Question

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlorine from 15.5 g of CF3Cl goes through 10 cycles of these reactions

Answer 1

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$.