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The difference rate law for the reaction `H_(2) +I_(2) to 2HI` isA. `- (d[H_(2)])/(dt) = -(d[I_(2)])/(dt) = + (1)/(2) (d[H])/(dt)`B. ` (d[H_(2)])/(dt) = (d[HI])/(dt) = (1)/(2) (d [HI])/(dt)`C. `(1)/(2) (d[H_(2)])/(dt) = (1)/(2) (d[I_(2)])/(dt) = - (d[HI])/(dt)`D. `-2 (d[H_(2)])/(dt) = - 2 (d[I_(2)])/(dt) = + (d [HI])/(dt)` |
Answer» Correct Answer - ad Concentration of reactants decreases while concentration of product increases. |
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