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InterviewSolution
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A sample of air consisting of `N_(2)` and `O_(2)` was heated to `2500 K` until the equilibrium `N_(2)(g)+O_(2)(g)hArr2NO(g)` was established with an equlibrium constant, `K_(c )=2.1xx10^(-3)`. At equilibrium, the `"mole"%` of `NO` was `1.8`. Eatimate the initial composition of air in mole fraction of `N_(2)` and `O_(2)`. |
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Answer» Let the total number of mole of `N_(2)` and `O_(2)` be `100` containing a mole of `N_(2)` initially. `N_(2)+O_(2)hArr2NO` `{:(a,(100-a),,0,"Initial mol"),((a-x),(100-a-x),,2x,"mole at equilibrium"):}` mole `%` of NO at equilibrium `=(2x)/((a-x)+(100-a-x)+2x)xx100=1.8` `:. x=0.9` Thus, at equilibrium, mole of `N_(2)=(a-0.9)` mole of `O_(2)=(100-a-0.9)=(99.1-a)` "mole" of `NO=2x=2xx0.9=1.8` `K_(c )=((2x//V)^(2))/(((a-0.9)/V)((99.1-a)/V))=((2x)^(2))/((a-0.9)(99.1-a))` `2.1xx10^(-3)=((1.8)^(2))/((a-0.9)(99.1-a))` `a=74.46%` and `20.54%` As we know that `%` of `N_(2)` in the air is more than that of `O_(2)`. |
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