Chem 113 1st Edition Lecture 3 Outline of Last Lecture I. The Third Law of ThermodynamicsII. Standard molar entropiesIII. Factors affecting entropya. Temperatureb. Physical statec. Formation of a solutiond. Atomic size and molecular complexityIV. Entropy and structureOutline of Current Lecture V. Comparing energy and entropyVI. Predicting the sign of ΔS°sysVII. Applications of the Second Law of ThermodynamicsVIII. Entropy changes in surroundingsIX. Calculating entropy changesX. Temperature at which heat is transferredCurrent LectureI. Comparing energy and entropya. The total energy of the universe remains constanti. ΔEsys + ΔEsurr = ΔEuniv= 0b. The total entropy of the universe increases for a spontaneous processi. For a spontaneous process: ΔSuniv>0ii. For a nonspontaneous process :ΔSuniv<01. Where ΔSuniv= ΔSsys+ ΔSsurrII. Predicting The Sign of ΔS°sysa. We can often predict the sign of S° for processes that involve a change in the number of moles of gasi. ΔS°sys is positive if the amount of gas increasesii. ΔS°sys is negative if the amount of gas decreasesiii. ΔS°sys is likely to be positive if a new structure forms that has more freedom of motion (ie changes in the state of matter)b. Example: CH4(g) +2O2(g) CO2(g) +2H2)(l); ΔHrxn= -890kJ/moli. ΔS°sys<0 indicating the entropy of the system decreasesIII. Applications of the Second Law of ThermodynamicsThese 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. The sign of ΔS°sys for a reaction does not, by itself, predict the direction of a spontaneous reactionb. If we consider both the system and the surroundings, we find that all real processes occur spontaneously in the direction that increases the entropy of the universec. For a process to be spontaneous, a decrease in the entropy of the system must be offset by a larger increase in the entropy of the surroundings or vic versaIV. Entropy changes in the surroundingsa. A decrease in the entropy of the system is outweighed by an increase in the entropy of the surroundingsb. The surroundings function as a heat source or heat sinkc. In an exothermic process, the surroundings absorb the heat released by the system, and Ssurr increasesi. qsys<0; qsurr>0 and ΔSsurr>0d. In an endothermic process the surroundings provide the heat absorbed by the system and Ssurr decreasesi. qsys>0; qsurr<0 and ΔSsurr<0V. Calculating entropy changesa. The standard entropy of a reaction ΔS°rxn is the entropy change that occurs whenall reactants and products are in their standard statei. Since entropy is a state function we can calculate the ΔS°sys in a similar way as calculating ΔHrxn1. State function meaning it’s based entirely on the chemical or physical state (or both), but not on how it reached that stateii. ΔS°rxn=ΣmS°products- ΣnS°reactants1. Where m and n are the amounts (mol) of products and reactants, given by the coefficients in the balanced equationVI. Temperature at which Heat is Transferreda. Since entropy depends on temperature, ΔS°surr is also affected by the temperatureat which heat is transferredb. For any reaction qsys=-qsurr; the heat transferred is specific for the reaction and is the same regardless of the temperature of the surroundingsc. The impact on the surroundings is larger when the surroundings are at lower temperature, because there is a greater relative change in Ssurri.ΔSsurr=−qsysT∨ΔS
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