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_2(g) The rate law for this reaction is rate of reaction = k[O_3] [NO] Given that k = 4.06 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 values [O_3] = 7.02 times 10^-6 M and [NO]_0 = 4.89 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

k = 4.06*10^6 1/Ms (second order)

given

[O3] = 7.02*10^-6

[NO]0 = 4.89*10^-5

substitutein rate

Rate = k*[O3][NO]

Rate = (4.06*10^6)*(7.02*10^-6)*(4.89*10^-5) = 0.00139 M/s

calculate number of moles of NO2 produced per hour and liter of air

the overall rate is

0.00139 M/s

in 1 liter --> 0.00139 moles/s

in 1 hour --> 0.00139 moles/s * 3600 / hr = 5.004 moles of NO2 per liter per hour

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