Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 12, Problem 12.63QA
Interpretation Introduction

To identify:

Which of the given reactions is spontaneous i) Only at low temperature ii) Only at high temperature iii) at all temperatures.

Expert Solution & Answer
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Answer to Problem 12.63QA

Solution:

a) 2 NOg+O2(g) 2 NO2(g); Spontaneous only at low temperature.

b) 2 NH3g+2 O2(g)  N2Og+3 H2O(g); Spontaneous only at low temperature.

c) NH4NO3s2 H2Og+N2Og; Spontaneous at all temperatures.

Explanation of Solution

To know if a reaction is spontaneous, we need to find out Hrxno, Srxno and Grxno.

a) 2 NOg+O2(g) 2 NO2(g)

i. Change in enthalpy of reaction i.e. Hrxno:

From the appendix 4, the values of standard molar enthalpies are,

Hfo for O2g=0.0 kJ/mol

Hfo for NOg=90.3  kJ/mol

Hfo for NO2g=33.2 kJ/mol

Hrxno(enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of nreactant moles of reactants and the standard molar enthalpies of nproduct moles of products,

Therefore,

Hrxno= nproduct  Hf (product)o- nreactant Hf (reactant)o

Hrxno= nNO2g H NO2go-n O2g  H O2go+n NOg  H NOgo

Hrxno=2 mol×33.2 kJ/mol-1 mol×0.0 kJ/mol+2 mol×90.3 kJ/mol

Hrxno=66.4 kJ-180.6 kJ

Hrxno= -114.2 kJ

ii. Change in entropy of reaction i.e. Srxno:

From the appendix 4, the values of standard molar entropies are,

So for O2g=205.0 J/mol.K

So for NOg=210.7 J/mol.K

So for NO2g=240.0 J/mol.K

Srxno(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of nreactant moles of reactants and the standard molar entropies of nproduct moles of products,

Therefore,

Srxno= nproduct  Sproducto- nreactant  Sreactanto

Srxno= nNO2g  SNO2go-n O2g  S O2go+n NOg  S NOgo

Srxno=2 mol×240.0 J/mol.K-1 mol×205.0 J/mol.K+2 mol×210.7 J/mol.K

Srxno=480.0 J/K -626.4 J/K

Srxno=-146.4 J/K

iii. The Gibb’s free energy change for reaction,Grxno:

Grxno=Hrxno-TSrxno

According to the above equation when both Hrxno and Srxno are negative Grxno<0 at low temperature and hence the reaction is spontaneous at low temperature.

b) 2 NH3g+2 O2(g)  N2Og+3 H2O(g)

i. Change in enthalpy of reaction i.e. Hrxno:

From the appendix 4, the values of standard molar enthalpies are,

Hfo for NH3g=-46.1kJ/mol

Hfo for O2g=0.0 kJ/mol

Hfo for N2Og=82.1 kJ/mol

Hfo for H2Og=-241.8 kJ/mol

Hrxno= nproduct  Hf (product)o- nreactant Hf (reactant)o

Hrxno= nN2Og H N2Ogo+nH2Og H H2Ogo

-n O2g  H O2go+n NH3g  H NH3go

Hrxno=1 mol×82.1 kJ/mol+3 mol×-241.8 kJ/mol

-2 mol×46.1 kJ/mol+2 mol×0.0 kJ/mol

Hrxno=-643.3 kJ-92.2kJ

Hrxno=-735.5 kJ

ii. Change in entropy of reaction i.e. Srxno:

From the appendix 4, the values of standard molar entropies are,

So for NH3g=192.5 J/mol.K

So for O2g =205.0 J/mol.K

So for N2Og=219.9 J/mol.K

So for H2Og=188.8 J/mol.K

Srxno= nproduct  Sproducto- nreactant  Sreactanto

Srxno= nN2Og S N2Ogo+nH2Og S H2Ogo

-n O2g  S O2go+n NH3g  S NH3go

Srxno=1 mol×219.9 J/mol.K+3 mol×188.8 J/mol.K

-2 mol×192.5 J/mol.K+2 mol×205.0 J/mol.K

Srxno=786.3 J/K-795 J/K

Srxno=-8.7 J/K

iii. The Gibb’s free energy change for reaction,Grxno:

Grxno=Hrxno-TSrxno

According to the above equation when both Hrxno and Srxno are negative Grxno<0 at low temperature and hence the reaction is spontaneous at low temperature.

c) NH4NO3s2 H2Og+N2Og

i. Change in enthalpy of reaction i.e. Hrxno:

From the appendix 4, the values of standard molar enthalpies are,

Hfo for NH4NO3s=-365.6 kJ/mol

Hfo for N2Og=82.1 kJ/mol

Hfo for H2Og=-241.8 kJ/mol

Hrxno= nproduct  Hf (product)o- nreactant Hf (reactant)o

Hrxno= nH2Og H H2Ogo+nN2Og H N2Ogo-nNH4NO3s  H NH4NO3so

Hrxno=2 mol×-241.8+1 mol×82.1 kJ/mol- mol× -365.6 kJ/mol

Hrxno= -401.5  kJ+365.6  kJ

Hrxno= -35.9 kJ

ii. Change in entropy of reaction i.e. Srxno:

From the appendix 4, the values of standard molar entropies are,

So for NH4NO3s=151.1 J/mol.K

So for N2Og=219.9 J/mol.K

So for H2Og=188.8 J/mol.K

Srxno= nproduct  Sproducto- nreactant  Sreactanto

Srxno= nH2Og S H2Ogo+nN2Og S N2Ogo-n NH4NO3s  S NH4NO3so

Srxno=2 mol×219.9 J/mol.K+1 mol×188.8 J/mol.K-1 mol×151.1 J/mol.K

Srxno=626.6 J/K-151.1 J/K

Srxno=477.5 J/K

iii. The Gibb’s free energy change for reaction,Grxno:

Grxno=Hrxno-TSrxno

According to the above equation when Hrxno is negative and Srxno is positive Grxno<0 at all temperatures and hence the reaction is spontaneous at all temperatures.

Conclusion:

Using change in enthalpy and entropy of reaction, sign of Gibb’s free energy can be found out.

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Chapter 12 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 12 - Prob. 12.13QACh. 12 - Prob. 12.14QACh. 12 - Prob. 12.15QACh. 12 - Prob. 12.16QACh. 12 - Prob. 12.17QACh. 12 - Prob. 12.18QACh. 12 - Prob. 12.19QACh. 12 - Prob. 12.20QACh. 12 - Prob. 12.21QACh. 12 - Prob. 12.22QACh. 12 - Prob. 12.23QACh. 12 - Prob. 12.24QACh. 12 - Prob. 12.25QACh. 12 - Prob. 12.26QACh. 12 - Prob. 12.27QACh. 12 - Prob. 12.28QACh. 12 - Prob. 12.29QACh. 12 - Prob. 12.30QACh. 12 - Prob. 12.31QACh. 12 - Prob. 12.32QACh. 12 - Prob. 12.33QACh. 12 - Prob. 12.34QACh. 12 - Prob. 12.35QACh. 12 - Prob. 12.36QACh. 12 - Prob. 12.37QACh. 12 - Prob. 12.38QACh. 12 - Prob. 12.39QACh. 12 - Prob. 12.40QACh. 12 - Prob. 12.41QACh. 12 - Prob. 12.42QACh. 12 - Prob. 12.43QACh. 12 - Prob. 12.44QACh. 12 - Prob. 12.45QACh. 12 - Prob. 12.46QACh. 12 - Prob. 12.47QACh. 12 - Prob. 12.48QACh. 12 - Prob. 12.49QACh. 12 - Prob. 12.50QACh. 12 - Prob. 12.51QACh. 12 - Prob. 12.52QACh. 12 - Prob. 12.53QACh. 12 - Prob. 12.54QACh. 12 - Prob. 12.55QACh. 12 - Prob. 12.56QACh. 12 - Prob. 12.57QACh. 12 - Prob. 12.58QACh. 12 - Prob. 12.59QACh. 12 - Prob. 12.60QACh. 12 - Prob. 12.61QACh. 12 - Prob. 12.62QACh. 12 - Prob. 12.63QACh. 12 - Prob. 12.64QACh. 12 - Prob. 12.65QACh. 12 - Prob. 12.66QACh. 12 - Prob. 12.67QACh. 12 - Prob. 12.68QACh. 12 - Prob. 12.69QACh. 12 - Prob. 12.70QACh. 12 - Prob. 12.71QACh. 12 - Prob. 12.72QACh. 12 - Prob. 12.73QACh. 12 - Prob. 12.74QACh. 12 - Prob. 12.75QACh. 12 - Prob. 12.76QACh. 12 - Prob. 12.77QACh. 12 - Prob. 12.78QACh. 12 - Prob. 12.79QACh. 12 - Prob. 12.80QACh. 12 - Prob. 12.81QACh. 12 - Prob. 12.82QACh. 12 - Prob. 12.83QACh. 12 - Prob. 12.84QACh. 12 - Prob. 12.85QACh. 12 - Prob. 12.86QA
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