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You have studied the gas-phase oxidation of HBr by O2:

4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g)

You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism:

HBr(g) + O2(g) HOOBr(g)
HOOBr(g) + HBr(g) 2 HOBr(g)
HOBr(g) + HBr(g) H2O(g) + Br2(g)

(a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

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Jarrod Robel
Jarrod RobelLv2
26 May 2020

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Related questions

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process. (b) HBr does not react with O2 at a measurable rate at room temperature under ordinary conditions. What can you infer from this about the magnitude of the activation energy for the rate-determining step? (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

Consider the following reaction:4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g)(a) The rate law for this reaction is first order in HBr(g) and first order in O2(g). What is the rate law for this reaction?Rate = k [HBr(g)] [O2(g)]Rate = k [HBr(g)]2 [O2(g)] Rate = k [HBr(g)] [O2(g)]2Rate = k [HBr(g)]2 [O2(g)]2Rate = k [HBr(g)] [O2(g)]3Rate = k [HBr(g)]4 [O2(g)](b) If the rate constant for this reaction at a certain temperature is 7750, what is the reaction rate when [HBr(g)] = 0.00344 M and [O2(g)] = 0.00687 M?Rate = M/s.(c) What is the reaction rate when the concentration of HBr(g) is doubled, to 0.00688 M while the concentration of O2(g) is 0.00687 M?Rate = M/s

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