22. Although the standard molar entropy, (AS), of a lot of chemicals are available, (AS) of most chemicals, especially the newly discovered chemicals are NOT available, you have to calculate the values of the (AS). The following table lists the standard molar entropy of a few compounds. The following reaction occurs when sucrose (cane sugar) is metabolized by the body. 0 C12H22O11(s) + 12 O₂(g) → 12 CO₂(g) + 11 H₂O(1) ASn = 512.01 J/K What is the standard molar entropy, sº of sucrose solid? Substance sº mol K Substance so (mol K CO(g) 197.9 CO₂(g) 213.6 H₂O(g) 188.83 MgO(s) 69.91 26.8 0₂(8) H₂O(1) Mg(OH)2 (s) C(graphite, s) 63.24 205.0 5.69 J mol K N₂(g) 191.50

22. Although the standard molar entropy, (AS), of a lot of chemicals are available, (AS) of most chemicals, especially the newly discovered chemicals are NOT available, you have to calculate the values of the (AS). The following table lists the standard molar entropy of a few compounds. The following reaction occurs when sucrose (cane sugar) is metabolized by the body. 0 C12H22O11(s) + 12 O₂(g) → 12 CO₂(g) + 11 H₂O(1) ASn = 512.01 J/K What is the standard molar entropy, sº of sucrose solid? Substance sº mol K Substance so (mol K CO(g) 197.9 CO₂(g) 213.6 H₂O(g) 188.83 MgO(s) 69.91 26.8 0₂(8) H₂O(1) Mg(OH)2 (s) C(graphite, s) 63.24 205.0 5.69 J mol K N₂(g) 191.50

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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A heat source generates heat at a rate of 71.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 27.3 °C? Assume the heat transfer is reversible. AS = J K· h
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Calculate the standard reacton entropy at 298K of the following: 1. CH3CHO (g) + O2 -----→ CH3COOH (l) 2. Sucrose [C12H22O11(s)] + O2(g) ----→ CO2 (g) + H2O (l) 3. Zn(s) + Cu2+ (aq) ---→ Zn2+ (aq) + Cu (s)
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The decomposition of nitrogen pentoxide in the reaction 2N₂O5 (s)→ 4NO2(g) + O2(g) is endothermic with AH = +110 kJ/mol. Even so, this reaction is spontaneous for temperatures above 230 K because the change of a solid into a gas is a large increase of entropy.
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20. Although the standard molar entropy, (AS), of a lot of chemicals are available, (AS) of most chemicals, especially the newly discovered chemicals are NOT available, you have to calculate the values of the (AS). The following table lists the standard molar entropy of a few compounds. Trinitrotoluene (C₂H5N3Q6), abbreviated as TNT is one of the most famous explosives. Please use the above table and the standard entropy of reaction of the detonation reaction of TNT to find the Sº of TNT. 0 2 C₂H5N306(5) → 3 N₂(g) + 5 H₂O(g) + 7 CO(g) + 7 C(graphite,s) ASrxn = 488.9 J/K so of TNT = Substance So mol K Substance sº (mlk) J mol K CO(g) 197.9 CO₂(g) 213.6 H₂O(g) 188.83 MgO(s) 69.91 26.8 0₂(g) H₂O(1) Mg(OH)2 (s) C(graphite, s) 63.24 205.0 5.69 N₂(g) 191.50
Chemistry
Consider the reaction: NH4Cl(aq) →→→→NH3(g) + HCl(aq) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of NH4Cl(aq) react at standard conditions. AS° surroundings = J/K
Consider the reaction HCI(g) + NH3(g)- →NH4Cl(s) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.34 moles of HCI(g) react at standard conditions. ASO surroundings = J/K