Consider the reaction: СаСО3{8) —СаO(9) + CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.96 moles of CaC03(s) react at standard conditions. AS° system J/K

Consider the reaction: СаСО3{8) —СаO(9) + CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.96 moles of CaC03(s) react at standard conditions. AS° system J/K

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter17: Chemcial Thermodynamics

Section: Chapter Questions

Problem 17.70QE: What is the sign of the standard Gibbs free-energy change at low temperatures and at high...

See similar textbooks

Related questions

Q: Consider the reaction: CaCO3(s)CaO(s) + CO2(g) Using standard thermodynamic data at 298K,…

A: Given Temperature = 298K Standard enthalpy of formation values ΔH°f (CaCO3(s)) = -1206.9 kJ/mol…

Q: Consider the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) Using standard thermodynamic data at 298K,…

Q: Consider the reaction:H2(g) + C2H4(g)C2H6(g)Using standard absolute entropies at 298K, calculate the…

A: Given the chemical equation - The entropy change for the reaction of 1 mole of H2(g) can be…

Q: Consider the reaction: 2H2O(l)2H2(g) + O2(g) Using standard absolute entropies at 298K, calculate…

A: Given: 2H2O(l)→2H2(g) + O2(g) Temperature=298 K Number of moles of H2O=1.98 moles To find: Entropy…

Q: Consider the reaction: Fe2O3(s) + 3H2(g)2Fe(s) + 3H2O(g) Using standard absolute entropies at…

A: The standard absolute entropies of different components of the given system can be stated as:

Q: The heat of vaporization AH, of heptane (C,H,6) is 31.2 kJ/mol. Calculate the change in entropy AS…

A: The change in entropy (in kJ/K) can be calculated from the heat of vaporization ∆Hv by the following…

Q: Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.29…

A: We have to find S°system when 2.29 moles of NO(g) react at standard conditions.

Q: Consider the reaction: 2H2O(l)2H2(g) + O2(g) Using standard absolute entropies at 298K, calculate…

Q: Consider the reaction: N2(g) + 3H2(g)–→2NH3(g) Using standard absolute entropies at 298K, calculate…

Q: Consider the reaction: 4NH3(g) + 502(g)4NO(g) + 6H2O(g) Using standard absolute entropies at 298K,…

A: In this question, we want to determine the value of ∆S° system when moles of NH3 gas is equal to…

Q: [References] Use the References to access important values if needed for this question. Consider the…

A: Given, NH4NO3(aq) ➝ N2O(g) + 2H2O(l) Standard entropies at 298 K are, NH4NO3(aq) = 259.8 J/mol.K…

Q: Consider the reaction: CO(g) + Cl₂(g)COCI₂(g) Using standard absolute entropies at 298K, calculate…

Q: Calculate the standard molar entropy of vaporization of water at 43.0 °C, given that its standard…

Q: Consider the reaction: 2SO2(g) + O2(g)2SO3(g) Using standard absolute entropies at 298K, calculate…

Q: Consider the reaction H2(g) + C2H4(g)C2H6(g) Using standard thermodynamic data at 298K, calculate…

Q: Given: H2O(l) → H2O(g) ΔH° = 40.7 kJ at 373K What is the entropy change in the system (ΔS) when…

A: The given chemical equation as follows --- H2O (l) → H2O (g) ΔH° = 40.7 kJ at 373 K…

Q: Consider the reaction: NH4NO3(aq)N2O(g) + 2H2O(l) Using standard thermodynamic data at 298K,…

A: Given: The reaction is : NH4NO3(aq) → N2O (g) + 2H2O(l) we know that,∆H∘f(NH4NO3)…

Q: Consider the reaction: NH4CI(aq)NH3(g) + HCI(aq) Using standard absolute entropies at 298K,…

Q: Consider the reactionCO(g) + 3H2(g)CH4(g) + H2O(g)Using standard thermodynamic data at 298K,…

A: The given reaction may be written as: The standard thermodynamic data for the given reactants is:

Q: Consider the reaction: Fe,03(s) + 3H2(9) 2Fe(s) + 3H20(g) Using standard absolute entropies at 298K,…

A: Fe2O3(s) +3 H2(g) --> 2Fe(s) + 3H2O(g) Number of moles = 2.19 mole Here may be answer slightly…

Q: Consider the reactionN2(g) + 3H2(g)2NH3(g)Using standard thermodynamic data at 298K, calculate the…

A: Answer :- The equation is N2 (g) + 3H2 (g) = 2NH3 (g) The entropy of the surroundings changes…

Q: Consider the reaction: 4HCI(g) + O2(g)- →2H2O(g) + 2CI2(g) Using standard absolute entropies at…

Q: Consider the reaction: 2Fe(s) + 3Cl2(g)2FeCl3(s) Using standard absolute entropies at 298K,…

A: The question is based on the concept of chemical thermodynamics. we have to calculate entropy change…

Q: The standard enthalpy of vaporization of Freon-12, CF2CI2, is 20.19 kJ/mol at its normal boiling…

A: ∆S =∆Hv T Where ∆S = Entropy change = ? J/mol K ∆Hv = 20.19 KJ/mol x 1000 J/1 KJ = 20190…

Q: 2C2H6(g) + 7O2(g)4CO2(g) ---------> 6H2O(g) Using standard absolute entropies at 298K, calculate…

A: Standard entropy value at 298K S°(298)H2O = 189 JK-1mol-1 , S°(298)CO2 =…

Q: Consider the reaction: NH4Cl(aq)NH3(g) + HCl(aq) Using standard thermodynamic data at 298K,…

A: Interpretation - To determine the entropy change for the surroundings when 1.76 moles of NH4Cl(aq)…

Q: Consider the reaction: 2BrF3(g)Br2(g) + 3F2(g)…

Q: Consider the reaction: Ca(OH),(aq) + 2HCI(aq)CaCly(s) + 2H,0() Using standard absolute entropies at…

Q: Consider the reaction 4HCI(g) + O2(g) 2H20(g) + 2C12(g) Using standard thermodynamic data at 298K,…

A: The standard enthalpy of formation of HCl(g) at 298 K is −92.3 kJ/mol. The standard enthalpy of…

Q: A-3.3. The aluminum block weighing 1.25 kg was cooled under constant pressure from the temperature…

Q: Calculate the entropy change for the following process: H₂O (1, 298 K, 1 bar)→ H₂O (g, 398 K, 0.8…

A: The change in entropy can be calculated as follows

Q: Consider the reaction: 2Fe(s) + 3Cl2(g)--------->2FeCl3(s) Using standard absolute entropies at…

A: Here I have explained using standard entropy values of reactant and product

Q: Consider the reaction: 2NO(g) + O2(g)–→2NO2(g) Using standard absolute entropies at 298K, calculate…

A: Answer

Q: Consider the reaction:4HCl(g) + O2(g)2H2O(g) + 2Cl2(g)Using standard absolute entropies at 298K,…

A: Given, 2.26 moles of HCl(g)

Q: Consider the reaction: I2(g) + Cl2(g)2ICl(g) Using standard absolute entropies at 298K, calculate…

Q: During their experimentation, a group of engineering students employed an unknown substance with a…

Q: Use the References to access important values if needed for this question. Consider the reaction:…

Q: Consider the reaction: CH4(g) + 2O2(g)------>CO2(g) + 2H2O(g) Using standard absolute entropies at…

A: Data taken from google; ∆SoCH4=186.3J/K/mol∆SoO2=205.1J/K/mol∆SoCO2=213.7J/K/mol∆SoH2O=188.1J/K/mol

Q: Consider the reaction: 2H2O2()2H,0(1) + O2(g) Using standard absolute entropies at 298K, calculate…

A: Given: A chemical reaction, Moles of H2O2 = 1.50 mol Temperature = 298K

Q: H2(g) + C2H4(g)- C2H6(g) Using standard absolute entropies at 298K, calculate the entropy change for…

A: Given-> Moles of H2 = 2.46 mole H2(g) + C2H4(g) ----> C2H6(g)

Q: Consider the reaction: NH4Cl(aq) → NH3(g) + HCl(aq) Using standard absolute entropies at 298 K,…

Q: The enthalpy of vaporization of methanol is 35.27 kJ mol1 at its normal boiling point of 64.1 °C.…

A: Enthalpy of vaporisation is 35.27 KJ mol-1 = 35270 Jmol-1 Temperature = 64.1°C = 64.1+273 =337.1K…

Q: Consider the reaction: 2NH3 (9) +202(g) → N₂O(g) + 3H₂O(l) Using standard thermodynamic data at 298…

A: The answer to the following question is-

Q: Consider the reaction: C,H4(g) + H,0(g)→CH3CH,OH(g) Using standard absolute entropies at 298K,…

A: The entropy change for the reaction of 1 mole of C2H4 (g) can be calculated by subtracting the…

Q: Consider the reaction: 2NO(g) + O2(g)2NO2(g) Using standard absolute entropies at 298K, calculate…

A: The reaction given is, => 2 NO (g) + O2 (g) → 2 NO2 (g) Given: Moles of NO reacting = 2.11 mol.…

Q: The heat of vaporization AH, of acetonitrile (CH,CN) is 29.8 kJ/mol. Calculate the change in entropy…

Q: An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS…

A: The given data is as follows: The temperature of the object = 400 K The object absorbs heat ∆H= 25.0…

Q: Consider the reaction CO(g) + H20(1)–CO2(g) + H2(g) Using standard thermodynamic data at 298K,…

Q: Consider the reaction: 2SO2(g) + O2(g)–→2S03(g) Using standard absolute entropies at 298K, calculate…

A: Solution we want to calculate the entropy for system ∆So system-given chemical…

Q: Consider the reaction: 2Na(s) + 2H2O(l)---->2NaOH(aq) + H2(g) Using standard absolute entropies at…

A: Standard absolute entropy:- ∆S°(Na) = 51J/K ∆S°(H2O)= 70J/K ∆S°(NaOH) = 50J/K ∆S°(H2) = 131J/K

Question

Consider the reaction:
CACO3(s)-
→CaO(s) + CO2(g)
Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.96 moles of CaCO3(s) react at standard conditions.
AS system
J/K

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Appendix J lists standard molar entropies S, not standard entropies of formation rS. Why is this possible forentropy but not for internal energy, enthalpy, or Gibbsfree energy?
Explain why absolute entropies can be measured.
What is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the decomposition of phosgene? COCl2(g)CO(g)+Cl2(g)
What is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the combustion of acetaldehyde? CH3CHO(l)+52O2(g)2CO2+2H2O(l)
10.69 If a sample of air were separated into nitrogen and oxygen molecules (ignoring other gases present), what would be the sign of for this process? Explain your answer. The next four questions relate to the following paragraph [Frank L. Lambert, Journal of Chemical Education, 76(10), 1999, 1385]. "The movement of macro objects from one location to another by an external agent involves no change in the objects' physical (thermodynamic) entropy. The agent of movement undergoes a thermodynamic entropy increase in the process."
A key component in many chemical engineering designs is the separation of mixtures of chemicals. (a) What happens to the entropy of the system when a chemical mixture is separated? (b) Are designs for chemical separation more likely to rely on spontaneous or nonspontaneous processes?
What is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the formation of hydrogen sulfide from the elements? H2(g)+18S8(s)H2S(g)
A 2.50-mol sample of gas is compressed isothermally from 20.0 L to 5.00 L under a constant external pressure of 10.0 atm. Calculate w, q, U, and H for the process.
For one day, keep a log of all the activities you undertake that consume Gibbs free energy. Distinguish betweenGibbs free energy provided by nutrient metabolism andthat provided by other energy resources.
The standard molar entropy of iodine vapor, I2(g), is 260.7 J Kl mol-1 and the standard molar enthalpy of formation is 62.4 kJ/mol. a) Calculate the entropy change for vaporization of 1 mol of solid iodine (use data from Table 16.1 or Appendix J). b) Calculate the enthalpy change for sublimation of iodine. c) Assuming that rSdoes not change with temperature, estimate the temperature at which iodine would sublime (change directly from solid to gas).
Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. Boilling point (1atm) Hvap Methane 112K 8.20 KJ/mol Hexane 342K 28.9 KJ/mol Compare the molar volume of gaseous methane at 112 K with that of gaseous hexane at 342 K. How do the differences in molar volume affect the values of Svap for these liquids?
The synthesis of glucose directly from CO2 and H2O and the synthesis of proteins directly from amino acids are both non-spontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?