Q17.104CP - Chemistry: Molecular Nature Of Matter And Change

Question

One mechanism for the synthesis of ammonia proposes that ${\text{N}}_2$ and ${\text{H}}_2$ molecules catalytically dissociate into atoms:

 $\begin{array}{l}{\text{N}}_2\left(g\right)\rightleftharpoons 2 \text{N}\left(g\right)log{K}_{\text{p}}=-43.10\\ {\text{H}}_2\left(g\right)\rightleftharpoons 2\text{H}\left(g\right)log{K}_{\text{p}}=-17.30\end{array}$

(a) Find the partial pressure of $\text{N}$ in ${\text{N}}_2$ at 1000 and  200atm.

(b) Find the partial pressure of  $\text{H}$ in ${\text{H}}_2$ at 1000 and 600atm.

(c) How many $\text{N}$ atoms and $\text{H}$ 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.

$\begin{array}{l}\text{N}\left(g\right)+\text{H}\left(g\right)\to \text{NH}\left(g\right)\\ {\text{N}}_2\left(g\right)+\text{H}\left(g\right)\text{NH}\left(g\right)+\text{N}\left(g\right)\end{array}$

Step-by-Step Solution

Answer

a)  .$P_{\text{N}}=4.0\times {10}^{-21}$

b)  .$P_{\text{H}}=5.5\times {10}^{-8}$

c)  $4.0\times {10}^{14}\text{H}$ Atoms per litre.

d) When comparing the two possible reaction steps, possible reaction step (2) appears to be the more reasonable option, as it maximises NH generation.

Step 1: Definition of pressure

Pressure is defined as a measure of the force applied over a unit area.

Step 2: Calculate the partial pressure of N

a)

In this problem, we are tasked to evaluate the given two-step reaction shown below.

 $\begin{array}{l}{\text{N}}_2\left(g\right)\rightleftharpoons 2 \text{N}\left(g\right)\log K_{p1}=-43.10\\ {\text{H}}_2\left(g\right)\rightleftharpoons 2\text{H}\left(g\right)\log K_{p2}=-17.30\end{array}$

First, solve for the $K_p$ of the reaction (1).

 $\begin{array}{c}\log K_{p1}=-43.10\\ K_{p1}={10}^{-43.10}\\ K_{p1}=7.9433\times {10}^{-44}.\end{array}$

Next, write the expression for the equilibrium constant of the reaction in terms of partial pressures, and solve for  $P_{\text{N}}$ .

 $\begin{array}{c}{K}_{p1}=\frac{P_{\text{products }}}{P_{\text{reactants }}}\\ K_{p1}=\frac{P_{\text{N}}^2}{P_{{\text{N}}_2}}{P}_{\text{N}}^2\\ =K_{p1}\cdot P_{{\text{N}}_2}{P}_{\text{N}}\\ =\sqrt{K_{p1}\cdot P_{{\text{N}}_2}}\\ =\sqrt{\left(7.4933\times {10}^{-44}\right)\left(200\right)}\\ P_{\text{N}}=4.0\times {10}^{-21}.\end{array}$

 

 

Therefore,  $P_{\text{N}}=4.0\times {10}^{-21}$.

Step 3: Calculate the partial pressure of H

b) 

We will do the same method in solving $P_{\text{N}}$ , but instead, we will solve for $P_{\text{H}}$ .

Solve for the  $K_p$ of reaction (2), first.

 $\begin{array}{l}{K}_{p^2}=-17.30\\ K_{p^2}={10}^{-17.30}\\ K_{p^2}=5.0119\times {10}^{-18}\end{array}$

Next, write the expression for the equilibrium constant of the reaction in terms of partial pressures, and solve for $P_{\text{H}}$ .

 $\begin{array}{c}{K}_{p2}=\frac{P_{\text{products }}}{P_{\text{reactants }}}\\ K_{p2}=\frac{P_{\text{H}}^2}{P_{{\text{H}}_2}}\\ P_{\text{H}}^2=K_{p2}\cdot P_{{\text{H}}_2}\\ P_{\text{H}}=\sqrt{K_{p2}\cdot P_{{\text{H}}_2}}\\ =\sqrt{\left(5.0119\times {10}^{-18}\right)\left(600\right)}\\ P_{\text{H}}=5.5\times {10}^{-8}\end{array}$

Therefore,  $P_{\text{H}}=5.5\times {10}^{-8}$.

Step 4: Calculate the atoms per litre

(c)

To solve for the number of atoms of $\text{N}$ and $\text{H}$ are present per litre, we will solve for the number of moles first using the ideal gas law equation. Identify the given values.

 $\begin{array}{l}{P}_{\text{N}}=4.0\times {10}^{-21} \text{atm}\\ P_{\text{H}}=5.5\times {10}^{-8} \text{atm}\\ V=1.0 \text{L}\\ T=100 \text{K}\\ R=0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\end{array}$

Solve for the number of moles of N

 $PV\&=nRT$

$\begin{array}{c}\frac{n_{\text{N}}}{V}=\frac{P_{\text{N}}}{lTT}\\ =\frac{\left(1.0\times {10}^{-21} \text{atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)(1000 \text{K})}\\ \frac{n_{\text{N}}}{V}=4.87\times {10}^{-23 \text{mol}/\text{L}}\end{array}$

 

Solve for the number of atoms per litre of N by using dimensional analysis.

 $\frac{4.87\times {10}^{-23} \text{mol}}{ \text{L}}\times \frac{6.022\times {10}^{23}\text{ atoms }}{1 \text{mol}}=29 \text{N atoms per liter }$

Next, solve for the number of moles of  $\text{H}$ .

 $PV=nRT$

 $\begin{array}{c}\frac{n_{\text{H}}}{V}=\frac{P_{\text{H}}}{RT}\\ =\frac{\left(5.5\times {10}^{-8} \text{atm}\right)}{\left(0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\right)(1000 \text{K})}\\ \frac{n_{\text{N}}}{V}=6.70\times {10}^{-10} \text{mol}/\text{L}\end{array}$

Solve for the number of atoms per litre of $\text{H}$ by using dimensional analysis.

 $\begin{array}{l}\frac{6.70\times {10}^{-10} \text{mol}}{ \text{L}}\times \frac{6.022\times {10}^{23}\text{ atoms }}{1 \text{mol}}\\ =4.0\times {10}^{14}\text{H atoms per liter .}\end{array}$

Therefore, the required value is  .$4.0\times {10}^{14}\text{H}$

Step 5: Which of the following is a more reasonable step

d)

Let us evaluate each possible step.

 .$\text{N}\left(g\right)+\text{H}\left(g\right)\to \text{NH}\left(g\right)$

Possible reaction step (1) has  $\text{N}$and  $\text{H}$as reactants. Noted from our previous reactions that  $\text{N}$has low yield with only 29 atoms per litre. With this amount, the reaction will also have a low yield.

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

Possible reaction step (2) has ${\text{N}}_2$  and $\text{H}$ as reactants. From our previous calculations,  ${\text{N}}_2$ has a high partial pressure (200 atm). With this amount, the reaction will also have a high yield.

Therefore, comparing the two possible reaction steps, possible reaction step (2) is the more reasonable step since it will maximize the production of NH.