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

The correct option is D 2d[H_2]dt= 2d[I_2]dt=d[HI]dt For the equation; H_2 + I_2→ 2HI Rate equation can be written as: 2d[H_2]dt = 2d[I_2]dt ...

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Standard XII

Chemistry

Rate of Reaction

The different...

Question

The differential rate law for the reaction $H_{2} + I_{2} \to 2 H I$ is:

- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = - \frac{d [H I]}{d t}

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\frac{d [H_{2}]}{d t} = \frac{d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}

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\frac{1}{2} \frac{d [H_{2}]}{d t} = \frac{1}{2} \frac{d [I_{2}]}{d t} = - \frac{d [H I]}{d t}

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- 2 \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}

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Solution

The correct option is D $- 2 \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$
For the equation; $H_{2} + I_{2} \to 2 H I$
Rate equation can be written as: $- 2 \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$

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I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}

\frac{- d [I_{2}]}{d t} = - \frac{1}{2} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

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\frac{- d [I_{2}]}{d t} = - 2 \frac{d [S_{2} O_{3}^{2 -}]}{d t}

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= \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

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

The correct option is D −2d[H2]dt=−2d[I2]dt=d[HI]dtFor the equation; H2+I2→2HIRate equation can be written as: −2d[H2]dt=−2d[I2]dt=d[HI]dt

The differential rate law for the reaction $H_{2} + I_{2} \to 2 H I$ is:

The correct option is D $- 2 \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$
For the equation; $H_{2} + I_{2} \to 2 H I$
Rate equation can be written as: $- 2 \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$