91影视

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

a)

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

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

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

$\begin{array}{c}\log K_{p1}=鈭�43.10\\ K_{p1}={10}^{鈭�43.10}\\ K_{p1}=7.9433脳{10}^{鈭�44}.\end{array}$

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

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

Therefore, $P_{\text{N}}=4.0脳{10}^{鈭�21}$.

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.

$\begin{array}{l}{K}_{p^2}=鈭�17.30\\ K_{p^2}={10}^{鈭�17.30}\\ K_{p^2}=5.0119脳{10}^{鈭�18}\end{array}$

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

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

Therefore, $P_{\text{H}}=5.5脳{10}^{鈭�8}$.

(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.

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

Solve for the number of moles of N

$PV\&=nRT$

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

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

$\frac{4.87脳{10}^{鈭�23}\text{mol}}{\text{L}}脳\frac{6.022脳{10}^{23}\text{atoms}}{1\text{mol}}=29\text{Natomsperliter}$

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

$PV=nRT$

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

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

$\begin{array}{l}\frac{6.70脳{10}^{鈭�10}\text{mol}}{\text{L}}脳\frac{6.022脳{10}^{23}\text{atoms}}{1\text{mol}}\\ =4.0脳{10}^{14}\text{Hatomsperliter.}\end{array}$

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

d)

Let us evaluate each possible step.

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

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.

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

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.