Reaction between `N_(2)` and `O_(2-)` takes place as follows :
`2N_(2)(g) + O_(2)(g) hArr 2N_(2)O (g)`
If a mixture of `0.482` mol `N_(2)` and `0.933` mol of `O_(2)` is placed in a `10 L` reaction vessel and allowed to form `N_(2)O` at a temperature for which `K_(c) = 2.0 xx 10^(-37)`, determine the composition of equilibrium mixutre.

Explanation

Suppose x moles of ` N_2 ` react till equilibrium is attained.
` {:(,, 2N_2(g) , +O_2(g) , hArr, 2N_2O(g) ),(" Innitial no. of moles ",, 0.482 ,, 0.933, ),(" Moles at equilibrium ",,0.482-x ,, 0.9333 - (x)/(2) , ):}`
Hence, at equilibrium,
` " "[N_2 (g) = (0.482 -x)/(10) , [O_2(g) ]=(0.933-(x)/(2)) /( 10)`
and ` " " [N_2O (g)] =(x)/(10) " " (because " volume " = 10 L) `
Since , the value of `K_c` for then reaction `(K_c= 2.0 xx 10^(-37))` is very small , the value of x must be very small. Therefore ,
` [N_2(g) = (0.482-x)/(10 ) ~~(0.482)/(10)= 0.0482 " mol " L^(-1)`

` [O_2(g)] =(0.933-(x)/(2))/(10 ) ~~ (0.933)/(10) = 0.0933 " mol " L^(-1) `
and ` " " [N_2O (g) ]=(x)/(10) `
` because " "K_c =( [N_2O(g)]^(2))/([N_2(g) ]^(2) [O_2(g) ]) `
we have ,`" " 2.0 xx 10^(-37) (((x)/(10))^(2))/((0.0482)^(2) xx 0.0933)`
which gives. ` " " x= 6.6 xx 10 ^(-20)`
` because [N_2O(g) ]=(x)/(10) =(6.6 xx 10^(-20))/(10) = 6.6 xx 10^(-21) " mol " L^(-1)`