Q22.63CP - Chemistry: Molecular Nature Of Matter And Change

Question

Step  1 of the Ostwald process for nitric acid production is 

  ${\mathbf{\text{4NH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + 5O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\stackrel{\mathbf{\text{Pt/Rh Catalyst}}}{\mathbf{\to }}{\mathbf{\text{4NO(g) + 6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}$

An unwanted side reaction for this step is 

  ${\mathbf{\text{4NH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + 3O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{2N}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + 6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}$

(a) Calculate  ${\mathbf{\text{K}}}_{\mathbf{p}}$ for these two  ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$ oxidations at  ${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$. 

(b) Calculate ${\mathbf{\text{K}}}_{\mathbf{p}}$  for these two ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$  oxidations at  ${\mathbf{\text{900}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.

(c) The  $\mathbf{\text{Pt/Rh}}$ catalyst is one of the most efficient in the chemical industry, achieving 96%  yield in   millisecond of contact with the reactants. However, at normal operating conditions ( 5 atm and ${\text{850}}^{\text{∘}}\text{C}$ ), about  175 mg of   is lost per metric ton ( $\mathit{t}$) of  ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$ produced. If the annual U.S. production of  ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$ is  $\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{t}$ and the market price of  $\mathit{P}\mathit{t}$ is  $\mathbf{\$}\mathbf{1260}\mathbf{/}\mathit{t}\mathit{r}\mathit{o}\mathit{y}\mathit{o}\mathit{z}$, what is the annual cost of the lost   ( $\mathbf{1}\mathit{k}\mathit{g}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{15}\mathit{t}\mathit{r}\mathit{o}\mathit{y}\mathit{o}\mathit{z}$)?

(d) Because of the high price of  $\mathit{P}\mathit{t}$, a filtering unit composed of ceramic fibre is often installed, which recovers as much as  75% of the lost  $\mathit{P}\mathit{t}$. What is the value of the  $\mathit{P}\mathit{t}$ captured by a recovery unit with 72% efficiency?

Step-by-Step Solution

Answer

(a) The ${\text{K}}_{\text{p}}$  for the two  ${\text{NH}}_{\text{3}}$ oxidations at  ${\text{25}}^{\text{o}}\text{C}$ is $\text{K = 1.38}\times {\text{10}}^{\text{168}}$  and  $\text{K = 7.125}\times {\text{10}}^{\text{228}}$.

(b) The  ${\text{K}}_{\text{p}}$ for the two  ${\text{NH}}_{\text{3}}$ oxidations at  ${\text{900}}^{\text{o}}\text{C}$ is $\text{K = 4.456}\times {\text{10}}^{\text{49}}$  and  $\text{K = 1.384}\times {\text{10}}^{\text{63}}$

(c) The annual cost of lost  $Pt$ is $\text{\$ 8.848}\times {\text{10}}^{\text{8}}$ .

(d) The cost of the saved, recovered  $\text{P}t$ is $\text{\$ 6.37}\times {\text{10}}^{\text{7}}$ .

Step 1: Concept Introduction

The Ostwald process is a cornerstone of contemporary chemical manufacturing, and it provides the primary raw material for the most prevalent type of fertiliser. It helps in the production of nitric acid ( ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$).

Step 2: Calculation for K p

a) 

The equilibrium constant, defined for partial pressures (${\text{K}}_{\text{p}}$ ) as all the reactants and products are in a gaseous state can be expressed for each reaction –

 $$\begin{gathered} {\text{K}}_{\text{p}}\text{ = }\frac{{\text{p[NO]}}^{\text{4}}·\text{p}{\left[{\text{H}}_{\text{2}}\text{O}\right]}^{\text{6}}}{\text{p}{\left[{\text{NH}}_{\text{3}}\right]}^{\text{4}}·\text{p}{\left[{\text{O}}_{\text{2}}\right]}^{\text{5}}}....\text{Step 1} \\ {\text{K}}_{\text{p}}\text{ = }\frac{\text{p}{\left[{\text{N}}_{\text{2}}\right]}^{\text{2}}·\text{p}{\left[{\text{H}}_{\text{2}}\text{O}\right]}^{\text{6}}}{\text{p}{\left[{\text{NH}}_{\text{3}}\right]}^{\text{4}}·\text{p}{\left[{\text{O}}_{\text{2}}\right]}^{\text{3}}}....\text{Step 2} \end{gathered}$$

To determine the equilibrium constant, the  ${\text{K}}_{\text{p}}$, can be calculated from   $∆G$. In this case, Gibb's energy can be calculated as –

 $$\begin{gathered} \Delta G=\Delta H-T·\Delta S \\ \Delta G=-R·T·\ln K \\ \ln K=\frac{\Delta G}{-R·T} \\ K=e^{\ln K} \end{gathered}$$

For the reaction  (1), calculating the enthalpy and entropy of the reaction using Appendix B –

 $$\begin{gathered} \Delta H_{rxn}=\sum \Delta H_{products}-\sum \Delta H_{reac\tan ts} \\ \Delta H_{rxn}=\left[4 mol\times \Delta H\left(NO\right)+6 mol\times \Delta H\left(H_{2}O\right)\right]-\left[4 mol\times \Delta H\left(N{H}_3\right)+5 mol\times \Delta H\left(O_2\right)\right] \\ \Delta H_{rxn}=[4 mol\times 90.29 kJ/mol+6 mol\times (-241.826)kJ/mol] \\ -[4 mol\times (-45.90)kJ/mol+5 mol\times 0.00 kJ/mol] \\ \Delta H_{rxn}=-906.196 kJ \end{gathered}$$

 $$\begin{gathered} \Delta S_{rxn}=\left[4 mol\times \Delta S\left(NO\right)+6 mol\times \Delta S\left(H_{2}O\right)\right]-\left[4 mol\times \Delta S\left(N{H}_3\right)+5 mol\times \Delta S\left(O_2\right)\right] \\ \Delta S_{rxn}=\left[4 mol\times 210.65\frac{J}{mol\times K}+6 mol\times 188.72\frac{J}{mol\times K}\right] \\ -\left[4 mol\times 193.00\frac{J}{mol\times K}+5 mol\times 205.00\frac{J}{mol\times K}\right] \\ \Delta S_{rxn}=177.92\frac{J}{K}=0.17792\frac{kJ}{K} \end{gathered}$$

Then, the Gibb's energy for ${\text{25}}^{\text{o}}\text{C}$  is –

 $$\begin{gathered} \Delta G_{25}=-906.196 kJ-(25+273)K\times 0.17792\frac{kJ}{K} \\ \Delta G_{25}=-959.216 kJ \end{gathered}$$

Finally, the  ${\text{K}}_{\text{p}}$, for step 2 , at  ${\text{25}}^{\text{o}}\text{C}$ is –

 $$\begin{gathered} \ln K_p=\frac{-959.216·{10}^3 J/mol}{-8.314\frac{J}{mol·K}·(25+273)K} \\ \ln K_p=387.159 \\ K=1.38\times {10}^{168} \end{gathered}$$

Similarly, for the reaction  (2), calculating the enthalpy and entropy of the reaction using Appendix B –

 $$\begin{gathered} \Delta H_{rxn}=\sum \Delta H_{products}-\sum \Delta H_{reac\tan ts} \\ \Delta H_{rxn}=\left[2 mol\times \Delta H\left(N_2\right)+6 mol\times \Delta H\left(H_{2}O\right)\right]-\left[4 mol\times \Delta H\left(N{H}_3\right)+3 mol\times \Delta H\left(O_2\right)\right] \\ \Delta H_{rxn}=[2 mol\times 0.0 kJ/mol+6 mol\times (-241.826)kJ/mol] \\ -[4 mol\times (-45.90)kJ/mol+3 mol\times 0.00 kJ/mol] \\ \Delta H_{rxn}=-1267.356 kJ \end{gathered}$$

 $$\begin{gathered} \Delta S_{rxn}=\left[2 mol\times \Delta S\left(N_2\right)+6 mol\times \Delta S\left(H_{2}O\right)\right]-\left[4 mol\times \Delta S\left(N{H}_3\right)+3 mol\times \Delta S\left(O_2\right)\right] \\ \Delta S_{rxn}=\left[2 mol\times 191.50\frac{J}{mol\times K}+6 mol\times 188.72\frac{J}{mol\times K}\right] \\ -\left[4 mol\times 193.00\frac{J}{mol\times K}+3 mol\times 205.00\frac{J}{mol\times K}\right] \\ \Delta S_{rxn}=128.32\frac{J}{K}=0.12832\frac{kJ}{K} \end{gathered}$$

Then, the Gibb's energy for  ${\text{25}}^{\text{o}}\text{C}$ is –

 $$\begin{gathered} \Delta G_{25}=-1267.356 kJ-(25+273)K\times 0.12832\frac{kJ}{K} \\ \Delta G_{25}=-1305.595 kJ \end{gathered}$$

Finally, the  $K_p$, for step  , at  ${\text{25}}^{\text{o}}\text{C}$ is calculated below

 $$\begin{gathered} \ln K_p=\frac{-1305.595\times {10}^3 J/mol}{-8.314\frac{J}{mol·K}\times (25+273)K} \\ \ln K_p=526.966 \\ K=7.125\times {10}^{228} \end{gathered}$$

Therefore, the values obtained are  $\text{K = 1.38}\times {\text{10}}^{\text{168}}$ and  $\text{K = 7.125}\times {\text{10}}^{\text{228}}$.

Step 3: Calculation for K p

(b)

To recalculate the equilibrium constant at  ${900}^{\text{o}}\text{C}$, Gibb's energy and then $K$  should be calculated.

Then, the Gibb's energy for step  1  is –

 $$\begin{gathered} \Delta G_{900}=-906.196 kJ-(900+273)K\times 0.17792\frac{kJ}{K} \\ \Delta G_{900}=-1114.896 kJ \end{gathered}$$

Finally, the  ${\text{K}}_{\text{p}}$, for step  1, at ${\text{900}}^{\text{o}}\text{C}$  is –

$$\begin{gathered} \ln K_p=\frac{-959.216\times {10}^3 J/mol}{-8.314\frac{J}{mol·K}\times (900+273)K} \\ \ln K_p=114.321 \\ K=4.456\times {10}^{49} \end{gathered}$$

Then, the Gibb's energy for step 2  is –

 $$\begin{gathered} \Delta G_{900}=-1267.356 kJ-(900+273)K\times 0.12832\frac{kJ}{K} \\ \Delta G_{900}=-1417.875 kJ \end{gathered}$$

Finally, the  , for step  2, at  is –

 $$\begin{gathered} \ln K_p=\frac{-1305.595\times {10}^3 J/mol}{-8.314\frac{J}{mol·K}\times (25+273)K} \\ \ln K_p=145.388 \\ K=1.384\times {10}^{63} \end{gathered}$$

Therefore, the values obtained are  $\text{K = 4.456}\times {\text{10}}^{\text{49}}$ and  $\text{K = 1.384}\times {\text{10}}^{63}$.

Step 4: Calculation for Cost

(c)

At first, calculate the loss of  $Pt$ per metric ton $\left(t\right)$  of  $HN{O}_3$ produced –

 $$\begin{gathered} (Pt{)}_{lost}=m\left(HN{O}_3\right)\times m(Pt{)}_{loss} \\ m(Pt{)}_{lost}=\left(1.01\times {10}^{7}t\right)\times 175\frac{mg}{t} \\ m(Pt{)}_{lost}=1767.5\times {10}^{6}mg \\ =1767.5 kg \end{gathered}$$

Knowing the price of   per troy oz, the cost of the lost   can be calculated as –

 $$\begin{gathered} Cost=1767.5 kg\times \frac{32.15troyoz}{1 kg}\times \$1557\times \frac{1}{troyoz} \\ Cost=\$88476719.63 \\ =\$8.848\times {10}^8 \end{gathered}$$

Therefore, the value for cost is obtained as $\text{\$ 8.848}\times {\text{10}}^{\text{8}}$ .

Step 5: Calculation for Cost

(d)

Consider that $72\%$  efficiency filter is used, thus the cost of the recovered  $Pt$ is –

 $$\begin{gathered} m(Pt{)}_{recov}=1767.5 kg\times \frac{72\%}{100\%} \\ m(Pt{)}_{recov}=1272.6 kg \\ Cost{}_{recov}=1272.6 kg\times \frac{32.15troyoz}{1 kg}\times \$1557·\frac{1}{troyoz} \\ Cost{}_{recov}=\$63703238.13 \\ Cost{}_{recov}=\$6.37\times {10}^7 \end{gathered}$$

Therefore, the value for cost is obtained as  $\text{\$ 6.37}\times {\text{10}}^{\text{7}}$.