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

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Answer 1

Question

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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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Answer 2

Question

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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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Answer 3

Question

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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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Answer 4

Question

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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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Answer 5

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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.

Answer 6

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

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) NH4NO3s→2 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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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.

Answer 7

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.

Answer 8

Expert Solution & Answer

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) NH4NO3s→2 H2Og+N2Og; Spontaneous at all temperatures.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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-T∆Srxno

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-T∆Srxno

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) NH4NO3s→2 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-T∆Srxno

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.

Answer 12

Ch. 12 - Prob. 12.1VP Ch. 12 - Prob. 12.2VP Ch. 12 - Prob. 12.4VP Ch. 12 - Prob. 12.5VP Ch. 12 - Prob. 12.6VP Ch. 12 - Prob. 12.8VP Ch. 12 - Prob. 12.9VP Ch. 12 - Prob. 12.10VP Ch. 12 - Prob. 12.11QA Ch. 12 - Prob. 12.12QA

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

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

Answer to Problem 12.63QA

Explanation of Solution

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Chapter 12 Solutions

To identify:

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