A reaction of importance in the formation of smog is that between ozone and nitr
ID: 516152 • Letter: A
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 = 4.01 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]_0 = 4.50 times 10^-6 M and [NO]_ 0 = 4.58 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
Solution:- From the given rate law the reaction first order with respect to each of the reactant and the over all reaction is second order.
rate = k [O3] [NO]
let's plug in the values in it...
rate = 4.01 x 106 M-1.s-1 [4.50 x 10-6 M] [4.58 x 10-5 M]
rate = 8.26 x 10-4 M.s-1
From the given equation......
rate = -d[O3]/dt = -d[NO]/dt = d[O2]/dt = d[NO2]/dt
where d[]/dt stands for the rates of these reactants and products. negative signs for the reactants indicates a decrease in their concentrations.
From this rate law...
d[NO2]/dt = rate = 8.26 x 10-4 M.s-1
M stands for molarity and it is mol/L or could also be written as mol.L-1. So,.....
d[NO2]/dt = 8.26 x 10-4 mol.L-1.s-1
This is the rate of formation of NO2 per second or the number of moles of NO2 formed per second per liter of air. let's calculate it for per hour. For this we will do the unit conversion from s to h that is second to hour.
d[NO2]/dt = 8.26 x 10-4 mol.L-1.s-1 x (60 s/1 min) x (60 min/1 h) = 2.97 mol.h-1.L-1
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