CH301 Worksheet 11: Thermodynamic Reaction Solution KeyCheck for explanation of solution key at bottom of page—assume room temperature for all calculations.Chemical ReactionDH(kJ)BE(kJ)Dngasw (kJ)=-DnRTDnsystemTDS(kJ)DG (kJ)CH4g + 2O2g ‡ CO2g + 2H2OgCombustion reaction, would predict large heat and spontaneousreaction. Dn = 0 means no work, small entropy change.-802-802000-1-8012H2g + O2g ‡ 2H2OgSpontaneous combustion reaction, would predict large heat. Dn =negative which means entropy decreases and work on system-484-482-1+2.5-1-27-4582H2Og ‡ 2H2g + O2gThis is the non-spontaneous reaction that only occurs when I throw abattery into water to produce H2 gas. Values are opposite samemagnitude but opposite signs of combustion reaction above.4844821-2.5127458C2H5OHl + 3O2g ‡ 2CO2g + 3H2OgSpontaneous combustion reaction, would predict large heat. Dn = ispositive so work on surroundings and positive entropy.-1234-12482-5166-1300C2H5OHl + 3O2g ‡ 2CO2g + 3H2OlCombustion reaction, would predict large heat and spontaneousreaction. But a liquid is formed, so there is a negative entropy change.-1366-1248-12.51-41-1325C3H8g + 5O2g ‡ 3CO2g + 4H2OgSpontaneous combustion reaction, would predict large heat. Dn = ispositive so there is work on surroundings and positive entropy change.-2043-20121-2.5131-20744H2Og + 3 CO2g ‡ C3H8g + 5O2gThis is the reverse of the combustion reaction. So we know that it hasthe opposite signs from the example above and is consequentlyendothermic and unfavorable entropy change.2043+2012-12.5-1-31+2074CCl4l ‡ Cs + 2Cl2gThis is a non-spontaneous process in which dry cleaning fluid would beexpected to decompose at room temperature into carbon and releasechlorine gas. Not a likely event. The reaction has a highly favorableentropy so it must be endothermic.1358602-527065Ba(OH)2(H2O)8s + 2NH4NO3 ‡ Ba(NO3)2s + 2NH3g + 10H2OlThis is the reaction involving two white solids I do all the time. Theprocess is spontaneous from experience. It is a highly endothermicprocess as we can tell from the cold temperature generated. The factthat ammonia is released as a gas indicates that the reaction spontaneityis driven by the large positive entropy.NAButpositiveNAButpositive2-510NANA2O3g ‡ 3O2gThis reaction is spontaneous. It should have a small exothermic valueand a favorable entropy change.-286NA1-2.5141-327H2Os ‡ H2OlPhase change from solid to gas. Expect temp. dependent spontaneity.From experience, spontaneous at room temperature. Reaction isendothermic, so the reaction spontaneity depends on positive entropy.NANA000NAnegativeat room tempCO2g ‡ CO2sThis is a deposition reaction as a gas becomes a solid. At roomtemperature we know this is not spontaneous. Since the reaction mustbe exothermic (heat leaves the system as the dry ice is formed), theentropy change must be sufficiently negative at room temperature tokeep the reaction from occurring.NANA-12.50NApositiveat room tempNH3g + HClg ‡ NH4ClsWe have seen that this is a spontaneous reaction that generates a whitesolid from the reaction of the gas vapors. Since the entropy change isclearly negative, a large negative DH must contribute to spontaneity.-176NA-25-1-85-912H2O2l ‡ 2H2Ol + O2 gFrom experience, this reaction is spontaneous. Reaction heat is notlarge but positive entropy contributes to spontaneity.-196-3521-2.5138-234Explanation of my solution key:Here is the answer key for the thermodynamic worksheet. Most of the calculations were done in a dark parking lot using the dome light of a 5 yearold Saturn, so take the answers with a grain of salt, and e-mail me if you find a problem. In the first column I have given qualitative explanations inwhich the sign of the thermodynamic process is predicted. As you can tell from the calculations, I was right every time!! In some cases the lack ofAppendix K data kept me from performing a calculation. Also note that when all species are in the gas phase, the BE calculation is pretty close, butwhen other phases are involved, the energy change for the transition is not considered and the BE calculation is poor. In cases where an ionic bondwas involved, no BE calculation was done. Also note that I didn’t simply calculate DS but also multiplied it by 300 to get TDS with units of kJ. Onemore thing, I use as few digits as possible to make the calculation simpler. So I use 300K, not 298K, and I never looked at the Appendix K dataafter the decimal. Amazingly (not), I still am within a kJ on everything. Finally, one great way to find out if you have done these calculationscorrectly is to compare the DGf calculation with a separate DH-TDS calculation of DG. If they are the same it means that even though you may havedone the work in a darkened parking lot, there is internal integrity in the calculation—makes you feel warm all
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