Consider the reaction:2NH3 (9) + 3N2O(g) → 4N2 (g) + 3H₂O(g)Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3(g) react atstandard conditions.AS⁰Substance AH (kJ/mol)NH3 (9)N₂O(g)N₂ (9)H₂O(g)surroundings-46.182.10.0-241.8J/K Hint: No calculations are required.For the reaction6CO₂(g) + 6H₂O(l)→→→C6H₁2O6 + 60₂(g)AH° = 2.80×10³ kJ and AS° = -259 J/KAt standard conditions, this reaction would be product favoredat relatively high temperatures.at no temperature.at all temperatures.Oat relatively low temperatures.

Answered: Image /qna-images/answer/c04fc48e-6d16-48ec-94c5-c7e6fc21fc85.jpg

Consider the reaction: 2NH3 (9) + 3N2O(g)→ 4N2(g) + 3H₂O(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3 (g) react at standard conditions. A.S⁰ Substance AH (kJ/mol) NH3 (9) -46.1 N₂O(g) 82.1 N₂ (9) 0.0 H₂O(g) -241.8 surroundings J/K

Consider the reaction: 2NH3 (9) + 3N2O(g)→ 4N2(g) + 3H₂O(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.54 moles of NH3 (g) react at standard conditions. A.S⁰ Substance AH (kJ/mol) NH3 (9) -46.1 N₂O(g) 82.1 N₂ (9) 0.0 H₂O(g) -241.8 surroundings J/K

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...

See similar textbooks

Similar questions

Consider the reaction 2NO(g) + O2(g)–→2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.82 moles of NO(g) react at standard conditions. AS surroundings ] J/K
Without doing any calculations, determine the sign of AS and ASu for each chemical reaction. In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction is spontaneous. a. 2 CO(g) + O2(g) 2 CO,(g) AHn = -566.0 kJ b. 2 NO,(g) 2 NO(g) + O,(g) AHan = +113.1 kJ c. 2 H,(g) + 0,(g) 2 H,0(g) d. CO,(g) - C(s) + O2(g) AHn = -483.6 kJ AHsn = +393.5 kJ Exn
Consider the reaction described by the chemical equation shown. 3 C₂H₂(g) → CH (1) AHxn=-633.1 kJ Use the data from the table of thermodynamic properties to calculate the value of ASixn at 25.0 °C. ASixn = Calculate AGixn. AGixn= In which direction is the reaction, as written, spontaneous at 25 °C and standard pressure? forward reverse both neither J.K-¹ kJ
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.
Consider the reaction: 2H₂(g) + O₂(g) → 2H₂O(g) Using standard thermodynamic data at 298 K, calculate the free energy change when 2.08 moles of H₂ (g) react at standard conditions. Substance AG (kJ/mol) H₂O(g) -228.6 H₂(g) 0.0 O₂(g) 0.0 AGO =-475.488 rxn kJ
Chemistry Above what temperature does the following reaction become non-spontaneous? 2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g) AH = -1036 kJ; AS = -153.2 J/K O 158.7 K This reaction is nonspontaneous at all temperatures. O 298 K This reaction is spontaneous at all temperatures O 6.762 × 103 K
Chemistry
QUESTION 12 Which of the following is false about the second law of thermodynamics? O It is not possible to construct a perpetual motion machine which can work endlessly without the expenditure of energy. O It is impossible to transfer heat from a heat source to a hotter body without doing some work. O t is not possible to take heat from a source and convert all of it into work without losing some of it to a colder reservoir. O The entropy fo the universe is always increasing.
eq M M IS M ots M M M # 3 E D Consider the reaction 2H₂O(l) 2H₂(g) + O₂(g) The standard free energy change for this reaction is 474.2 kJ. The free energy change when 1.88 moles of H₂O(1) react at standard condition is kJ. 20 C What is the maximum amount of useful work that the reaction of 1.88 moles of H₂O(l) is capable of producing in the surroundings under standard conditions? If no work can be done, enter none. kJ Submit Answer $ 4 R F > % 5 T [References] Use the References to access important values if needed for this question. Retry Entire Group 9 more group attempts remaining G Cengage Learning Cengage Technical Support 6 B MacBook Pro Y Topics] H & 7 N U J ▶II * 00 8 M I ( 9 K. O < 96 ) L P V t Previous Email Instructor Save and Exit { + 11 [ Next ? } 1 dele
Methanol is burned as fuel in some race cars. This makes it clear that the reaction is spontaneous once methanol is ignited. Yet the entropy change for the reaction 2 CH3 OH (l) + 3 O2 (g) to 2 CO2 (g) + 4 H2 O(l) is negative. Why doesn't this violate the second law of thermodynamics?
2 NH3 (g) + 2 O2 (g) → N20 (g) + 3 H2O (g) Is this reaction spontaneous only at low temperatures, only at high temperatures, at all temperatures? AH°r, kJ/mol s°, J/mol·K NH3(g) -46.1 192.5 0:(g) 205.0 N 20 (g) 82.1 219.9 H2O(g) |-241.8 188.7 AH°i, kJ/mol s°, J/mol·K What is the lowest temperature at which the following reaction is spontaneous? H2O (g) + C (s) → H2 (g) + CO (g) Co(g) H2(g) C (s) -110.5 | 197.7 130.6 5.7 H2O(g) |-241.8 188.7
Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction A B с observations Crystallization of a pure compound is spontaneous only below 23. °C. This reaction is spontaneous only above 31. °C but proceeds at a faster rate above 83. °C. The reverse of this reaction is always spontaneous but proceeds faster at temperatures above 108. °C. AH is AS is AH is AS is conclusions ΔΗ is AS is (pick one) (pick one) (pick one) (pick one) (pick one) (pick one) V