MISSED THIS? Read Section19.8 (Pages 871-877); Watch IWE 19.6.Liquid methanol burns in oxygen to form gaseouscarbon dioxide and gaseous water.Substance H (kJ/mol) S° (J/(mol · K))CH, OH (1)-238.6126.8O₂(g)0205.2CO₂(g)-393.5213.8CO2(aq) -413.8117.6H₂O(g) -241.8188.8H₂O(1)-187.870.0▼Part Dwhere no represents the stoichiometric coefficients of the products, n, represents the stof the reactants, and Sº represents the standard entropies.AGxn=Calculate AG at 25°C.Express your answer in kilojoules to three significant figures.Submit=Provide Feedback15] ΑΣΦRequest AnswerPart E Complete previous part(s)[2(S° co₂(g)) + 4(S° H₂O(g))]- [2(S° CH₂OH()) + 3(S° 0₂(E[2(213.8 J/K)+4(188.8 J/K)] - [2(126.8 J/K) + 3(2314 J/KMacBook ProP Pearson?kJCopyright © 2023 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions Contac Gibbs free energy (G) is a measure of thespontaneity of a chemical reaction. It is thechemical potential for a reaction, and is minimizedat equilibrium. It is defined asG=H-TSwhere H is enthalpy, T is temperature, and Sisentropy.The chemical reaction that causes iron to corrodein air is given by4Fe +302 2Fe2O3in which at 298 K, AH = -1684 kJ andAS-543.7 J/K.AGrxn = 302 kJSubmitPart C✓ CorrectThis positive value for AG means that, at extremely high temperatures, the corrosion reaction will actually greverse, converting iron(III) oxide into iron and oxygen.Teq =Previous AnswersAt what temperature Teq do the forward and reverse corrosion reactions occur in equilibrium?Express your answer as an integer and include the appropriate units.► View Available Hint(s)SubmitProvide FeedbackμÃValueP PearsonUnitsReview | Constants I Periodic Ta?5 of 20Copyright © 2023 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | Contact Us |

Since part C and D contains different questions, we solve only part C which is first question…

MISSED THIS? Read Section 19.8 (Pages 871-877); Watch IWE 19.6. Liquid methanol burns in oxygen to form gaseous carbon dioxide and gaseous water. Substance H° (kJ/mol) S° (J/(mol-K)) CH, OH (1) -238.6 126.8 O₂(g) 0 205.2 213.8 117.6 CO₂(g) -393.5 CO,(aq) -413.8 H₂O(g) -241.8 H₂O(1) -187.8 188.8 70.0 ▼ Part D where no represents the stoichiometric coefficients of the products, n, repres of the reactants, and Sº represents the standard entropies. Calculate AG at 25°C. Express your answer in kilojoules to three significant figures. AGixn= Submit 5. ΑΣΦ 1412 314 J/K Request Answer Part E Complete previous part(s) [2(126.8 J ? kJ

MISSED THIS? Read Section 19.8 (Pages 871-877); Watch IWE 19.6. Liquid methanol burns in oxygen to form gaseous carbon dioxide and gaseous water. Substance H° (kJ/mol) S° (J/(mol-K)) CH, OH (1) -238.6 126.8 O₂(g) 0 205.2 213.8 117.6 CO₂(g) -393.5 CO,(aq) -413.8 H₂O(g) -241.8 H₂O(1) -187.8 188.8 70.0 ▼ Part D where no represents the stoichiometric coefficients of the products, n, repres of the reactants, and Sº represents the standard entropies. Calculate AG at 25°C. Express your answer in kilojoules to three significant figures. AGixn= Submit 5. ΑΣΦ 1412 314 J/K Request Answer Part E Complete previous part(s) [2(126.8 J ? kJ

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...

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Using the following information, calculate the ΔH∘rsn Δ H r s n ° for the reaction between ammonia gas and oxygen. 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l). NH3 -45.9 kJ/mol O2 0.0 kJ/mol NO 90.3 kJ/mol H2O -285.8 kJ/mol
Which one of the following reactions cannot proceed spontaneously as written at any temperature? Assume that ΔH°rxn does not depend on temperature. A. 2Cl2O(g) → 2Cl2(g) + O2(g) ΔH°rxn = –151 kJ B. 3O2(g) → 2O3(g) ΔH°rxn = +284 kJ C. COCl2(g) → CO(g) + Cl2(g) ΔH°rxn = +112 kJ D. 2C(s) + 2S(s) + O2(g) → 2COS(g) ΔH°rxn = –137 kJ E. N2O4(g) → 2NO2(g) ΔH°rxn = +171 kJ
1. A reaction with AH° = -64.0 kJ/ mole has K = 1.00 X 108 at 25.0 °C. What would be the value of K at 0.00 °C? 2. A reaction with AH° = 618.0 kJ/mole has K = 1.00 X 1060 at 25.0 °C. What would be the value of K at 500.0 °C?
Using H° data, calculate the standard enthalpy change for the reaction at 298K. 2 NH3(g) HNO3(aq) --> NO2(g) 2 H20(1) AH° (KJ/mol) -46.19 -206.6 33.84 -285.83 (Note: the table AH values are located directly below the corresponding compound, this is the number found in the Appendix)
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Calculate the ΔH°rxn for the following reaction. SiO2(s) + 4HCl(g) → SiCl4(g) + 2H2O(g) ΔH°f [SiO2(s)] = –910.9 kJ/mol; ΔH°f [SiCl4(g)] = –657.0 kJ/mol; ΔH°f [HCl(g)] = –92.3 kJ/mol; ΔH°f [H2O(g)] = –241.8 kJ/mol Group of answer choices –139.5 kJ –137.4 kJ –104.4 kJ 104.4 kJ 139.5 kJ
Calcium ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) Ca (s) 0 0 41.4 Ca (g) 178.2 144.3 158.9 Ca2+ (g) 1925.9 CaC2 (s) -59.8 -64.9 70.0 CaCO3 (s, calcite) -1206.9 -1128.8 92.9 CaCl2 (s) -795.8 -748.1 104.6 CaF2 (s) -1219.6 -1167.3 68.9 CaH2 (s) -186.2 -147.2 42.0 CaO (s) -635.1 -604.0 39.8 CaS (s) -482.4 -477.4 56.5 Ca(OH)2 (s) -986.1 -898.5 83.4 Ca(OH)2 (aq) -1002.8 -868.1 -74.5 Ca3(PO4)2 (s) -4126.0 -3890.0 241.0 CaSO4 (s) -1434.1 -1321.8 106.7 CaSiO3 (s) -1630.0 -1550.0 84.0 Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) C (s, graphite) 0 0 5.7 C (s, diamond) 1.9 2.9 2.4 C (g) 716.7 671.3 158.1 CCl4 (l) -135.4 -65.2 216.4 CCl4 (g) -102.9 -60.6 309.9 CHCl3 (l) -134.5 -73.7 201.7 CHCl3 (g) -103.1 -70.3 295.7 CH4 (g) -74.8 -50.7 186.3 CH3OH (g)…
What is the energy required to evaporate two moles of liquid water given the following equations? 2H₂(g) + O₂(g) → 2H₂O(g) 2H2(g) + O₂(g) → 2H₂O(1) -88,0 kJ -1055.2 KJ 1055.2 kJ O 44.0 kJ O88.0 kJ ΔΗ = -483.6 kJ ΔΗ = -571,6 kJ
Given the following at 25°C and 1.00 atm: 1/2 N2 (g) + O2 (g) → NO2 (g) AHO = 33.2 kJ N2 (g) + 2 02 (g) → N204 (g) AH: = 11.1 kJ Calculate the AHº for the reaction below at 25°C. 2 NO2 (g) – N204 (g) +44.3 kJ +11.0 kJ O -55.3 kJ +55.3 kJ
Use the AH°f and AHO rxn information provided to calculate AH°f for IF: kJ IF7(g) IF5(g) AHᵒf (kJ/mol) -900. -850. IF7(g) + 12(g) →→ IF5(g) + 2 IF(g) AH rxn = -87
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
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