1. Tardigrade
  2. Question
  3. Chemistry
  4. In the formation of HBr from H 2 Br 2, following mechanism is observed. 1. Br 2 leftharpoons 2 Br bullet Equilibrium step 2. H 2+ Br bullet arrow HBr + H bullet Slow step 3. H bullet+ Br 2 arrow HBr + Br bullet Fast step Calculate the rate of reaction, if concentration of hydrogen is twice that of bromine and the rate constant is equal to 1 M -1 / 2 Sec -1 . Concentration of bromine is 1 M

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. In the formation of $HBr$ from $H _{2}\, \&\, Br _{2},$ following mechanism is observed.
1. $Br _{2} \rightleftharpoons 2 Br ^{\bullet} $ Equilibrium step
2. $H _{2}+ Br ^{\bullet} \rightarrow HBr + H ^{\bullet}$ Slow step
3. $H ^{\bullet}+ Br ^{2} \rightarrow HBr + Br ^{\bullet} $ Fast step
Calculate the rate of reaction, if concentration of hydrogen is twice that of bromine and the rate constant is equal to 1 $M ^{-1 / 2} Sec ^{-1} .$ Concentration of bromine is $1\, M$

Chemical Kinetics

Solution:

Rate $= k \left[ H _{2}\right]^{1}\left[ Br _{2}\right]^{1 / 2}$

Rate $=1[2]^{1}[1]^{1 / 2}=2\, M\, \operatorname{Sec}^{-1}$