In the following gaseous pjhase equlibrium at constant temperature the concentration of `[SO_(2)]=3.0xx10^(-3)M.[O_(2)]=3.5xx10^(-3)M.[SO_(3)]=5.0xx10^(-2)M.` Calculate equlibarium constant for both the directions.
`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`

Explanation

For forward direction, `2SO_(2((g)+O_(2)(g)hArr2SO_(3)(g)`
`K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=((5xx10^(-2))^(2))/((3xx10^(-3))^(2)(3.5xx10^(-3)))=7xx^(-4)M^(-1)`
For backward direction.
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
`K_(c)=(1)/(K_(c))=12.6xx10^(-6)M`