CHEM-1112 Lecture Support Chapter XI Entropy Calculations Jan 11/25 PDF
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This document is a set of lecture notes on thermodynamics, specifically focusing on entropy calculations for various chemical processes. It explores the concepts in detail using examples such as the melting of ice, freezing of water, and boiling of water, and provides calculations for the changes in entropy of the water and the related apparatus. The document is well-organized and includes clear explanations along with calculations to demonstrate the processes.
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## CHEM-1112 Lecture Support, Chapter XI ### Example 11-4: Will Ice Spontaneously Melt **The entropy change for the process:** H₂O(s) -> H₂O(l) = 22.1J/K This requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at -10.00°C? + 10.00°C? #### Solution...
## CHEM-1112 Lecture Support, Chapter XI ### Example 11-4: Will Ice Spontaneously Melt **The entropy change for the process:** H₂O(s) -> H₂O(l) = 22.1J/K This requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at -10.00°C? + 10.00°C? #### Solution We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔS<sub>univ</sub> is positive, then the process is spontaneous. At both temperatures, ΔS<sub>sys</sub> = 22.1 J/K **At -10.00ºC (263.15 K):** ΔS<sub>surr</sub> = -6.00kJ/T ΔS<sub>univ</sub> = ΔS<sub>sys</sub> + ΔS<sub>surr</sub> = ΔS<sub>sys </sub>+ q<sub>surr</sub>/T = 22.1J/K + (-6.00 x10^3 J/263.15K) = -0.7J/K ΔS<sub>univ</sub> < 0, so melting is nonspontaneous at -10.00°C **At 10.00ºC (283.15 K):** ΔS<sub>univ</sub> = ΔS<sub>sys</sub> + ΔS<sub>surr</sub> = ΔS<sub>sys</sub> + q<sub>surr</sub>/T = 22.1J/K + (-6.00 x 10^3 J/283.15K) = +0.9J/K ΔS<sub>univ</sub> > 0, so melting is spontaneous at 10.00°C ### Entropy Change for Melting of Ice Courtesy Dr. Desiree Vanderwell 2024 & the Web Example 5.4.2: Entropy change for melting ice **Sample Calculation** Ice melts at 0.0°C (273.15 K) with a ΔH<sub>fus</sub> = 6.01 kJ/mol. What is the entropy change for the melting of ice when 1.0 mol of ice melts at 0.0°C? **NOTE:** ΔH<sub>fus</sub> = "Enthalpy of fusion" = "Heat of Fusion" * The amount of heat transferred from the surroundings when a substance undergoes a phase change from solid to liquid (ie, "melts"), * The opposite of "heat-of-fusion" = "heat of solidification" **Entropy change for a phase change at constant pressure:** ΔS = q/T = ΔH<sub>phase</sub>/T ΔS = (1 mol) (6010J/mol) / 273.15 K = 22 J/K ### Entropy Change for Water & Freezer What is the total entropy change when 5.00 mol of water (ΔH<sub>fus</sub> = 6.01kJ/mol) freezes at 0.0°C in a freezer compartment whose temperature is held at -10°C? #### Solution Total Entropy change includes: * ΔS for the water * ΔS for the freezer compartment When water freezes, heat flows from water to surroundings (ie, the freezer compartment) Thus, q = (-ve) for water (& its entropy decreases) At the same time: q = (+ve) for the freezer compartment (& its entropy increases) Identify T for water: (Water-ice mix; T=0.0°C = 273.15 K). Constant while the phase transition occurs Eventually, the ice will cool to -10°C. But the question is only for the freezing process. The freezer is at constant T = 10°C (± 263.15 K) Given: * n = 5.00 mol * ΔH<sub>fus</sub> = 6.01 x 10^3 J/mol = Amount of heat required for the phase conversion of 1 mol of substance. For the phase change: q = nΔH<sub>fus</sub> Then calculate the entropy changes for these constant-temperature processes: ΔS = q/T ΔH<sub>fus</sub> = (+ve) & melting of Ice **NOTE:** T = Kelvin Water-ice mixture releases heat as water freezes: q for water has (-ve) value q<sub>H₂O</sub> = (5.00 mol) (6.01 x 10^3 J/mol) = -30,050 J ΔS<sub>H₂O</sub> = q<sub>H₂O</sub> / T<sub>H₂O</sub> = -30,050J / 273.15K = -110. 0 J/K Freezer compartment absorbs heat released by the water (but does so at its temp = -10°C + 273 = 263.15 K Because freezer absorbs heat: q<sub>freezer</sub> = -q<sub>H₂O</sub> = + 30,050 J ΔS<sub>freezer</sub> = q<sub>freezer</sub> / T<sub>freezer</sub> = 30,050 J / 263.15 K = 114.19 J/K #### Total Entropy Change ΔS<sub>total </sub>= ΔS<sub>H₂O</sub> + ΔS<sub>freezer</sub> = (-110.0 J/K) + (114.19 J/K) = +4.19 J/K Indicates spontaneous Entropy change of ice-water mix (-110.0 J/K) has (-ve) sign because water is more dispersed in the liquid than in the solid state. Entropy change for the freezer (114.19 J/K) has (+ve) sign because heat is being absorbed by the freezer. This increases the dispersal of energy. **NOTE:** ΔS<sub>total</sub> > 0. This must be the case for a spontaneous process. ### Changes in Entropy for Boiling Water, Pot, Hot Plate, & Overall A pot on a hot plate at 300°C boils off 5.0 mL water. Calculate the changes in entropy for the water and for the hot plate, & the overall entropy Given: * Enthalpy of vaporization of water = 40.79 kJ/mol * Boiling point of water = 100°C #### Solution: ΔS = q/T For vaporization of H₂O Enthalpy q = nΔH<sub>vap</sub> = (5.0 mol) (40.79 kJ/mol) = 203.95 kJ T<sub>vap</sub> = 100°C = (100°C + 273.15) K = 373.15 K ΔS<sub>vap</sub> = 203.95 kJ / 373.15 K = 0.547 kJ/K **Hot plate supplies the heat**: q<sub>hot plate</sub> = -(203.95 kJ) ΔS<sub>hot plate</sub> = -203.95 kJ/(300°C + 273.15) K = -203.95 kJ/573.15 K = -0.356 kJ/K ΔS<sub>total</sub> = (0.547 - 0.356) kJ/K = 0.191 kJ/K ### Calculate the ΔS<sub>rxn</sub> for the Synthesis of NH₃ N<sub>2</sub>(g) + 3H<sub>2</sub>(g) -> 2NH<sub>3</sub>(g) **Solution**: ΔS<sub>rxn</sub> = ∑νS<sub>products </sub>- ∑νS<sub>reactants</sub> = 2mol(S<sub>NH₃</sub>) - [1mol(S<sub>N₂</sub>)+ 3mol(S<sub>H₂</sub>)] = 2mol(192.80 J/mol/K) - [1mol (191.61 J/mol/K) + 3mol (130.68 J/mol/K)] = 2(192.8 J/K) - [1(191.61 J/K) + 3(130.68 J/K)] = 385.6 J/K - [191.61 J/K + 392.04 J/K] = 385.6 J/K - 583.65 J/K = -198.05 J/K
