The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction in which it is oxidized by oxygen, i. e., 2NO + O2- --> 2NO2 is 2 x 10^38 molecule2 cm s. For typical concentrations encountered in morning smog episodes, namely 40 ppb for ozone and 80 ppb for nitric oxide, deduce the rates of these two reactions and decide which one is the dominant process.

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.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: