Q105CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$, 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

Answer

a) The partial pressure of ${\mathrm{P}}_{\mathrm{total}}$ is 174 atm, ${\mathrm{P}}_{\mathrm{H}2}$is 50 atm, and ${\mathrm{P}}_{\mathrm{N}2}$is 31 atm.

b) The partial pressure of ${\mathrm{P}}_{\mathrm{total}}$ is 179 atm, ${\mathrm{P}}_{\mathrm{H}2}$is 111 atm, and ${\mathrm{P}}_{\mathrm{N}2}$is 18 atm.

Step1: (a) Calculate the partial pressures of reactant and P total .

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

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

Step2: (b) Calculate the partial pressure of each reactant and P total under the conditions

Now,

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