2. Standar... 1req Question O Question Question 3. AS surrou 1req 4. AG° =A... 1req 5. AG: Pre... 1req Consider the reaction: P4010 (8) + 6H2O(l) → 4H3PO4(aq) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.80 moles of P4010 (8) react at standard conditions. J 6. AG: Ent... 1req Substance STO K.mol 7. AG fro... 1req H3PO4(aq) 158.0 P4010(s) 228.9 8. Calculat... 1req H2O(l) 69.9 9. Calculat... 1req AS° = system J/K 10. Calcul... 1req 2. Standar... 1req For the reaction 3. AS surrou... 1req 4. AG° =A... 1req Question Question Question 5. AG: Pre... 1req Fe2O3 (8) + 3H2(g) → 2Fe(s) + 3H2O(g) AH 98.8 kJ and AS° = 142.5 J/K The standard free energy change for the reaction of 1.74 moles of Fe2O3(s) at 330 K, 1 atm would be KJ. This reaction is favored under standard 6. AG: Ent... 1req conditions at 33 ✓ 7. AG fro... 1req Assume that AH° and 8. Calculat... 1req As are indepen reactant product 9. Calculat... 1req 10. Calcul... 1req Submit Answer Retry Entire Group S more group attempts Temammmg

2. Standar... 1req Question O Question Question 3. AS surrou 1req 4. AG° =A... 1req 5. AG: Pre... 1req Consider the reaction: P4010 (8) + 6H2O(l) → 4H3PO4(aq) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.80 moles of P4010 (8) react at standard conditions. J 6. AG: Ent... 1req Substance STO K.mol 7. AG fro... 1req H3PO4(aq) 158.0 P4010(s) 228.9 8. Calculat... 1req H2O(l) 69.9 9. Calculat... 1req AS° = system J/K 10. Calcul... 1req 2. Standar... 1req For the reaction 3. AS surrou... 1req 4. AG° =A... 1req Question Question Question 5. AG: Pre... 1req Fe2O3 (8) + 3H2(g) → 2Fe(s) + 3H2O(g) AH 98.8 kJ and AS° = 142.5 J/K The standard free energy change for the reaction of 1.74 moles of Fe2O3(s) at 330 K, 1 atm would be KJ. This reaction is favored under standard 6. AG: Ent... 1req conditions at 33 ✓ 7. AG fro... 1req Assume that AH° and 8. Calculat... 1req As are indepen reactant product 9. Calculat... 1req 10. Calcul... 1req Submit Answer Retry Entire Group S more group attempts Temammmg

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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The equilibrium constant of the reaction COCl2(g) CO(g) + Cl2(g) was found to fit the expression ln K = + 1.75 ln T(K) – 2.667 Find the degree of dissociation of COCl2(g) at 227°C when the total pressure is 2 bar. Calculate the standard reaction enthaply and standard reaction entropy at 227°C.
Consider the reaction 3Fe,O3(s) + H2(g)- →2FE3O4(s) + H20(g) for which AH° = -6.000 kJ and AS° = 88.70 J/K at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.575 moles of Fe,O3(s) 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'.
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For the reaction S(s,rhombic) + 2co(g)----> so2(g) +2c(s, graphite) Delta H =75kj and delta s= -167.6j/k The equilibrium constant for this reaction at 298.0 k is ----- Assume that delta h and delta s are independent of temperature.
Consider each of the following reactions and decide whether the standard entropy change for the reaction will be, (a) Negative; or (b) Positive; or (c) Very close to Zero. Cu (s) + 2 H2S04 (aq) = CuSO4 (aq) + 2 H20 (1) + SO2 (8) 2 Na (I) + H2 (g) = 2 NaH (s) 1. negative Mg (s) + CO2 (g) = MgO (s) + CO (g) 2. positive 2 H20 (I) = 2 H2 (g) + O2 (g) 3. very close to zero N2 (g) + 3 H2 (g) = 2 NH3 (g) NH3 (g) + HCI (g) = NH4CI (s) > > >
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Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical reaction: P4010 (s) + 6H₂O (1)→ 4H₂PO4(s) Round your answer to zero decimal places. x10 0 -/-/ K X ? Ś
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