Estimate the [O3]/[O] ratio in Chapman mechanism at 30 km with p=10 Torr, T=250K using the J values and rate constant given below. Hint: The mixing ratio of O2 is 0.21, which mean [O2]=0.21[M]
Estimate the [O3]/[O] ratio in Chapman mechanism at 30 km with p=10 Torr, T=250K using the J values and rate constant given below. Hint: The mixing ratio of O2 is 0.21, which mean [O2]=0.21[M]
Chapter12: Spectrochemical Methods
Section: Chapter Questions
Problem 11P
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A. Estimate the [O3]/[O] ratio in Chapman mechanism at 30 km with p=10 Torr, T=250K using the J values and rate constant given below. Hint: The mixing ratio of O2 is 0.21, which mean [O2]=0.21[M]
B. Calculate the expected [O3] in ppm at 30 km, with p=10Torr, T=250K, using the J values and rate constant given below:
![Chapman mechanism:
1. O₂+ hv→ 0+0 (λ<242 nm)
2.0 + O₂ + M → 0₂ + M* k₂
-
3. 03 + hv→ 0₂ +0 (λ<320 nm)
4.0 +03 → 0₂ + O₂
Joz 10-11 s¹ at 30 km
9x10-34 cm³/(molec²s) at 250 K
Jos 10³ s² at 30 km
k₁2x10-15 cm³/(molecs) at 250 K
"Odd Oxygen"
The rate of interconversion between 0 and O3 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] = Jos/(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, [O]/[03] <<1.
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[O]/dt = 2Joz[0₂] - 2k4[0][03] - 0
Joz[0₂]=k4[0][03] = K4J03[03]²/(k₂[0₂][M])
[03]²=J0₂k₂[0₂]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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb4fb731-3ce7-4d4b-8230-1bfc9860d0f0%2Fa8e52b62-95b9-4551-966e-25b5da8dda0e%2F9cletek_processed.png&w=3840&q=75)
Transcribed Image Text:Chapman mechanism:
1. O₂+ hv→ 0+0 (λ<242 nm)
2.0 + O₂ + M → 0₂ + M* k₂
-
3. 03 + hv→ 0₂ +0 (λ<320 nm)
4.0 +03 → 0₂ + O₂
Joz 10-11 s¹ at 30 km
9x10-34 cm³/(molec²s) at 250 K
Jos 10³ s² at 30 km
k₁2x10-15 cm³/(molecs) at 250 K
"Odd Oxygen"
The rate of interconversion between 0 and O3 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] = Jos/(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, [O]/[03] <<1.
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[O]/dt = 2Joz[0₂] - 2k4[0][03] - 0
Joz[0₂]=k4[0][03] = K4J03[03]²/(k₂[0₂][M])
[03]²=J0₂k₂[0₂]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.
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