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. What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Pretend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at 500° C is 6.9 x 10^-5. (Hint: You'l have to solve a quadratic equation.)

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.

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

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 $K=6.9\times {10}^{-5}$so,

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

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)

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

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

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

Solving the above quadratic equation, we get

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

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

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

Substitute into (6) to get

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

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