Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 17, Problem 17.58QP

(a)

Interpretation Introduction

Interpretation: The reactions only at low temperatures; those that are spontaneous only at high temperatures and those that are spontaneous at all temperatures are to be identified.

Concept introduction: A process is spontaneous at all temperature when the values of,

ΔHrxnο=NegativeΔSrxnο=Positive

A process is spontaneous at low temperature when the values of,

ΔHrxnο=NegativeΔSrxnο=Negative

A process is spontaneous at high temperature when the values of,

ΔHrxnο=PositiveΔSrxnο=Positive

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

(a)

Expert Solution
Check Mark

Answer to Problem 17.58QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

2H2S(g)+3O22H2O(g)+2SO2 (1)

The enthalpy of formation of H2S(g) (ΔHf,H2S(g)ο) is 20.17kJ/mol .

The enthalpy of formation of H2O(g) (ΔHf,H2O(g)ο) is 241.8kJ/mol .

The enthalpy of formation of SO2(g) (ΔHf,SO2(g)ο) is 296.8kJ/mol .

The enthalpy of formation of O2(g) (ΔHf,O2(g)ο) is 0.0kJ/mol .

The standard molar entropy of H2S(g) (SH2S(g)ο) is 205.6Jmol1K1 .

The standard molar entropy of SO2(g) (SSO2(g)ο) is 248.2Jmol1K1

The standard molar entropy of O2(g) (SO2(g)ο) is 205.0Jmol1K1

The standard molar entropy of H2O(g) (SH2O(g)ο) is 188.8Jmol1K1

A process is spontaneous at all temperature when the values of,

ΔHrxnο=NegativeΔSrxnο=Positive (2)

A process is spontaneous at low temperature when the values of,

ΔHrxnο=NegativeΔSrxnο=Negative (3)

A process is spontaneous at high temperature when the values of,

ΔHrxnο=PositiveΔSrxnο=Positive (4)

The enthalpy change for a reaction (ΔHrxnο) is calculated by the formula,

ΔHrxnο=n×ΔHfο(Products)m×ΔHfο(Reactants) (5)

Where,

  • ΔHfο(Products) is standard enthalpy of formation of the products.
  • ΔHfο(Reactants) is of standard enthalpy of formation of the reactants.
  • n is number of moles of products.
  • m is number of moles of reactants.

In chemical equation (1) the,

  • Number of moles of product H2O(g) is 2 .
  • Number of moles of product SO2(g) is 2 .
  • Number of moles of reactant H2S(g) is 2 .
  • Number of moles of reactant O2(g) is 3 .

The n×ΔHfο(Products) for the chemical reaction (1) is expressed as,

n×ΔHfο(Products)=2mol×ΔHf,H2O(g)ο+2mol×ΔHf,SO2(g)ο (6)

Substitute the values of ΔHf,H2O(g)ο and ΔHf,SO2(g)ο in equation (6).

n×ΔHfο(Products)=2mol×241.8kJ/mol+2mol×296.8kJ/mol=483.6kJ593.6kJ=1077.2kJ

The m×ΔHfο(Reactants) for the balanced chemical reaction (1) is expressed as,

m×ΔHfο(Reactants)=2mol×ΔHf,H2S(g)ο+3mol×ΔHf,O2(g)ο (7)

Substitute the values of ΔHf,H2S(g)ο and ΔHf,O2(g)ο in equation (7).

m×ΔHfο(Reactants)=2mol×20.17kJ/mol+1mol×0kJ/mol=40.34kJ

Substitute the values of n×ΔHfο(Products) and m×ΔHfο(Reactants) in equation (5).

ΔHrxnο=1077.2kJ(40.34kJ)=1036.86kJ

The standard entropy change (ΔSο) is calculated by the formula,

ΔSrxnο=nproducts×Sο(products)mreactants×Sο(reactants) (8)

Where,

  • nproducts is the number of moles of products.
  • mreactants is the number of moles of reactants.
  • Sο(products) is the standard molar entropy of products.
  • Sο(reactants) is the standard molar entropy of reactants.

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=2mol×SH2O(g)ο+2mol×SSO2(g)ο (9)

Substitute the value of SH2O(g)ο and SSO2(g)ο in equation (9).

nproducts×Sο(products)=2mol×188.8Jmol1K1+2mol×248.2Jmol1K1=377.6JK-1+496.4JK-1=874JK-1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=2mol×SH2S(g)ο+3mol×SO2(g)ο (10)

Substitute the values of SH2S(g)ο and SO2(g)ο in equation (10).

mreactants×Sο(reactants)=(2mol×205.6Jmol1K1)+(3mol×205.0Jmol1K1)=1026.2JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (8).

ΔSrxnο=874JK11026.2JK1=152.2JK1

The values of ΔHrxnο and ΔSrxnο satisfies the equation (3).

Hence the given reaction is spontaneous at low temperature.

(b)

Interpretation Introduction

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

(b)

Expert Solution
Check Mark

Answer to Problem 17.58QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

SO2(g)+2H2O2(l)H2SO4(l) (11)

The enthalpy of formation of H2SO4(l) (ΔHf,H2SO4(l)ο) is 814.0kJ/mol .

The enthalpy of formation of SO2(g) (ΔHf,SO2(g)ο) is 296.8kJ/mol .

The enthalpy of formation of H2O2(l) (ΔHf,H2O2(l)ο) is 187.8kJ/mol .

The standard molar entropy of SO2(g) (SSO2(g)ο) is 248.2Jmol1K1

The standard molar entropy of H2O2(l) (SH2O2(l)ο) is 109.6Jmol1K1 .

The standard molar entropy of H2SO4(l) (SH2SO4(l)ο) is 156.9Jmol1K1

In chemical equation (11) the,

  • Number of moles of product H2SO4(l) is 1 .
  • Number of moles of reactant SO2(g) is 1 .
  • Number of moles of reactant H2O2(l) is 1 .

The n×ΔHfο(Products) for the chemical reaction (11) is expressed as,

n×ΔHfο(Products)=1mol×ΔHf,H2SO4(l)ο (12)

Substitute the value of ΔHf,H2SO4(l)ο in equation (12).

n×ΔHfο(Products)=1mol×814.0kJ/mol=814.0kJ

The m×ΔHfο(Reactants) for the balanced chemical reaction (11) is expressed as,

m×ΔHfο(Reactants)=1mol×ΔHf,SO2(g)ο+1mol×ΔHf,H2O2(l)ο (13)

Substitute the values of ΔHf,H2O2(l)ο and ΔHf,SO2(g)ο in equation (13).

m×ΔHfο(Reactants)=1mol×296.8kJ/mol+1mol×187.8kJ/mol=484.6kJ

Substitute the values of n×ΔHfο(Products) and m×ΔHfο(Reactants) in equation (5).

ΔHrxnο=814.0kJ(484.6kJ)=329.4kJ

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=1mol×SH2SO4(l)ο (14)

Substitute the values of SH2SO4(l)ο in equation (14).

nproducts×Sο(products)=1mol×156.9Jmol1K1=156.9JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=1mol×SSO2(g)ο+1mol×SH2O2(l)ο (15)

Substitute the values of SSO2(g)ο and SH2O2(l)ο in equation (15).

mreactants×Sο(reactants)=1mol×109.6Jmol1K1+1mol×248.2Jmol1K1=357.8JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (8).

ΔSrxnο=156.9JK1357.8JK1=200.9JK1

The values of ΔHrxnο and ΔSrxnο satisfies the equation (3).

Hence the given reaction is spontaneous at low temperature.

(c)

Interpretation Introduction

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

(c)

Expert Solution
Check Mark

Answer to Problem 17.58QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

S(g)+O2(g)SO2(g) (16)

The enthalpy of formation of SO2(g) (ΔHf,SO2(g)ο) is 296.8kJ/mol .

The enthalpy of formation of O2(g) (ΔHf,O2(g)ο) is 0.0kJ/mol .

The enthalpy of formation of S(g) (ΔHf,S(g)ο) is 102.3kJ/mol .

The standard molar entropy of S(g) (SS(g)ο) is 167.8Jmol1K1 .

The standard molar entropy of SO2(g) (SSO2(g)ο) is 248.2Jmol1K1

The standard molar entropy of O2(g) (SO2(g)ο) is 205.0Jmol1K1

In chemical equation (16) the,

  • Number of moles of product SO2(g) is 1 .
  • Number of moles of reactant S(g) is 1 .
  • Number of moles of reactant O2(g) is 1 .

The n×ΔHfο(Products) for the chemical reaction (16) is expressed as,

n×ΔHfο(Products)=1mol×ΔHf,SO2(g)ο (17)

Substitute the value of ΔHf,SO2(g)ο in equation (17).

n×ΔHfο(Products)=1mol×296.8kJ/mol=296.8kJ

The m×ΔHfο(Reactants) for the chemical reaction (16) is expressed as,

m×ΔHfο(Reactants)=1mol×ΔHf,S(g)ο+1mol×ΔHf,O2(g)ο (18)

Substitute the values of ΔHf,S(g)ο and ΔHf,O2(g)ο in equation (18).

m×ΔHfο(Reactants)=1mol×102.3kJ/mol+1mol×0.0kJ/mol=102.3kJ

Substitute the values of n×ΔHfο(Products) and m×ΔHfο(Reactants) in equation (5).

ΔHrxnο=296.8kJ102.3kJ=399.1kJ

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=1mol×SSO2(g)ο (19)

Substitute the value of SSO2(g)ο in equation (19).

nproducts×Sο(products)=1mol×248.2Jmol1K1=248.2JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=1mol×SS(g)ο+1mol×SO2(g)ο (20)

Substitute the value of SS(g)ο and SO2(g)ο in equation (20).

mreactants×Sο(reactants)=1mol×167.8Jmol1K1+1mol×205.0Jmol1K1=372.8JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (8).

ΔSrxnο=248.2JK1372.8JK1=124.6JK1

The values of ΔHrxnο and ΔSrxnο satisfies the equation (3).

Hence the given reaction is spontaneous at low temperature.

Conclusion

  1. a. The given reaction is spontaneous at low temperature.
  2. b. The given reaction is spontaneous at low temperature.
  3. c. The given reaction is spontaneous at low temperature

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Chapter 17 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 17 - Prob. 17.3VPCh. 17 - Prob. 17.4VPCh. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9QPCh. 17 - Prob. 17.10QPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.27QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85QPCh. 17 - Prob. 17.86QPCh. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91APCh. 17 - Prob. 17.92APCh. 17 - Prob. 17.93APCh. 17 - Prob. 17.94APCh. 17 - Prob. 17.95APCh. 17 - Prob. 17.96APCh. 17 - Prob. 17.97APCh. 17 - Prob. 17.98APCh. 17 - Prob. 17.99APCh. 17 - Prob. 17.100APCh. 17 - Prob. 17.101APCh. 17 - Prob. 17.102APCh. 17 - Prob. 17.103APCh. 17 - Prob. 17.104APCh. 17 - Prob. 17.105AP
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