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Which of the following diagrams best rep-resents the final state of this system afterequilibrium is achieved?1.2.3.4.correct5.Explanation:Only the solvent goes through the mem-brane thus increasing the volume on the solu-tion side and decreasing volume on the solventside.The solution side is therefore dilutedslightly and the shade is lightened somewhat.The solute itself (the darker color) cannotpass through the membrane so the right sideMUST also stay the same color of pure solvent(the lighest shade shown).005(part 1 of 1) 10 pointsThe bond breaking process has ΔH(greater,less) than zero and ΔS(greater, less) thanzero; thus it is favorable at (high, low) tem-peratures.1.less; greater; low.2.greater; less; high.3.less; greater; high.4.greater; greater; high.correct5.less; less; low.6.greater; less; low.7.greater; greater; low.8.less; less; high.Explanation:Energy input (positive ΔH, endothermic)is required to break bonds.Because you in-evitably end up with more individual particles(atoms) than what you start with (molecules),the entropy of the system increases.Processes are spontaneous when ΔGis neg-ative.ΔG= ΔH-TΔS= (+)-T(+)
Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe4= (+)-TΔGwill be negative only at high tempera-tures.006(part 1 of 3) 10 pointsStart with a liquid of molar fraction 0.3. Boilit, catch the vapor, then cool the vapor untilit condenses into a new liquid.01120◦65◦What is the composition of the new liquid?1.0.85correct2.0.033.0.34.895.686.115Explanation:0.85Each horizontal unit is 0.1.007(part 2 of 3) 10 pointsWhat is the boiling point of the new con-densed liquid?1.0.32.0.853.68correct4.895.1156.0.03Explanation:
Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe568◦Each vertical unit is120◦-65◦10= 5.5◦.008(part 3 of 3) 0 pointsAt what temperature would this solution boil?1.0.32.0.853.0.034.1155.89correct6.68Explanation:89◦009(part 1 of 1) 10 pointsCalculate the standard enthalpy change forthe reaction2 HCl(g) + F2(g)→2 HF(‘) + Cl2(g)given4 HCl(g) + O2(g)→2 H2O(‘) + 2 Cl2(g)ΔH0=-202.4 kJ/mol rxn12H2(g) +12F2(g)→HF(‘)ΔH0=-600.0 kJ/mol rxnH2(g) +12O2(g)→H2O(‘)ΔH0=-285.8 kJ/mol rxn1.ΔH0= +1116.6 kJ/mol rxn2.ΔH0=-1088.2 kJ/mol rxn3.ΔH0=-1015.4 kJ/mol rxncorrect4.ΔH0= +1088.2 kJ/mol rxn5.ΔH0=-516.6 kJ/mol rxn6.ΔH0=-1587.2 kJ/mol rxn
Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe67.ΔH0= +516.6 kJ/mol rxn8.ΔH0= +1587.2 kJ/mol rxn9.ΔH0=-1116.6 kJ/mol rxn10.ΔH0= +1015.4 kJ/mol rxnExplanation:The first equation needs to be multiplied by12in order to get the equation we’re interestedin. Thus its ΔH0is multiplied by12as well.The second equation needs to be multipliedby two in order to get the equation we’reinterested in.We also multiply its ΔH0bytwo.The third equation needs to be reversed, sothe sign of its ΔH0should change.Then we add the equations to get the equa-tion we’re interested in. We also add the ad-justed ΔH0values to get the answer,-1015.4kJ/mol rxn.