Reaction between nitrogen and oxygen takes place as follows : 2 N_(2) (g) + O_(2) (g) hArr 2 N_(2) O (g) If a mixture of 0*482 mol of N_(2) and 0*933 mol of O_(2) is placed in a reaction vessel of volume 10 L and allowed to form N_(2)Oat a temperature for which K_(c) = 2*0 xx 10^(-37) . Determine the composition of the equilibrium mixture.

Reaction between nitrogen and oxygen takes place as follows : 2 N_(2)

1.	Reaction between nitrogen and oxygen takes place as follows : 2 N_(2) (g) + O_(2) (g) hArr 2 N_(2) O (g) If a mixture of 0482 mol of N_(2) and 0933 mol of O_(2) is placed in a reaction vessel of volume 10 L and allowed to form N_(2)Oat a temperature for which K_(c) = 2*0 xx 10^(-37) . Determine the composition of the equilibrium mixture.
Answer» Solution :` {:(,2 N_(2) (g) ,+,O_(2)(g),hArr,2 N_(2)O(g)),("Intial",0482 "mol",,0933"mol",,),("At eqm.",0482 "mol",,0.933-x//2,,x),("Molar conc." ,(0482-x)/10,,(0933-x//2)/10,,x/10):}`, As K ` = 20xx 10 ^(-37)` is very small , this means that the amount of `N_(2) and O_(2) ` reacted (x)is very very small . Hence , at equilibrium , we have `[N_(2)] = 0* 0482 "mol"L^(-1), [O_(2)] = 0* 0933 "mol" L^(-1) , [N_(2)O]= 01x ` ` :. K_(c) = (01 x)^(2) /((00482)^(2) (00933)) = 2* 0 xx 10^(-37) ` (GIVEN) On solving, this GIVES `x = 66XX 10^(20)` `:. [N_(2)O ] = 01x 6*6xx 10^(-21)"mol"L^(-1)`