Problem 54 - Chemistry

Question

At a particular temperature, 8.0 moles of \(\mathrm{NO}_{2}\) is placed into a 1.0 -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is 2.0 \(\mathrm{M}\) . Calculate \(K\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant, K, for this reaction is calculated as follows. Using an ICE table and the given equilibrium concentration of NO, we determine the equilibrium concentrations of all species involved. Finally, we plug these equilibrium concentrations into the equilibrium constant expression, resulting in \(K = 0.111\).

Step 1: Calculate the initial concentration of NO2

We are given the initial amount of NO2 (8.0 moles) and the volume of the container (1.0 L). We can find the initial concentration by dividing the moles by the volume: Initial concentration of NO2 = (Moles of NO2) / (Volume of the container) \(= \frac{8.0\,\text{moles}}{1.0\,\text{L}} = 8.0\,\text{M}\)

Step 2: Setup the ICE table

We can set up an ICE (Initial, Change, and Equilibrium) table to keep track of the concentrations of each species as the reaction proceeds. | | NO2 | NO | O2 | |------|------|-----|----| | I | 8.0 | 0 | 0 | | C | -2x | +2x | +x | | E | 8.0-2x | 2x | x | Here, x is the amount of NO2 that dissociates at equilibrium.

Step 3: Find the value of x using the given equilibrium concentration of NO

At equilibrium, the concentration of NO is given as 2.0 M. From the ICE table, we know that the equilibrium concentration of NO is 2x. So: \(2x = 2.0\,\text{M}\) To find x, we can divide both sides by 2: \(x = \frac{2.0\,\text{M}}{2} = 1.0\,\text{M}\)

Step 4: Calculate the equilibrium concentrations of all species

Now, we can use the value of x to find the equilibrium concentrations of all species from the ICE table: For NO2: Equilibrium concentration = \(8.0 - 2x = 8.0 - 2(1.0) = 6.0\,\text{M}\) For NO: Equilibrium concentration = \(2x = 2(1.0) = 2.0\,\text{M}\) For O2: Equilibrium concentration = \(x = 1.0\,\text{M}\)

Step 5: Calculate K using the equilibrium concentrations of all species

The equilibrium constant, K, for the given reaction can be expressed as: \[K = \frac{[\mathrm{NO}]^2[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^2}\] Substituting the equilibrium concentrations of all species into the expression: \(K = \frac{(2.0\,\text{M})^2 (1.0\,\text{M})}{(6.0\,\text{M})^2} \) Calculate K: \(K = \frac{4.0}{36} = \frac{1}{9} = 0.111\) The equilibrium constant, K, for this reaction is 0.111.

Key Concepts

ICE table

The ICE table is a helpful tool in chemistry to understand changes in concentrations during a chemical reaction. "ICE" stands for Initial, Change, and Equilibrium. This table allows us to organize and track the concentration changes of each species in a reaction as it approaches equilibrium.

To start, we first input the initial concentrations. In this case, we only have \( ext{NO}_2\) with an initial concentration of \( ext{8.0 M}\) in a 1.0-liter container, while \( ext{NO}\) and \( ext{O}_2\) start at 0 M because they are products.

Next, we consider the change (C) in concentration, noted as \(x\), using reaction stoichiometry. For every unit of change in the dissociation of \( ext{NO}_2\), the change in \(\text{NO}\) and \( ext{O}_2\) is represented as plus 2x and plus x respectively.

Finally, the equilibrium (E) row shows the concentration when the reaction reaches equilibrium. By applying values, we solve for \(x\) and find out all equilibrium concentrations.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of reactants and products remain constant over time. This concept is crucial when calculating the equilibrium constant, \(K\).

In our dissociation reaction of \( ext{2 NO}_2 ightleftharpoons 2 ext{NO} + ext{O}_2\), equilibrium means the formation and dissociation of \(\text{NO}_2\) and its products occur at an equal rate.

Equilibrium does not imply that reactants and products are equal in concentration. Instead, it's the state at which their rates are balanced.

Understanding this helps in using the equilibrium constant expression, which reflects the ratio of product concentrations to reactant concentrations at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It is essential in determining how amounts change as the reaction proceeds toward equilibrium.

In the reaction \(2 ext{NO}_2 ightleftharpoons 2 ext{NO} + ext{O}_2\), stoichiometry is fundamental in setting up the "Change" line in our ICE table.

From the balanced equation, the dissociation of \(2 ext{NO}_2\) produces \(2 ext{NO}\) and \( ext{O}_2\), indicating their mole ratio. Thus, if \(x\) moles of \(\text{O}_2\) is produced, \(2x\) moles of \(\text{NO}_2\) and \(\text{NO}\) are involved.

This understanding helps calculate changes in concentration and helps predict the system's behavior at equilibrium.

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into its component parts. In our problem, \(\text{2 NO}_2\) dissociates into \(\text{2 NO}\) and \(\text{O}_2\). This type of reaction is common in equilibrium chemistry and requires understanding how various factors affect the dissociation process.

Upon reaching equilibrium, the system is balanced as the breakdown and formation of \(\text{NO}_2\) occur at the same rate. This dynamic process continues but with constant overall concentrations.

This type of reaction is ideal for using the equilibrium constant, \(K\), which quantifies the extent of dissociation and helps us understand how products form from reactants. By measuring and calculating these equilibrium concentrations, we gain insights into how dissociation impacts the overall chemical system.