For the reaction H_2 + I_2 rightarrow 2HI,sta | vedclass

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(D) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by: $Rate = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d

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$\frac{-d[H_2]}{dt} = \frac{-d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$

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(D) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by: $Rate = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$ For the given reaction $H_2 + I_2 \rightarrow 2HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively. Therefore,the rate of reaction is: $Rate = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$ Multiplying by $2$,we get: $2 \left( -\frac{d[H_2]}{dt} \right) = 2 \left( -\frac{d[I_2]}{dt} \right) = \frac{d[HI]}{dt}$ Alternatively,the expression can be written as: $-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$ Comparing this with the given options,option $D$ represents the correct relationship.

For the reaction $H_2 + I_2 \rightarrow 2HI$,state the correct differential rate law.

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