At a certain temperature, the equilibrium constant `(K_(c ))` is `16` for the reaction:
`SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)`
If we take one mole of each of the four gases in one litre container then what will be the equilibrium concentration of `NO` and `NO_(2)`?

Options

A. `1.6 mol L^(-1)`
B. `0.8 mol L^(-1)`
C. `0.4 mol L^(-1)`
D. `0.6 mol L^(-1)`

Answer

C

Explanation

`{:(SO_(2)(g),+,NO_(2)(g),hArr,SO_(3)(g),+,NO(g),),(1,,1,,1,,1,"Initial conc"),(1-x,,1-x,,1+x,,1+x,"At Eq."):}`
`K=([SO_(3)][NO])/([SO_(2)][NO_(2)])=((1+x)(1+x))/((1-x)(1-x))`
`16=((1+x)^(2))/((1-x)^(2))rArr((1+x))/((1-x))=4` or `x=0.6`
`[NO_(2)]=1-x=1-0.6=0.4 "mol" L^(-1)`