for the reaction of the formation of HBr the experimental rate law established f

Question

for the reaction of the formation of HBr the experimental rate law established for this reaction is r=K[H2][Br2] the following mechanism was proposed: Step 1 Br2<->2Br fast step equilibrium Second Step 2Br+H2->2HBr slow step Demonstrate this is an aaceptable mechanism for the reaction of the formation of HBr the experimental rate law established for this reaction is r=K[H2][Br2] the following mechanism was proposed: Step 1 Br2<->2Br fast step equilibrium Second Step 2Br+H2->2HBr slow step Demonstrate this is an aaceptable mechanism for the reaction of the formation of HBr the experimental rate law established for this reaction is r=K[H2][Br2] the following mechanism was proposed: Step 1 Br2<->2Br fast step equilibrium Second Step 2Br+H2->2HBr slow step Demonstrate this is an aaceptable mechanism

Explanation / Answer

for the reaction of the formation of HBr the experimental rate law established for this reaction is r=K[H2][Br2] the following mechanism was proposed: Step 1 Br2<->2Br fast step equilibrium Second Step 2Br+H2->2HBr slow step Demonstrate this is an aaceptable mechanism

H2 + Br2 2HBr

r = k [H ][Br ]^1/2 / 1+ k '[HBr] / [Br 2]

At t=0

[HBr]=0

R= k[Br2]^ 1/2[H2]

but at t=

r=k[Br2] ^3/2[H2]/[HBr]

aacep

1

Br2 k 1 2Br

initiate reaction

2

Br• + H2 k 2 HBr + H

Propagation of reaction

3

H• + Br2 k 3 HBr + Br

4

HBr + H• k 2 Br• + H2

Inhibition------reverse of 2nd step

5

Br• + Br• k 1 Br2

terminate------reverse of 2n first step

r = k [H ][Br ]^1/2 / 1+ k '[HBr] / [Br 2]

1

Br2 k 1 2Br

initiate reaction

2

Br• + H2 k 2 HBr + H

Propagation of reaction

3

H• + Br2 k 3 HBr + Br

4

HBr + H• k 2 Br• + H2

Inhibition------reverse of 2nd step

5

Br• + Br• k 1 Br2

terminate------reverse of 2n first step

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