Equilibrium concentration of `SO_(2)Cl_(2)` formed from 3 moles of `SO_(2)` and 4 moles of `Cl_(2)` present in a litre flask is 1.5 mole/L. Calculate `K_(c )`.

Explanation

`SO_(2)+Cl_(2) hArr SO_(2)Cl_(2)`
`{:("Initial NO. of moles",3,4,0),("At equilibrium",,3-x,4-x x):}`
Concentration of `SO_(2)Cl_(2)` at equilibrium is 1.5 moles.
x=1.5 moles/L
Concentration of `SO_(2)` at equilibrium =3-1.5=1.5 moles/L
`therefore K_(c )=([SO_(2)Cl_(2)])/([SO_(2)][Cl_(2)])=(1.5)/(1.5xx2.5)=0.4 " mole" // L`