Equilibrium constant for the given reaction is `K=10^(20)` at temperature 300 K
`A(s)+2B(aq.)hArr2C(s)+D(aq.) K=10^20`
The equilibrium conc. of B starting with mixture of 1 mole of A and `1//3` mole/litre of B at 300 K is

Options

A. `~4xx10^(-11)`
B. `~2xx10^(-10)`
C. `~2xx10^(-11)`
D. `~10^(-11)`

Answer

A

Explanation

`{:(,A(s)+,2B(aq)" "hArr,2C(s)+,D(aq)),("Initial",1,1/3,0,0),(At_(eq),1-x,underset(~~a)(1/3-x),2x,x):}`
`x~~1//3`
`10^(20)=(1/3)/[B]^2 " " implies 10^20=(1/3)/a^2" " implies a^2=1/(3xx10^20)=10^(-20)/3`
`a=10^(-10)/sqrt3~~4xx10^(-11)M`