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The differential rate law for the reaction $H_{2} + I_{2} ⇌ 2 HI is$

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

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

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

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

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detailed solution

Correct option is D

The chemical equilibrium reaction is

$H_{2}_{(g)} + I_{2}_{(g)} ⇌ {2HI}_{(g)}$

$The differential rate law = - \frac{{d[H}_{2}]}{dt} = - \frac{{d[I}_{2}]}{dt} = + \frac{1}{2} \frac{d[HI]}{dt}$

The modified form of equation is

$Differential rate law = - 2 \frac{{d[H}_{2}]}{dt} = - 2 \frac{{d[I}_{2}]}{dt} = + \frac{d[HI]}{dt}$

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The differential rate law for the reaction H2+I2⇌2HI is