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. col1 Time(us) 0 0.6 1.0 1.4 1.8 2.4col2 [HO2](uM) 8.5 5.1 3.6 2.6 1.1 1.1(a) Determine the rate law for the reaction.

Rate =

(b) Determine the value of the rate constant at 298 K.

us-1

Rate = k [HO2]

Rate constant = 0.8456us-1

Explanation:

Time(us) 0 0.6 1.0 1.4 1.8 2.4

[HO2](uM) 8.5 5.1 3.6 2.6 1.1 1.1

The rate law is given as;

Rate = k [HO2]^x

Where x signify the order of reaction.

For an order of reaction, the rate constant is constant for all concentrations. We are going to use this to obtain the order of reaction.

Zero Order:

[A] = [A]o -kt

5.1 = 8.5 - k(0.6)

-k (0.6) = 5.1 - 8.5

k = 5.67

3.6 = 5.1 - k(0.4)

-k (0.4) = 3.6 - 5.1

k = 3.75

The fact that the rate constant was not constant means the reaction is not a zero order reaction.

First Order:

ln[A] = ln[A]o -kt

(5.1) = ln(8.5) - k(0.6)

-k (0.6) = ln(5.1) - ln(8.5)

k = 0.8524

ln(3.6) = ln(5.1) - k(0.4)

-k (0.4) = ln(3.6) - ln(5.1)

k = 0.8708

ln(2.6) = ln(3.6) - k (0.4)

-k (0.4) = ln(2.6) - ln(3.6)

k = 0.8136

From the three calculations we see that the value of the rate constant is fairly constant in the range of 0.8 This means our reaction is a first order reaction.

The rate law is given as;

Rate = k [HO2]

We can represent the rate constant as the average of the three rate constants calculated above;

Rate constant = (0.8136 + 0.8708 + 0.8524 / 3)

Rate constant = 0.8456us-1

the dissolution of polymer can be thermodynamically described in 2 steps.