Question

The first step in \(\mathrm{HNO}_{3}\) production is the catalyzed oxidation of \(\mathrm{NH}_{3}\). Without a catalyst, a different reaction predominates: $$ 4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ When \(0.0150 \mathrm{~mol}\) of \(\mathrm{NH}_{3}(g)\) and \(0.0150 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) are placed in a 1.00 - \(\mathrm{L}\) container at a certain temperature, the \(\mathrm{N}_{2}\) concentration at equilibrium is \(1.96 \times 10^{-3} M .\) Calculate \(K_{\mathrm{c}}\)

Short answer

The equilibrium constant \( K_c \) is approximately 1.81.

Step-by-step solution

Write Down the Balanced Equation

The balanced equation for the given reaction is: \[ 4 \mathrm{NH}_{3}(g) + 3 \mathrm{O}_{2}(g) \leftrightarrows 2 \mathrm{N}_{2}(g) + 6 \mathrm{H}_{2}O(g) \]

Define Initial Concentrations

Initial concentration of \( \mathrm{NH}_{3} \) and \( \mathrm{O}_{2} \) are both 0.0150 mol in a 1.00 L container: \[ \text{[NH}_3]_0 = \frac{0.0150 \text{ mol}}{1.00 \text{ L}} = 0.0150 \text{ M} \] \[ \text{[O}_2]_0 = \frac{0.0150 \text{ mol}}{1.00 \text{ L}} = 0.0150 \text{ M} \]

Determine the Change in Concentration

Let the change in the concentration of \(\mathrm{NH}_{3}\) be \( -4x \), \(\mathrm{O}_{2}\) be \( -3x \), \(\mathrm{N}_{2}\) be \( 2x \) and \( \mathrm{H}_2O \) be \( 6x \). Given the equilibrium concentration of \[ \text{[N}_2]_\text{eq} = 1.96 \times 10^{-3} \text{ M} \] equates to \( 2x \): \[ 2x = 1.96 \times 10^{-3} \text{ M} \] Solving for \( x \): \[ x = \frac{1.96 \times 10^{-3}}{2} = 0.98 \times 10^{-3} \text{ M} \]

Calculate Equilibrium Concentrations

Find the equilibrium concentrations for \( \text{NH}_3 \) and \( \text{O}_2 \): \[ \text{[NH}_3]_\text{eq} = \text{[NH}_3]_0 - 4x \ = 0.0150 \text{ M} - 4 \times 0.98 \times 10^{-3} \text{ M} = 0.01108 \text{ M} \] \[ \text{[O}_2]_\text{eq} = \text{[O}_2]_0 - 3x \ = 0.0150 \text{ M} - 3 \times 0.98 \times 10^{-3} \text{ M} = 0.0121 \text{ M} \]

Write the Expression for the Equilibrium Constant \( K_c \)

For the reaction: \[ K_c = \frac{[\text{N}_2]^2 [\text{H}_2O]^6}{[\text{NH}_3]^4 [\text{O}_2]^3} \]

Substitute Equilibrium Concentrations into \( K_c \) Expression

Substitute the known equilibrium concentrations back into the \( K_c \) expression: \[ K_c = \frac{(1.96 \times 10^{-3})^2 (6 \times 0.98 \times 10^{-3})^6}{(0.01108)^4 (0.0121)^3} \]

Calculate the Value of \(K_c\)

Perform the calculations: \[ K_c = \frac{(1.96 \times 10^{-3})^2 \times (5.88 \times 10^{-3})^6}{(0.01108)^4 \times (0.0121)^3} \approx 1.81 \]

Key concepts

Equilibrium Concentrations

In chemical reactions, not all reactants are fully converted to products. Instead, they reach a state where both reactants and products are present. This state is called chemical equilibrium. At equilibrium, the concentrations of reactants and products remain constant. To determine these equilibrium concentrations, we use reaction stoichiometry and the known initial concentrations.
In the given problem, we start with initial concentrations of \( \text{NH}_3 \) and \( \text{O}_2 \). As the reaction proceeds, their concentrations change, represented by \( -4x \) and \( -3x \) respectively. The product concentrations increase by \( 2x \) for \( \text{N}_2 \) and \( 6x \) for \( \text{H}_2O \). Using the provided \( \text{N}_2 \) equilibrium concentration (1.96 x 10^{-3} M), we solve for \( x \) and find it to be 0.98 x 10^{-3} M.
We then calculate the equilibrium concentrations of \( \text{NH}_3 \) and \( \text{O}_2 \) using the equations:
\[ \text{[NH}_3]_\text{eq} = \text{[NH}_3]_0 - 4x \]
\[ \text{[O}_2]_\text{eq} = \text{[O}_2]_0 - 3x \]
Plugging in the values, we find:
\[ \text{[NH}_3]_\text{eq} = 0.0150 - 4(0.98 x 10^{-3}) = 0.01108 \]
\[ \text{[O}_2]_\text{eq} = 0.0150 - 3(0.98 x 10^{-3}) = 0.0121 \]
This process helps us understand how equilibrium concentrations are derived from initial conditions and changes in concentration.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, establishing a balance between reactants and products. At this point, their concentrations do not change over time. The position of equilibrium depends on the reaction conditions, such as temperature and pressure, and is described by the equilibrium constant \( K_c \).
The given reaction can be written as:
\[ 4 \text{NH}_3(g) + 3 \text{O}_2(g) \rightleftharpoons 2 \text{N}_2(g) + 6 \text{H}_2_O(g) \]
To calculate \( K_c \), we use the equilibrium concentrations of the reactants and products. The expression for \( K_c \) is:
\[ K_c = \frac{[\text{N}_2]^2 [\text{H}_2O]^6}{[\text{NH}_3]^4 [\text{O}_2]^3} \]
By substituting the equilibrium concentrations into this formula, we can determine the value of \( K_c \). This constant indicates the extent to which a reaction proceeds towards the products at equilibrium.
Understanding chemical equilibrium helps us predict the behavior of reactions under different conditions and is fundamental in fields like chemistry and chemical engineering.

Reaction Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate how the amount of one substance changes in relation to another.
In the given exercise, the balanced equation is:
\[ 4 \text{NH}_3 + 3 \text{O}_2 \rightarrow 2 \text{N}_2 + 6 \text{H}_2O \]
This equation tells us the ratio of reactants and products. For example, 4 moles of \( \text{NH}_3 \) react with 3 moles of \( \text{O}_2 \) to produce 2 moles of \( \text{N}_2 \) and 6 moles of \( \text{H}_2O \).
We can use stoichiometric coefficients to determine how the concentrations change. In this problem, if the change in \( \text{NH}_3 \) is \( -4x \), then the change in \( \text{N}_2 \) is \( +2x \), reflecting the ratio in the balanced equation.
In summary, understanding reaction stoichiometry allows us to relate changes in reactant and product amounts and is crucial for calculating concentrations at equilibrium.

Catalysis

Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process. They work by lowering the activation energy required for the reaction to proceed. This can significantly affect both the rate at which equilibrium is achieved and the overall efficiency of industrial processes.
In the exercise, the production of \( \text{HNO}_3 \) involves a catalyzed oxidation reaction. Without a catalyst, a different reaction predominates, resulting in different products. This highlights the importance of catalysts in directing the desired reaction pathway.
Catalysis is essential in many industrial processes, such as the synthesis of ammonia in the Haber process, where catalysts help achieve equilibrium faster and at lower temperatures. Understanding how catalysts work helps in the design and optimization of chemical processes, leading to more sustainable and cost-effective production methods.