MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 20, Problem 20.98P

(a)

Interpretation Introduction

Interpretation:

The molar entropy So of given standard formation reaction has to be founded at 267.1J/molK 

Concept Introduction:

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process.  For all spontaneous process in an isolated system there will be an increase in entropy.  Entropy is represented by the letter ‘S’. It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy of one mole of substance at a given standard state is called standard molar entropy (So).

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

Explanation of Solution

The formation reaction for propylene is,

  3C(s)+3H2(g)CH3CH=CH2(g)

Equal molar of carbon C and hydrogen (H) produced to one mole of propylene molecule.

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.

Sfo Formation of values,

  C(s)=5.686J/molKH2(g)=130.6J/molKCH3CHCH2(g)=267.1J/molK

Calculate the molar entropy for given reaction,

  Sfo=[(1molCH3CH=CH2)(SoofCH3CH=CH2)][(3molC)(SoofC)+(3molH2)(SoofH2)]=[(1mol)(267.1J/molK)][(3molC)(5.686J/molK)+(3molH2)(130.6J/molK)]=[(1mol)(267.1J/molK)][(17.058J/molK)+(391.8J/molK)]=[(267.1J/molK)][(408.858J/molK)]=141.758J/KSfo=-141.8J/K

The (S°f) of the reaction is -141.8J/K_

The molar entropy change is negative sign for (S°f) indicates the formation of propylene is disfavored.

(b)

Interpretation Introduction

Interpretation:

For the propylene formation reaction ΔGfo value should be found.

Concept introduction:

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 equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

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=ΔHoTΔSo

Where,

  T is the temperature

  ΔG is the free energy

  ΔGo, ΔHo and ΔSo is standard free energy, enthalpy and entropy values.

(b)

Expert Solution
Check Mark

Explanation of Solution

The formation reaction for propylene is,

  3C(s)+3H2(g)CH3CH=CH2(g)

Standared Free energy change equation is,

  ΔGrxno = ΔΗrxno- TΔSrxno

Free energy change ΔGfo

Calcualted enthalpy and entropy  values are

  ΔΗrxno=20.4kJ/molΔGrxno=141.758J/K

These values are plugging above standard free energy equation,

   ΔGfo=(20.4kJ/mol)(298K)(141.758J/K)(103J/1kJ)ΔGfo=62.643884kJ/molΔGfo=62.6kJ/mol

Hence the free energy (ΔGfo) of the reaction is 62.6kJ/mol_.

(c)

Interpretation Introduction

Interpretation:

For the dehydrogenation reaction the enthalpy ΔΗrxno and free energy ΔGrxno values have to be calculated.

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

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 equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

(c)

Expert Solution
Check Mark

Explanation of Solution

The dehydrogantion  reaction is as follows,

  CH3CH2CH3(g)CH3CH=CH2(g)+H2(g)

Standard enthalpy change is,

Let us find enthalpy change for the reaction,

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

  ΔH°rxn =[(1molCH3CH=CH2)(ΔHfoofCH3CH=CH2)+(2molH2)(ΔHfoofH2)][(1molCH3CH2CH3)(ΔHfoofCH3CH2CH3)]ΔH°rxn =[(1mol)(20.4kJ/mol)+(1mol)(0kJ/mol)][(1mol)(105kJ/mol)]ΔH°rxn =125.4=125kJ

Hence, the enthalpy (ΔH°rxn) value is 125kJ_

Standared Free energy change equation iss,

  ΔGrxno = ΔΗrxno- TΔSrxno

Free energy change ΔGfo

Calculated enthalpy and entropy  values are

  ΔΗrxno=125.4kJ/molΔGrxno=62.643884kJ/mol

These values are plugging above standard free energy equation,

ΔG°rxn =[(1molCH3CH=CH2)(ΔGfoofCH3CH=CH2)+(1molH2)(ΔGfoofH2)][(1molCH3CH2CH3)(ΔGfoofCH3CH2CH3)]ΔG°rxn =[(1mol)(62.643884kJ/mol)+(1mol)(0kJ/mol)][(1mol)(24.5kJ/mol)]ΔG°rxn =87.1438874=87.1kJ

The free energy (ΔG°rxn) of the reaction is 87.1kJ_.

(d)

Interpretation Introduction

Interpretation:

For the formation of propylene theoretical yield has to be calculated at 580oC.

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

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 equation given below helps us to calculate the change in free energy in a system.

  ΔGo = ΔΗo- TΔSo

(d)

Expert Solution
Check Mark

Explanation of Solution

The formation reaction for propylene is,

  CH3CH2CH3(g)CH3CH=CH2(g)+H2(g)

Entropy change ΔS°system

Calculate the change in entropy for this reaction as follows,

  ΔS°rxn = S°Products- nS°reactants

  ΔS°rxn =[(1molCH3CH=CH2)(SoofCH3CH=CH2)+(1molH2)(SoofH2)][(1molCH3CH2CH3)(SoofCH3CH2CH3)]ΔS°rxn =[(1mol)(267.1J/molK)+(1mol)(130.6J/molK)][(1mol)(269.9J/molK)]ΔS°rxn =127.8kJ

Standared Free energy change equation is,

  ΔGfo = ΔΗfo- TΔSfo

Free energy change ΔGfo

Calcualted enthalpy and entropy  values are

  ΔΗrxno=1125.4kJΔGrxno=127.8kJ

These values are plugging above standard free energy equation,

   ΔGfo=(125.4kJ/mol)(273+580K)(127.8J/K)(103J/1kJ)ΔGfo=16.3866kJ/mol

Calculation for equilibrium pressureKp

The equilibrium equation is,

  ΔGrxno=RTln(Kp)

Rearrange the above equation,

  lnK=ΔGo-RT=(16.3866kJ/mol(8.314J/molK)(853K))(103J1kJ)lnK=2.3106267K=e2.3106267K=0.099199K=0.0992

The equilibrium reaction is,

  CH3CH2CH3(g)CH3CH=CH2(g)+H2(g)

  K=(PCH3CH=CH2)(PH2)(PCH3CH2CH3)

Here,

  x=pressureofCH3CH=CH2andH21.00xispressureofCH3CH2CH3

So,

  0.099199=(x)(x)(1.00x)x2+0.099199x0.099199=0

Solve the above quadratic equation,

  x=b±b24ac2ax=(0.099199)±(0.099199)24(1)(0.099199)2(1)x=0.2692408x=0.269atm

Hence, the theoretical yield of propylene is 27%.

(e)

Interpretation Introduction

Interpretation:

Identify whether there is any yield change if the reactor wall were preamble to hydrogen H2.

Concept introduction:

Theoretical yield: The amount of product formed, assuming complete reaction of the limiting reagent.

Actual yield: The amount of product actually formed in a reaction.

Percent yield: The percentage of the theoretical yield actually obtained from a chemical reaction.

Percent yield =ActualyieldTheoretical yield×100

(e)

Expert Solution
Check Mark

Explanation of Solution

The formation reaction for propylene,

  CH3CH2CH3(g)CH3CH=CH2(g)+H2(g)

If hydrogen could escape through the reactor walls, the reaction would be shifted to the right side and improving the yield.

(f)

Interpretation Introduction

Interpretation:

The temp at which the dehydrogenation spontaneous has to be identified, provided all substances in the standard state.

Concept introduction:

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process. For all spontaneous process in an isolated system there will be an increase in entropy. Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy changes associated with a phase transition reaction can be found by the following equation.

  =ΔΗoΔSo

Where,

  ΔΗo is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔSo is the change in entropy in the system

(f)

Expert Solution
Check Mark

Explanation of Solution

The formation reaction for propylene,

  CH3CH2CH3(g)CH3CH=CH2(g)+H2(g)

Standared Free energy change equation is,

  ΔGo = ΔΗo- TΔSoΔGo =0=ΔΗo- TΔSo[1]ΔΗo=TΔSo[2]

Rearrange the equation (2) to calculate temprature T,

T=ΔΗoΔSo[3]

Hence,

Enthalpy and entropy values are

  ΔΗfo=125.4kJΔGrxno=0.1278kJ/K

  T=125.4kJ0.1278J/K(1031kJ)=981.22066T=981K707.85oC=708oC_

Hence, the propylene formation founded temperature value is T=708oC.

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Chapter 20 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

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? 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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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