Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 20, Problem 20.95P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the ΔSrxno value has to be calculated at 298K, using the given equilibrium K value.

Concept introduction:

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants.  (ΔS°rxn) can be calculated by the following equation.

  ΔS°rxn = S°Products- nS°reactants

Where,

  S°reactants is the standard entropy of the reactants

  S°Products is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same.

(a)

Expert Solution
Check Mark

Answer to Problem 20.95P

The standard entropy changes of the given reaction value is ΔSrxno=121.5J/K_.

Explanation of Solution

The given equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Entropy change  ΔS°system

ΔSrxno Formations of values are,

  NO(g)=210.65J/KmolBr2(g)=245.38J/KmolNOBr(g)=272.6J/Kmol

Calculate the change in entropy for this reaction as follows,

  ΔS°rxn = S°Products- nS°reactants

Calculate the change in entropy for this reaction as follows

  ΔSrxno =[(2molNO)(SoofNO)+(1molBr2)(SoofBr2)][(2molofNOBr)(SoofNOBr)]

Substituting the values of standard entropy (So) of individual species in the above equation

  ΔSrxno =[(2molof NO)(210.65J/molK)+(1molof Br2)(245.38J/molK)][(2molofNOBr)(272.6J/molK)]ΔSrxno =121.48=121.5J/K.

The standard entropy of the reaction ΔSrxno=121.5J/K_ and the entropy change is positive.

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the ΔGrxno value has to be calculated at 373K, using the given data’s.

Concept introduction:

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

  ΔG =ΔGo+RTln(K)ΔGo=- RTln(K)

  Where,

  T is the temperature

  ΔG is the free energy

  ΔGo is standard free energy change.

(b)

Expert Solution
Check Mark

Answer to Problem 20.95P

Given reaction the standard free ΔG°rxn energy value is 2.7×103J/mol.

Explanation of Solution

The given equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Solving for ΔGrxno using following free energy equation,

  ΔG=ΔGo+RTln(Q)

Calculate the ΔGrxno of given equilibrium reaction is as follows,

     ΔG=0=ΔGrxno+RTlnKΔGrxno=RTlnK=(8.314J/mol×K)(373K)In(0.42)=(8.314J/mol×K)(373K)×(0.867500)=2690.223J/molΔGrxno=2.7×103J/mol.

Therefore, the given reaction (ΔGrxno) value is 2.7×103J/mol.

(c)

Interpretation Introduction

Interpretation:

For the following reaction, the ΔHrxno value has to be calculated at 373K, using the given data’s.

Concept introduction:

Free energy 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,

  ΔΗo is the change in enthalpy of the system

  ΔGo is the standard change in free energy of the system

  T is the absolute value of the temperature

(c)

Expert Solution
Check Mark

Answer to Problem 20.95P

Given reaction enthalpy (ΔΗrxno) changes is 4.80×104J/mol

Explanation of Solution

The given equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Calculation for enthalpy change  ΔHrxno

Standared Free energy change equation is,

  ΔΗrxno = ΔGrxno- TΔSrxno

Calcualted entropy ΔSrxno and  ΔGrxno values are

  ΔSrxno=121.48J/K

  ΔGrxno=2690.225J/mol

These values are plugging following above equation,

  ΔΗrxno=2690.225J/mol+(373K)(121.48J/K)=4.8002265×104ΔΗrxno=4.80×104J/mol.

Therefore, the given reaction (ΔΗrxno) value is 4.80×104J/mol.

(d)

Interpretation Introduction

Interpretation:

For the following reaction, the ΔHfo value has to be calculated at 298K.

Concept introduction:

Enthalpy is the amount energy absorbed or released in a process.

The enthalpy change in a system Ηsys) can be calculated by the following equation.

  ΔHrxn = ΔH°produdcts-ΔH°reactants 

Where,

  ΔHfo(reactants) is the standard enthalpy of the reactants

  ΔHfo(produdcts) is the standard enthalpy of the products

(d)

Expert Solution
Check Mark

Answer to Problem 20.95P

Given reaction standard enthalpy changes is ΔH°rxn=81.7kJ/mol_.

Explanation of Solution

Consider the equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Standard enthalpy change is,

The equation for the standard enthalpy of the above reaction is,

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

Calculate the change in entropy for this reaction as follows

  ΔHrxno =[(2molNO)(ΔHfoofNO)+(1molBr2)(ΔHfoofBr2)][(2molofNOBr)(ΔHfoofNOBr)]

Substituting the values of standard enthalpies of individual species in the above equation

  ΔHrxno =[(2molof NO)(90.29kJ/mol)+(1molof Br2)(30.91kJ/mol)][(2molofNOBr)(4.8002265×104J)(1kJ/103J)]ΔHrxnoofNOBr =81.7438675=81.7kJ/mol.

Hence, the standard enthalpy of the reaction ΔH°rxn=81.7kJ/mol_.

(e)

Interpretation Introduction

Interpretation:

For the following reaction, the ΔGrxno 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

(e)

Expert Solution
Check Mark

Answer to Problem 20.95P

The ΔGrxno value of the reaction is 1.18×104J/mol_.

Explanation of Solution

Consider the equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Calculate the Free energy change  ΔGrxno

Standared Free energy change equation is,

  ΔGrxno = ΔΗrxno- TΔSrxno

Calcualted enthalpy and entropy values are

  ΔΗrxno=4.8002265×104J/mol

  ΔSrxno=121.48kJ/K

These values are plugging following standard free energy equation,

  ΔGrxno =4.8002265×104J/mol-(298K)(121.48J/molK)=1.1801225×105ΔGrxno=1.18×104J/mol.

Therefore, the ΔGrxno value of the reaction is 1.18×104J/mol_.

(f)

Interpretation Introduction

Interpretation:

For the following reaction, the NOBr ΔGfo value has to be calculated at 298K.

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=mΔGf°(Products)-nΔGf°(Reactants)

Where,

  nΔGf°(Reactants) is the standard entropy of the reactants

  mΔGf°(products) is the standard free energy of the products

(f)

Expert Solution
Check Mark

Answer to Problem 20.95P

The standard free energy value is ΔGfo=82.3kJ/mol.

Explanation of Solution

Given equilibrium reaction is,

  2NOBr(g)2NO(g)+Br2(g)K=0.42at373K

Free energy changes ΔG°rxn

Gibbs free energy equation is,

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

Free energy change for the reaction is calculated as follows,

  ΔG°rxn =[(2molof NO)(ΔGfoof NO)+(1molofBr2)(ΔGfoof Br2)][(2molofNOBr)(ΔGfoofNOBr)]ΔG°rxn =[(2molof NO)(86.60kJ/mol)+(1molof Br2)(3.13kJ/mol)][(2molofNOBr)(1.1801225×104J)(1kJ/103J)]ΔGfo ofNOBr=82.264=82.3kJ/mol.

Hence, the standard free energy value is ΔGfo=82.3kJ/mol.

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Chapter 20 Solutions

Loose Leaf 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
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