Question

The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.0 x 10-5 s-1 at 10 km above the planet’s surface. Consider a laboratory experiment in which a vessel of ozone is exposed to UV radiation at an intensity chosen to mimic the conditions at that altitude. If the initial O3 concentration is 5.0 x 10-3M, what will the concentration be after 10.0 day?

Answer 1

[O₃]= 8.84x10⁻⁷M

Step-by-step explanation:

The photodissociation of ozone by UV light is given by:

O₃ + hν → O₂ + O (1)

The first-order reaction of the equation (1) is:

`rate = k [O_(3)] = - k (\Delta [O_(3)])/(\Delta t)` (2)

where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration

We can get the following expression of the first-order integrated law of the reaction (1), by resolving the equation (2):

`[O_(3)]_(t) = [O_(3)]_(0) \cdot e^(-kt)` (3)

where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration

We can calculate the initial ozone concentration using equation (3):

`[O_(3)]_(t) = 5.0 \cdot 10^(-3)M \cdot e^{-(1.0\cdot 10^(-5)s^(-1)) ((10d \cdot 24h \cdot 3600 s)/(1d \cdot 1h))} = 8.84 \cdot 10^(-7)M`

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!