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13.8 g of N_2O_4 was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium, N_2O_(4(g)) hArr 2NO_(2(g)) The total pressure at equilibrium was found to be 9.15 bar. Calculate K_c, K_p and partial pressure at equilibrium.

Answer» <html><body><p></p>Solution :Molecular mass of `N_2O_4`=2(N)+4(O) <br/> =2(14)+4(16) = 92 g `"mol"^(-1)` <br/> Mole of `N_2O_4 = n = "<a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a>"/"Molecular mass"= "13.8 g"/"92 g mol"^(-1)`<br/> = 0.15 <br/> Temperature T=<a href="https://interviewquestions.tuteehub.com/tag/400-315233" style="font-weight:bold;" target="_blank" title="Click to know more about 400">400</a> K <br/> Gas constant = <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>= 0.083 bar L `mol^(-1) K^(-1)` <br/> pV=nRT where , p=Partial pressure `N_2O_4` <br/> `therefore p=(nRT)/V` <br/> `therefore p=((0.1 "mol")(0.083 "bar L mol"^(-1) K^(-1))(400 K))/(1 L)` <br/> = 4.98 bar partial pressure of `N_2O_4`<br/> `{:("Equilibrium reaction :",N_2O_(4(g)) hArr, 2NO_(2(g))),("<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> pressure :", "4.98 bar","0 bar"),("Pressure change :","-x bar" , "+<a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a> bar"),("Partial pressure at equilibrium :","(4.98-x)bar","2x bar"):}` <br/> `therefore` Total pressure at equilibrium = Addition of partial pressure <br/> `p_"total"=p_(N_2O_4)+p_(NO_2)` <br/> `therefore` 9.15=(4.98-x)+2x <br/> `therefore` 9.15=4.98 +x <br/> `therefore` x=(9.15-4.98)=4.17 bar <br/> So, partial pressure at equilibrium <br/> `p_(NO_2)`= 2x=2(4.17)=8.34 bar<br/> `p_(N_2O_4)`=(4.98-x)=(4.98-4.17)=0.81 bar <br/> The expression of `K_p`, <br/> `K_p=(p_(NO_2))^2/(p_(N_2O_4))="(8.34 bar)"^2/"0.81 bar"`=85.87 bar <br/> The calculation of `K_c` from `K_p`: <br/> `K_p=K_c(RT)^(Deltan)` So, `K_c=K_p/(RT)^(Deltan_(g))` <br/> where , `K_p`=85.87 bar, R=0.083 L bar `mol^(-1) K^(-1)`<br/> T=400 K , `Deltan_((g))`=(2-1)=+1 <br/> `therefore K_c=(85.87)/(400xx0.083)^1=2.5864 approx` 2.586 M</body></html>


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