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For a reversible first order reaction, ...

For a reversible first order reaction,

`K_(f) =10^(-2) sec^(-1) and (B_(eq))/(A_(eq))=4," If "A_(0) =0.01 ML^(-1) and B_(0) =0`, what will be concentration of B after 30 sec ?

Text Solution

Verified by Experts

`A_(0) rarr B`
`0.01" "0`
`0.01-X_(eq)" "X_(eq)`
`([B]_(eq))/([A]_(eq))=(10^(-2))/(K_(b))=4=([X]_(eq))/(0.01-[X]_(eq))`
`therefore K_(b) =0.25xx10^(-2) and X_(eq) =(0.04)/(5)=0.008`
`t=(2.303)/((K_(f)+K_(b)))log ""([X]_(eq))/([X]_(eq)-X)`
`30=(2.303)/(1.25xx10^(-2))log""(0.008)/((0.008-x)`
`therefore (0.008)/(0.008-x)=1.455`
`therefore " "x=2.50 xx10^(-3)`
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