Question

The Ostwald process for the commercial production of nitric acid involves three steps:

$4N{H}_3+5O_2\left(g\right)\stackrel{Pt}{\to }4NO\left(g\right)+6H_{2}O$

$\mathbf{2}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

$\mathbf{3}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{H}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}\mathit{N}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

1. Calculate $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{,}\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{,}\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathit{K}$ for each of the tree steps in the Ostwald process
2. Calculate the equilibrium constant for the first step at${\mathbf{825}}^{\mathbf{o}}\mathit{C}$ . Assume that$\mathit{\Delta }\mathit{H}\mathit{a}\mathit{n}\mathit{d}\mathit{\Delta }\mathit{S}$ are temperature independent
3. Is there a thermodynamic reason for the high temperature in the first step assuming standard conditions?

Short answer

a.$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-908\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-181\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-962\text{\hspace{0.17em}kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-112\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-147\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-68\text{\hspace{0.17em}kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-74\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-267\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-6\text{\hspace{0.17em}kJ/mol}\end{array}$

b.$\text{K}=4.6\times {10}^{52}$

c.

Enthalpy change is negative and entropy change is positive. Temperature should be lowered to shift the equilibrium to the right side. But here, the purpose of using higher temperature is to increase the rate of the reaction.

Step-by-step solution

Step 1: Definition of Standard free energy, Standard enthalpy, and Standard entropy

The free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under standard conditions is defined as standard free energy.

A measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements is defined as Standard enthalpy.

The standard entropy is at 1 atm pressure.

Calculations

a.$4{\text{NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\text{(g)}\stackrel{Pt}{\to }4\text{NO(g)}+{\text{6H}}_{\text{2}}\text{O}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-908\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-181\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-962\text{\hspace{0.17em}kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}=-\text{RTlnK}\\ \text{K}=4.33\times {10}^{168}\end{array}$

$2\text{NO(g)}+{\text{O}}_{\text{2}}\text{(g)}\to 2{\text{NO}}_{\text{2}}\text{(g)}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-112\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-147\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-68\text{\hspace{0.17em}kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}=-\text{RTlnK}\\ \text{K}=8.3\times {10}^{11}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}=-74\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔS}}^{\text{o}}=-267\text{\hspace{0.17em}J/K.mol}\\ {\text{ΔG}}^{\text{o}}=-6\text{\hspace{0.17em}kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}=-\text{RTlnK}\\ \text{K}=9.0\times {10}^{-2}\end{array}$

b.The equilibrium constant

$\begin{array}{l}{\text{ΔG}}^{\text{0}}={\text{ΔH}}^0-{\text{TΔS}}^0\\ {\text{ΔG}}^{\text{0}}=-1107\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔG}}^{\text{0}}=-\text{RTlnK}\\ \text{K}=4.6\times {10}^{52}\end{array}$

c.

1. There is no thermodynamic reason for using higher temperature in step of the Ostwald process.Enthalpy change is negative and entropy change is positive, temperature should be lowered to shift the equilibrium to the right side.But here, the purpose of using higher temperature is to increase the rate of the reaction.