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In the formation of HBr form H(2) and Br...

In the formation of `HBr` form `H_(2)` and `Br_(2)`, following mechanism is observed:
I. `Br_(2) hArr 2Br^(*)` (Equilibrium step)
(II) `H_(2) + Br^(*)rarr HBr+H^(*)` (Slow step)
(III) `H^(*) + Br_(2) rarr HBr + H^(*)` (Fast step)
The rate law for the above reaction is

A

`2xx10^6` dps

B

`10xx10^9` dps

C

`20.2xx10^10` dps

D

`4xx10^2` dps

Text Solution

Verified by Experts

The correct Answer is:
A
1-Rutherford=`10^6` dps
Rate =`k[H_2]^1[Br_2]^(1//2)`
Rate = `10^6[2]^1[1]^(1//2)=2xx10^6` dps
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