At `700 K`, the equilibrium constant `K_(p)` for the reaction
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
is `1.80xx10^(-3) kPa`. What is the numerical value of `K_(c )` in moles per litre for this reaction at the same temperature?

Options

A. `3.09xx10^(-7) mol//L`
B. `5.07xx10^(-8)mol//L`
C. `8.18xx10^(-9)mol//L`
D. `9.24xx10^(-10)mol//L`

Answer

A

Explanation

`{:(2SO_(3),hArr,2SO_(2),+,O_(2)),(2,,3,,):}`
`Delta n= 3-2 = +1` , `K_(p) = 1.80xx10^(-3)`
`[RT]^(Delta n) = (8.314xx700)^(1)`
`K_(c) = (K_(p))/((RT)^(Delta n)) = (1.8xx10^(-3))/((8.314xx700)^(1))`
`=3.09xx10^(-7)` mol -L.