Chapter 22, Problem 22.61P

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant ${\text{K}}_{\text{p}}$ for the two given ammonia oxidations has to be calculated.

Explanation of Solution

For the first reaction (1):

    $\begin{array}{l}{\text{4NH}}_{\text{3(g)}}{\text{ + 5O}}_{\text{2(g) }}\underset{}{\overset{\text{pt/Rh}\text{catalyst}}{\to }}{\text{4NO}}_{\text{(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \\ {\text{ΔG}}_{\text{25}}^{\text{o}}\text{=}\text{-906}\text{.196 kJ}\text{-[(273 + 25)K]( 0}\text{.17792 kJ/K)}\text{=}\text{-959}\text{.216 kJ}\\ \\ {\text{ΔG}}_{\text{900}}^{\text{o}}\text{=}\text{-906}\text{.196 kJ}\text{-[(273 + 900)K]( 0}\text{.17792 kJ/K)}\text{=}\text{-1114}\text{.896kJ}\end{array}$

For the first reaction (2):

    $\begin{array}{l}{\text{4NH}}_{\text{3(g)}}{\text{ + 3O}}_{\text{2(g) }}\underset{}{\overset{}{\to }}2{\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{ + 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \\ {\text{ΔG}}_{\text{25}}^{\text{o}}\text{=}\text{-1267}\text{.356 kJ}\text{-[(273 + 25)K]( 0}\text{.12832 kJ/K)}\text{=}\text{-1305}\text{.595 kJ}\\ \\ {\text{ΔG}}_{\text{900}}^{\text{o}}\text{=}\text{-1267}\text{.356 kJ}\text{-[(273 + 900)K]( 0}\text{.12832 kJ/K)}\text{=}\text{-1417}\text{.875 kJ}\text{.}\end{array}$

At ${\text{25}}^{\text{o}}\text{C}$, the equilibrium constant is calculated as,

    $\begin{array}{l}{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{-}{\text{RTlnK}}_{\text{c}}\\ \\ {\text{lnK}}_{\text{25}}\text{(1)}\text{=}\frac{{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}}{\text{-RT}}\text{=}\frac{\text{-959}\text{.216 kJ/mol}}{\text{(8}\text{.314 K/mol}\text{.K)(298k)}}\left(\frac{\text{1000J}}{\text{1KJ}}\right)\text{=}\text{387}\text{.15977}\\ \\ {\text{K}}_{\text{25}}\text{(1)}\text{=}{\text{1x10}}^{\text{168}}\\ \\ {\text{lnK}}_{\text{25}}\text{(2)}\text{=}\frac{{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}}{\text{-RT}}\text{=}\frac{\text{-1305}\text{.595 kJ/mol}}{\text{(8}\text{.314 K/mol}\text{.K)(298k)}}\left(\frac{\text{1000J}}{\text{1KJ}}\right)\text{=}\text{526}\text{.9655}\\ \\ {\text{K}}_{\text{25}}\text{(2)}\text{=}{\text{7x10}}^{\text{228}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant ${\text{K}}_{\text{p}}$ for the two given ammonia oxidations has to be calculated.

Explanation of Solution

At ${900}^{\text{o}}\text{C}$, the equilibrium constant is calculated as,

    $\begin{array}{l}{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{-}{\text{RTlnK}}_{\text{c}}\\ \\ {\text{lnK}}_{\text{900}}\text{(1)}\text{=}\frac{{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}}{\text{-RT}}\text{=}\frac{\text{-1114}\text{.896 kJ/mol}}{\text{(8}\text{.314 K/mol}\text{.K)(298k)}}\left(\frac{\text{1000J}}{\text{1KJ}}\right)\text{=}\text{114}\text{.3210817}\\ \\ {\text{K}}_{\text{900}}\text{(1)}\text{=}\text{4}{\text{.6x10}}^{\text{49}}\\ \\ {\text{lnK}}_{\text{900}}\text{(2)}\text{=}\frac{{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}}{\text{-RT}}\text{=}\frac{\text{-1417}\text{.875 kJ/mol}}{\text{(8}\text{.314 K/mol}\text{.K)(298k)}}\left(\frac{\text{1000J}}{\text{1KJ}}\right)\text{=}\text{145}\text{.388}\\ \\ {\text{K}}_{\text{900}}\text{(2)}\text{=}\text{1}{\text{.4x10}}^{\text{63}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The annual cost of the lost platinum has to be calculated.

Explanation of Solution

The annual cost of the lost platinum is calculated as,

    $\begin{array}{l}\text{Cost (\$) Pt =}\left(\text{1}{\text{.01x10}}^{\text{7}}{\text{ t HNO}}_{\text{3}}\right)\left(\frac{\text{175 mg Pt}}{\text{1}{\text{t HNO}}_{\text{3}}}\right)\left(\frac{\text{0}\text{.001g}}{\text{1mg}}\right)\left(\frac{\text{1kg}}{\text{1000g}}\right)\\ \\ \left(\frac{\text{32}\text{.15 Troy Oz}}{\text{1kg}}\right)\left(\frac{\text{\$1557}}{\text{1 troy oz}}\right)\\ \\ \text{=}\$\text{8}{\text{.8x10}}^{\text{7}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The value of the platinum captured by a recovery unit with 72% efficiency has to be calculated.

Explanation of Solution

The value of the platinum captured by a recovery unit is calculated as,

    $\begin{array}{l}\text{Cost (\$) Pt =}\left(\text{1}{\text{.01x10}}^{\text{7}}{\text{ t HNO}}_{\text{3}}\right)\left(\frac{\text{175 mg Pt}}{\text{1}{\text{t HNO}}_{\text{3}}}\right)\left(\frac{\text{72\%}}{\text{100\%}}\right)\left(\frac{\text{0}\text{.001g}}{\text{1mg}}\right)\left(\frac{\text{1kg}}{\text{1000g}}\right)\\ \\ \left(\frac{\text{32}\text{.15 Troy Oz}}{\text{1kg}}\right)\left(\frac{\text{\$1557}}{\text{1 troy oz}}\right)\\ \\ \text{=}\text{\$6}{\text{.4x10}}^{\text{7}}\end{array}$