For the reaction
`Cl_(2)(g)+F_(2)(g)hArrClF(g),K_(c)=19.9` What will happen in a mixture originally containing `[Cl_(2)]=0.04 "mol"L^(-)`,
`[F_(2)]=0.2"mol"L^(-1)` and `[ClF]=7.3 "mol" L^(-)`?

Explanation

Given reaction
`Cl_(2_((g)))+F_(2_((g))) hArr 2Cl_((g)),K_(c)=19.9`
Given `[Cl_(2)]=0.4` mole/lit `[F_(2)]=0.2` mole/lit
[ClF]=7.3 mole/lit
`:.Q_(c)=([ClF]^(2))/([Cl_(2)][F_(2)])=([7.3]^(2))/(0.4xx0.2)=(53.29)/(0.8)=66.6125`
`Q_(c) gt K_(c)`
`:.` The reaction proceeds in the backward direction: (Reactants side, Reverse Reaction)