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For the reaction Cl(2)(g)+F(2)(g)hArrC...

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^(-)`?

Text Solution

Verified by Experts

The reaction quotient
`Q=([ClF]^(2))/([Cl_(2)][F_(2)])=((7.3)^(2))/(0.04xx0.2)=6.66xx10^(3)`
Since the Q value is very high than the `K_(c)` value 19.9 the reaction takes place in backward direction.
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