1. Given the thermodynamic data in the table below, calculate G° (at 298.15 K) for the reaction:2S02 (g) + 02 (g) 2S03 (g)Substance | AH (kJ/mol) | S° (J/mol · K)SO2 (g)02 (g)SO3 (g)-297249205-395256Note: HºUXIН (prods) -H° (reacts)S°(prods) - S° reacts)Go- TXnНо —S°

1. Given the thermodynamic data in the table below, calculate G° (at 298.15 K) for the reaction: 2SO2 (g) + 02 (g) 2S03 (g) Substance | AH (kJ/mol) | S° (J/mol · K) SO2 (g) 02 (g) SO3 (g) -297 249 205 -395 256 Note: H°, НО (prods) - Но (reacts) S°xn = S°(prods) - S° (reacts) H° - S° TXI

1. Given the thermodynamic data in the table below, calculate G° (at 298.15 K) for the reaction: 2SO2 (g) + 02 (g) 2S03 (g) Substance | AH (kJ/mol) | S° (J/mol · K) SO2 (g) 02 (g) SO3 (g) -297 249 205 -395 256 Note: H°, НО (prods) - Но (reacts) S°xn = S°(prods) - S° (reacts) H° - S° TXI

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol) H2O2(l) H2O2(g) Use the following thermodynamic information at 298 K to estimate this temperature. Substance: H2O2(l) H2O2(g) ΔH°f (kJ/mol): –187.7 –136.3 ΔG°f (kJ/mol): –120.4 –105.6 S° (J/K • mol): 109.6 232.7
Predict the normal boiling point of hydrogen peroxide (H2O2) from the following data: ΔHºf, kJ/mol Sº, J/(mol • K) H2O2(ℓ) –187.8 110 H2O2(g) –136.3 233 Tb = Based on the estimated boiling point, is H2O2 a liquid or a gas at room temperature (25°C)? Why?Choose one:A liquid, because its estimated boiling point is less than 25°C.A gas, because its estimated boiling point is less than 25°C.A liquid, because its estimated boiling point is greater than 25°C.A gas, because its estimated boiling point is greater than 25°C.
Calculate the standard free energy change (kJ/mol) for the reaction shown at a temperature of 298.15 K using the information provided. Reaction ΔHrxn (kJ/mol) ΔSrxn (J/mol-K) 2 N2(g) + 3 O2(g) → 2 N2O3(g) 167.4 -374.5
Ozone (O3) in the atmosphere is subject to reaction with nitric oxide (NO) as O3(g) + NO(g) -> NO2(g) + O2(g) From the following data, calculate the ΔGo (KJ) for the reaction at 25oC ΔHo = -199 kJ and ΔSo = -4.1 J/K
Calculate the ∆H°rxn for the reaction C(graphite) + 2H2(g) → CH4(g)from the following enthalpy changes: - Cgraphite + O2(g) → CO2(g) ∆H = -393.5 kJ - 2H2(g) + O2(g) → 2H2O(l) ∆H = -571.6 kJ - CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = -890.4 kJ
The synthesis of the pesticide thionyl chloride (OSCI2) proceeds according to the following reaction: SO3(g) + SCI2(e) = OSCl2(e) + SO2(g) If AG° is -106 kJ/mol for this reaction at 298K, what is the Keg at this temperature? Recall that R is 8.314 J mol-1 K-1. A) 7.21 x 1012 B 3.50 x 1014 c) 3.81 x 1018 D) 1.19 x 1020
Species So NH3 111.3 O2 205.0 CH4 186.1 HCN 201.7 H2O 70.0 Calculate the ΔSo for the reaction: 2 NH3 + 3 O2 + 2 CH4 ⟶ 2 HCN + 6 H2O T = 25o CExpress the answer to 1 decimal point.
Calculate the change of the entropy of the system in J/K for the following phase transition: condensation of 64.1 g methanol The molar mass of methanol (CH3OH) is 32.04 g/mo%; the enthalpy of vaporization of methanol is 37.6 kJ/mol; the boiling point of methanol is 64.7°C A) -3640 B) -1130 C) 1.13446677 D) -223 E) 0.22258
The element oxygen was prepared by Joseph Priestley in 1774 by heating mercury(II) oxide: HgO(s) → Hg() + 1/2O2(g), AH = 90.84 kJ/mol. Estimate the temperature at which this reaction will become spontaneous under standard state conditions. S'(Hg) = 76.02 J/K mol S(O2) = 205.0 J/K mol C S(HgO) = 70.29 J/K mol 11
Given the below thermodynamic data, calculate ΔΔH for the reaction CH4 (g) + NH3 (g) ⟶⟶ HCN (g) + 3 H2 (g): N2 (g) + 3 H2 (g) ⟶⟶ 2 NH3 (g) ΔΔH = -85.6 kJ C (s) + 2 H2 (g) ⟶⟶ CH4 (g) ΔΔH = -75.2 kJ H2 (g) + 2 C (s) + N2 (g) ⟶⟶ 2 HCN (g) ΔΔH = +271.4 kJ
Calculate the ΔGo for X, in kJ, from the data given below at 327 K: 2X(g) → X2(g) So (J/mol-K) ΔHo (kJ/mol) X(g) 51.2 77.4 X2(g) 61.4 171.1
Given the data in the table below, AH° rxn IF5 (g) + F₂ (g) →→ IF7 (g) f Substance (kJ/mol) IF 5 (g) -840 IF7(g) -941 kJ. more information is needed 1801 -121 -101 121 is ΔΗ° for the reaction