Question

Estimate the lifetime of odd oxygen (i.e. ozone) at 20 km and 45 km. It may be assumed that the photolysis rates of O3 (reaction 3 from Chapman Mechanism) at these two altitudes are: k3 = 7.0 x 10^-4 s^-1 at 20 km k3 = 6.0 x 10^-3 s^-1 at 45 km k2 = 6 x 10^-34 s^-1 (T/300)^-2.4 cm^6 molecules^-2 s^-1 k4 = 8.0 x 10^-12 exp(-2060/T) cm^3 molecules^-1 s^-1 and the O3 and M concentrations at the two altitudes are: [O3] = 2 x 10^12 molecules cm^-3 and [M] = 1.4 x 10^18 cm^-3 at 20 km (T = 217 K) [O3] = 0.2 x 10^12 molecules cm^-3 and [M] = 3.6 x 10^16 cm^-3 at 45 km (T = 265 K) Hint: you will need to calculate k2 and k4 rate constants from the Chapman Mechanism. k3 was given in this problem.

          Estimate the lifetime of odd oxygen (i.e. ozone) at 20 km and 45 km. It may be assumed that the photolysis rates of O3 (reaction 3 from Chapman Mechanism) at these two altitudes are:
k3 = 7.0 x 10^-4 s^-1 at 20 km
k3 = 6.0 x 10^-3 s^-1 at 45 km
k2 = 6 x 10^-34 s^-1 (T/300)^-2.4 cm^6 molecules^-2 s^-1
k4 = 8.0 x 10^-12 exp(-2060/T) cm^3 molecules^-1 s^-1
and the O3 and M concentrations at the two altitudes are:
[O3] = 2 x 10^12 molecules cm^-3 and [M] = 1.4 x 10^18 cm^-3 at 20 km (T = 217 K)
[O3] = 0.2 x 10^12 molecules cm^-3 and [M] = 3.6 x 10^16 cm^-3 at 45 km (T = 265 K)
Hint: you will need to calculate k2 and k4 rate constants from the Chapman Mechanism. k3 was given in this problem.

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Added by Victoria M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Supreeta N

11/06/2023

Step 1

Calculate k2 and k4 at both altitudes using the given formulas and temperatures: At 20 km (T = 217 K): k2 = 6 x 10^-34 s^-1 (217/300)^-2.4 cm^6 molecules^-2 s^-1 k4 = 8.0 x 10^-12 exp(-2060/217) cm^3 molecules^-1 s^-1 At 45 km (T = 265 K): k2 = 6 x 10^-34 s^-1 Show more…

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Estimate the lifetime of odd oxygen (i.e. ozone) at 20 km and 45 km. It may be assumed that the photolysis rates of O3 (reaction 3 from Chapman Mechanism) at these two altitudes are: k3 = 7.0 x 10^-4 s^-1 at 20 km k3 = 6.0 x 10^-3 s^-1 at 45 km k2 = 6 x 10^-34 s^-1 (T/300)^-2.4 cm^6 molecules^-2 s^-1 k4 = 8.0 x 10^-12 exp(-2060/T) cm^3 molecules^-1 s^-1 and the O3 and M concentrations at the two altitudes are: [O3] = 2 x 10^12 molecules cm^-3 and [M] = 1.4 x 10^18 cm^-3 at 20 km (T = 217 K) [O3] = 0.2 x 10^12 molecules cm^-3 and [M] = 3.6 x 10^16 cm^-3 at 45 km (T = 265 K) Hint: you will need to calculate k2 and k4 rate constants from the Chapman Mechanism. k3 was given in this problem.

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Transcript

00:01 Hello students, so the question is to estimate the lifetime of odd oxygen at 20 km and at 45 km.

00:08 So, first we will see for the 20 km where the temperature is 217 kelvin.

00:14 So here we have been given the k3 which is 7 into 10 raised to the power minus 4 per second.

00:20 Now by the chapman mechanism we will calculate the k2.

00:25 So here, so the k2 is given this in the question.

00:29 So by putting the value of t over here we can calculate the k2 value.

00:33 So as we know the t value is 217 kelvin.

00:38 So here the k2 is equal to 6 into 10 raised to the power minus 34 into 217 divided by 300 whole raised to the power minus 2 .4 centimeter whole 6 molecules per cube per square per second.

01:07 So here by calculating this whole equation we will get the k2 and it is equal to 1 .17 into 10 raised to the power minus 33 centimeter whole 6 molecules per square per second.

01:29 So similarly we will calculate for the k4.

01:32 So the k4 is thus equal to as it is given as so this is the equation for the k4.

01:38 So by putting the value of t over here we will get the k4 and it is equal to 1 .26 into 10 raised to the power minus 10 centimeter cube molecules per molecules per second.

01:56 So now we know the o3 concentration which is 2 into 10 raised to the power 12 molecules per centimeter cube and the concentration of the m is equal to 1 .4 into 10 raised to the power 18 centimeter per cube...

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