For the reaction , `Cl_(2)+2I^(-) rarr I_(2)+2Cl^(-)`, the initial concentration of `I^(-)` was `0.20 "mol lit"^(-1)` and the concentration after 20 minutes was `0.18 "mol lit"^(-1)`. Then the rate of formation of `I_(2)` in `"mol lit"^-1min^-1` would be

Options

A. `1xx10^(-4)`
B. `5xx10^(-4)`
C. `1xx10^(-3)`
D. `5xx10^(-3)`

Answer

C

Explanation

For `Cl_(2) + 2I^(-) rarr I_(2) +2CI^(-)`
`r=k[I^(-)]^(2)=2.5xx10^(-2) xx (0.2)^(2) =2.5xx0.04xx10^(-2)=1xx10^(-3)` mol/L sec
But `(d[I_(2)])/(dt) =r=1xx10^(-3)`