Nitrogen Oxides, NOx (mixture of NO and NO2).
The NOx in the atmosphere is slowly broken down to N2 and O2 in a first-order reaction. The average half-life of NOx in the smokestack emissions in a large city during daylight is 3.9 hours.
(a) Starting with 1.50 mg in an experiment, what quantity of NOx remains after 5.25 hours?
(b) How many hours of daylight must have elapsed to decrease 1.50 mg of NOx to 2.50X10^-6 mg?
a).
k = 0.693/t1/2
ln(No/N) = kt
No = 1.5mg
N = unknown mg
k = from above.
t = 5.25 hours.
b)
Same equation as above.
No = 1.50 mg
N = 2.50E-6 mg
k from above
solve for t in hours.
Thank you Bob
To solve these problems, we can use the concept of half-life and the first-order reaction.
(a) Starting with 1.50 mg of NOx, we want to find the quantity of NOx that remains after 5.25 hours. First, let's determine the number of half-lives that have elapsed during this time.
Half-life (t1/2) is the time it takes for half of the reactant to degrade. In this case, the half-life of NOx is given as 3.9 hours.
To find the number of half-lives elapsed, we can divide the total time (5.25 hours) by the half-life of NOx:
Number of half-lives = 5.25 hours / 3.9 hours ≈ 1.35
Now, we can use the formula for exponential decay to calculate the remaining quantity of NOx. The formula is:
[N] = [N0] * (1/2)^(t/t1/2)
Where:
[N] = Quantity of NOx remaining after time t
[N0] = Initial quantity of NOx
t = Time elapsed
t1/2 = Half-life of NOx
Plugging in the values:
[N] = 1.50 mg * (1/2)^(1.35)
Using a calculator, evaluate (1/2)^(1.35) ≈ 0.407
[N] ≈ 1.50 mg * 0.407 ≈ 0.6105 mg
Therefore, approximately 0.6105 mg of NOx remains after 5.25 hours.
(b) In this problem, we need to determine the time required for the quantity of NOx to decrease from 1.50 mg to 2.50 × 10^-6 mg.
Using the same formula as above, we can rearrange it to solve for time (t):
t = t1/2 * log([N]/[N0]) / log(1/2)
Plugging in the values:
t = 3.9 hours * log(2.50 × 10^-6 mg / 1.50 mg) / log(1/2)
Using a calculator, evaluate log(2.50 × 10^-6 mg / 1.50 mg) / log(1/2) ≈ -14.92
t ≈ 3.9 hours * -14.92 ≈ -58.2 hours
Since time cannot be negative, we can conclude that it would take more than 58.2 hours of daylight to decrease 1.50 mg of NOx to 2.50 × 10^-6 mg.