Solid iodine, I2(s), has an enthalpy of sublimation AHsubl = 62.4 kJ mol. Suppose 2.61 g of iodine at a constanttemperature of 29.5 °C is allowed to sublime into the atmosphere at a temperature of 44.9 °C.Your AnswerCorrect Answer (Used)-1Calculate the entropy change of the system (i.e. the iodine) in J K2.120JK!Calculate the entropy change of the surroundings in J K.JK1iSave for LaterAttempts: 0 of 3 usedSubmit Answer

Solid iodine, I2(s), has an enthalpy of sublimation AHsubl = 62.4 kJ mol. Suppose 2.61g of iodine at a constant temperature of 29.5 °C is allowed to sublime into the atmosphere at a temperature of 44.9 °C. Your Answer Correct Answer (Used) Calculate the entropy change of the system (i.e. the iodine) in J K!. JK! 2.120 Calculate the entropy change of the surroundings in JK | JK1. Save for Later Attempts: 0 of 3 used Submit Answer

Solid iodine, I2(s), has an enthalpy of sublimation AHsubl = 62.4 kJ mol. Suppose 2.61g of iodine at a constant temperature of 29.5 °C is allowed to sublime into the atmosphere at a temperature of 44.9 °C. Your Answer Correct Answer (Used) Calculate the entropy change of the system (i.e. the iodine) in J K!. JK! 2.120 Calculate the entropy change of the surroundings in JK | JK1. Save for Later Attempts: 0 of 3 used Submit Answer

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 these two reactions: Reaction 1: 20F2 (g) →02(g) + 2 F2(g) AH= -49.4 kJ Reaction 2: N2 (g) + 202(9) → 2NO2(g) AH= +66.4 kJ Explain which reaction is spontaneous and justify your answer in terms of change in enthalpy and change in entropy. Edit View Insert Format Tools Table hp $ & 4 7 8. 9. r y f j k
i need help with these pleasee Consider the reaction: 2NH3(g) +202(g) → N₂O(g) + 3H₂O(1) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.64 moles of NH3(g) react at standard conditions. ΔS Substance AH (kJ/mol) NH3(g) As O₂(g) N₂O(g) H₂O (1) surroundings Consider the reaction: system -46.1 0.0 82.1 -285.8 H₂CO(g) + O2(g) → CO₂(g) + H₂O(1) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.63 moles of H₂CO(g) react at standard conditions. J/K J/K Substance Sº (J/K mol) H₂CO(g) 219.0 0₂ (9) 205.1 CO₂(g) 213.7 H₂O(l) 69.9 SUVIL
4) The heat capacity of CO2(g) between 30 C and 100 C is given by the following equation Cp = 26.65+ 42.3x10-3 T/K-142.4X10 -7 T 2/K 2 in J K -1 mol -1 . a )Calculate the enthalpy change and b) the entropy change between 30C and 100 C.
The heat of vaporization AH, of heptane (C-H₁6) is 31.2 kJ/mol. Calculate the change in entropy AS when 736. g of heptane condenses at 98.4 °C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits. x10 ロ・ロ X H ■
# Consider the reaction 20 E D 2NH3(g) + 3N₂O(g) → 4N2(g) + 3H₂O(g) for which AH = -879.5 kJ and AS = 288.1 J/K at 298.15 K. 1. Calculate the entropy change of the UNIVERSE when 1.642 moles of NH3 (9) react under standard conditions at 298.15 K. $ Submit Answer 4 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'. 000 000 R F do L % [Review Topics] [References] Use the References to access important values if needed for this question. 5 O Retry Entire Group T Cengage Learning Cengage Technical Support G ‹ 6 8 more group attempts remaining Y H & 7 U 8 J 1 ( 9 K O O L P > Previous Email Instructor Save and Exit + 11 ( Next ? 11 } ] d
Mark the correct statements. (Select all that apply.) O [The reaction 2H20(g) → 2H2(g) + O2(g) is expected to have a positive value of ASº.] O [If a sample of pure ozone changed to oxygen at constant T and P, the entropy of the sample would decrease.] O [For the process H20(1) → H20(s) occurring at 0°C and 1 atm, the sign of AS is negative.] O [The value of AS for the process CHC13(1) – CHC13(g) is negative.] O [The reaction CH4(g) + 202(g) → CO2(g) + 2H2O(g) has a negative value of ASsurr-]
The heat of fusion AH, of acetic acid (HCH₂CO₂) is 11.7 kJ/mol. Calculate the change in entropy AS when 5.7 g of acetic acid melts at 17.0 °C. Be sure your answer contains a unit symbol. Round your answer to 2 significant digits.
A chem.. ngineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 52.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. AH = 181. kJ J AS = 557. K N, (g) + 0, (g) → 2NO (g) AG = N, (g) + 0,(8) kJ Which is spontaneous? this reaction the reverse reaction neither ΔΗ - - 1353. kJ J AS = K 2CH,OH (g) + 30,(g)…
Hint: No calculations are required. For the reaction 2H₂O(1) 2H₂(g) + O₂(g) AH° 572 kJ and AS° = 327 J/K At standard conditions, this reaction would be product favored O at all temperatures. at no temperature. O at relatively low temperatures. O at relatively high temperatures.
M Preq 3 Consider the reaction 00 E D 4HCI(g) + O₂(g) 2H₂O(g) + 2Cl₂(8) for which AH° = -114.4 kJ and AS° = -128.9 J/K at 298.15 K. J/K (1) Calculate the entropy change of the UNIVERSE when 1.775 moles of HCI(g) react under standard conditions at 298.15 K. ASuniverse (2) Is this reaction reactant or product favored under standard conditions? Submit Answer (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'. 4 000 000 R F DO LO V % 5 [Review Topics] [References] Use the References to access important values if needed for this question. T Retry Entire Group 8 more group attempts remaining Cengage Learning Cengage Technical Support G 6 B Y H & 7 N U 1 * 8 J M 9 K बैं O < L P Previous Email Instructor Save and Exit { + 11 Next ? } ]
Try Again Your answer is wrong. • Second reaction: Your answer for A Sis incorrect. A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 88.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 2Fe₂O3(s) 4Fe(s) + 30₂ (g) ΔΗ = 1648. kJ AS = 4631. AG = J K -24 kJ Which is spontaneous? this…