Hi Albert, let's look at the Popp question again. Popp is the opposing atmospheric pressure. When a gas forms in a reaction, it has to expand against the opposing atmospheric pressure. It takes work to do this, called "pressure-volume" work, w = –PoppΔV, where ΔV is the change in volume of the gases in the reaction.
What I assumed incorrectly was that ΔV is 20 mL (the volume of NO2 formed) but I forgot that in this reaction, we are starting with gases too, which will be consumed as the NO2 product forms. Therefore, ΔV of this reaction will be the volume of the gaseous product minus the volume of the gaseous reactants.
What was the original volume of the NO and O2 reactants? We can tell from the balanced reaction, because the volumes will be related to each other according to the coefficients in the reaction.
2 NO (g) + O2 (g) ––�?gt; 2 NO2 (g)
Since the coefficients for NO and NO2 are the same, the moles of NO reacted and NO2 formed are the same. This also means that their volumes are the same. Since 20 mL of NO2 formed, we must have started with 20 mL of NO. Similarly, we can see from the reaction that the moles of O2 is 1/2 the moles of NO, and therefore half the volume also, which is 10 mL. So, the total volume of the reactants is 20 mL of NO plus 10 mL of O2 = 30 mL.
Now we can get ΔV: ΔV = Vfinal �? Vinitial = Vproducts �? Vreactants = 20 mL �? 30 mL = �?0 mL.
And now we can calculate the work: w = –PoppΔV = �?9.86923 atm)(�?.010 L) = +0.0986923 L atm
Converting L atm to joules gives the work as +10 J.
Hopefully, that is one of the answers! 
Your other two questions involve the reaction of a strong acid with a strong base and a strong acid with a weak base. I'll look at those in the next post.
Free Forum Hosting
Important Announcement
Prev Message
