3.1 mol `Fe^(3+)` and 3.2 mol `"SCN"^(Theta)` present in 1 L solution. At equilibrium 3 mol `FeSCN^(2+)` are formed The equilibrium constant `K_(c)` for the reaction `Fe^(3+)+SCN^(Theta)hArrFeSCN^(2+)` will

Options

A. `6.66 xx 10^(-3)`
B. 0.3
C. 3.3
D. 150

Explanation

`Fe^(3+) +SCN^(-) Leftrightarrow FeSCN^(2+)`
`{:(,t_(0),3.1,3.2,0),(,t_(eq),(0.1)/(1),(0.2)/(1),3/1):}`
`K_(c)=([FeSCN^(2+)])/([Fe^(3+)] [SCN^(-)]) =(3)/(0.1 xx 0.2)=150`