The equilibrium constant of the reaction `SO_2(g)+1/2O_2(g) hArr SO_3(g) "is " 4xx10^(-3) "atm"^(-1//2)`.
The equilibrium constant of the reaction `2SO_3(g)hArr2SO_(2)(g)+O_(2)(g)` would be :

Options

A. 250 atm
B. `4 xx 10^3` atm
C. `0.25xx10^4` atm
D. `6.25 xx 10^4` atm

Answer

D

Explanation

`SO_2(g)+1/2 O_2(g) harr SO_3(g)" "K_p=4xx10^(-3)`
`SO_3 harr SO_2(g)+1/2 O_2(g)" "K_p^(1)=(1)/(K_p)`
`K_p^(1)=((1)/(4xx10^(-3)))`
`2SO_3 harr 2SO_2+O_2(g)`
`K_p^(n)=(K_p^(1))^2=[(1)/(4xx10^(-3))]^2=[(1000)/(4)]^2`
`=6250=625xx10^(2)=6.25xx10^(4)` atm