CHEM 102 1nd Edition Lecture 11 Outline of Last Lecture I. Determining ΔG, ΔS, and ΔHII. Temperature Range of Spontaneity III. Entropy of SurroundingsIV. Predicting Boiling PointsV. Entropy Phase ChangesVI. KineticsOutline of Current Lecture I. Entropy in Solution and SurroundingsII. Boiling PointsIII. Reaction RatesCurrent LectureI. Entropy in Solution and Surroundingsa. In the dissolution of ammonium nitrate, does the entropy of the reaction increase, decrease, remain the same, or more information is needed to determine?i. NH4NO2 (s) NH4+ (aq) + NO2- (aq)ii. The solution’s entropy increases because the number of molecules in the reaction will increase. Before the reaction occurred, there was 11 mole of ammonium nitrate, after the reaction, there were each 1 mole of ammonium and nitrate.b. In the same reaction, does the entropy of the surroundings increase, remain the same, decrease, or more information is needed?i. More information is needed because if we were to plug the known information into the equation ΔSuniv = ΔSsys + ΔSsurr, where the entropy of the universe is always positive, and the entropy of this system is positive, then the equation would look like: + = (+) + ΔSsurrii. The entropy of the surroundings could be negative but not negative enough to make the final answer negative; however, the entropy of the surroundings could also be positive and make the entropy of the universemore positive. II. Boiling PointsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. When will the solution of 8.77 e-4 of a nonelectrolyte in 10.0g benzene boil?i. Kb of benzene = 2.53 c/m and the boiling point is 80.10 °C. Answer choicesare: 80.0, 79.0, 81.0, 78.0 (all answer choices are in Celsius) ii. There are two ways of determining, to predict the answer, you can recall that when a solute is added to a solvent, the solution’s boiling point will be higher than that of the original; therefore, the only answer that is logical is 81.0.iii. To determine algebraically: ΔT =kb×miv. m= moles/kg solvent = 8.77 e -4 / 0.0100 kg = 8.77 e -2v.ΔT = kb×m=2.53 × 0.0877=0.222 ° Cvi.ΔT =80.10+ 0.222=80.32vii. You want a temperature higher than 80.32 in order for the solution to boilIII. Reaction Ratesa. Rxn rate = −1a×ΔAΔt=1b×ΔBΔt, where aA bBb. For example, for the equation H2O 2H2 + O2 , −11×Δ H 20Δt=12×Δ H
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