Question

At a particular temperature, a 3.50 -L flask contains 1.16 moles of \(\mathrm{NH}_{3}, 2.40\) moles of \(\mathrm{H}_{2},\) and 1.14 moles of \(\mathrm{N}_{2}\) in equilibrium. Calculate the value of \(K\) for the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$

Short answer

The equilibrium constant, K, for the given reaction 3H₂(g) + N₂(g) ⇌ 2NH₃(g) with the specified equilibrium concentrations in a 3.50-L flask can be calculated using the expression \( K = \frac{[\mathrm{NH}_3]^2}{[\mathrm{H}_2]^3 \times [\mathrm{N}_2]} \). Substituting the equilibrium concentrations, the value of K is found to be approximately 0.0145.

Step-by-step solution

Calculate equilibrium concentrations

First, we need to find the equilibrium concentrations of each substance involved in the reaction. We do this by dividing the moles by the volume of the flask. \[ [\mathrm{NH}_3] = \frac{1.16\,\text{moles}}{3.50\,\text{L}} \] \[ [\mathrm{H}_2] = \frac{2.40\,\text{moles}}{3.50\,\text{L}} \] \[ [\mathrm{N}_2] = \frac{1.14\,\text{moles}}{3.50\,\text{L}} \]

Calculate equilibrium constant expression

Next, we need to write the equilibrium constant expression for the given reaction. The general form of the expression is: \[ K = \frac{[\text{products}]^{\text{stoichiometric coefficients}}}{[\text{reactants}]^{\text{stoichiometric coefficients}}} \] For our reaction, the equilibrium constant expression is: \[ K = \frac{[\mathrm{NH}_3]^2}{[\mathrm{H}_2]^3 \times [\mathrm{N}_2]} \]

Substitute equilibrium concentrations into expression

Now we will substitute the equilibrium concentrations from Step 1 into the expression from Step 2: \[ K = \frac{\left(\frac{1.16\,\text{moles}}{3.50\,\text{L}}\right)^2}{\left(\frac{2.40\,\text{moles}}{3.50\,\text{L}}\right)^3 \times \left(\frac{1.14\,\text{moles}}{3.50\,\text{L}}\right)} \]

Simplify and solve for K

Finally, we will simplify the expression and solve for K: \[ K = \frac{\left(\frac{1.16}{3.50}\right)^2}{\left(\frac{2.40}{3.50}\right)^3 \times \left(\frac{1.14}{3.50}\right)} \] \[ K \approx 0.0145 \] Thus, the equilibrium constant K for the reaction is approximately 0.0145.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry that describes a state where a chemical reaction and its reverse reaction occur at the same rate. This means there is no net change in the concentrations of reactants and products. When a system reaches equilibrium, it doesn't mean the reactants and products are equal in concentration, but rather their rates of formation are equal, so their concentrations remain constant over time.

In the context of our exercise, we're dealing with the equilibrium state of the reaction where hydrogen gas ( \(3 \text{H}_2 (g)\)), nitrogen gas ( \(\text{N}_2 (g)\)), and ammonia ( \(2 \text{NH}_3 (g)\)) are involved. They interact in a closed system until the point where their concentrations no longer change. This state is what we compute as the equilibrium for this given temperature.

Remember, achieving chemical equilibrium in a reaction can be influenced by factors like temperature, pressure, and concentration, referred to as Le Chatelier’s Principle. Changes in these conditions can shift the equilibrium position and result in different concentrations of reactants and products.

Equilibrium Concentrations

To understand equilibrium concentrations, we must first calculate how the concentration of each reactant and product changes over time until equilibrium is achieved. In our example, the concentration of each substance is initially determined by dividing the number of moles by the volume of the container.

Let's break down this calculation:

• [NHequilibrium Concentrations] = \(\frac{1.16 \text{ moles}}{3.50 \text{ L}}≈ 0.331 \text{ mol/L}\)
• [Hequilibrium Concentrations] = \(\frac{2.40 \text{ moles}}{3.50 \text{ L}}≈ 0.686 \text{ mol/L}\)
• [Nequilibrium Concentrations] = \(\frac{1.14 \text{ moles}}{3.50 \text{ L}}≈ 0.326 \text{ mol/L}\)

Equilibrium concentrations are crucial because they allow us to apply the equilibrium constant expression, making it possible to determine the extent to which a reaction will proceed under a given set of conditions.

Understanding these concentrations is also essential for predicting how the system will react to changes in conditions, which is particularly useful in industrial and laboratory settings.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions, playing a vital role in predicting the outcomes of reactions. It involves the use of a balanced chemical equation to determine the relationships between the amounts of reactants and products.

In the balanced reaction given: \(3 \text{H}_2 (g) + \text{N}_2 (g) \rightleftharpoons 2 \text{NH}_3 (g)\), stoichiometry helps us understand the stoichiometric coefficients, which indicate the proportion of each substance involved in the reaction. In this case, the coefficients 3, 1, and 2 correspond to hydrogen, nitrogen, and ammonia, respectively.

These coefficients are crucial in constructing the equilibrium constant expression, \(K\). The stoichiometric coefficients become exponents in the equilibrium constant expression, thus making them vital for calculating \(K\).

• For \(NH_3\), the expression is raised to the power of 2.
• For \(H_2\), raise it to the power of 3.
• For \(N_2\), the power is 1.

Understanding stoichiometry can help accurately predict how much product will be formed from given reactants and provide insights into how changing conditions affect the reaction's direction and yield. This knowledge is not only useful for academic purposes but also has practical applications in industrial settings where chemical production efficiency is desired.