A mechanism for the reaction H_2 + Br_2 rightarrow 2HBr is as follows: Br_2 reve

Question

A mechanism for the reaction H_2 + Br_2 rightarrow 2HBr is as follows: Br_2 reversiblearrow 2Br. Br + H_2 rightarrow k_2 HBr + H. H +Br_2 rightarrow HBr +Br. The HBr + H rightarrow k_4 H_2 +Br. The rate law predicted by this mechanism is: d[HBr]/dt = 2k_2 squareroot k_1/k_-1 [Br_2]^1/2 [H_2]/1 +k_4[HBr]/k_3[Br_2]. (It is not necessary to derive this rate law!) Consider the reaction initiated with only Br_2 and H_2 present. By evaluating the rate law above, show that at time t = 0 this rate aw reduces to the expression:[d[HBr]/dt]_i = 0 =-2k_2 (k _1/k_2]^1/2 [H_2]_0[Br_2]^1/2_0.

Explanation / Answer

At t=0 the concentration of reactants is 100%, the reaction will not get intiated at t=0
therefore the concentration of products as well as intermediates will be zero.
Concentration of reactant [H2] at t=0 will be [H2]0
concentration of reactant [Br2] at t=0 will be [Br2]0
[HBr] = 0, substitute this in the R.H.S of the given rate law
[d [HBr] /dt ]t=0 = { 2k2 (k1/k-1)1/2 [H2]0 [Br2]01/2 } / {1+ ( k4[0] / k3[Br2] ) }

[d [HBr] /dt ]t=0 = 2k2 (k1/k-1)1/2 [H2]0 [Br2]01/2

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