Answer:
The NO + O3 is the dominant reaction.
Explanation:
First of all, let's convert to molecules/cm³;
For O3;
O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT
Thus, plugging in the relevant values to get;
n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)
So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =
9.84 x 10^(11) molecules/cm³
But from the question, NO has 2 moles, and thus concentration is;
2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³
For O2;
Following the same pattern for O3, we obtain;
(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³
Now, for NO and O3 reaction the rate is; k[NO] [O3]
Thus rate;
= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s
For 2NO + O2 → 2NO2 reaction, rate = k[NO]2
[O2]
Thus, rate;
= (2 x 10^(-38) cm^(6)/molec².s
)( 1.968 x 10^(12) molec/cm³)
²
(5.167 x 1018 molec/cm³)
= 40,000 molec/cm³.s
Observing the two rates, it's clear that the NO + O3 is the dominant reaction.
