A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by
O3(g)+NO(g) --> O2(g)+NO2(g)
The rate law for this reaction is:
rate=k[O3][NO]
Given that k = 2.87 × 106 M–1·s–1 at a certain temperature, calculate the initial reaction rate when [O3] and [NO] remain essentially constant at the values [O3]0 = 2.56 × 10–6 M and [NO]0= 7.14 × 10–5 M, owing to continuous production from separate sources.
________ M·s–1
Calculate the number of moles of NO2(g) produced per hour per liter of air.
________ mol·h-1·L-1
M,s-1 is a zero order; i.e., it is mols/L/s = mols/Ls.
Let's call your answer for the first part y, then
y mols/Ls x (60 s/min) x (60 min/hr) = 3600(y) mols/L.hr. Check my thinking
i know how to do the first part, I just need the second part of the problem.
To calculate the initial reaction rate, we can substitute the given values into the rate law equation.
rate = k[O3][NO]
Given:
k = 2.87 × 10^6 M–1·s–1
[O3]0 = 2.56 × 10–6 M
[NO]0 = 7.14 × 10–5 M
Substituting these values into the rate law equation:
rate = (2.87 × 10^6 M–1·s–1)(2.56 × 10–6 M)(7.14 × 10–5 M)
Simplifying:
rate = 2.87 × 10^6 M–1·s–1 × 2.56 × 10–6 M × 7.14 × 10–5 M
≈ 0.051 M·s–1
Therefore, the initial reaction rate is approximately 0.051 M·s–1.
To calculate the number of moles of NO2(g) produced per hour per liter of air, we need to convert the reaction rate to mol·h-1·L-1.
Given:
Volume of air = 1 L
Time = 1 hour
The reaction rate is already in mol·s-1, so we can convert it to mol·h-1 by multiplying by the number of seconds in an hour (3600 seconds).
Moles of NO2(g) produced per hour per liter of air = rate × (3600 s/h)
Substituting the value of the rate calculated earlier:
Moles of NO2(g) produced per hour per liter of air = 0.051 M·s–1 × 3600 s/h
= 183.6 M·h–1
Therefore, the number of moles of NO2(g) produced per hour per liter of air is 183.6 mol·h-1·L-1.
To calculate the initial reaction rate, you can substitute the given values into the rate law equation.
Given:
k = 2.87 × 10^6 M–1·s–1
[O3]0 = 2.56 × 10^–6 M
[NO]0 = 7.14 × 10^–5 M
Substituting these values into the rate law equation, we get:
rate = k[O3][NO]
rate = (2.87 × 10^6 M–1·s–1)(2.56 × 10^–6 M)(7.14 × 10^–5 M)
Now, multiply the numbers together to find the rate:
rate = 2.57 × 10^–11 M·s–1
So, the initial reaction rate is 2.57 × 10^–11 M·s–1.
Now, to calculate the number of moles of NO2(g) produced per hour per liter of air, we need to use stoichiometry and consider the molar ratio between NO and NO2.
From the balanced reaction equation:
1 mole of NO reacts to form 1 mole of NO2.
Given that the initial concentration of NO is [NO]0 = 7.14 × 10^–5 M, we can directly say that the moles of NO2 produced per liter of air per second is also 7.14 × 10^–5 mol·L^–1·s^–1.
To convert the rate to hours, we need to multiply it by the conversion factor of 3600 seconds per hour:
(7.14 × 10^–5 mol·L–1·s–1) × (3600 s/h) = 0.2574 mol·L–1·h–1
So, the number of moles of NO2(g) produced per hour per liter of air is 0.2574 mol·L–1·h–1.