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For the reaction Cl(2)+2I^(-)toI(2)+2CI^...

For the reaction `Cl_(2)+2I^(-)toI_(2)+2CI^(-)`, the initial concentration of `I^(-)` was 0.20 mol `"lit"^(-1)` and the concentration after 20 min was 0.18 mol `"lit"^(-1)`. If the rate of formation of `I_(2)` in mol `"lit"^(-1) "min"^(-1)` is `x xx 10^(-4)` then the value of x is

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The correct Answer is:
5
`(r_(I^(-)))/2=(r_(I_(2)))/1,r_(I_(2))=0.001/2=5xx10^(-4),r_(I^(-))=(0.20-0.18)/20=0.02/20=0.001`
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