91Ó°ÊÓ

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 ( ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$).

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 role="math" localid="1663400587610" $∆G$. In this case, Gibb's energy can be calculated as –

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

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

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

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

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

$$\begin{gathered} \mathrm{Δ}{G}_{25}=-906.196kJ-(25+273)K×0.17792\frac{kJ}{K} \\ \mathrm{Δ}{G}_{25}=-959.216kJ \end{gathered}$$

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

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

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

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

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

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

$$\begin{gathered} \mathrm{Δ}{G}_{25}=-1267.356kJ-(25+273)K×0.12832\frac{kJ}{K} \\ \mathrm{Δ}{G}_{25}=-1305.595kJ \end{gathered}$$

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

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

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

(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 –

$$\begin{gathered} \mathrm{Δ}{G}_{900}=-906.196kJ-(900+273)K×0.17792\frac{kJ}{K} \\ \mathrm{Δ}{G}_{900}=-1114.896kJ \end{gathered}$$

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

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

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

$$\begin{gathered} \mathrm{Δ}{G}_{900}=-1267.356kJ-(900+273)K×0.12832\frac{kJ}{K} \\ \mathrm{Δ}{G}_{900}=-1417.875kJ \end{gathered}$$

Finally, the , for step 2, at is –

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

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

(c)

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

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

Knowing the price ofper troy oz, the cost of the lostcan be calculated as –

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

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

(d)

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

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

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