Question

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}.\)

Short answer

The total pressure, \(P_{Total}\), of the system at equilibrium is obtained by evaluating \(\sqrt[5]{4.31 \times 10^{-4}}\).

Step-by-step solution

Interpreting the Given Information

You need to understand the given information and set up your strategy. You are given that 1 mole of \(N_{2}\) and 3 moles of \(H_{2}\) are placed in a flask to react via the equation: \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \) . The reaction reaches equilibrium at 397°C, the mole fraction of NH3 is given as 0.21, and \(K_p = 4.31 \times 10^{-4} \).

Calculating the Partial Pressure of \(NH_3\)

The mole fraction of \(NH_3\) equals the partial pressure of \(NH_3\), \(x_{NH_3}\), over the total pressure \(P_{Total}\): \(x_{NH_3} = \frac{P_{NH_3}}{P_{Total}} \). We can rearrange this to calculate the partial pressure of \(NH_3\): \(P_{NH_3} = x_{NH_3} \times P_{Total} \). Using this, the fact that we need the total pressure, and the given \(K_p\) value, we can form the equation: \(K_p = \frac{P_{NH_3}^{2}}{P_{N_2} P_{H_2}^{3}} = \frac{(x_{NH_3} \times P_{Total})^{2}}{(0.79 \times P_{Total}) (0.79^{2} \times P_{Total}^{3})} = \frac{(0.21 \times P_{Total})^{2}}{(0.79 \times P_{Total}) (0.79^{3} \times P_{Total}^{3})} \). Note that the mole fractions of \(N_{2}\) and \(H_{2}\) are \(1-0.21 = 0.79 \) because at equilibrium the sum of all mole fractions equals 1.

Solving for the Total Pressure

Upon rearranging the equation from step 2, we get \(P_{Total} = \sqrt[5]{K_p} \), where the root is 5 because the denominator in step 2 has \(P_{Total}\) raised to the power 5. Substituting the given value for \(K_p\), we find \(P_{Total} = \sqrt[5]{4.31 \times 10^{-4}}\).

Final Calculation

Calculate \(P_{Total}\) to get the final answer.

Key concepts

Partial Pressure

In chemistry, the concept of partial pressure is essential when dealing with gas mixtures in reactions. Each gas in a mixture exerts pressure as if it alone occupied the entire volume. This pressure is referred to as its partial pressure. In the context of the reaction involving nitrogen (\(N_{2}\)) and hydrogen (\(H_{2}\)) to form ammonia (\(NH_{3}\)), the partial pressure of each gas plays a crucial role in solving equilibrium problems.
To simplify, consider the partial pressure of ammonia, \(P_{NH_3}\), in our exercise. It can be related to the mole fraction, \(x_{NH_3}\), and the total pressure of the system, \(P_{Total}\), through this formula:

• \(P_{NH_3} = x_{NH_3} \times P_{Total}\)

This relationship helps in determining one unknown if the others are given.
By understanding partial pressures, you can predict how changes in conditions affect the concentration of the products and reactants at equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant for gases, denoted as \(K_p\), provides insight into the extent of a reaction at equilibrium. It is a ratio that compares the partial pressures of the derived products and original reactants, each raised to the power of their respective coefficients in the balance equation. For our ammonia synthesis reaction, the equilibrium constant is defined as:

• \[K_p = \frac{P_{NH_3}^{2}}{P_{N_2} P_{H_2}^{3}}\]

In this case, the given \(K_p\) value is \(4.31 \times 10^{-4}\). This value indicates a small tendency for the reaction to form products under standard conditions.
Understanding \(K_p\) is crucial for predicting the direction of the reaction upon any disturbance of equilibrium, such as pressure or composition changes. An equilibrium constant only changes with temperature, aiding in the anticipation of reaction shifts in different conditions.

Mole Fraction

The mole fraction is an essential concept when dealing with mixtures. It represents the ratio of the number of moles of a particular component to the total number of moles in the mixture. In our ammonia synthesis reaction, each gas's mole fraction helps in understanding its relative abundance in the reaction system.
The mole fraction of ammonia, \(x_{NH_3}\), is calculated and given as 0.21. It is essential for determining the partial pressure. The sum of the mole fractions of all components in a system always equals 1. Therefore, the mole fraction for the nitrogen and hydrogen is \(1 - x_{NH_3} = 0.79\).

• Mole fraction of \(NH_3\): \(x_{NH_3} = 0.21\)
• Mole fraction of other gases: \(x = 0.79\)

Understanding these fractions aids in calculating equilibrium concentrations and predicting the system’s behavior under different conditions.

Ammonia Synthesis Reaction

The ammonia synthesis reaction is a crucial industrial process, famously known as the Haber process. It involves combining nitrogen (\(N_{2}\)) and hydrogen (\(H_{2}\)) to form ammonia (\(NH_{3}\)) under specific conditions of temperature and pressure. The balanced equation for the reaction is:

• \[N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\]

This reversible reaction reaches equilibrium, where the forward and reverse rates equalize, resulting in constant concentrations of products and reactants.
The synthesis of ammonia is vital for manufacturing fertilizers and other chemical products. Understanding this reaction involves applying principles of chemical equilibrium, such as using the equilibrium constant (\(K_p\)) and partial pressures, to optimize the production under varying conditions.