Example: Estimate Keg at 400°C for• AS-198.3 J/KΟ ΔΗ° = -92.38 ΚJ• AG-33.32 kJrxneqN₂+ 3H2 → 2NH3• Keg at 298 K = 6.9 × 105eq

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Answered: Example: Estimate Keg at 400°C for •…

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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Consider the reaction 2NaOH(aq) + H₂(g)—2Na(s) + 2H₂O(1) for which AH° = 368.6 kJ and AS° = 15.30 J/K at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 2.290 moles of NaOH(aq) react under standard conditions at 298.15 K. ASuniverse = (2) Is this reaction reactant or product favored under standard conditions? J/K (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'.
Use the values you measured for Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) ΔH°1: -452kJ, and MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) ΔH°2: -133kJ, with the known ΔH°3 = ΔH°f(H2O, l) = –286 kJ/mol, to find your experimental ΔH°f(MgO), as: [f MgO] Mg(s) + ½O2(g) → MgO(s) ΔH°f(MgO) = ΔH°1 – ΔH°2 + ΔH°3 = ______ kJ/mol
Using values from Appendix C of your textbook, calculate the value of Keq at 298 K for each of the following reactions: (a) 6 C(graphite) + 6 H₂O(1) C6H1206(5) Keq=1.1E- (b) 2 NO(g) + Cl₂(g) = 2 NOCI(g) Kea (c) 2 N₂Ha(g) + 2 H₂O(1) = 4 NHy(9) + O₂(g) Kea
Calculate ΔS (for the system) when the state of 4.00 mol diatomic ideal gas molecules, for which Cp,m = 7/2 R, is changed from 27°C and 3.0 atm to 157°C and 10.00 atm? R=8.314 J/Kmol
21. Given the following two equations CH5OH(1) + 3 O2(8) 2 СO2(g) + 3 Н20(1) AH = -1370 kJ С-Н(8) + 3 О2(s) → 2 CO2(g) + 2 H2O(1) AH = -1410 kJ Calculate AH for the reaction C2H4(8) + H2O(1) CH5OH(I) ΔΗ- a. -2780 kJ b. 2780 kJ с. 40 kJ d. 40 kJ
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reaction АТР(aq) + H,О() — ADP(aq) + HPO, (aq) for which AGan = -30.5 kJ/mol at 37.0 °C and pH 7.0. Calculate the value of AGxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.10 mM, and [HPO?-] = 5.0 mM. AGrxn = kJ/mol Is the hydrolysis of ATP spontaneous under these conditions? yes no
Gaseous SO2Cl2 reacts with liquid water to form hydrogen chloride gas and liquid sulfuric acid. Using data tabulated in Appendix G of your text, calculate the enthalpy change, in units of kJ/mol, for the formation of 2 mol of liquid sulfuric acid. The answer is no -139.
Given the information A+ B → 2 D AH° = -776.1 kJ AS = 353.0 J/K С — D AH° = 493.0 kJ AS° = -222.0 J/K calculate AG at 298 K for the reaction А + В — 2С AG° = kJ
Give a clear handwritten answer with explanation
Using data found in Appendix E of your textbook calculate the nonstandard emf for each of the following reactions if the concentration of each of the lons in these reactions is 0.0003 molar and everything else is standard (use 298 K for the temperature, R = 8.314 J/mol-K, and F = 96,485 C/mol): (a) 1 Sn4+ (aq) + 1 Hg(1) --> 1 Sn²+ (aq) + 1 Hg²+ (aq) (b) 2 Hg2+ (aq) + 1 Co(s) --> 1 Hg₂2+ (aq) + 1 Co²+ (aq) (c) 3 F₂(g) + 2 Cr(s) --> 6 F (aq) + 2 Cr³+ (aq) (d) 1 Ni²+ (aq) + 1 Ca(s) --> 1 NI(s) + 1 Ca²+ (aq) E = E = 11 E = E = V
Calculate ΔH°298 for the process Co3O4(s)⟶3Co(s) + 2O2(g)from the following information:Co(s) + (1)/(2)O2(g)⟶CoO(s) ΔH°298 = −237.9kJ3CoO(s) +(1)/(2)O2(g)⟶Co3O4(s) ΔH°298 = −177.5kJ
Consider the reactions that produce acetic acid: CHA (9) + CO2(g) → CH; CO2H(1) CH; OH(9) + CO(g) → CH3CO2H(1) Substance AH¡ (kJ/mol) S° (J/K • mol) and State CH4 (9) -75 186 CO2(9) -394 214 CH,ОН() -201 240 CO(g) -111 198 CH; CO,H(1) -484 160

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Example: Estimate Keg at 400°C for • AS-198.3 J/K Ο ΔΗ° = -92.38 ΚJ • AG-33.32 kJ rxn eq N₂+ 3H2 → 2NH3 • Keg at 298 K = 6.9 × 105 eq