Estimate the [03]/[O] ratio in Chapman mechanism at 30 km with p = 10 Torr, T = 250 K, using the J values and rate constants given. Hint: The mixing ratio of O₂ is 0.21, which means [0₂] = 0.21 [M]. The j values and rate constants are given below Chapman mechanism:
Estimate the [03]/[O] ratio in Chapman mechanism at 30 km with p = 10 Torr, T = 250 K, using the J values and rate constants given. Hint: The mixing ratio of O₂ is 0.21, which means [0₂] = 0.21 [M]. The j values and rate constants are given below Chapman mechanism:
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter32: Radiochemical Methods
Section: Chapter Questions
Problem 32.12QAP
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Question
![Estimate the [03]/[O] ratio in Chapman mechanism at 30 km with p = 10 Torr, T = 250 K, using the J
values and rate constants given.
Hint: The mixing ratio of O2 is 0.21, which means [0₂] = 0.21 [M].
The j values and rate constants are given below
Chapman mechanism:
1. O₂+ hv→ 0+0 (λ<242 nm) Joz 10-11 s¹ at 30 km
2.0 + O₂ + M → 03 + M*
k₂9x10-34 cm5/(molec²s) at 250 K
3. 0₂+ hv→ O₂ +O (A<320 nm) Jos 10-¹³ s¹ at 30 km
4.0 +03 → 0₂ + O₂
k₁2x10-15 cm³/(molecs) at 250 K
"Odd Oxygen
The rate of interconversion between 0 and O₂ is MUCH faster than either the
production of O from O₂ photolysis or the loss of ozone via reaction with O. It is
therefore useful to define a" family" of odd oxygen species:
0x = 0+03
The partitioning between O and O3 is set by reactions 2 and 3 (steady-state):
k₂[0][0₂][M] =J03 [03]/
[O]/[03] = J03/(k₂[0₂][M])
The timescale for establishing this steady-state is quite fast: Even at 50 km,
k₂[0₂][M] is = 0.05 s1. Jos is much smaller than this and so, [0]/[0₂] <<1.
4
We can now calculate the concentration of ozone at steady state from Chapman's
mechanism. We assume that the rate of production of odd oxygen is balanced by
the rate of loss of odd oxygen:
d[0]/dt = 2Jo₂[0₂] - 2k4[0][03] - 0
Joz[0₂]=K4[0][03] = K4J03[03]²/(k₂[0₂][M])
|[03]²=J0₂k₂[O₂]²[M]/(K4J03)
Homework: Estimate the [03]/[O] ratio.
Calculate the expected [03] in ppm at 30
km using the J values and rate constants
given on the previous page.
***Updated***
Assume:
P= 10 Torr
T= 250 K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb4fb731-3ce7-4d4b-8230-1bfc9860d0f0%2Fc425e57a-163d-40e2-99fe-861172c818de%2Ftl9wr0q_processed.png&w=3840&q=75)
Transcribed Image Text:Estimate the [03]/[O] ratio in Chapman mechanism at 30 km with p = 10 Torr, T = 250 K, using the J
values and rate constants given.
Hint: The mixing ratio of O2 is 0.21, which means [0₂] = 0.21 [M].
The j values and rate constants are given below
Chapman mechanism:
1. O₂+ hv→ 0+0 (λ<242 nm) Joz 10-11 s¹ at 30 km
2.0 + O₂ + M → 03 + M*
k₂9x10-34 cm5/(molec²s) at 250 K
3. 0₂+ hv→ O₂ +O (A<320 nm) Jos 10-¹³ s¹ at 30 km
4.0 +03 → 0₂ + O₂
k₁2x10-15 cm³/(molecs) at 250 K
"Odd Oxygen
The rate of interconversion between 0 and O₂ is MUCH faster than either the
production of O from O₂ photolysis or the loss of ozone via reaction with O. It is
therefore useful to define a" family" of odd oxygen species:
0x = 0+03
The partitioning between O and O3 is set by reactions 2 and 3 (steady-state):
k₂[0][0₂][M] =J03 [03]/
[O]/[03] = J03/(k₂[0₂][M])
The timescale for establishing this steady-state is quite fast: Even at 50 km,
k₂[0₂][M] is = 0.05 s1. Jos is much smaller than this and so, [0]/[0₂] <<1.
4
We can now calculate the concentration of ozone at steady state from Chapman's
mechanism. We assume that the rate of production of odd oxygen is balanced by
the rate of loss of odd oxygen:
d[0]/dt = 2Jo₂[0₂] - 2k4[0][03] - 0
Joz[0₂]=K4[0][03] = K4J03[03]²/(k₂[0₂][M])
|[03]²=J0₂k₂[O₂]²[M]/(K4J03)
Homework: Estimate the [03]/[O] ratio.
Calculate the expected [03] in ppm at 30
km using the J values and rate constants
given on the previous page.
***Updated***
Assume:
P= 10 Torr
T= 250 K
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