Chapter 12, Problem 12.65QA

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 to Problem 12.65QA

Solution:

a) $2 NO\left(g\right)+O_2\left(g\right) \to 2 {NO}_2\left(g\right)$; Spontaneous only at low temperature.

b) $2 {NH}_3\left(g\right)+2 O_2\left(g\right) \to N_{2}O\left(g\right)+{3 H}_{2}O\left(g\right)$; Spontaneous only at low temperature.

c) ${NH}_4{NO}_3\left(s\right)\to {2 H}_{2}O\left(g\right)+N_{2}O\left(g\right)$; Spontaneous at all temperatures.

Explanation of Solution

To know if a reaction is spontaneous, we need to find out $∆H_{rxn}^o, ∆S_{rxn}^o$ and $∆G_{rxn}^o$.

a) $2 NO\left(g\right)+O_2\left(g\right) \to 2 {NO}_2\left(g\right)$

i. Change in enthalpy of reaction i.e. $∆H_{rxn}^o$:

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

${∆H}_f^o for O_2\left(g\right)=0.0 kJ/mol$

${∆H}_f^o for NO\left(g\right)=90.3 kJ/mol$

${∆H}_f^o for {NO}_2\left(g\right)=33.2 kJ/mol$

$∆H_{rxn}^o$(enthalpy change for the reaction) can be calculated from the difference between the standard molar enthalpies of $n_{reactant}$ moles of reactants and the standard molar enthalpies of $n_{product}$ moles of products,

Therefore,

$∆H_{rxn}^o= \sum n_{product} { ∆H}_{f \left(product\right)}^o- \sum n_{reactant} ∆H_{f \left(reactant\right)}^o$

$∆H_{rxn}^o= \left[n_{{NO}_2\left(g\right)} {∆H}_{ {NO}_2\left(g\right)}^o\right]-\left[\left(n_{ O_2\left(g\right)} { ∆H}_{ O_2\left(g\right)}^o\right)+\left(n_{ NO\left(g\right)} { ∆H}_{ NO\left(g\right)}^o\right)\right]$

$∆H_{rxn}^o=\left[\left(2 mol\times 33.2 kJ/mol\right)\right]-\left[\left(1 mol\times 0.0 kJ/mol\right)+\left(2 mol\times 90.3 kJ/mol\right)\right]$

$∆H_{rxn}^o=66.4 kJ-180.6 kJ$

$∆H_{rxn}^o= -114.2 kJ$

ii. Change in entropy of reaction i.e. $∆S_{rxn}^o$:

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

$S^o for O_2\left(g\right)=205.0 J/mol.K$

$S^o for NO\left(g\right)=210.7 J/mol.K$

$S^o for {NO}_2\left(g\right)=240.0 J/mol.K$

$∆S_{rxn}^o$(entropy change for the reaction) can be calculated from the difference between the standard molar entropies of $n_{reactant}$ moles of reactants and the standard molar entropies of $n_{product}$ moles of products,

Therefore,

$∆S_{rxn}^o= \sum n_{product} { S}_{product}^o- \sum n_{reactant} { S}_{reactant}^o$

$∆S_{rxn}^o= \left[n_{{NO}_2\left(g\right)} { S}_{{NO}_2\left(g\right)}^o\right]-\left[\left(n_{ O_2\left(g\right)} { S}_{ O_2\left(g\right)}^o\right)+\left(n_{ NO\left(g\right)} { S}_{ NO\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(2 mol\times 240.0 J/mol.K\right)\right]-\left[\left(1 mol\times 205.0 J/mol.K\right)+\left(2 mol\times 210.7 J/mol.K\right)\right]$

$∆S_{rxn}^o=480.0 J/K -626.4 J/K$

$∆S_{rxn}^o=-146.4 J/K$

iii. The Gibb’s free energy change for reaction,$∆G_{rxn}^o$:

$∆G_{rxn}^o=∆H_{rxn}^o-T∆S_{rxn}^o$

According to the above equation when both $∆H_{rxn}^o$ and $∆S_{rxn}^o$ are negative $∆G_{rxn}^o<0$ at low temperature and hence the reaction is spontaneous at low temperature.

b) $2 {NH}_3\left(g\right)+2 O_2\left(g\right) \to N_{2}O\left(g\right)+{3 H}_{2}O\left(g\right)$

i. Change in enthalpy of reaction i.e. $∆H_{rxn}^o$:

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

${∆H}_f^o for {NH}_3\left(g\right)=-46.1kJ/mol$

${∆H}_f^o for O_2\left(g\right)=0.0 kJ/mol$

${∆H}_f^o for N_{2}O\left(g\right)=82.1 kJ/mol$

${∆H}_f^o for H_{2}O\left(g\right)=-241.8 kJ/mol$

$∆H_{rxn}^o= \sum n_{product} { ∆H}_{f \left(product\right)}^o- \sum n_{reactant} ∆H_{f \left(reactant\right)}^o$

$∆H_{rxn}^o= \left[\left(n_{N_{2}O\left(g\right)} {∆H}_{ N_{2}O\left(g\right)}^o\right)+\left(n_{H_{2}O\left(g\right)} {∆H}_{ H_{2}O\left(g\right)}^o\right)\right]$

$-\left[\left(n_{ O_2\left(g\right)} { ∆H}_{ O_2\left(g\right)}^o\right)+\left(n_{ {NH}_3\left(g\right)} { ∆H}_{ {NH}_3\left(g\right)}^o\right)\right]$

$∆H_{rxn}^o=\left[1 mol\times \left(82.1 kJ/mol\right)+3 mol\times \left(-241.8 kJ/mol\right)\right]$

$-\left[2 mol\times \left(46.1 kJ/mol\right)+2 mol\times \left(0.0 kJ/mol\right)\right]$

$∆H_{rxn}^o=-643.3 kJ-92.2kJ$

$∆H_{rxn}^o=-735.5 kJ$

ii. Change in entropy of reaction i.e. $∆S_{rxn}^o$:

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

$S^o for {NH}_3\left(g\right)=192.5 J/mol.K$

$S^o for O_2\left(g\right) =205.0 J/mol.K$

$S^o for N_{2}O\left(g\right)=219.9 J/mol.K$

$S^o for H_{2}O\left(g\right)=188.8 J/mol.K$

$∆S_{rxn}^o= \sum n_{product} { S}_{product}^o- \sum n_{reactant} { S}_{reactant}^o$

$∆S_{rxn}^o= \left[\left(n_{N_{2}O\left(g\right)} S_{ N_{2}O\left(g\right)}^o\right)+\left(n_{H_{2}O\left(g\right)} S_{ H_{2}O\left(g\right)}^o\right)\right]$

$-\left[\left(n_{ O_2\left(g\right)} { S}_{ O_2\left(g\right)}^o\right)+\left(n_{ {NH}_3\left(g\right)} { S}_{ {NH}_3\left(g\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(1 mol\times 219.9 J/mol.K\right)+\left(3 mol\times 188.8 J/mol.K\right)\right]$

$-\left[\left(2 mol\times 192.5 J/mol.K\right)+\left(2 mol\times 205.0 J/mol.K\right)\right]$

$∆S_{rxn}^o=786.3 J/K-795 J/K$

$∆S_{rxn}^o=-8.7 J/K$

iii. The Gibb’s free energy change for reaction,$∆G_{rxn}^o$:

$∆G_{rxn}^o=∆H_{rxn}^o-T∆S_{rxn}^o$

According to the above equation when both $∆H_{rxn}^o$ and $∆S_{rxn}^o$ are negative $∆G_{rxn}^o<0$ at low temperature and hence the reaction is spontaneous at low temperature.

c) ${NH}_4{NO}_3\left(s\right)\to {2 H}_{2}O\left(g\right)+N_{2}O\left(g\right)$

i. Change in enthalpy of reaction i.e. $∆H_{rxn}^o$:

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

${∆H}_f^o for {NH}_4{NO}_3\left(s\right)=-365.6 kJ/mol$

${∆H}_f^o for N_{2}O\left(g\right)=82.1 kJ/mol$

${∆H}_f^o for H_{2}O\left(g\right)=-241.8 kJ/mol$

$∆H_{rxn}^o= \sum n_{product} { ∆H}_{f \left(product\right)}^o- \sum n_{reactant} ∆H_{f \left(reactant\right)}^o$

$∆H_{rxn}^o= \left[\left(n_{H_{2}O\left(g\right)} {∆H}_{ H_{2}O\left(g\right)}^o\right)+\left(n_{N_{2}O\left(g\right)} {∆H}_{ N_{2}O\left(g\right)}^o\right)\right]-\left[\left(n_{{NH}_4{NO}_3\left(s\right)} { ∆H}_{ {NH}_4{NO}_3\left(s\right)}^o\right)\right]$

$∆H_{rxn}^o=\left[2 mol\times \left(-241.8\right)+1 mol\times \left(82.1 kJ/mol\right)\right]-\left[ mol\times \left(-365.6 kJ/mol\right)\right]$

$∆H_{rxn}^o= -401.5 kJ+365.6 kJ$

$∆H_{rxn}^o= -35.9 kJ$

ii. Change in entropy of reaction i.e. $∆S_{rxn}^o$:

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

$S^o for {NH}_4{NO}_3\left(s\right)=151.1 J/mol.K$

$S^o for N_{2}O\left(g\right)=219.9 J/mol.K$

$S^o for H_{2}O\left(g\right)=188.8 J/mol.K$

$∆S_{rxn}^o= \sum n_{product} { S}_{product}^o- \sum n_{reactant} { S}_{reactant}^o$

$∆S_{rxn}^o= \left[\left(n_{H_{2}O\left(g\right)} S_{ H_{2}O\left(g\right)}^o\right)+\left(n_{N_{2}O\left(g\right)} S_{ N_{2}O\left(g\right)}^o\right)\right]-\left[\left(n_{ {NH}_4{NO}_3\left(s\right)} { S}_{ {NH}_4{NO}_3\left(s\right)}^o\right)\right]$

$∆S_{rxn}^o=\left[\left(2 mol\times 219.9 J/mol.K\right)+\left(1 mol\times 188.8 J/mol.K\right)\right]-\left[\left(1 mol\times 151.1 J/mol.K\right)\right]$

$∆S_{rxn}^o=626.6 J/K-151.1 J/K$

$∆S_{rxn}^o=477.5 J/K$

iii. The Gibb’s free energy change for reaction,$∆G_{rxn}^o$:

$∆G_{rxn}^o=∆H_{rxn}^o-T∆S_{rxn}^o$

According to the above equation when $∆H_{rxn}^o$ is negative and $∆S_{rxn}^o$ is positive $∆G_{rxn}^o<0$ 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.