Problem 124 - Chemistry The Molecular Science

Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ has the value \(5.97 \times 10^{-2}\) at \(500 .{ }^{\circ} \mathrm{C}\). If \(1.00 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{gas}\) and \(1.00 \mathrm{~mol} \mathrm{H}_{2}\) gas are heated to \(500 .{ }^{\circ} \mathrm{C}\) in a \(10.00-\mathrm{L}\) sealed flask together with a catalyst, calculate the percentage of \(\mathrm{N}_{2}\) converted to \(\mathrm{NH}_{3}\). (Hint: Assume that only a very small fraction of the reactants is converted to products. Obtain an approximate answer and use it to obtain a more accurate result.)

Step-by-Step Solution

Answer

Approximately 0.0122% of \(\text{N}_2\) is converted to \(\text{NH}_3\).

Step 1: Write the Expression for Kc

The equilibrium constant expression for the given reaction is: \[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] where \([\text{NH}_3]\), \([\text{N}_2]\), and \([\text{H}_2]\) are the equilibrium concentrations of ammonia, nitrogen, and hydrogen, respectively.

Step 2: Initial Concentrations

Determine the initial concentrations of the gases. Since the volume of the flask is 10.00 L, the initial concentration of \(\text{N}_2\) is \(\frac{1.00}{10.00} = 0.100\, \text{mol/L}\), and for \(\text{H}_2\), it is also \(\frac{1.00}{10.00} = 0.100\, \text{mol/L}\). There's initially no \(\text{NH}_3\) in the flask.

Step 3: Define Changes in Concentrations

Let the change in concentration of \(\text{N}_2\) be \(x\). According to the reaction stoichiometry, the change in \(\text{NH}_3\) will be \(+2x\) and in \(\text{H}_2\) will be \(-3x\). Thus, at equilibrium, the concentrations will be:- \([\text{N}_2] = 0.100 - x\),- \([\text{H}_2] = 0.100 - 3x\),- \([\text{NH}_3] = 2x\).

Step 4: Set Up the Equilibrium Expression

Substitute the equilibrium concentrations into the expression for \(K_c\):\[ 5.97 \times 10^{-2} = \frac{(2x)^2}{(0.100 - x)(0.100 - 3x)^3} \]

Step 5: Approximate and Solve for x

Assume \(x\) is small compared to the initial concentrations due to the given hint, so \(0.100 - x \approx 0.100\) and \(0.100 - 3x \approx 0.100\). Simplifying:\[ 5.97 \times 10^{-2} = \frac{4x^2}{(0.100)^4} \]Solving for \(x\):\[ 4x^2 = 5.97 \times 10^{-2} \times 10^{-8} \]\[ x^2 = 1.4925 \times 10^{-10} \]\[ x = \sqrt{1.4925 \times 10^{-10}} \approx 1.221 \times 10^{-5} \].

Step 6: Calculate Percentage Conversion

The percentage of \(\text{N}_2\) converted to \(\text{NH}_3\) is given by:\[ \text{Percentage conversion} = \left(\frac{x}{0.100}\right) \times 100\% \]Substituting the value of \(x\):\[ \text{Percentage conversion} = \left(\frac{1.221 \times 10^{-5}}{0.100}\right) \times 100\% \approx 0.0122\% \]

Key Concepts

Equilibrium Constant (Kc)

In any reversible chemical reaction, a state of balance or equilibrium is reached where the rates of the forward and reverse reactions are equal. This balance is characterized by the equilibrium constant, represented as \( K_c \). The value of \( K_c \) indicates the extent to which reactants are converted into products at equilibrium conditions. For the equilibrium reaction \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \), the expression for \( K_c \) is given by:\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] Here, \([\text{NH}_3]\), \([\text{N}_2]\), and \([\text{H}_2]\) are the molar concentrations of ammonia, nitrogen, and hydrogen at equilibrium.The magnitude of \( K_c \) provides insight into the position of equilibrium:

• If \( K_c \) is large, the equilibrium position is product-favored.
• If \( K_c \) is small, the equilibrium position is reactant-favored.

In our example with \( K_c = 5.97 \times 10^{-2} \), the relatively small value suggests that under these conditions, the reaction significantly favors the reactants, and only a tiny amount of \( \text{N}_2 \) and \( \text{H}_2 \) transforms into \( \text{NH}_3 \). This understanding helps us predict the reaction's behavior and the minor changes in concentration characteristics at equilibrium.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict how much products will be formed from given reactants or how much of the reactants are required to form desired products.In the reaction \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \), stoichiometry guides us in understanding the ratio in which each species interacts:

• For every 1 mole of \( \text{N}_2 \) used, 3 moles of \( \text{H}_2 \) are needed.
• From this, 2 moles of \( \text{NH}_3 \) are produced.

Using stoichiometry, we can define changes in concentration during the attainment of equilibrium. If we denote \( x \) as the change in concentration of \( \text{N}_2 \), then the corresponding changes for \( \text{H}_2 \) and \( \text{NH}_3 \) are \(-3x\) and \(+2x\) respectively, due to their stoichiometric coefficients in the balanced equation.This relationship is instrumental when calculating equilibrium concentrations as it simplifies the setup of the reaction's dynamic equilibrium. Understanding and utilizing stoichiometry allows for accurate predictions of the extent of chemical reactions and efficient calculation of unknown quantities in chemical processes.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products in a reaction mixture when it has reached a state of equilibrium. These concentrations are crucial for calculating the equilibrium constant and determining the extent of the reaction.Upon setting up the initial concentrations of \( \text{N}_2 \) and \( \text{H}_2 \) in a 10 L flask — both at \(0.100 \, \text{mol/L}\), and acknowledging that \( \text{NH}_3 \) starts at zero, we move through the reaction:

• Let \( x \) be the amount of \( \text{N}_2 \) reacting.
• The equilibrium concentration of \( \text{N}_2 \) becomes \( 0.100 - x \).
• For \( \text{H}_2 \), with triple its stoichiometric requirement, the equilibrium is \( 0.100 - 3x \).
• And, produced \( \text{NH}_3 \) arises to \( 2x \).

Plugging these values into the expression for \( K_c \), we substitute to find \( x \). Often approximations are made by assuming \( x \) is small relative to the initial concentrations, streamlining calculations. This specific approach is warranted here due to the hint that only a small fraction of reactants is converted, further validated by the low \( K_c \) value. As such, these equilibrium concentrations are pivotal for understanding the system's specific balance at 500 °C.