Problem 103 - Chemistry

Question

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas are placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to \(\mathrm{NO}\) and \(\mathrm{O}_{2} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 \(\mathrm{mol} / \mathrm{L}\) . Calculate the value of \(K\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

Step 1: Calculating initial concentrations

First, let's calculate the initial concentration of \(\mathrm{NO}_{2}\) by dividing the given moles (8.1 mol) by the volume of the container (3.0 L): \[\mathrm{NO}_{2} \text{ initial concentration} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \frac{\text{mol}}{\text{L}}\]

Step 2: Setting up changes in concentration

Let's use the stoichiometric coefficients in the balanced chemical equation to understand the changes in concentrations during the reaction. At equilibrium, let the decrease in concentration of \(\mathrm{NO}_{2}\) be \(x\). Since two moles of \(\mathrm{NO}_{2}\) are required to produce two moles of \(\mathrm{NO}\) and one mole of \(\mathrm{O}_{2}\), the concentration of \(\mathrm{NO}_{2}\) will decrease by \(x\) molar, while the concentration of \(\mathrm{NO}\) will increase by \(x\) molar, and the concentration of \(\mathrm{O}_{2}\) will increase by \(\frac{x}{2}\) molar.

Step 3: Calculating x

Given the equilibrium concentration of \(\mathrm{NO}(g)\) as 1.4 M, we can find the value of \(x\): \(x = \text{equilibrium concentration of } \mathrm{NO} = 1.4 \frac{\text{mol}}{\text{L}}\)

Step 4: Finding equilibrium concentrations

Now that we have found the value of \(x\), we can find the equilibrium concentrations for all reactants and products: \[\text{Equilibrium concentration of } \mathrm{NO}_{2} = 2.7 \frac{\text{mol}}{\text{L}} - x = 2.7 - 1.4 = 1.3 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{NO} = x = 1.4 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{O}_{2} = \frac{x}{2} = \frac{1.4}{2} = 0.7 \frac{\text{mol}}{\text{L}}\]

Step 5: Calculating the equilibrium constant, K

Using the equilibrium concentrations, we can write the expression for the equilibrium constant, \(K\): \[K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}}\] Now, substitute the equilibrium concentrations into the expression: \[K = \frac{(1.4)^{2}(0.7)}{(1.3)^{2}} = \frac{1.586}{1.69}\] After performing the calculation: \[K \approx 0.939\]

Step 6: Final Answer

The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

Key Concepts

Chemical Equilibrium

Chemical equilibrium describes the state where the concentrations of reactants and products remain constant over time in a closed system. This happens when the forward and reverse reactions occur at the same rate. In our reaction \( 2 \mathrm{NO}_{2} \rightleftharpoons 2 \mathrm{NO} + \mathrm{O}_{2} \), equilibrium is reached when the rate at which \( \mathrm{NO}_{2} \) decomposes matches the rate at which \( \mathrm{NO} \) and \( \mathrm{O}_{2} \) combine to reform \( \mathrm{NO}_{2} \).

At equilibrium, the concentrations don’t change, but that doesn’t mean they are equal. Instead, the relative concentrations help us determine the equilibrium constant \( K \), a powerful tool in predicting how a system will react to changes.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to relate the amounts of reactants and products. It's essential in understanding how changes in one part of a reaction affect the rest.

In this case, the stoichiometry of the reaction \( 2 \mathrm{NO}_{2} \rightarrow 2 \mathrm{NO} + \mathrm{O}_{2} \) tells us that two moles of \( \mathrm{NO}_{2} \) produce two moles of \( \mathrm{NO} \) and one mole of \( \mathrm{O}_{2} \). This allows us to calculate the changes in concentration as the reaction progresses.

• \( \mathrm{NO}_{2} \) concentration decreases by \( x \) moles per liter.
• \( \mathrm{NO} \) concentration increases by \( x \) moles per liter.
• \( \mathrm{O}_{2} \) concentration increases by \( \frac{x}{2} \) moles per liter.

Understanding stoichiometry helps ensure that these calculations are precise, ensuring we accurately calculate equilibrium concentrations.

Concentration Calculations

Concentration calculations are crucial for finding the equilibrium constant. We begin by determining the initial concentration of \( \mathrm{NO}_{2} \). We do this by dividing the number of moles by the volume:

\[ \mathrm{NO}_{2} \text{ initial concentration} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \text{ mol/L} \]

Next, we calculate the equilibrium concentrations using given data. For \( \mathrm{NO}(g) \), the equilibrium concentration is provided as 1.4 mol/L. With this information, we can calculate the other concentrations:

• \[ \text{\( \mathrm{NO}_{2} \) equilibrium concentration} = 2.7 - 1.4 = 1.3 \text{ mol/L} \]
• \[ \text{\( \mathrm{O}_{2} \) equilibrium concentration} = \frac{1.4}{2} = 0.7 \text{ mol/L} \]

These calculations lay the foundation for finding the equilibrium constant, \( K \).

Gas Laws

Gas laws might not seem directly connected, but they help us understand reactions involving gases. In this exercise, knowing the relationships between pressure, volume, and temperature can enable us to predict how reaction rates might change with different conditions.

Although specific gas laws like Boyle’s Law or Charles' Law aren't directly used in the calculation of \( K \), understanding that gas concentrations can change with pressure and temperature is crucial in practical applications.

This insight allows chemists to manipulate conditions to favor the production of desired products, which is a valuable tool in industries such as pharmaceuticals and manufacturing.