Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 20, Problem 20.91P

(a)

Interpretation Introduction

Interpretation:

For magnesite decomposition the balanced equation has to be written.

Concept introduction:

Balanced equation: A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides, is known as balanced equation.

Reactant: In a chemical reaction the species that present at left is denoted as reactant which undergoes chemical change and result to given new species called product.

Product: In a chemical reaction the species that present in right side is denoted as product that results from the reactant.

(a)

Expert Solution
Check Mark

Answer to Problem 20.91P

Balance chemical eqaution is,

  MgCO3(s)MgO(s)+CO2(g)

Explanation of Solution

Balance chemical eqaution is,

  MgCO3(s)MgO(s)+CO2(g)

The MgCO3 decomposed at 1200oC to forms Magnesia and carbon dioxide the balanced equation is shows above. The above reaction is balanced one since both reactant and the product side contains same type and same number of elements.

(b)

Interpretation Introduction

Interpretation:

For the given decomposition reaction the free energy ΔGo value has to be calculated at 298K using the enthalpy and entropy values.

Concept introduction:

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous processes are associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

Where,

  ΔGo is the standard change in free energy of the system

  ΔΗo is the standard change in enthalpy of the system

  T is the absolute value of the temperature

  ΔSo is the change in entropy in the system

(b)

Expert Solution
Check Mark

Answer to Problem 20.91P

For given decomposition reaction standard free energy values is ΔGrxno=65.2kJ_

Explanation of Solution

Given decomposition reaction is,

  MgCO3(s)MgO(s)+CO2(g)

Calculate the change in Gibb’s free energy at 298K as follows,

Standard enthalpy change is,

ΔHo Formation of values are shown as follows,

  MgO(g)=601.2kJ/molCO2(g)=393.5kJ/molMgCO3(g)=1112.0kJ/mol

Enthalpy values referred from Appendix Table. 

The enthalpy change for the reaction is calculated as follows,

  ΔH°rxn = ΔH°f(Products)- nΔH°f(reactants)

  ΔH°rxn =[(1molMgO)(ΔH°fofMgO)+(1molCO2)(ΔH°fofCO2)][(1molMgCO3)(ΔH°fofMgCO3)]ΔH°rxn =[(1molMgO)(601.2kJ/mol)+(1molCO2)(393.5kJ/mol)][(1molMgCO3)(1112kJ/mol)]ΔH°rxn =117.3kJ

The enthalpy change is positive. Hence, the enthalpy (ΔH°rxn) value is 117.3kJ_

Entropy change  ΔS°system

Calculate the change in entropy for this reaction as follows,

  ΔS°rxn = S°Products- nS°reactants

Where, (m)and(n) are the stoichiometric co-efficient.

  ΔS°rxn =[(1molMgO)(SoofMgO)+(1molCO2)(SoofCO2)][(1molMgCO3)(SoofMgCO3)]ΔS°rxn =[(1mol)(26.9J/mol×K)+(1mol)(213.7J/mol×K)][(1mol)(65.86J/mol×K)]ΔS°rxn =174.74J/K

The (ΔS°rxn) of the reaction is 174.74J/K_

The enthalpy and entropy changes are positive sign for ΔS°rxn indicates the formation of methanol is favored at given temperature.

Calculate the Free enrgy change ΔGrxno

Standared Free energy change equation is ΔGrxno = ΔΗrxno- TΔSrxno

Calcualted enthalpy and entropy values are

  ΔΗrxno=117.3kJΔSrxno=174.74J/KT1=273+25=298K

Plugging these values into above standard free energy equation,

  ΔGrxno=117.3kJ-[(298K)(174.74J/K)(1kJ/103)]=65.22748kJΔGrxno=65.2kJ

Therefore, for the given decomposition reaction standard free energy value is ΔGrxno=65.2kJ_.

(c)

Interpretation Introduction

Interpretation:

The minimum temperature at which the given reaction is spontaneous has to be identified.

Concept introduction:

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

Where,

  ΔGo is the standard change in free energy of the system

  ΔΗo is the standard change in enthalpy of the system

  T is the absolute value of the temperature

  ΔSo is the change in entropy in the system

(c)

Expert Solution
Check Mark

Answer to Problem 20.91P

The required minimum temperature is T=671K.

Explanation of Solution

Given decomposition reaction is,

  MgCO3(s)MgO(s)+CO2(g)

Calculate the change in Gibb’s free energy at 298K as follows,

The reaction becomes spontaneous below the temprature where ΔGrxno=0

Consider the follwing free energy equation,

  ΔGrxno =0ΔΗrxno- TΔSrxno[1]ΔΗrxno=TΔSrxno[2]

Rearrange equation (2) to calculate temprature T,

T=ΔΗrxnoΔSrxno

Hence,

  T=117.3kJ0.17474kJ/K=671.283T=671K

At temprature above 671K, this reaction is spontaneous. Because both enthalpy ΔΗrxno and entropy ΔSrxno values are positve hence the reaction becomes spontaneous above this temprature. 

(d)

Interpretation Introduction

Interpretation:

For the MgCO3 decomposition reaction, the equilibrium constant PCO2 value has to be calculated at 298K.

Concept introduction:

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

Where,

  ΔGo is the standard change in free energy of the system

  ΔΗo is the standard change in enthalpy of the system

  T is the absolute value of the temperature

  ΔSo is the change in entropy in the system

Free energy changeΔG: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant K

  ΔG =ΔG0+RTln(K)ΔG0=ΔH0TΔS0

  Where,

  T is the temperature

  ΔG is the free energy

  ΔG0, ΔH0 and ΔS0 is standard free energy, enthalpy and entropy values.

(d)

Expert Solution
Check Mark

Answer to Problem 20.91P

For given reaction, the equilbrium pressure PCO2 value is K=3.68×10-12atm_.

Explanation of Solution

Given,

Free energy equation is,

  MgCO3(s)MgO(s)+CO2(g)

Solving for PCO2 using following free energy equation,

  ΔG=ΔGoRTln(K)[1](or)ΔGo=RTln(K)

Rearrange the above equation as shown below,

  InK=ΔGoRT[2]

Then,

  InK=ΔGoRT=65.22748kJ/mol(8.314J/mol×K)(298K)(1000J1kJ)InK=26.327178K=e26.327178=3.6834253×1012K=3.68×10-12atmofPCO2

Therefore, the given equilbrium pressure PCO2 value is K=3.68×10-12atm_

(e)

Interpretation Introduction

Interpretation:

For the following MgCO3 decomposition reaction, the equilibrium constant PCO2 value has to be calculated at 1200K.

Concept introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change (ΔG°rxn) is the difference in free energy of the reactants and products in their standard state.

ΔG°rxn=nΔGf°(Products)-nΔGf°(Reactants)

Free energy changeΔG: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant K

  ΔG =ΔG0+RTln(K)ΔG0=ΔH0TΔS0

  Where,

  T is the temperature

  ΔG is the free energy

  ΔG0, ΔH0 and ΔS0 is standard free energy, enthalpy and entropy values.

(e)

Expert Solution
Check Mark

Answer to Problem 20.91P

For given reaction the equilbrium pressure PCO2 value is K=1.0512×104atm_

Explanation of Solution

Given,

Free energy equation is,

  MgCO3(s)MgO(s)+CO2(g)

Calculate the Free enrgy change  ΔGrxno

Standared Free energy change equation iss,

  ΔGrxno = ΔΗrxno- TΔSrxno

Calcualted enthalpy and entropy  values are

  ΔΗrxno=117.3kJΔSrxno=174.74J/KT1=273+25=298K

These values are plugging above standard free energy equation,

  ΔGrxno=117.3kJ-[(1200K)(174.74J/K)(1kJ/103)]=92.388kJΔGrxno=-92.38kJ

Next, we solving for PCO2 using following free energy equation at 1200K

  ΔG=ΔGoRTln(K)[1](or)ΔGo=RTln(K)

Rearrange the above equation,

  InK=ΔGoRT[2]

Then,

  InK=ΔGoRT=92.388kJ/mol(8.314J/mol×K)(1200K)(1000J1kJ)InK=9.26028K=e9.26028=1.0512×104K=1.0512×104atmofPCO2

Therefore, the given equilbrium pressure PCO2 value is K=1.0512×104atm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 20 Solutions

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change

Ch. 20.3 - Prob. 20.6AFPCh. 20.3 - Prob. 20.6BFPCh. 20.3 - Prob. 20.7AFPCh. 20.3 - Prob. 20.7BFPCh. 20.3 - Prob. 20.8AFPCh. 20.3 - Prob. 20.8BFPCh. 20.3 - Prob. B20.1PCh. 20.3 - Nonspontaneous processes like muscle contraction,...Ch. 20.4 - Use Appendix B to find K at 298 K for the...Ch. 20.4 - Use the given value of K to calculate ΔG° at 298 K...Ch. 20.4 - Prob. 20.10AFPCh. 20.4 - Prob. 20.10BFPCh. 20.4 - At 298 K, ΔG° = −33.5 kJ/mol for the formation of...Ch. 20.4 - Prob. 20.11BFPCh. 20 - Prob. 20.1PCh. 20 - Distinguish between the terms spontaneous and...Ch. 20 - State the first law of thermodynamics in terms of...Ch. 20 - State qualitatively the relationship between...Ch. 20 - Why is ΔSvap of a substance always larger than...Ch. 20 - Prob. 20.6PCh. 20 - Prob. 20.7PCh. 20 - Which of these processes are spontaneous? (a)...Ch. 20 - Prob. 20.9PCh. 20 - Which of these processes are spontaneous? (a)...Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Predict which substance has greater molar entropy....Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - Without consulting Appendix B, arrange each group...Ch. 20 - In the reaction depicted in the molecular scenes,...Ch. 20 - Describe the equilibrium condition in terms of the...Ch. 20 - Prob. 20.32PCh. 20 - For each reaction, predict the sign and find the...Ch. 20 - For each reaction, predict the sign and find the...Ch. 20 - Find for the combustion of ethane (C2H6) to...Ch. 20 - Find for the combustion of methane to carbon...Ch. 20 - Find for the reaction of nitrogen monoxide with...Ch. 20 - Find for the combustion of ammonia to nitrogen...Ch. 20 - Find for the formation of Cu2O(s) from its...Ch. 20 - Find for the formation of HI(g) from its...Ch. 20 - Find for the formation of CH3OH(l) from its...Ch. 20 - Find for the formation of N2O(g) from its...Ch. 20 - Sulfur dioxide is released in the combustion of...Ch. 20 - Oxyacetylene welding is used to repair metal...Ch. 20 - What is the advantage of calculating free energy...Ch. 20 - Given that ΔGsys = −TΔSuniv, explain how the sign...Ch. 20 - Is an endothermic reaction more likely to be...Ch. 20 - Explain your answers to each of the following for...Ch. 20 - With its components in their standard states, a...Ch. 20 - How can ΔS° for a reaction be relatively...Ch. 20 - Calculate ΔG° for each reaction using ...Ch. 20 - Calculate ΔG° for each reaction using ...Ch. 20 - Prob. 20.53PCh. 20 - Prob. 20.54PCh. 20 - Consider the oxidation of carbon...Ch. 20 - Consider the combustion of butane gas: Predict...Ch. 20 - For the gaseous reaction of xenon and fluorine to...Ch. 20 - For the gaseous reaction of carbon monoxide and...Ch. 20 - One reaction used to produce small quantities of...Ch. 20 - A reaction that occurs in the internal combustion...Ch. 20 - As a fuel, H2(g) produces only nonpolluting H2O(g)...Ch. 20 - The U.S. government requires automobile fuels to...Ch. 20 - If K << 1 for a reaction, what do you know about...Ch. 20 - How is the free energy change of a process related...Ch. 20 - The scenes and the graph relate to the reaction of...Ch. 20 - What is the difference between ΔG° and ΔG? Under...Ch. 20 - Calculate K at 298 K for each reaction: MgCO3(s) ⇌...Ch. 20 - Calculate ΔG° at 298 K for each reaction: 2H2S(g)...Ch. 20 - Calculate K at 298 K for each reaction: HCN(aq) +...Ch. 20 - Calculate ΔG° at 298 K for each reaction: 2NO(g) +...Ch. 20 - Use ΔH° and ΔS° values for the following process...Ch. 20 - Use ΔH° and ΔS° values to find the temperature at...Ch. 20 - Prob. 20.73PCh. 20 - Use Appendix B to determine the Ksp of CaF2. Ch. 20 - For the reaction I2(g) + Cl2(g) ⇌ 2ICl(g),...Ch. 20 - For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g),...Ch. 20 - The Ksp of PbCl2 is 1.7×10−5 at 25°C. What is ΔG°?...Ch. 20 - Prob. 20.78PCh. 20 - The equilibrium constant for the...Ch. 20 - The formation constant for the reaction Ni2+(aq) +...Ch. 20 - Prob. 20.81PCh. 20 - Prob. 20.82PCh. 20 - High levels of ozone (O3) cause rubber to...Ch. 20 - A BaSO4 slurry is ingested before the...Ch. 20 - According to advertisements, “a diamond is...Ch. 20 - Prob. 20.86PCh. 20 - Prob. 20.87PCh. 20 - Prob. 20.88PCh. 20 - Is each statement true or false? If false, correct...Ch. 20 - Prob. 20.90PCh. 20 - Prob. 20.91PCh. 20 - Prob. 20.92PCh. 20 - Prob. 20.93PCh. 20 - Write a balanced equation for the gaseous...Ch. 20 - Prob. 20.95PCh. 20 - Hydrogenation is the addition of H2 to double (or...Ch. 20 - Prob. 20.97PCh. 20 - Prob. 20.98PCh. 20 - Prob. 20.99PCh. 20 - Prob. 20.100PCh. 20 - From the following reaction and data, find (a) S°...Ch. 20 - Prob. 20.102PCh. 20 - Prob. 20.103PCh. 20 - Prob. 20.104PCh. 20 - Prob. 20.105PCh. 20 - Prob. 20.106PCh. 20 - Prob. 20.107PCh. 20 - Consider the formation of ammonia: N2(g) + 3H2(g)...Ch. 20 - Kyanite, sillimanite, and andalusite all have the...Ch. 20 - Prob. 20.110PCh. 20 - Prob. 20.111P
Knowledge Booster
Background pattern image
Chemistry
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY