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InterviewSolution
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The `K_(p) " for the reaction," N_(2) O_(4) hArr 2NO_(2) ` is 640 mm at 775 K. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of 160 mm. At what pressure the dissociation will be 50% ? |
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Answer» Suppose intially `N_(2)O_(4)` taken = 1 mole and its degree of dissociation = `alpha` ` {:(,N_(2)O_(4),hArr,2NO_(2),),("Intial ",1 "mole",,,),("At eqm.",1-alpha,2alpha,"Total " =1-alpha + 2 alpha = 1+alpha,):}` If P is the total pressure at equilibrium , then ` p_(N_(2)O_(4)) = (1-alpha)/(1+alpha) xxP and p_(NO_(2)) = (2 alpha)/(1+alpha) xxP ` Now, ` K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= ((2 alpha)/(1+alpha).P)^(2)/((1-alpha)/(1+alpha) *P)= (4 a^(2))/((1+alpha)(1-alpha)) = (4a^(2))/(1-alpha^(2)) xx P` Putting ` K_(p) = 640 " mm (Given ) and equilibrium pressure , P+ 160 mm, we get "` ` 640 = (4a^(2))/(1-alpha)^(2) xx 160 or (alpha^(2))/(1-alpha)^(2) = or alpha^(2) = 1 - alpha^(2) or 2 alpha^(2) = 1 or alpha^(2) = 0*5 or alpha = 0* 707 = 70* 7 % ` For dissociation to be ` 50 % , alpha = 0*50 , K_(p) = 640` mm ( constant ) ` :. 640 = (4 ( 0*5)^(2))/(1-(0*5)^(2)) xx P or 640 = 1/(1-1/4) P= 4/3 P or P =480 "mm"` . |
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