How much energy, in joules, must be absorbed if all the O_3 molecules in the sample of air are to dissociate? Assume that each photon absorbed causes one O_3 molecule to dissociate, and that the wavelength of the radiation is 254 nm.

Information provided with question Ozone, ${O}_{3}$, absorbs ultraviolet radiation and dissociates into ${O}_{2}$ molecules and $O$ atoms: O_3+h\nu→O_2+O. A 1.85-$L$ sample of air at 22°C and 748 $m m H g$ contains 0.25 ppm of ${O}_{3}$. My thoughts I assume "parts per million" (ppm) is "$\frac{m L}{L}$"? Also, I don't really know what section I should be posting this in, sorry.

Nov 20, 2016

8.9 mJ

Explanation:

This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.

The outline of how we are going to solve the problem is this:
1) figure out how much energy, in Joules, a single photon of 254 nm light has.
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or $\mu$l/l. Fortunately, the convention for gas is $\mu$l/l, so we will go with that.
3) multiply the two together to get the answer.

So, step 1: we will use E = hc/ $\lambda$ where E is in Joules, h is Plank's constant ($6.62607 \times {10}^{- 34} J \cdot s$), c is the speed of light in meters/sec (299,792,458 m/sec), and $\lambda$ is 254 nm ($254 \times {10}^{- 9}$ meters). Each photon thus has an energy of $7.82065 \times {10}^{- 19}$ J.

Reality check: is this number reasonable?
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)

Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present.

${m}_{g a s} = \frac{{P}_{a t m} \cdot {V}_{l}}{{R}_{\frac{l a t m}{m o l K}} \cdot {T}_{K}}$

We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.

${m}_{g a s} = \frac{\left(\frac{748 m m H g}{760 m m H g}\right) \cdot 1.85 l i t e r s}{{0.082057}_{\frac{l a t m}{m o l K}} \cdot \left(22 + 273.15\right) K} = 7.5180 \times {10}^{- 2} m$

Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm.

So far, so good!

So how many molecules of ozone do we have?

$\text{molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules}$

Now to multiply the part 1 answer by the part 2 answer:

$7.82065 \times {10}^{- 19} \text{ J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules}$

For this question, the limiting number of significant figures is 2 from the 0.25 ppm ozone concentration. Thus the final answer is

8.9 mJ.

Nov 29, 2016

WARNING! Long answer! The energy absorbed is 8.9 mJ.

Explanation:

Step 1: Calculate the number of molecules in the container.

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange the formula to get

$n = \frac{P V}{R T}$

In this problem,

P = 748 color(red)(cancel(color(black)("mmHg"))) × ("1 atm")/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9842 atm"

$V = \text{1.85 L}$

$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$

$T = \text{22 °C" = "295.15 K}$

n = (0.9842 color(red)(cancel(color(black)("atm"))) × 1.85 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 295.15 color(red)(cancel(color(black)("K")))) = "0.075 18 mol"

Step 2. Calculate the number of gas molecules

"Number of gas molecules" = "0.075 18" color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 color(white)(l)"molecules")/(1 color(red)(cancel(color(black)("mol"))))

= 4.527 × 10^22 color(white)(l)"molecules"

Step 3. Calculate the number of ${\text{O}}_{3}$ molecules.

$\text{Number of"color(white)(l) "O"_3 color(white)(l)"molecules" = 4.527 × 10^22 color(red)(cancel(color(black)("gas molecules"))) × ("0.25 O"_3 color(white)(l)"molecule")/(10^6 color(red)(cancel(color(black)("gas molecules")))) = 1.13 × 10^16 color(white)(l)"O"_3 color(white)(l)"molecules}$

Step 4. Calculate the energy absorbed by 1 ${\text{O}}_{3}$ molecule.

The formula for the energy $E$ of a quantum is

color(blue)(bar(ul(|color(white)(a/a) E = (hc)/λ color(white)(a/a)|)))" "

where

$h$ = Planck's constant
$c$ = the speed of light
λ = the wavelength of the light

E = (6.626 × 10^"-34" "J"·color(red)(cancel(color(black)("s"))) × 2.998 × 10^8 color(red)(cancel(color(black)("m·s"^"-1"))))/(254 × 10^"-9" color(red)(cancel(color(black)("m")))) = 7.821 × 10^"-19"color(white)(l) "J"

Step 5. Calculate the energy absorbed by all the ${\text{O}}_{3}$ molecules

$\text{Total energy" = 1.13 × 10^16 color(red)(cancel(color(black)("O"_3color(white)(l) "molecules"))) × (7.821 × 10^"-19"color(white)(l) "J")/(1 color(red)(cancel(color(black)("O"_3color(white)(l) "molecule")))) = 8.9 × 10^"-3"color(white)(l) "J" = "8.9 mJ}$