A reaction of importance in the formation of smog is that between ozone and nitr

Question

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O_3(g) + NO(g) rightarrow O_2(g) + NO_(g) The rate law for this reaction is rate of reaction = k[O_3] [NO] Given that k = 2.66 times 10^6 M^-1 s^-1 at a certain temperature, calculate the initial reaction rate when [O_3] and [NO] remain essentially constant at the value [O_3]_0 = 5.34 times 10^-6 M and [NO]_0 = 5.24 times 10^-5 M, owing to continuous production from separate sources. Calculate the number of moles of NO_2(g) produced per hour per liter of air.

Explanation / Answer

NO (g) + O3 (g) NO2 (g) + O2 (g)

Rate of reaction = k [O3] [NO]j

Given

K = 2.66 x 106 M-1s-1

[O3]0 = 5.34 x 10-6 M

[NO]0 = 5.24 x 10-5 M

1. Rate of reaction = k [O3] [NO]

                             = 2.66 x 106 M-1S-1* [5.34 x 10-6 M] [5.24 x 10-5 M]

= 0.00074431M.s-1

= 7.44 x 10 -4Ms-1

2. Number of moles of NO2 (g) produced per hour per liter of air

Rate of reaction = 7.44 x 10-4M /s

                             =7.44 x 10-4moles/L /s

NO2 (g) per hour per liter of air = (7.44 ×10^-4 moles/L) (60s/1min) (60min/1hour)

                                                                                                = 2.678moles/ L h

NO2 per hour per liter of air = 2.6784moles

2.678 moles/ L h NO2 (g) produced per hour per liter of air

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