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The following reaction was carried out in water `Cl_(2) + 2I^(-) to 2Cl^(-) + I_(2)` The initial concentration of `I^(-)` was `0.50 mol L^(-1)` and concentration after 10 minutes was 0.46 mol `L^(-1)`. Calculate the rate of disappearance of `I^(-)` and rate of appearance of `I_(2)`. |
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Answer» The chemicaal reaction is: `Cl_(2) + 2I^(-) to 2Cl^(-) +I_(2)` `Delta[I^(-)]=(0.046 -0.50) = -0.04 mol L^(-1)`, `Deltat=10min`. Rate of disappearance of `I^(-)` `-(Delta[I^(-)])/(Deltat) = -(0.04 mol L^(-1))/(10 "min")=0.004 mol L^(-1)min^(-1)` Rate of appearance of `I_(2)`. `(Delta[I_(2)])/(Deltat) = 1/2(Delta[I^(-)])/(Deltat) = 0.04/2 mol L^(-1)min^(-1) = 0.002 mol L^(-1)min^(-1)`. |
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