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Rate law and rate constant

For the reaction, {H}_{2}+{I}_{2} 2 {HI}, the differential ra

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

Name: Video solution 1: For the reaction, H2+I2⟶2HI, the differential rate law is
Uploaded: 2022-09-09T18:34:57.000Z
Duration: 5 min
Description: Video solution 1: For the reaction, H2+I2⟶2HI, the differential rate law is

- \frac{d [ H _{2} ]}{d t} = - \frac{d [ I _{2} ]}{d t} = 2 \frac{d [ HI ]}{d t}

- 2 \frac{d [ H _{2} ]}{d t} = - 2 \frac{d [ I _{2} ]}{d t} = \frac{d [ HI ]}{d t}

- \frac{d [ H _{2} ]}{d t} = - \frac{d [ I _{2} ]}{d t} = \frac{d [ HI ]}{d t}

- \frac{d [ H _{2} ]}{2 d t} = - \frac{d [ I _{2} ]}{2 d t} = \frac{d [ HI ]}{d t}

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Updated on: Sep 17, 2022

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Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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JEE Mains 2002

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For the reaction,

H_{2} + I_{2} ⟶ 2 HI

, the differential rate law is

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Text solutionVerified

H_{2} + I_{2} ⟶ 2 HI

Rate of reaction = \frac{- d [ H _{2} ]}{d t} = \frac{- d [ I _{2} ]}{d t} = \frac{1}{2} \frac{d [ HI ]}{d t}

or \frac{- 2 d [ H _{2} ]}{d t} = \frac{- 2 d [ I _{2} ]}{d t} = \frac{d [ HI ]}{d t}

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Question Text	For the reaction, $H_{2} + I_{2} ⟶ 2 HI$ , the differential rate law is
Updated On	Sep 17, 2022
Topic	Chemical Kinetics
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	202
Avg. Video Duration	14 min

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

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