Exp No. 1 2 3 Br₂ + H₂ [Br₂] (molecules/cm³) The reaction is.... 1.00E18 4.00E18 4.00E18 zero order in both [Br₂] and [H₂] 2HBr 1/2 order in [Br₂], 2nd order in [H₂] [H₂] (molecules/cm³) 1.00E19 1.00E19 2.00E19 1st order in [Br₂], 2nd order in [H₂], 3rd order overall Initial Rate (molecules/cm³/s) 4.71E15 9.42E15 1.88E16 1st order in [Br₂], 1st order in [H₂], 2nd order overall ½ order in [Br₂], 1st order in [H₂]

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Exp No.
1
2
3
Br₂ + H₂ → 2HBr
[Br₂]
[H₂]
(molecules/cm³) (molecules/cm³)
The reaction is ....
1.00E18
4.00E18
4.00E18
zero order in both [Br₂] and [H₂]
1.00E19
1/2 order in [Br₂], 2nd order in [H₂]
1.00E19
2.00E19
1st order in [Br₂], 2nd order in [H₂], 3rd order overall
Initial Rate
(molecules/cm³/s)
4.71E15
9.42E15
1.88E16
1st order in [Br₂], 1st order in [H₂], 2nd order overall
½ order in [Br₂], 1st order in [H₂]
Transcribed Image Text:Exp No. 1 2 3 Br₂ + H₂ → 2HBr [Br₂] [H₂] (molecules/cm³) (molecules/cm³) The reaction is .... 1.00E18 4.00E18 4.00E18 zero order in both [Br₂] and [H₂] 1.00E19 1/2 order in [Br₂], 2nd order in [H₂] 1.00E19 2.00E19 1st order in [Br₂], 2nd order in [H₂], 3rd order overall Initial Rate (molecules/cm³/s) 4.71E15 9.42E15 1.88E16 1st order in [Br₂], 1st order in [H₂], 2nd order overall ½ order in [Br₂], 1st order in [H₂]
Expert Solution
Step 1

Reaction given:-

Br2 + H2 => 2HBr

•We have to find out oreder of reaction regarding Br2 and H2.

•To find out degree regarding we use above given table.

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