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At certain temperature and under a pressure of 4 atm , PCl_(5) is 10 % dissociated . Calculate the pressure at which PCl_(5) will be 20 % dissociated at temperature remaining constant . |
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Answer» Solution :Calculation of `K_(P)` `{:(PCl_(5) (g) , to , PCl_(3) (g)+ Cl_(2) (g)) , (1 , ," "0 "" 0) , ((1 - alpha) , , ""alpha "" alpha):}` Total no. of moles in the EQUILIBRIUM mixture = `1 - alpha + alpha + alpha = (1 + alpha)` mol . Let the total pressure of equilibrium mixture = p atm Partial pressure of `PCl_(5) , P_(PCl_(5)) = (1 - alpha)/(1 + alpha)xx p` atm Partial pressure of `PCl_(3) = (alpha)/(1 + alpha) xx p` atm Partial pressure of `Cl_(2) , PCl_(2) = (alpha)/(1 + alpha) xx p` atm `K_(p) = (P_(PCl_(2)) xx P_(Cl_(2)))/(P_(PCl_(3))) = (((alpha)/(1 + alpha) p " atm" ) xx ((alpha)/(1 + alpha) p atm))/((1-alpha)/(1 + alpha) p atm) = (alpha^(2) p)/(1 - alpha^(2))` P = 4 atm and `alpha = 10% = (10)/(100) = 0.1` `K_(p) = ((0.1) xx (0.1) xx (4 atm))/(1 - (0.1)^(2)) = (0.04)/(0.99) = 0.04` atm Calculation of P under NEW condition `alpha = 0.2 , K_(p) = 0.04` atm `K_(p) = (alpha^(2) p)/(1 - alpha^(2)) ` or p = `(K_(p) ( 1 - alpha^(2)))/(alpha^(2))` `= ((0.04 atm) [(1- (0.2)^(2))])/((0.2)^(2)) = (0.04 atmxx 0.96)/(0.04)` `= 0.96` atm |
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