  Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes into H2O2 and O2. The following data was obtained at 298 K. Time (μμs)[HO2] (μμM)0.008.500.605.101.003.601.402.601.801.802.401.10  Determine the rate law for the reaction. Do not add multiplication symbols to your answer.Rate= Determine the value of the rate constant at 298 K.     μs-1

Question
Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes into H2O2 and O2. The following data was obtained at 298 K.

 Time (μμs) [HO2] (μμM) 0.00 8.50 0.60 5.10 1.00 3.60 1.40 2.60 1.80 1.80 2.40 1.10

Determine the rate law for the reaction. Do not add multiplication symbols to your answer.

Rate=

Determine the value of the rate constant at 298 K.
μs-1
Step 1

The rate law of a reaction is expressed as: help_outlineImage Transcriptionclose>C+ D A B Rate k [A]a [B]b Where, k rate constant [A], [B] concentration of reactants a b order of the reaction fullscreen
Step 2

Firstly, we need to find the order of the reaction. This can be done by plotting graphs of time versus concentration. If the time vs [conc] gives a straight line it will be zero order reaction.  If the time vs ln[conc] gives a straight line it will be first order reaction. If the time vs 1/[conc] gives a straight line it will be second order reaction. help_outlineImage Transcriptionclose[HO2 In [HO2 1/[HO2 Time [НO:] 2.14 0.12 8.50 0.00 0.20 1.63 0.60 5.10 0.28 1.28 1.00 3.60 0.96 0.38 1.40 2.60 0.59 0.56 1.80 1.80 0.90 0.095 2.40 1.10 Time vs [conc Time vs In [conc] Time vs 1/[conc] 10 2.5 1 0.8 0.6 6 1.5 4 0.4 1 0.2 2 0.5 0 1 2 3 1 2 3 1 2 3 Time Time Time [conc] In [conc] en 1/[conc] 생 명:명 fullscreen
Step 3

Since, time vs ln [ HO2] gives a straight line. Therefore, the reaction is first order.

The rat...

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Chemical Kinetics 