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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. |
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Answer» Solution :Molecular mass of `N_2O_4`=2(N)+4(O) =2(14)+4(16) = 92 g `"mol"^(-1)` Mole of `N_2O_4 = n = "WEIGHT"/"Molecular mass"= "13.8 g"/"92 g mol"^(-1)` = 0.15 Temperature T=400 K Gas constant = R= 0.083 bar L `mol^(-1) K^(-1)` pV=nRT where , p=Partial pressure `N_2O_4` `therefore p=(nRT)/V` `therefore p=((0.1 "mol")(0.083 "bar L mol"^(-1) K^(-1))(400 K))/(1 L)` = 4.98 bar partial pressure of `N_2O_4` `{:("Equilibrium reaction :",N_2O_(4(g)) hArr, 2NO_(2(g))),("INITIAL pressure :", "4.98 bar","0 bar"),("Pressure change :","-x bar" , "+2X bar"),("Partial pressure at equilibrium :","(4.98-x)bar","2x bar"):}` `therefore` Total pressure at equilibrium = Addition of partial pressure `p_"total"=p_(N_2O_4)+p_(NO_2)` `therefore` 9.15=(4.98-x)+2x `therefore` 9.15=4.98 +x `therefore` x=(9.15-4.98)=4.17 bar So, partial pressure at equilibrium `p_(NO_2)`= 2x=2(4.17)=8.34 bar `p_(N_2O_4)`=(4.98-x)=(4.98-4.17)=0.81 bar The expression of `K_p`, `K_p=(p_(NO_2))^2/(p_(N_2O_4))="(8.34 bar)"^2/"0.81 bar"`=85.87 bar The calculation of `K_c` from `K_p`: `K_p=K_c(RT)^(Deltan)` So, `K_c=K_p/(RT)^(Deltan_(g))` where , `K_p`=85.87 bar, R=0.083 L bar `mol^(-1) K^(-1)` T=400 K , `Deltan_((g))`=(2-1)=+1 `therefore K_c=(85.87)/(400xx0.083)^1=2.5864 approx` 2.586 M |
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