91Ó°ÊÓ

A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of 500° C.

The gases behave ideally despite the very high pressure.

The equilibrium constant at 500° C is 6.9 x 10^-5

Consider the following reaction: one part nitrogen and three parts hydrogen are heated to 500 degrees Celsius with a final pressure of 400 atmospheres.

${\mathrm{N}}_2+3{\mathrm{H}}_2→2{\mathrm{NH}}_3\left(1\right)$

In terms of partial pressure, the equilibrium constant for this reaction may be represented as:

$K=\frac{{\left(P_{{\mathrm{NH}}_3}/P_0\right)}^2}{\left(P_{{\mathrm{N}}_2}/P_0\right){\left(P_{{\mathrm{H}}_2}/P_0\right)}^3}$

Where,

$P_o$is atmospheric pressure

$K=\frac{P_{{\mathrm{NH}}_3}^2}{P_{{\mathrm{N}}_2}{P}_{{\mathrm{H}}_2}^3}$

At a temperature of 500 C, the constant is $K=6.9×{10}^{-5}$so,

$\frac{P_{{\mathrm{NH}}_3}^2}{P_{{\mathrm{N}}_2}{P}_{{\mathrm{H}}_2}^3}=6.9×{10}^{-5}\left(2\right)$

The total final pressure is:

$P_{{\mathrm{NH}}_3}+P_{{\mathrm{N}}_2}+P_{{\mathrm{H}}_2}=400\mathrm{atm}\left(3\right)$

Because every part of N2 requires three parts of H2, the ratio of partial derivatives reactants remains constant during the reaction, we may write:

$$\begin{gathered} \frac{P_{{\mathrm{N}}_2}}{P_{{\mathrm{H}}_2}}=\frac{1}{3} \\ {P}_{{\mathrm{H}}_2}=3P_{{\mathrm{N}}_2}\left(4\right) \end{gathered}$$

So we have three equations to solve for the three pressures: (2), (3), and (4). To eliminate P_H2, first substitute from (4) to (3), as follows:

$$\begin{gathered} P_{{\mathrm{NH}}_3}+P_{{\mathrm{N}}_2}+3P_{{\mathrm{N}}_2}=400\mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{NH}}_3}+4{\mathrm{P}}_{{\mathrm{N}}_2}=400\text{atm}\left(5\right) \end{gathered}$$

Substitute from from (4) into (2)

$$\begin{gathered} \frac{P_{{\mathrm{NH}}_3}^2}{27P_{{\mathrm{N}}_2}^4}=6.9×{10}^{-5} \\ {P}_{{\mathrm{NH}}_3}^2=\left(1.863×{10}^{-3}\right)P_{{\mathrm{N}}_2}^4 \\ {P}_{{\mathrm{NH}}_3}=\sqrt{1.863×{10}^{-3}}{P}_{{\mathrm{N}}_2}^2\left(6\right) \end{gathered}$$

Substitute from (6) into (5), to eliminate $P_{{\mathrm{NH}}_3}$

$$\begin{gathered} \sqrt{1.863×{10}^{-3}}{P}_{{\mathrm{N}}_2}^2+4P_{{\mathrm{N}}_2}=400\mathrm{atm} \\ \sqrt{1.863×{10}^{-3}}{\mathrm{P}}_{{\mathrm{N}}_2}^2+4{\mathrm{P}}_{{\mathrm{N}}_2}-400\mathrm{atm}=0 \end{gathered}$$

Solving the above quadratic equation, we get

$P_{{\mathrm{N}}_2}=\frac{-4Â\pm \sqrt{16+4\sqrt{1.863×{10}^{-3}}(400\mathrm{atm})}}{2\sqrt{1.863×{10}^{-3}}}$

Since the presSure cannot be negative, the accepted solution is:

$P_{{\mathrm{N}}_2}=60.50\mathrm{atm}$

Substitute into (4) to get,

$P_{{\mathrm{H}}_2}=3(60.50\mathrm{atm})=181.5\mathrm{atm}$

Substitute into (6), we get

$P_{{\mathrm{NH}}_3}=\sqrt{1.863×{10}^{-3}}(60.50\mathrm{atm}{)}^2=158\mathrm{atm}$

Substitute into (6) to get

$P_{{\mathrm{NH}}_3}=\sqrt{1.863×{10}^{-3}}(60.50\mathrm{atm}{)}^2=158\mathrm{atm}$

We must use the ideal gas law so: to convert these partial pressures to number of molecules.

$$\begin{gathered} N=\left(\frac{V}{kT}\right)P \\ N=CP \end{gathered}$$

Where C is constant

role="math" localid="1647207370400" $$\begin{gathered} \text{The number of}{\mathrm{N}}_2\text{molecules is}60.5C \\ \text{The number of molecules of}{\mathrm{NH}}_3\text{is}158C \\ The\text{number of atoms of}{\mathrm{N}}_2\text{is}60.5C×2=121C \\ \text{The number of atoms of}{\mathrm{NH}}_3\text{is}158 \\ \text{The number of nitrogen atoms is}121C+158C=279C \\ Hence,\text{the fraction of the nitrogen atom which converted to ammonia is}158C/279C=0.5663 \\ =56.6\% \end{gathered}$$