https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://www.ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://www.ejercicios-fyq.com/Calculation-of-equilibrium-constants-and-degree-of-dissociation-of-N2O4-8413 https://www.ejercicios-fyq.com/Calculation-of-equilibrium-constants-and-degree-of-dissociation-of-N2O4-8413 2025-03-14T05:49:47Z text/html es F_y_Q SOLVED Equilibrium constant Degree of dissociation <p>At and 1 atm, dissociates into by according to the following equilibrium: <br class='autobr' /> <br class='autobr' /> Calculate: <br class='autobr' /> a) The values of the equilibrium constants and at this temperature. <br class='autobr' /> b) The percentage of dissociation at and a total pressure of 0.1 atm. <br class='autobr' /> Data: .</p> - <a href="https://www.ejercicios-fyq.com/Chemical-reactions-270" rel="directory">Chemical reactions</a> / <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a>, <a href="https://www.ejercicios-fyq.com/Equilibrium-constant" rel="tag">Equilibrium constant</a>, <a href="https://www.ejercicios-fyq.com/Degree-of-dissociation" rel="tag">Degree of dissociation</a> <div class='rss_texte'><p>At <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/e377e6d6fa467c168c039e4ec52eff00-8b3d5.png?1732957639' style='vertical-align:middle;' width='55' height='42' alt="30\ ^oC" title="30\ ^oC" /> and 1 atm, <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L38xH15/2275ea8c97ef884a19d0b29c49b74f82-d567c.png?1732971950' style='vertical-align:middle;' width='38' height='15' alt="\ce{N2O4}" title="\ce{N2O4}" /> dissociates into <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L31xH15/3bff4ff64128b7961af6d9893d7df955-4b7e7.png?1732958360' style='vertical-align:middle;' width='31' height='15' alt="\ce{NO2}" title="\ce{NO2}" /> by <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L41xH19/241c990a7dee7378dc0bd3d60c2585e1-bdb86.png?1733051788' style='vertical-align:middle;' width='41' height='19' alt="20\ \%" title="20\ \%" /> according to the following equilibrium:</p> <p> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L163xH18/dc13eca0b7410c9e801460f082fa8b72-1ffe1.png?1732972179' style='vertical-align:middle;' width='163' height='18' alt="\ce{N2O4(g) <=> 2NO2(g)}" title="\ce{N2O4(g) <=> 2NO2(g)}" /></p> </p> <p>Calculate:</p> <p>a) The values of the equilibrium constants <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L21xH15/d1eabe2ec5e8b66c30d676030f0ed0f1-cc54d.png?1732956653' style='vertical-align:middle;' width='21' height='15' alt="\ce{K_P}" title="\ce{K_P}" /> and <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L21xH16/a0ec985ab1c8f83ce4abf56d2d6fd75a-a9ca3.png?1732956653' style='vertical-align:middle;' width='21' height='16' alt="\ce{K_C}" title="\ce{K_C}" /> at this temperature.</p> <p>b) The percentage of dissociation at <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/e377e6d6fa467c168c039e4ec52eff00-8b3d5.png?1732957639' style='vertical-align:middle;' width='55' height='42' alt="30\ ^oC" title="30\ ^oC" /> and a total pressure of 0.1 atm.</p> <p>Data: <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L145xH27/a4bb6a710452a0da15b10aedf4d6bd13-bff24.png?1741931447' style='vertical-align:middle;' width='145' height='27' alt="R = 0.082\ \textstyle{atm \cdot L \over K \cdot mol}" title="R = 0.082\ \textstyle{atm \cdot L \over K \cdot mol}" />.</math></p></div> <hr /> <div <div class='rss_ps'><p>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 (<img src='https://www.ejercicios-fyq.com/local/cache-TeX/7b7f9dbfea05c83784f8b85149852f08.png' style="vertical-align:middle;" width="18" height="30" alt="\alpha" title="\alpha" />): <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/5fd5ff3d40492d8725773c12b3f0fe93.png' style="vertical-align:middle;" width="163" height="31" alt="\ce{\underset{n_0(1-\alpha)}{\ce{N2O4(g)}}} \ce{<=> \underset{2n_0\alpha}{\ce{2NO2(g)}}}" title="\ce{\underset{n_0(1-\alpha)}{\ce{N2O4(g)}}} \ce{<=> \underset{2n_0\alpha}{\ce{2NO2(g)}}}" /></p> <br/> Since the degree of dissociation is <img src='https://www.ejercicios-fyq.com/local/cache-TeX/05612f0a57949a7c3d1cf278dddd1049.png' style="vertical-align:middle;" width="52" height="12" alt="\alpha = 0.2" title="\alpha = 0.2" />, the moles can be written as: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/52cf66dd18afb2c9c34b61307bba2def.png' style="vertical-align:middle;" width="163" height="29" alt="\ce{\underset{0.8n_0}{\ce{N2O4(g)}}} \ce{<=> \underset{0.4n_0}{\ce{2NO2(g)}}}" title="\ce{\underset{0.8n_0}{\ce{N2O4(g)}}} \ce{<=> \underset{0.4n_0}{\ce{2NO2(g)}}}" /></p> <br/> The total number of moles at equilibrium is the sum of all species: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/8e1e3b96379a36aa0d2aec01d4a3f66b.png' style="vertical-align:middle;" width="258" height="16" alt="n_T = 0.8n_0 + 0.4n_0\ \to\ \color[RGB]{2,112,20}{\bm{n_0 = 1.2n_0}}" title="n_T = 0.8n_0 + 0.4n_0\ \to\ \color[RGB]{2,112,20}{\bm{n_0 = 1.2n_0}}" /> <br/> <br/> Next, calculate the mole fraction for each substance: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/9639dcb8e5266665feaead120115462c.png' style="vertical-align:middle;" width="135" height="39" alt="x_{\ce{N2O4}} = \frac{0.8\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{2}{3}}}" title="x_{\ce{N2O4}} = \frac{0.8\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{2}{3}}}" /> <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/c0ea6e6347d34920e7bc4047748a7a92.png' style="vertical-align:middle;" width="129" height="39" alt="x_{\ce{NO2}} = \frac{0.4\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{1}{3}}}" title="x_{\ce{NO2}} = \frac{0.4\cancel{n_0}}{1.2\cancel{n_0}} = \color[RGB]{0,112,192}{\bm{\frac{1}{3}}}" /> <br/> <br/> a) The equilibrium constant in terms of partial pressures is: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/7bb1a5ae1ada4ae1a36684478625349c.png' style="vertical-align:middle;" width="95" height="43" alt="\color[RGB]{2,112,20}{\bm{K_p = \frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}}}" title="\color[RGB]{2,112,20}{\bm{K_p = \frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}}}" /> <br/> <br/> Substitute the values into this equation to determine the constant: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/9ea30ddc8f45f9e83979a2523afb9021.png' style="vertical-align:middle;" width="369" height="46" alt="\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}}}" title="\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}}}" /></p> <br/> The equilibrium constant in terms of concentrations is calculated as: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/80977f2607a0b2eeeee8ad9f720e4c5c.png' style="vertical-align:middle;" width="143" height="20" alt="\color[RGB]{2,112,20}{\bm{K_c = K_p(RT)^{-\Delta n}}}" title="\color[RGB]{2,112,20}{\bm{K_c = K_p(RT)^{-\Delta n}}}" /> <br/> <br/> The change in the number of moles of gas is one, so substitute the values to find the constant: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/f7ff6fd4a9c39b9974d8bbd0dcb23cc0.png' style="vertical-align:middle;" width="489" height="43" alt="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}}}" title="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}}}" /></p> <br/> <br/> b) When the total pressure of the system changes to 0.1 amt, the equilibrium shifts to favor greater dissociation of <img src='https://www.ejercicios-fyq.com/local/cache-TeX/2275ea8c97ef884a19d0b29c49b74f82.png' style="vertical-align:middle;" width="38" height="15" alt="\ce{N2O4}" title="\ce{N2O4}" />. The same expression for <img src='https://www.ejercicios-fyq.com/local/cache-TeX/a869f115bfd1bcf8582c86f84e8d04f4.png' style="vertical-align:middle;" width="18" height="18" alt="\ce{K_p}" title="\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: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/a0f881dc56c5467d04858d56f02b1163.png' style="vertical-align:middle;" width="506" height="88" alt="\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)}}}" title="\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)}}}" /> <br/> <br/> For simplicity, work with the denominator and substitute to make solving <img src='https://www.ejercicios-fyq.com/local/cache-TeX/7b7f9dbfea05c83784f8b85149852f08.png' style="vertical-align:middle;" width="18" height="30" alt="\alpha" title="\alpha" /> easier: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/4aaf3d54e847786b16b0cd1934b8c589.png' style="vertical-align:middle;" width="398" height="37" alt="\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\%}}}" title="\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\%}}}" /></p> </math></p></div>