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

Answer:

[O₃]= 8.84x10⁻⁷M  

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:

$rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t}$ (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):  

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

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

I hope it helps you!