Question

One mechanism for the synthesis of ammonia proposes that ${}^{{\mathbf{H}}_{\mathbf{2}}}$molecules catalytically dissociate into atoms:

$$\begin{gathered} {\mathbf{N}}_{\mathbf{2}\left(g\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{N}}_{\left(g\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{-}\mathbf{43}\mathbf{.}\mathbf{10} \\ {\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{H}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{30} \end{gathered}$$

(a) Find the partial pressure of N in${}^{{\mathbf{N}}_{\mathbf{2}}}$at 1000. K and 200. atm.

(b) Find the partial pressure of H in${}^{{\mathbf{H}}_{\mathbf{2}}}$at 1000. K and 600. atm.

(c) How many N atoms and H atoms are present per litre?

(d) Based on these answers, which of the following is a more reasonable step to continue the mechanism after the catalytic dissociation? Explain

$$\begin{gathered} {\mathbf{N}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\mathbf{\to }\mathbf{}{\mathbf{NH}}_{\left(g\right)} \\ {\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{}{\mathbf{NH}}_{\left(g\right)}\mathbf{+}\mathbf{N}\left(g\right) \end{gathered}$$

Short answer

1. The partial pressure of N is ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}}$.

2. The partial pressure of H is ${}^{\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}$.

3. The number of N atoms in a liter is 29.5. the number of H atoms in a liter is ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}}$.

4. The number of molecules of ${}^{{\mathbf{N}}_{\mathbf{2}}}$is quite high (due to its high partial pressure) and it can react with atoms of H to produce a considerable amount of NH. Thus, the step is more reasonable.

localid="1655189421474" ${}^{{\mathbf{N}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\mathbf{\to }\mathbf{}{\mathbf{NH}}_{\left(g\right)}\mathbf{+}\mathbf{}{\mathbf{N}}_{\left(g\right)}}$

Step-by-step solution

(a) Find the partial pressure of N in  at 1000. K and 200. atm.

The reaction for thedissociation of nitrogenmolecule into its element is given below:

${}^{{\mathbf{N}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{N}}_{\left(g\right)}}$

The expression for the equilibrium constant $\left({}^{{\mathbf{K}}_{\mathbf{p}}}\right)$of the reaction is,

${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{{\mathbf{P}}^{\mathbf{2}}}_{\mathbf{N}}}{{\mathbf{P}}_{{\mathbf{N}}_{\mathbf{2}}}}$

The value of ${}^{\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{p}}}$for the reaction is -43.10. Then calculate the value of ${}^{{\mathbf{K}}_{\mathbf{p}}}$as follows:

$$\begin{gathered} \mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{43}\mathbf{.}\mathbf{10} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{43}\mathbf{.}\mathbf{10}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{44}} \end{gathered}$$

The partial pressure of role="math" localid="1655098427719" ${}^{{\mathbf{N}}_{\mathbf{2}}}$at 1000 K is 200 atm.

Substitute the value in relation (1), and calculate ${}^{{\mathbf{P}}_{\mathbf{N}}}$:

$$\begin{gathered} \mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{44}}\mathbf{=}\frac{{{\mathbf{P}}^{\mathbf{2}}}_{\mathbf{N}}}{\mathbf{200}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{P}}_{\mathbf{N}}\mathbf{=}\sqrt{\left(7.9\times {10}^{-44}\right)\left(200\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}} \end{gathered}$$

Thus, the partial pressure of N is ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}}$.

(b) Find the partial pressure of H in  at 1000. K and 600 atm

The reaction for the dissociation of hydrogen molecule into its element is given below:

${}^{{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{H}}_{\left(g\right)}}$

The expression for the equilibrium constant $\left({}^{{\mathbf{K}}_{\mathbf{p}}}\right)$of the reaction is,

${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{{\mathbf{P}}^{\mathbf{2}}}_{\mathbf{H}}}{{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}}$

The value of ${}^{\mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{p}}}$for the reaction is -17.30. Then calculate the value of ${}^{{\mathbf{K}}_{\mathbf{p}}}$as follows:

$$\begin{gathered} \mathbf{log}\mathbf{}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{30} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{30}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}} \end{gathered}$$

The partial pressure of ${\mathrm{H}}_2$at 1000 K is 600 atm.

Substitute the values in relation (1), and calculate ${}^{{\mathbf{P}}_{\mathbf{H}}}$:

$$\begin{gathered} \mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{=}\frac{{{\mathbf{P}}^{\mathbf{2}}}_{\mathbf{H}}}{\mathbf{600}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{P}}_{\mathbf{H}}\mathbf{=}\sqrt{\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{)}\mathbf{\left(}\mathbf{600}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}} \end{gathered}$$

Thus, the partial pressure of H is${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}$

(c) How many N atoms and H atoms are present per litre?

Ideal gas law is given below:

PV=nRT

Here, P is pressure, V is volume, n is number of moles of gas, R is gas constant and T is absolute temperature.

Calculate moles of N formed from its partial pressure ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{21}}\mathbf{}\mathbf{atm}}$and volume 1.0 L as follows:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{N}}=\frac{{\mathrm{P}}_{\mathrm{N}}\mathrm{V}}{\mathrm{RT}} \\ =\frac{(4.0\times {10}^{-21}\mathrm{atm})(1.0\mathrm{L})}{(0.08206\mathrm{L}.\mathrm{atm}.\mathrm{mol}-1\mathrm{K}-1)(1000\mathrm{K})} \\ =4.9\times {10}^{-23}\mathrm{mol} \end{gathered}$$

Then multiply the moles of H with Avogadro number ${}^{\left(6.023\times {10}^{-23}{\mathrm{mol}}^{-1}\right)}$, to obtain atoms of H.

localid="1655190390654" $$\begin{gathered} 4.9\times {10}^{-23}\mathrm{mol}\mathrm{N}\times \frac{6.023\times {10}^{23}\mathrm{N}\mathrm{atoms}}{\mathrm{mol}\mathrm{N}} \\ =29.5\mathrm{N}\mathrm{atoms} \end{gathered}$$

Thus, the number of N atoms in a litre is 29.5.

Calculate moles of H formed from its partial pressure ${}^{\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{atm}}$and volume 1.0 L as follows:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{H}}=\frac{{\mathrm{P}}_{\mathrm{H}}\mathrm{V}}{\mathrm{RT}} \\ =\frac{(5.5\times {10}^{-8}\mathrm{atm})(1.0\mathrm{L})}{(0.08206\mathrm{L}.\mathrm{atm}.\mathrm{mol}-1\mathrm{K}-1)(1000\mathrm{K})} \\ =6.7\times {10}^{-10}\mathrm{mol} \end{gathered}$$

Then multiply the moles of H with Avogadro number ${}^{(6.023\times {10}^{-23}{\mathrm{mol}}^{-1})}$, to obtain atoms of H.

$$\begin{gathered} 6.7\times {10}^{-10}\mathrm{mol}\mathrm{H}\times \frac{6.022\times {10}^{23}\mathrm{H}\mathrm{atoms}}{\mathrm{mol}\mathrm{H}} \\ =4.0\times {10}^{24}\mathrm{H}\mathrm{atoms} \end{gathered}$$

Thus, the number of H atoms in a litre is ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}}$

(d) Which of the following is a more reasonable step?

Consider the step as follows:

${\mathrm{N}}_{\left(\mathrm{g}\right)}+{\mathrm{H}}_{\left(\mathrm{g}\right)}\to {\mathrm{NH}}_{\left(\mathrm{g}\right)}$

The number of atoms of N is only 29, where as number of atoms of H. So, the production of NH is very low. Thus, the step is not reasonable.

Consider the other step as follows:

${{\mathrm{N}}_2}_{\left(\mathrm{g}\right)}+{\mathrm{H}}_{\left(\mathrm{g}\right)}\to {\mathrm{NH}}_{\left(\mathrm{g}\right)}+{\mathrm{N}}_{\left(\mathrm{g}\right)}$

The number of molecules of ${}^{{\mathbf{N}}_{\mathbf{2}}}$is quite high (due to its high partial pressure) and it can react with atoms of H to produce considerable amount of NH. Thus, the step is more reasonable.