Question

One mole of \(\mathrm{N}_{2}\) and three moles of \(\mathrm{H}_{2}\) are placed in a flask at \(375^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is 0.21 . The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}\).

Short answer

The total pressure at equilibrium is approximately calculated from the expression using the equilibrium mole fractions and the equilibrium constant value.

Step-by-step solution

Write the Balanced Equation and Initial Setup

The balanced chemical equation for the formation of ammonia is \( \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \). We start with one mole of \( \mathrm{N}_2 \) and three moles of \( \mathrm{H}_2 \). At equilibrium, the mole fraction of \( \mathrm{NH}_3 \) is given as 0.21.

Define Mole Fractions

Let \( x \) be the number of moles of \( \mathrm{NH}_3 \) at equilibrium. The mole fraction, which is \( \frac{2x}{1 + 3 + 2x} \), is given as 0.21. Set up the equation: \( \frac{2x}{4 + 2x} = 0.21 \). Solve for \( x \).

Solve for \( x \)

Rearrange the equation \( \frac{2x}{4 + 2x} = 0.21 \) to \( 2x = 0.21(4 + 2x) \). Simplify to get \( 2x = 0.84 + 0.42x \), which simplifies further to \( 1.58x = 0.84 \). Solving for \( x \) gives \( x \approx 0.532 \) moles.

Calculate Equilibrium Moles of Reactants and Products

From \( x = 0.532 \):\[ \begin{align*} \text{Moles of} & \ \mathrm{NH}_3 = 2x = 1.064 \, \text{moles}, \ \text{Moles of} & \ \mathrm{N}_2 = 1 - x = 0.468 \, \text{moles}, \ \text{Moles of} & \ \mathrm{H}_2 = 3 - 3x = 1.404 \, \text{moles}. \end{align*} \]

Calculate Total Moles and Partial Pressures

Total moles at equilibrium is \( 0.468 + 1.404 + 1.064 = 2.936 \). Mole fractions: \( y_{\mathrm{NH}_3} = \frac{1.064}{2.936} \), \( y_{\mathrm{N}_2} = \frac{0.468}{2.936} \), \( y_{\mathrm{H}_2} = \frac{1.404}{2.936} \). Use partial pressures: \( P_{\mathrm{NH}_3} = y_{\mathrm{NH}_3} P_{\text{total}} \), etc.

Apply Equilibrium Constant Expression

Using the equation \( K_p = \frac{(P_{\mathrm{NH}_3})^2}{P_{\mathrm{N}_2}(P_{\mathrm{H}_2})^3} = 4.31 \times 10^{-4} \), substituting partial pressures \( P_{\mathrm{NH}_3} = y_{\mathrm{NH}_3} P_{\text{total}} \), solve for \( P_{\text{total}} \).

Solve for Total Pressure

Plug the mole fractions into the expression for \( K_p \): \[ \left(\frac{1.064}{2.936} \cdot P_{\text{total}}\right)^2 = 4.31 \times 10^{-4} \left( \frac{0.468}{2.936} \cdot P_{\text{total}} \right) \left( \frac{1.404}{2.936} \cdot P_{\text{total}} \right)^3 \]. Solve this equation to find \( P_{\text{total}} \).

Solution Verification

Ensure that the calculated \( P_{\text{total}} \) satisfies the conditions of the equilibrium constant \( K_p \) within the context of the given mole fractions and initial conditions.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions happen at equal rates. It results in constant concentrations of reactants and products over time. This doesn't mean the amounts are equal, but that their ratios remain unchanged. In this example, the formation of ammonia (\( \mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3\) ) reaches equilibrium at a certain pressure and temperature. Once equilibrium is established, the concentrations of nitrogen, hydrogen, and ammonia remain stable.

Understanding equilibrium in chemical reactions helps explain many natural phenomena. It's essential in optimizing industrial processes, like ammonia production via the Haber process. Recognizing when a reaction reaches equilibrium allows chemists to control conditions for maximum yield.

Mole Fraction

Mole fraction is a way of expressing the composition of a mixture using ratios. It represents the ratio of the number of moles of a component to the total moles in the mixture. In mathematical terms, the mole fraction, denoted as \( y_i \), is calculated using: \[y_i = \frac{n_i}{n_{\text{total}}}\]where \( n_i \) is the moles of the component, and \( n_{\text{total}} \) is the total moles of all components in the system.

For the given problem, the mole fraction of ammonia was provided as 0.21. This means that for every mole of mixed gases present at equilibrium, 0.21 moles are ammonia. Understanding and calculating mole fractions are crucial for determining partial pressures and applying equilibrium constant expressions. In essence, it gives a snapshot of the composition of a mixture at equilibrium.

Equilibrium Constant Expression

The equilibrium constant expression is a mathematical formula that represents the relationship between the concentrations (or pressures) of reactants and products in an equilibrium state. For reactions involving gases, we often use the equilibrium constant \( K_p \), which is based on partial pressures rather than concentrations. For the reaction \( \mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3 \), the equilibrium constant expression is:\[K_p = \frac{(P_{\mathrm{NH}_3})^2}{P_{\mathrm{N}_2}(P_{\mathrm{H}_2})^3}\]Here, \( P_{\mathrm{NH}_3} \), \( P_{\mathrm{N}_2} \), and \( P_{\mathrm{H}_2} \) represent the partial pressures of ammonia, nitrogen, and hydrogen, respectively.

The given \( K_p \) value helps determine how pressures at equilibrium relate to each other, guiding adjustments to system conditions and predicting shifts in the equilibrium position. It’s a crucial part of solving equilibrium problems in chemistry.

Partial Pressures

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. Each gas contributes to the total pressure in proportion to its mole fraction. The relationship between the mole fraction and partial pressure is given by:\[P_i = y_i \times P_{\text{total}}\]where \( P_i \) is the partial pressure of gas \( i \), \( y_i \) is its mole fraction, and \( P_{\text{total}} \) is the total pressure of the system.

In the equilibrium system, knowing the mole fractions allows us to calculate the partial pressures of nitrogen, hydrogen, and ammonia. These partial pressures are essential when plugging values into the equilibrium constant expression. Calculating partial pressures helps in understanding how each component affects the overall system pressure and achieving a meaningful interpretation of equilibrium conditions.