]^2 [\mathrm{NO}]} \]

Step 4: Substitute Concentrations into the Expression

Substitute the concentrations into the equilibrium expression:

\[ K_c = \frac{(0.05474)^3 \times 4.889}{(0.1381)^2 \times 0.1259} \]

Step 5: Calculate $K_c$

Calculate the value of $K_c$:

\[ K_c = \frac{(0.05474)^3 \times 4.889}{(0.1381)^2 \times 0.1259} = \frac{0.0001641 \times 4.889}{0.01907 \times 0.1259} = \frac{0.0008023}{0.002400} = 0.3343 \]