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Calculation of equilibrium constants and degree of dissociation of N2O4 (8413)

Viernes 14 de marzo de 2025, por F_y_Q

At $30\ ^oC$ and 1 atm, $\ce{N2O4}$ dissociates into $\ce{NO2}$ by $20\ \%$ according to the following equilibrium:

$\ce{N2O4(g) <=> 2NO2(g)}$

Calculate:

a) The values of the equilibrium constants $\ce{K_P}$ and $\ce{K_C}$ at this temperature.

b) The percentage of dissociation at $30\ ^oC$ and a total pressure of 0.1 atm.

Data: $R = 0.082\ \textstyle{atm \cdot L \over K \cdot mol}$ .

The first step is to determine the number of moles of each substance at equilibrium, based on the initial moles and the degree of dissociation ( $\alpha$ ):

$\ce{\underset{n_0(1-\alpha)}{\ce{N2O4(g)}}} \ce{<=> \underset{2n_0\alpha}{\ce{2NO2(g)}}}$

Since the degree of dissociation is $\alpha = 0.2$ , the moles can be written as:

$\ce{\underset{0.8n_0}{\ce{N2O4(g)}}} \ce{<=> \underset{0.4n_0}{\ce{2NO2(g)}}}$

The total number of moles at equilibrium is the sum of all species:

$n_T = 0.8n_0 + 0.4n_0\ \to\ \color[RGB]{2,112,20}{\bm{n_0 = 1.2n_0}}$

Next, calculate the mole fraction for each substance:

$x_{\ce{N2O4}} = \frac{0.8\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{2}{3}}}$

$x_{\ce{NO2}} = \frac{0.4\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{1}{3}}}$

a) The equilibrium constant in terms of partial pressures is:

$\color[RGB]{2,112,20}{\bm{K_p = \frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}}}$

Substitute the values into this equation to determine the constant:

$\ce{K_p} = \frac{x_{\ce{NO2}}^2\cdot P_T\cancel{^2}}{x_{\ce{N2O4}}\cdot \cancel{P_T}} = \frac{(\frac{2}{3})^2\cdot 1\ atm}{\frac{1}{3}}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{K_p = \frac{1}{6}\ atm}}}$

The equilibrium constant in terms of concentrations is calculated as:

$\color[RGB]{2,112,20}{\bm{K_c = K_p(RT)^{-\Delta n}}}$

The change in the number of moles of gas is one, so substitute the values to find the constant:

$K_c = \frac{1}{6}\ \cancel{\cancel{atm}}\left(0.082\ \frac{\cancel{atm}\cdot L}{mol\cdot \cancel{K}}\cdot 303\ \cancel{K}\right)^{-1}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{K_c = 6.68\cdot 10^{-3}\ M}}}$

b) When the total pressure of the system changes to 0.1 amt, the equilibrium shifts to favor greater dissociation of $\ce{N2O4}$ . The same expression for $\ce{K_p}$ holds, and it is used to calculate the new degree of dissociation under these conditions. Be cautious to express the mole fractions in terms of the initial moles:

$\ce{K_p} = \frac{x_{\ce{NO2}}^2\cdot P_T}{x_{\ce{N2O4}}}\ \to\ K_p = \frac{\dfrac{4\cancel{n_0^2}\alpha^2}{\cancel{n_0^2}(1+\alpha)\cancel{^2}}\cdot P_T}{\dfrac{\cancel{n_0}(1-\alpha)}{\cancel{n_0}\cancel{(1+\alpha)}}}\ \to\ \color[RGB]{2,112,20}{\bm{K_c = \frac{4\alpha^2\cdot P_T}{(1-\alpha)(1+\alpha)}}}$

For simplicity, work with the denominator and substitute to make solving $\alpha$ easier:

$\frac{1}{6} = \frac{0.4\alpha^2}{1-\alpha^2}\ \to\ 0.415 = 1.415\alpha\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{\alpha = 0.54 = 54\%}}}$

Calculation of equilibrium constants and degree of dissociation of N2O4 (8413)

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