Tuesday, December 8, 2009

Here's a problem for you

When I took over this teaching gig, my predecessor left a fold behind labeled "Diabolical Organic Problems." I love the word "Diabolical" but some may argue that that title is redundant with "Organic."

I have on occasion given my students diabolical problems to hone their skills. Today in preparation for the cumulative final exam, I gave my General Chemistry class the following question:

A 21.0 cm3 piece of dry ice is placed into a sealed 5.0 L container at STP. The dry ice is allowed to sublime while the temperature is held constant at 0°C. The container also contains 12.7 grams of sodium hydroxide. Sodium hydroxide reacts with carbon dioxide to form solid sodium carbonate and liquid water. What will the volume of the dry ice be when the pressure in the container is 2.5 atm? The density of dry ice is 1.5 g/cm3. Assume the volume of the container is 5.0L.

I'm kind of proud of this question as it incorporates several concepts we covered in the first half of the semester. I'll give anyone who answers this 50 extra credit points.


the iNDefatigable mjenks said...

This question is tricky, as it depends on how badly you need the dry ice. Just about to do a metal-halogen exchange? There will be no dry ice. Rotovap cold finger getting a little empty? There will be a slight dusting in the corners of the box. Got a methanol fire that needs quenching? Time to update that resume, buddy, because there's nary a bit of dry ice to be found and the safety committee is doing their monthly inspection.

the iNDefatigable mjenks said...

You should give extra credit to any of your students who write "I'm more of a 'think outside of the box' type".

Adam L. said...

The correct answer would be "I used liquid nitrogen, so what do I care about dry ice?"

Liberal Arts Chemist said...

I must be missing something obvious here (twouldn’t be the first time). The information on the CO2 / NaOH rxn seems superfluous.

Assuming that the Ptot = PCO2 (not correcting for PH20 at STP) and that as a primary iteration of the question assuming that Vgas = V contain (assuming V H2O, VCO2 and VNaOH/Na2CO3 ~ 0) valid for the # sig. figs allowed in the question) we would calc.

21.0 cm3 CO2 - - > 31.5 g CO2 - - > 0.716 mol CO2

PV=nRT - - > nCO2 = 0.56 mol CO2 required to create 2.5 atm in chamber

nCO2 solid = nCO2 init – n CO2 gas = 0.16 mol CO2 must still be solid

0.16 mol CO2 - - > 6.95 g CO2 - - > 4.63 mL CO2 solid

You would then re-calculate the numbers using the volume in the chamber as 5.0 - VCO2 solid = 5.0 - 0.00463 which is not significant given the sig figs.

So what is with the CO2 / NaOH chemistry?

Chemgeek said...

Matt- Of course, but this is Gen Chem. We don't deal with the real world.

Adam- You gave me an idea. On the final exam I'm going to write a question something like: "Liquid N2 is placed in a container that will explode if the pressure reaches 3 atm. What is the max amount of liquid N2 that one can put in the container without exploding it?"

LAC- the NaOH reacts with some gaseous CO2 and thus "removes" it from the container. The dry ice sublimes, the CO2(g) reacts with the NaOH until the NaOH is gone. Then the dry ice continues to sublime until the gas pressure is 2.5 atm. The Ptot = PCO2 + Pair. They were told to ignore the contribution from the vapor pressure of water. There of course are some simplifications that are made.

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