In the Chemistry and the Environment box on free radicals in this chapter, we discussed the importance of the hydroxyl radical in reacting with and eliminating many atmospheric pollutants. However, the hydroxyl radical does not clean up everything. For example, chlorofluorocarbons—which destroy stratospheric ozone—are not attacked by the hydroxyl radical. Consider the hypothetical reaction by which the hydroxyl radical might react with a chlorofluorocarbon: OH( g) + CF2Cl2( g)¡HOF( g) + CFCl2( g) Use bond energies to explain why this reaction is improbable. (The C¬F bond energy is 552 kJ>mol.)

Video transcript

Welcome back everyone in this example. We have a seedling in nature formed from the thermal decomposition of long chain hydrocarbons at high temperatures up to 1700 Kelvin, we're told that a seedling is a hydrocarbon composed of two carbon and two hydrogen atoms. And the decomposition composition reaction for a sibling is given below here. So we need to use bond energies to explain why this reaction is improbable and were given the bond energies for each bond making up our compound a seedling below here. So this is our settling here. So to make use of these bond energies, we want to recall how to calculate the change in entropy delta H of our reaction. And we should recall that to calculate this. We're going to take the sum of the entropy or change in entropy of the reactant or bonds that are broken subtracted from the sum of the entropy change for the bonds formed or the products. And just so it's clear, let's actually write this part in a different color. So we'll say mine in purple minus the sum in entropy of the bonds formed. So they give us our reaction with the molecular formulas above in the prompt. But we want to translate this into a structural formula equation. So writing out our first reactant, we have a seedling and we have in a seedling to hydrogen carbon bonds that are single bonds. So we have two carbons as well and the carbons in the center are bonded by a triple bond because we should recall that carbon should have four bonds to be stable and to have a full octet. So this is going to be our structure for a ceiling here. And then in our reaction, this is going through a thermal decomposition to form two moles of carbon and this is just solid carbon. Whereas our acetylene is a gaseous reactant and then added onto this product. We have a second product which is going to be our one mole of hydrogen gas. And because hydrogen is given as a diatonic molecule, we will draw it out with a structural formula with two hydrogen single bonded to one another. And again, this is a gas. So here we have our bonds that are formed and then on the reactive side we have our bonds that are broken which is our ceiling. So we're going to utilize this information to calculate the entropy change of the reaction here, which will give us an idea of why this reaction is actually improbable. So calculating from above, we're going to say that the entropy change of our reaction is equal to first beginning with the some of the bonds broken of our reactant, a sideline. We take the some of the entropy change of each type of bond in our molecule of a ceiling. So we can see we have a carbon carbon triple bond. We just have one of these bonds. So we would say we have the bond energy of the carbon carbon triple bond. We just have one of these. This is then going to be added on to the second type of bond making up our sideline, which is our carbon hydrogen single bond. And we have two of these. So we would say plus two of our two moles of our bond energy of our carbon hydrogen single bond. So this is our only reactant. And now we can just subtract this from the bond energies of our products. And so we're going to begin that by taking first our first type of bond here on our product side, and we're only going to be considering our gaseous products. So we are going to disregard our two moles of solid carbon here. We just want to focus on this bond here, which is our hydrogen hydrogen single bond. We just have one of these. So we're going to say minus the entropy change of our hydrogen hydrogen single bond. And so now we have this outline, we're going to plug in our bond or bond energies. So making more space below, we will say that our entropy change of our reaction is equal to We have the bond energy of our carbon carbon triple bond given above as 839 kg joules per mole. So we'll plug that in. This is then added to the second type of bond making up our sideline, which is our carbon hydrogen single bond. We have two moles of this. So we're going to say plus two moles of our carbon hydrogen single bonds multiplied by its bond energy of 413 kg joules per mole given in the prompt. So times 413 kg joules per mole. So this will complete our some of our bond energies for our reactant. And now we will go ahead and subtract this from our bond energies of our products. So beginning with our only gaseous product, which we said is the hydrogen hydrogen single bond from H2 gas. We have a bond energy of 432 kg per mole. So we'll plug that in And in our calculators, we're going to type in our above equation. And we would get that the entropy change of our reaction is equal to a value of positive 1,233 killah jewels per mole. And so because we have a positive value here, it's not negative. We can say that therefore energy is needed to break the reactant bonds for our reaction to take place. And so this would be understood as then being a endo thermic reaction. And because this reaction is endo thermic and needs energy in order for our reactant bonds to break to form our products, we would say thus the reaction is improbable. And so for our final answer, we have our entropy change of our reaction as this value here in kilograms per mole. And we have the fact that energy is needed to break our reactant bonds which explained as to why this reaction above, given in the prompt, is improbable. So what's highlighted in yellow represents our final answers. I hope that everything I went through was clear, But if you have any questions, please leave them down below, and I will see everyone in the next practice video.