You are a member of a research team of chemists discussing the plans to operate an ammonia processing plant: ${\mathbf{N}}_{\mathbf{2}\left(g\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}\left(g\right)}\mathbf{}\mathbf{2}{\mathbf{NH}}_{\mathbf{3}\left(g\right)}$

(a) The plant operates at close to 700 K, at which ${\mathbf{K}}_{\mathbf{p}}$is role="math" localid="1654929481926" $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$, and employs the stoichiometric 1/3 ratio of ${\mathbf{N}}_{\mathbf{2}}\mathbf{/}{\mathbf{H}}_{\mathbf{2}}$. At equilibrium, the partial pressure of ${\mathbf{NH}}_{\mathbf{3}}$is 50atm. Calculate the partial pressures of each reactant and ${\mathbf{P}}_{\mathbf{total}}$.

(b) One member of the team suggests the following: since the partial pressure of ${\mathbf{H}}_{\mathbf{2}}$is cubed in the reaction quotient, the plant could produce the same amount of ${\mathbf{NH}}_{\mathbf{3}}$if the reactants were in a 1/6 ratio of ${\mathbf{N}}_{\mathbf{2}}\mathbf{/}{\mathbf{H}}_{\mathbf{2}}$and could do so at a lower pressure, which would cut operating costs. Calculate the partial pressure of each reactant and ${\mathbf{P}}_{\mathbf{total}}$under these conditions, assuming an unchanged partial pressure of 50. atm for ${\mathbf{NH}}_{\mathbf{3}}$. Is the suggestion valid?

Step by step solution

01

(a) Calculate the partial pressures of reactant and Ptotal. 

For the reaction; ${\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}2{\mathrm{OH}}_{3\left(\mathrm{g}\right)}$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{{{\mathrm{P}}^2}_{\mathrm{NH}3}}{{\mathrm{P}}_{{\mathrm{N}}_2}\times {\mathrm{P}}^3}\left(1\right) \\ \frac{{\mathrm{N}}_2}{{\mathrm{H}}_2}=\frac{1}{3} \\ \mathrm{We}\mathrm{have}, \\ {\mathrm{P}}_{\mathrm{H}2}=3{\mathrm{P}}_{\mathrm{N}2} \\ \mathrm{From}\left(1\right) \\ {\mathrm{K}}_{\mathrm{p}=}\frac{{{\mathrm{P}}^2}_{\mathrm{NH}3}}{{\mathrm{P}}_{{\mathrm{N}}_2}\times {\left({\mathrm{P}}_{\mathrm{H}2}\right)}^3} \\ =\frac{{{\mathrm{P}}^2}_{\mathrm{NH}3}}{27{{\mathrm{P}}^4}_{\mathrm{N}2}} \\ 1.00\times {10}^{-4}=\frac{{\left(50\right)}^2}{27{{\mathrm{P}}^4}_{\mathrm{N}2}} \\ {\mathrm{P}}_{\mathrm{N}=31\mathrm{atm}} \\ \mathrm{Hence}, \\ {\mathrm{P}}_{\mathrm{H}2}=3{\mathrm{P}}_{\mathrm{N}2} \\ =3\times 31\mathrm{atm} \\ =93\mathrm{atm} \\ {\mathrm{AndP}}_{\mathrm{NH}3}=50\mathrm{atm} \\ \mathrm{Thus},\mathrm{total}\mathrm{pressure}: \\ {\mathrm{P}}_{\mathrm{total}}=31\mathrm{atm}+93\mathrm{atm}+50\mathrm{atm} \\ =174\mathrm{atm} \\ \mathrm{Thus},\mathrm{the}\mathrm{partial}\mathrm{pressure}{\mathrm{ofP}}_{\mathrm{total}}\mathrm{is}174\mathrm{atm},{\mathrm{P}}_{\mathrm{H}2}\mathrm{is}50\mathrm{atm},\mathrm{and}{\mathrm{P}}_{\mathrm{N}2}\mathrm{is}31\mathrm{atm}. \end{gathered}$$

02

(b) Calculate the partial pressure of each reactant and Ptotal under the conditions

Now,

$$\begin{gathered} \frac{N_2}{H_2}=\frac{1}{6} \\ {H}_2=6N_2 \\ Hence,O{P}_{H_2}=6P_{N_2} \\ From\left(1\right), \\ {\mathrm{K}}_{\mathrm{p}=}\frac{{{\mathrm{P}}^2}_{{\mathrm{NH}}_3}}{{\mathrm{P}}_{{\mathrm{N}}_2}\times {\left({\mathrm{P}}_{\mathrm{H}2}\right)}^3} \\ =\frac{{{\mathrm{P}}^2}_{\mathrm{NH}3}}{216{{\mathrm{P}}^4}_{{\mathrm{N}}_2}} \\ 1.00\times {10}^{-4}=\frac{{\left(50\right)}^2}{216{{\mathrm{P}}^4}_{\mathrm{N}2}} \\ {\mathrm{P}}_{{\mathrm{N}}_2=18\mathrm{atm}} \\ {\mathrm{P}}_{\mathrm{H}2}=6{\mathrm{P}}_{\mathrm{N}2} \\ =6\times 18.44\mathrm{atm} \\ \approx 93\mathrm{atm} \\ {\mathrm{PNH}}_3=50\mathrm{atm} \\ \mathrm{Thus}, \\ {\mathrm{P}}_{\mathrm{total}}={\mathrm{P}}_{{\mathrm{NH}}_3}+{\mathrm{P}}_{{\mathrm{N}}_2}+{\mathrm{P}}_{{\mathrm{H}}_2} \\ =50\mathrm{atm}+18\mathrm{atm}+111\mathrm{atm} \\ =179\mathrm{atm} \\ \mathrm{Thus},\mathrm{the}\mathrm{partial}\mathrm{pressure}{\mathrm{ofP}}_{\mathrm{total}}\mathrm{is}179\mathrm{atm},{\mathrm{P}}_{\mathrm{H}2}\mathrm{is}111\mathrm{atm},\mathrm{and}{\mathrm{P}}_{\mathrm{N}2}\mathrm{is}18\mathrm{atm}. \end{gathered}$$

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