(a)
Interpretation:
The value of
Concept Introduction :
The formula to calculate the change in entropy at constant pressure:
Here
Tf = Final Temperature
Ti = Initial Temperature
Cp,m = molar heat capacity at constant pressure
n = number of moles
(b)
Interpretation:
The value of
for the transformation H2O(s, 0 0C): H2O(g, 1000C) needs to be determined.
Concept Introduction :
The relation for change in entropy for phase change from solid to vapor state is as follows:
Here
Tfusion = Fusion Temperature
Tboiling = Boiling temperature
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Thermodynamics, Statistical Thermodynamics, & Kinetics
- Calculate 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_forwardDetermine the standard Gibbs free energy change, rG, for the reactions of liquid methanol, of CO(g), and ofethyne, C2H2(g), with oxygen gas to form gaseous carbondioxide and (if hydrogen is present) liquid water at298 K. Use your calculations to decide which of thesesubstances are kinetically stable and which are thermodynamically stable: CH3OH(), CO(g), C2H9(g), CO2(g),H2O().arrow_forwardWhat is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the synthesis of ammonia? 3H2(g) + N2(g) 2NH3(g)arrow_forward
- From data in Appendix J, estimate (a) the boiling point of titanium(IV) chloride. (b) the boiling point of carbon disulfide, CS2, which is a liquid at 25 C and 1 bar.arrow_forwardFor a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction C₂H(g) + H₂(g) = 2 CH₂(g) the standard change in Gibbs free energy is AG° = -69.0 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are PC₂H₂ = 0.350 atm, PH₂ = 0.500 atm, and PCH, = 0.950 atm? AG = kJ/molarrow_forwardAt 25 ∘C25 ∘C, the equilibrium partial pressures for the reaction 3A(g)+4B(g)↽−−⇀2C(g)+3D(g) were found to be PA=5.55 atm, PB=5.46 atm, PC=5.18 atm, and PD=5.93 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C? kJ/molarrow_forward
- 7. 2 KC1O3(s) → 2 KC1(s) + 3 O₂(g) 0 AGTxn= Please calculate the standard Gibbs free energy of reaction of the following chemical equation at 298 K. Substance AHO KJ mol 5⁰ ( mol K Substance J kJ Fe₂O3(s) CO₂(g) AH(K) -393.5 mol Smol K -822.16 89.96 sº (mlk) Substance SiC14 (1) kJ AH, mol 213.6 -640.1 239.3 H₂O(g) 0₂(g) -241.82 188.83 69.91 H₂O(l) K₂CO3(s) KC1O3(s) 205.0 -285.83 -1150.18 -391.2 -910.9 50.29 Please provider your answer below. 155.44 N₂(g) SiO₂ (s) CaCO3(s) CaCl₂(s) 191.50 -1207.1 92.88 143.0 -795.8 104.6 Fe(s) 27.15 KCl(s) -435.9 82.7 SO₂(g) -296.9 248.5 Zn(s) KCIO3(aq) NaCl(aq) 41.63 -124.73 310.0 SO3(g) C4H10(g) NH4NO3(s) N₂O(g) -395.2 -349.5 256.2 265.7 -365.6 151 HCl(g) -92.30 -407.1 186.69 115.5 81.6 220.0 ZnO(s) -348.0 43.9arrow_forwardFor a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction C2H6(g)+H2(g)↽−−⇀2CH4(g) the standard change in Gibbs free energy is ΔG°=−32.8 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are PC2H6=0.500 atm , PH2=0.100 atm , and PCH4=0.800 atm ?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_forward
- A box contains a gas composed of 12 atoms each corresponding to the different stable isotopes of Tin (Sn). Initially, there are 10 atoms in the right compartment and 2 in the left compartment. After a certain amount of time passes, the system attains equilibrium with 6 atoms on each compartment. What is the change in entropy of the system? Use the constant KB= 1.3 x 10-23 [J/K*Molecule] in the solutionarrow_forwardThe equilibrium constant for the following reaction is KP=10 at 960 K. Note that R=8.314 J/mol K 2SO2 (g) + O2 (g) ↔ 2SO3 (g) Calculate ΔrG° Calculate Q for the conditions PSO2=1e-3 bar, PO2=0.20 bar, PSO3=1e-4 bar. Calculate ΔrG for the conditions in part (b). Based on part (c), indicate which direction (forward or backward) will proceed spontaneouslyarrow_forwardFor a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO(g) + 0,(g) =2 NO,(g) the standard change in Gibbs free energy is AG° = -32.8 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are PNO = 0.500 atm, Po, = 0.350 atm, and PNO, = 0.800 atm?arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning