Answer to Question #106477 in General Chemistry for Beverlie

Question

1. 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 (μs) [HO2] (μM)
0.00 8.50
0.60 5.10
1.00 3.60
1.40 2.60
1.80 1.80
2.40 1.10
a) Determine the rate law for the reaction. Do not add multiplication symbols to your answer.
b) Determine the value of the rate constant at 298 K. answer in μs-1.

Answer 1

HO2 + HO2 → H2O2 + O2 - decomposition

a) The rate law: rate = k [HO₂]^x

If we take ln [HO2] of all given concentrations we receive:

2,14 1,63 1,28 0,96 0,59 0,10

Finding the differences between the values gives:

0,51 0,35 0,33 0,37 0,49

which are close values, that proves that the order is 1

So: The rate law: rate = k [HO2]¹ - 1st order reaction.

b) rate 1 = (8.5-5.1)/(0.6-0) = 5.67

rate1 = k [HO2]

5.67 = k 8.5

k = 0.67 (μs^-1)