What should be the value of `K_(c)` for the reaction `2SO_(2_((g)))+O_(2_((g))hArr2SO_(3_((g)))`. If the amount are `SO_(3)=48g`. `SO_(2)=12.8 "and" O_(2)=9.6` at equilibrium and the volume of the container is one litre?

Options

A. `64`
B. `30`
C. `42`
D. `8.5`

Answer

B

Explanation

`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`
`K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])` Concentration in gram mole//litre. Therefore
`[SO_(3)](48)/(80xx1)` (Where `80` is molecular weight of `SO_(3)`
`[SO_(2)]=(126)/(64xx1)` (Where `64` is molecular weight of `SO_(2)`
`[O_(2)]=(9.6)/(32xx1)` (Where `32` is molecular weight of `O_(2))`
Thus, `K_(c)=((48)/(80))^(2)/(((12.8)/(64))^(2)((9.6)/(32)))=0.30`