In the polluted air moving over a city, the following concentrations were measur

ID: 1771133 • Letter: I

Question

In the polluted air moving over a city, the following concentrations were measured on a sunny late morning during a field experiment:

[CO] = 1.0 * 1013 molecules cm-3 (380 ppb)

[CH4] = 5.3 * 1013 molecules cm-3 (2000 ppb)

[HCHO] = 1.4 * 1010 molecules cm-3 (0.5 ppb)

The largest sinks for CO and CH4 are via reactions with the hydroxyl radical. HCHO is a significant sink for OH.

The reactions and reaction rate constants at 298K are given below:

(1) CO+OH CO2 +H (k1=1.51013cm3 molecule1 s1)

(2) CH4+OHCH3+H2O (k2=6.31015cm3molecule1s1)

(3) HCHO + OH CHO +H2O (k3=9.41012cm3 molecule1 s1)

(4) CHO+O2 CO+HO2

(a) (15 Pts) Determine the lifetime of OH in the next town's air, considering reactions 1-3.

(b) (10 Pts) Which of the three reactions is the dominant sink for OH? Hint: What are the individual lifetimes for CO, CH4, and HCHO?

These reactions (1,2,3) are of second order so overall rate of disappearance of OH is given by:

d (COH)/dt = K1* CCO*COH + K2*CCH4*COH + K3*CHCHO*COH

(a) For Lifetime of reaction 1:

d (COH)/dt = K1* CCO*COH

For Lifetime of reaction 2:

d (COH)/dt = K2*CCH4*COH

For Lifetime of reaction 3:

d (COH)/dt = K3*CHCHO*COH

We know the values of rate constants and concentrations of reactants.

In order to calculate time required to calculate a Finite amount of OH, We need concentration of OH available.

Then we can integrate these equation individually to calculate overall time by simply adding individually time required in a reaction.

(b) If we can calculate individually lifetime of reactions so whichever will take less time will be the dominant sink for OH.

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