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 17.57QP

(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 17.57QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

2NO(g)+O2(g)2NO2(g) (1)

The enthalpy of formation of NO(g) (ΔHf,NO(g)ο) is 90.3kJ/mol .

The enthalpy of formation of NO2(g) (ΔHf,NO2(g)ο) is 33.2kJ/mol .

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

The standard molar entropy of NO(g) (ΔSf,NO(g)ο) is 210.7Jmol1K1 .

The standard molar entropy of NO2(g) (ΔSf,NO2(g)ο) is 240.0Jmol1K1

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

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 NO2(g) is 2 .
  • Number of moles of reactant NO(g) is 2 .
  • Number of moles of reactant O2(g) is 1 .

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

n×ΔHfο(Products)=2mol×ΔHf,NO2(g)ο (6)

Substitute the value of ΔHf,NO2(g)ο in equation (6).

n×ΔHfο(Products)=2mol×33.2kJ/mol=66.4kJ

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

m×ΔHfο(Reactants)=2mol×ΔHf,NO(g)ο+1mol×ΔHf,O2(g)ο (7)

Substitute the values of ΔHf,NO(g)ο and ΔHf,O2(g)ο in equation (7).

m×ΔHfο(Reactants)=2mol×90.3kJ/mol+1mol×0kJ/mol=180.6kJ

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

ΔHrxnο=66.4kJ180.6kJ=114.2kJ

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×SNO2(g)ο (9)

Substitute the value of SNO2(g)ο in equation (9).

nproducts×Sο(products)=2mol×240.0Jmol1K1=480JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=(2mol×SNO(g)ο)+(1mol×SO2(g)ο) (10)

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

mreactants×Sο(reactants)=(2mol×210.7Jmol1K1)+(1mol×205.0Jmol1K1)=421.4JK1+205.0JK1=626.4JK1

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

ΔSrxnο=480JK1626.4JK1=146.4JK1

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 17.57QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

2NH3(g)+2O2N2O(g)+3H2O (11)

The enthalpy of formation of NH3(g) (ΔHf,NH3(g)ο) is 46.1kJ/mol .

The enthalpy of formation of N2O(g) (ΔHf,N2O(g)ο) is 82.1kJ/mol .

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

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

The standard molar entropy of NH3(g) (SNH3(g)ο) is 192.5Jmol1K1 .

The standard molar entropy of N2O(g) (SN2O(g)ο) is 219.9Jmol1K1 .

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

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

In chemical equation (11) the,

  • Number of moles of product N2O(g) is 1 .
  • Number of moles of product H2O(g) is 3 .
  • Number of moles of reactant NH3(g) is 2 .
  • Number of moles of reactant O2(g) is 2 .

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

n×ΔHfο(Products)=1mol×ΔHf,N2O(g)ο+3mol×ΔHf,H2O(g)ο (12)

Substitute the values of ΔHf,N2O(g)ο and ΔHf,H2O(g)ο in equation (12).

n×ΔHfο(Products)=1mol×82.1kJ/mol+3mol×241.8kJ/mol=643.3kJ

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

m×ΔHfο(Reactants)=2mol×ΔHf,NH3(g)ο+2mol×ΔHf,O2(g)ο (13)

Substitute the values of ΔHf,NH3(g)ο and ΔHf,O2(g)ο in equation (13).

m×ΔHfο(Reactants)=2mol×46.1kJ/mol+1mol×0kJ/mol=92.2kJ

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

ΔHrxnο=643.3kJ(92.2kJ)=551.3kJ

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=1mol×SN2O(g)ο+3mol×SH2O(g)ο (14)

Substitute the values of SN2O(g)ο and SH2O(g)ο in equation (14).

nproducts×Sο(products)=1mol×219.9Jmol1K1+3mol×188.8Jmol1K1=786.3JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=2mol×SNH3(g)ο+2mol×ΔSO2(g)ο (15)

Substitute the values of SNH3(g)ο and SO2(g)ο in equation (15).

mreactants×Sο(reactants)=2mol×192.5Jmol1K1+2mol×205.0Jmol1K1=795JK1

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

ΔSrxnο=786.3JK1795JK1=8.7JK1

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 17.57QP

Solution

The given reaction is spontaneous at all temperature.

Explanation of Solution

Explanation

The given reaction is,

NH4NO3(s)2H2O(g)+N2O(g) (16)

The enthalpy of formation of NH4NO3(s) (ΔHf,NH4NO3(s)ο) is 365.6kJ/mol .

The enthalpy of formation of N2O(g) (ΔHf,N2O(g)ο) is 82.1kJ/mol .

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

The standard molar entropy of NH4NO3(s) (SNH4NO3(s)ο) is 151.1Jmol1K1 .

The standard molar entropy of N2O(g) (SN2O(g)ο) is 219.9Jmol1K1 .

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

In chemical equation (16) the,

  • Number of moles of product N2O(g) is 1 .
  • Number of moles of product H2O(g) is 3 .
  • Number of moles of reactant NH3(g) is 2 .
  • Number of moles of reactant O2(g) is 2 .

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

n×ΔHfο(Products)=1mol×ΔHf,N2O(g)ο+2mol×ΔHf,H2O(g)ο (17)

Substitute the values of ΔHf,N2O(g)ο and ΔHf,H2O(g)ο in equation (17).

n×ΔHfο(Products)=1mol×82.1kJ/mol+2mol×241.8kJ/mol=401.5kJ

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

m×ΔHfο(Reactants)=1mol×ΔHf,NH4NO3(s)ο (18)

Substitute the value of ΔHf,NH4NO3(s)ο in equation (18).

m×ΔHfο(Reactants)=1mol×365.6kJ/mol=365.6kJ

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

ΔHrxnο=401.5kJ(365.6kJ)=35.9kJ

The nproducts×Sο(products) is expressed as,

nproducts×Sο(products)=1mol×SN2O(g)ο+2mol×SH2O(g)ο (19)

Substitute the values of SN2O(g)ο and SH2O(g)ο in equation (19).

nproducts×Sο(products)=1mol×219.9Jmol1K1+2mol×188.8Jmol1K1=597.5JK1

The mreactants×Sο(reactants) is expressed as,

mreactants×Sο(reactants)=1mol×SNH4NO3(s)ο (20)

Substitute the value of SNH4NO3(s)ο in equation (20).

mreactants×Sο(reactants)=1mol×151.1Jmol1K1=151.1JK1

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

ΔSrxnο=597.5JK1151.1JK1=446.4JK1

The values of ΔHrxnο is negative and ΔSrxnο is positive. Thus it satisfies the equation (2).

Hence the given reaction is spontaneous at all 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 all 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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