You are monitoring air quality for your city and around noontime you measure NO2

Question

You are monitoring air quality for your city and around noontime you measure NO2 and NO concentrations of 38.2 and 4.6 ppbv, respectively. Neglecting hydrocarbons and assuming the concentrations of NO2 and NO are controlled by: NO2 + h --> NO + O (a) O + O2 --> O3 (b) NO + O3 --> NO2 + O2 (c) where ka is the noontime photolysis rate constant, ka = 0.0027 s-1, and kc = 2.2 * 10-12 * exp(-1430 Kelvin/T) in units of cm3 molecule-1 s-1. From this, please estimate the steady-state concentration of ozone. Give your answer in molecule cm-3. Assume a temperature of 15 C.

Assume steady state for [O] and derive it from reactions (a) and (b)

Thanks a lot

Explanation / Answer

rate of reaction (a), ra = ka*[NO2]
rate of reaction (c), rc = kc*[NO]*[O3]
ka = 0.0027 s-1
kc = 2.21E-12*exp(-1430 Kelvin/T) cm3 molecule-1 s-1
T = 15C = 288.15K
kc = 2.21E-12*exp(-1430 Kelvin/288.15) cm3 molecule-1 s-1=1.546E-10 cm3 molecule-1 s-1

Assume steady state for [O]
O is formed in the reaction(a) and consumed in reaction (b).
So rate of reaction of O from first two reactions, rO = ra-rb = 0
So, rb = ra = ka*[NO2]

rO3 = rb - rc = ra -rc = ka*[NO2]-kc*[NO]*[O3] = 0 at steady state
Steady state ozone concentration [O3] = ka/kc*[NO2]/[NO]

microg/m3 = ppbv*MW/(0.08205*T)
1 microg/m3 = 1E6*g/microg*microg/m3*1E-6*m3/cm3 = 1.0 g/cm3
g/cm3 =   ppbv*MW/(0.08205*T)

1g/cm3 = g*MW* mol/g *1/NA*molecule/mol*1/cm3= MW/NA molecule/cm3
molecule/cm3 = NA/MW g/cm3 = NA/MW*ppbv*MW/(0.08205*T) = NA*ppbv/(0.08205*T)

T = 15C = 288.15K
molecule/cm3 = 6.02E23*ppbv/(0.08205*288.15) = 2.55E22*ppbv
[NO2] = 38.2 ppbv =38.2*2.55E22 = 0.974E24 molecule/cm3
[NO] = 4.6 ppbv = 4.6*2.55E22 = 0.117E24 molecule/cm3

ka = 0.0027 s-1
kc = 1.546E-10 cm3 molecule-1 s-1

Steady state ozone concentration [O3] = ka/kc*[NO2]/[NO]
[O3] = 0.0027/1.546E-10*0.974E24/0.117E24 = 1.454E8 molecule/cm3

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