For the reaction: H_{2} + I_{2} longrightarro | vedclass

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

Vedclass · Accepted Answer

$-\frac{d[H_{2}]}{dt} = -\frac{d[I_{2}]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$

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(B) For a general reaction $aA + bB \longrightarrow cC + dD$,the rate of reaction is given by: $	ext{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 reaction $H_{2} + I_{2} \longrightarrow 2 HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively.Therefore,the differential rate law is: $	ext{Rate} = -\frac{d[H_{2}]}{dt} = -\frac{d[I_{2}]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.

For the reaction: $H_{2} + I_{2} \longrightarrow 2 HI$,the differential rate law is

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