(a)
Interpretation: The standard entropy of liquid lead needs to be detremined at 725 oC.
Concept Introduction:
For liquid, the molar entropy can be calculated under constant pressure as follows:
Here,
(b)
Interpretation: The change in enthalpy for the given transformation needs to be determined.
Concept Introduction:The change in enthalpy can be calculated in terms of molar heat capacity of liquid and solid lead as follows:
Here,
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Thermodynamics, Statistical Thermodynamics, & Kinetics
- What is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the explosive decomposition of TNT? Use your knowledge of TNT and the chemical equation, particularly the phases, to answer this question. (Thermodynamic data for TNT are not in Appendix G.) 2C7H5N3O6(s) 3N2(g) + 5H2O() + 7C(s) + 7CO(g)arrow_forwardCalculate the standard Gibbs free-energy change when SO3 forms from SO2 and O2 at 298 K. Why is sulfur trioxide an important substance to study? (Hint: What happens when it combines with water?)arrow_forwardAt 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 2 B(g) = C(g) + 2 D(g) were found to be PA = 4.25 atm, Pg = 5.05 atm, Pc = 4.13 atm, and Pp 4.94 atm. %3D What is the standard change in Gibbs free energy of this reaction at 25 °C? kJ molarrow_forward
- Lithium bromide has several applications, including air-conditioning systems, organic synthesis, and early 20th century sedatives. Lithium bromide vapour dissociates according to LiBr(g) →Li(g) + ½ Br2(g). At what temperature does the partial pressure of Li reach the value of 105 atm when the gas is heated at a constant total pressure of 1 atm? The expression for Gibbs free energy is AG (J) = 333,900-42.09T.arrow_forwardThe molar volume of a certain solid is 161.0 cm3 mol−1 at 1.00 atm and 350.75 K, its melting temperature. The molar volume of the liquid at this temperature and pressure is 163.3 cm3 mol−1. At 100 atm the melting temperature changes to 351.26 K. Calculate the enthalpy and entropy of fusion of the solid. E4B.5(b) The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 K, its melting temperature. The molar volume of the liquid at this temperature and pressure is 152.6 cm3 mol−1. At 1.2 MPa the melting temperature changes to 429.26 K. Calculate the enthalpy and entropy of fusion of the solid.arrow_forwardCalculate the entropy of vaporization for 3 moles of water at 25 °C given that Cp, m (H2O(l)) = 75.29 J/Kmol, ΔHvap Tb = 4.07 × 104 J/mol and Cp, m (H2O(g)) = 33.58 J/K. Give your answer to one decimal point and DO NOT indicate the units.arrow_forward
- The standard molar entropy of liquid water at 273.15 K is 65 J K−1 mol−1 , and that of ice at the same temperature is 43 J K−1 mol−1 . Calculate the change in chemical potential of liquid water and of ice when the temperature is decreased by 1.5 K from the normal melting point. Giving your reasons, explain which phase is thermodynamically the more stable at the new temperature.arrow_forwardAt 25 °C, the equilibrium partial pressures for the reaction 3 A(g) + 2 B(g) = C(g) + 2 D(g) were found to be PA 5.61 atm, PB = 4.31 atm, Pc = 5.66 atm, and P = 5.84 a %3D %3D What is the standard change in Gibbs free energy of this reaction at 25 °C? kJ AGixn molarrow_forwardConsider the reaction 6CO2 (g) + 6H₂O(l) → C6H12O6 + 602 (9) for which AH° = 2801 kJ and AS⁰ = − 259.0 J/K at 298.15 K. 1. Calculate the entropy change of the UNIVERSE when 2.021 moles of CO₂ (g) react under standard conditions at 298.15 K. ASuniverse J/K 2. Is this reaction reactant or product favored under standard conditions? 3. If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'.arrow_forward
- Estimate the change in the molar entropy of N2(g) when the temperature is lowered from 298 K to 273 K, given that Cp,m(N2) = 29.125 J K−1 mol−1 at 298 K.arrow_forwardCalculate the enthalpy of vaporization of SO2 at –25°C if the same at its boiling point (i.e. –10°C) be 5950 cal mol–1. Given for SO2, molar heat capacities in liquid and vapor phase are 206 cal K–1 mol–1 and 9.3 cal K–1 mol–1 respectively.arrow_forwardThe molar enthalpy of fusion of ice at 273.15 K and one atm is ΔfusHm (H2O)=6.01 kJ mol-1, andthe molar entropy of fusion under the same conditions is ΔfusSm (H2O)=22.0 J K-1 mol-1. Show that(a) ΔfusGm (H2O)=0 at 273.15 K and one atm, (b) ΔfusG,m (H2O) < 0 when the temperature is greaterthan 273.15 K, and (c) ΔfusGm (H2O) > 0 when the temperature is less than 273.15 K.arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co