Question

nitrogen dioxide is one of the many oxides of nitrogen (often collectively called
ox\) that are of interest to atmospheric chemistry. it can react with itself to form another form of nox, dinitrogen tetroxide.
a chemical engineer studying this reaction fills a 25 l tank with 5.2 mol of nitrogen dioxide gas. when the mixture has come to equilibrium he determines that it contains 2.4 mol of nitrogen dioxide gas.
the engineer then adds another 2.6 mol of nitrogen dioxide, and allows the mixture to come to equilibrium again. calculate the moles of dinitrogen tetroxide after equilibrium is reached the second time. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(2NO_2(g)
ightleftharpoons N_2O_4(g)\)

Step2: Calculate the initial equilibrium concentrations

Initial moles of \(NO_2 = 5.2\space mol\), at first equilibrium moles of \(NO_2=2.4\space mol\).
Moles of \(NO_2\) reacted \(=5.2 - 2.4=2.8\space mol\)
From the stoichiometry (\(2\space mol\space NO_2\) produce \(1\space mol\space N_2O_4\)), moles of \(N_2O_4\) at first equilibrium \(n_1=\frac{2.8}{2}=1.4\space mol\)
Concentration of \(NO_2\) at first equilibrium \(C_{NO_2,1}=\frac{2.4}{25}=0.096\space M\), concentration of \(N_2O_4\) at first equilibrium \(C_{N_2O_4,1}=\frac{1.4}{25}=0.056\space M\)
Equilibrium constant \(K_c=\frac{C_{N_2O_4,1}}{C_{NO_2,1}^2}=\frac{0.056}{(0.096)^2}\approx6.1\)

Step3: Analyze the second - stage change

After adding \(2.6\space mol\) of \(NO_2\), moles of \(NO_2\) before second - stage equilibrium \(n_{NO_2,initial}=2.4 + 2.6=5.0\space mol\), concentration \(C_{NO_2,initial}=\frac{5.0}{25}=0.2\space M\)
Let \(x\) be the change in concentration of \(N_2O_4\) at second - stage equilibrium. Then concentration of \(NO_2=(0.2 - 2x)\space M\) and concentration of \(N_2O_4=(0.056 + x)\space M\) (since volume is constant, we can use moles per liter or just moles for the ratio as \(K_c\) is a ratio of concentrations)
Using \(K_c = 6.1=\frac{0.056 + x}{(0.2 - 2x)^2}\)
\(6.1(0.04-0.8x + 4x^2)=0.056+x\)
\(0.244-4.88x+24.4x^2=0.056+x\)
\(24.4x^2-5.88x + 0.188 = 0\)
Using the quadratic formula \(x=\frac{5.88\pm\sqrt{5.88^2-4\times24.4\times0.188}}{2\times24.4}\)
\(x=\frac{5.88\pm\sqrt{34.5744 - 18.3424}}{48.8}=\frac{5.88\pm\sqrt{16.232}}{48.8}=\frac{5.88\pm4.03}{48.8}\)
We take the positive root \(x=\frac{5.88 - 4.03}{48.8}\approx0.038\) (the other root gives a negative concentration for \(NO_2\) when substituted back)

Step4: Calculate moles of \(N_2O_4\) at second - stage equilibrium

Moles of \(N_2O_4\) at second - stage equilibrium \(n=(0.056 + x)\times25\) (since \(C=\frac{n}{V}\), \(n = C\times V\)). Substituting \(x = 0.038\)
\(n=(0.056+0.038)\times25=2.35\approx2.4\space mol\)

Answer:

\(2.4\space mol\)