The experimentally observed rate law for the reaction of H 2 ( g ) with Br 2 ( g

Question

The experimentally observed rate law for the reaction of H2 (g) with Br2 (g) to form HBr(g) is:

where k and m are constants. The following mechanism for this process has been proposed:

Classify the steps as being initiation, propagation, or termination, and show that the rate law derived from this mechanism (using any reasonable assumptions) is consistent with the observed rate law. Express the observed constants k and m in terms of the rate constants of the steps in the mechanism.

The experimentally observed rate law for the reaction of H2 (g) with Br2 (g) to form HBr(g) is: where k and m are constants. The following mechanism for this process has been proposed: Classify the steps as being initiation, propagation, or termination, and show that the rate law derived from this mechanism (using any reasonable assumptions) is consistent with the observed rate law. Express the observed constants k and m in terms of the rate constants of the steps in the mechanism.

Explanation / Answer

initiation ;

Br2 ---> 2Br.

Propagation ;

Br. + H2 -> Hbr + H.

H. + Br2 --> HBr + Br,

Termination ;

H. + HBr --> Br. + H2

2Br. ---> Br2

from the reactions

the final reaction is H2 + Br2 --> HBr

  

so the rate equation should only contain H2, Br2 ,HBr

from the above equations

d[ Br]/dt =2 k1 [ Br2] - k2 [ H2] [ Br] + k3 [ H] [ Br2] - ( k_2) [ HBr] [ H] - 2 x( k_1)[Br]2

So

2 k1 [ Br2] - k2 [ H2] [ Br] + k3 [ H] [ Br2] - ( k_2) [ HBr] [ H] - 2 x( k_1)[Br]2 =0

d[H] /dt = k2 [ H2] [ Br] - k3 [ H] [ Br2] - ( k_2) [ HBr] [ H] =0

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