1.

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 and1//3 mole/litre of B at 300 K is

Answer»

`~4XX10^(-11)`
`~2xx10^(-10)`
`~2xx10^(-11)`
`~10^(-11)`

Solution :`{:(,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`


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