Chem 113 1st Edition Lecture 22 Outline of Last Lecture I. The Simplifying AssumptionII. Correlating Reaction Direction with ΔG°III. Manipulating the relationship between ΔG and QOutline of Current Lecture IV. ΔG, Q, and KV. ΔG, Equilibrium, and Reaction DirectionVI. ΔG and the Equilibrium ConstantVII. The Effect of a Change in TemperatureVIII. Temperature and KCurrent LectureI. ΔG, Q, and Ka.ΔG=RTlnQK=RTlnQ−RTlnKb. If Q and K are very different, ΔG has a very large value (positive or negative). The reaction releases or absorbs a large amount of free energyc. If Q and K are nearly the same, ΔG has a very small value (positive or negative). The reaction releases or absorbs very little free energyII. ΔG, Equilibrium, and Reaction Directiona. A reaction proceeds spontaneously to the right if Q<K; lnQK<0 and ΔG<0b. A reaction proceeds spontaneously to the left if Q>K; lnQK>0 and ΔG>0c. A reaction is at equilibrium if Q=K; QK=1 so lnQK=0 and ΔG=0III. ΔG and the Equilibrium ConstantThese 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. For standard state conditions, Q=1 and ΔG° = -RT lnKb. A small change in ΔG° causes a large change in K, due to their logarithmic relationshipc. As ΔG° becomes more positive, K becomes smallerd. As ΔG° becomes more negative, K becomes largerIV. The Effect of a Change in Temperaturea. Heat is a product in an exothermic reaction (ΔH°rxn < 0).b. Heat is a reactant in an endothermic reaction (ΔH°rxn > 0).c. An increase in temperature adds heat, which favors the endothermic reaction, and shifts equilibrium to the right.d. A decrease in temperature removes heat, which favors the exothermic reaction, and shifts equilibrium to the right.V. Temperature and Ka. The only factor that affects the value of K for a given equilibrium system is temperatureb. For a reaction with ΔH°rxn>0 (endothermic), an increase in temperature will causethe K to increasec. For a reaction with ΔH°rxn<0 (exothermic) an increase in temperature will cause the K to decreased. Van’t Hoff Equationi.lnK2K1=−∆ H °rxnR(1T2−1T2)ii. K1 is the equilibrium constant at T1e. Combining equations: -RTlnK = ΔH° - TΔS°i. In a plot of lnK vs. 1/T, the slope of the line is –
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