Problem 48 - Chemistry An Atoms First Approach

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 the given reaction can be calculated using the equilibrium concentrations of the reactants and products: \([NO]_{eq}=2.0\mathrm{M}\), \([O_{2}]_{eq}=0.5\mathrm{M}\), and [NO\(_{2}\)]\(_{eq}\)=\(7.0\mathrm{M}\). Using the formula \(K=\frac{[\text{NO}]_{eq}^{2}[\text{O}_{2}]_{eq}}{[\text{NO}_{2}]_{eq}^{2}}\), we get \(K \approx 0.041\).

Step 1: Calculate initial concentrations

Since we are given the moles of \(\mathrm{NO}_{2}\) and its volume, we can calculate the initial concentration of \(\mathrm{NO}_{2}\) by dividing moles by volume: Initial concentration of \(\mathrm{NO}_{2}=\frac{8.0 \text{ moles}}{1.0 \text{ L}}=8.0 \mathrm{M}\). We are not given any information about \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) initially, so we can assume their initial concentrations to be zero. Initial concentrations: [N\(_{2}\)]\(_{0}\) = \(8.0\mathrm{M}\) [NO]\(_{0}\) = \(0\mathrm{M}\). [O\(_{2}\)]\(_{0}\) = \(0\mathrm{M}\).

Step 2: Determine the change in concentration for each species

In order to find the equilibrium concentrations, first, we need to know how the concentrations of the reactants and products change as the reaction proceeds to equilibrium. In this reaction, 2 moles of NO and 1 mole of O\(_{2}\) are formed for every 2 moles of N\(_{2}\) that dissociates. Hence, the decrease in the concentration of \(\mathrm{NO}_{2}\) at equilibrium will be half of the increase in the concentration of \(\mathrm{NO}\). Since the equilibrium concentration of \(\mathrm{NO}\) is given as 2.0 M, and its initial concentration was 0 M, the change in its concentration is given by: \(\Delta[\text{NO}] = 2.0\mathrm{M}\,-0\,\text{M} = 2.0\,\mathrm{M}\). The increase in the concentration of \(\mathrm{NO}_{2}\) refers to the moles dissociated. For every 2 moles of \(\mathrm{NO}_{2}\) dissociated, 2 moles of \(\mathrm{NO}\) are formed. Hence, \(\Delta[\text{NO}_{2}] = -\frac{1}{2} \Delta[\text{NO}] = -\frac{1}{2} \times 2.0\mathrm{M}= -1.0\,\mathrm{M}\). Since 1 mole of \(\mathrm{O}_{2}\) is formed for every 2 moles of \(\mathrm{NO}_{2}\) that dissociates, we have: \(\Delta[\text{O}_{2}] = -\frac{1}{2} \Delta[\text{NO}_{2}] = -\frac{1}{2} \times -1.0\,\mathrm{M}= 0.5\,\mathrm{M}\).

Step 3: Calculate equilibrium concentrations

Now, we can find the equilibrium concentrations of all species by adding the change in concentrations to their initial concentrations: [NO\(_{2}\)]\(_{eq}\) = [NO\(_{2}\)]\(_{0}\) + \(\Delta[\text{NO}_{2}] = 8.0\,\mathrm{M} -1.0\,\mathrm{M} = 7.0\,\mathrm{M}\). [NO]\(_{eq}\) = [NO]\(_{0}\) + \(\Delta\)[NO] = \(0\,\mathrm{M} + 2.0\,\mathrm{M} = 2.0\,\mathrm{M}\). [O\(_{2}\)]\(_{eq}\) = [O\(_{2}\)]\(_{0}\) + \(\Delta\)[O\(_{2}\)] = \(0\,\mathrm{M} + 0.5\,\mathrm{M} = 0.5\,\mathrm{M}\).

Step 4: Calculate the equilibrium constant, K

Now we will use the equilibrium concentrations to calculate the equilibrium constant, K. The equilibrium constant is given by the ratio of the concentrations of the products to the power of their stoichiometric coefficients to that of reactants : \(K=\frac{[\text{NO}]_{eq}^{2}[\text{O}_{2}]_{eq}}{[\text{NO}_{2}]_{eq}^{2}}\) Substituting the equilibrium concentrations obtained in step 3: \(K=\frac{(2.0\,\mathrm{M})^{2}(0.5\,\mathrm{M})}{(7.0\,\mathrm{M})^{2}}\) Finally, calculate the value of K: \(K=\frac{2.0}{49}\approx 0.041\) Thus, the equilibrium constant K for this reaction is approximately 0.041.

Key Concepts

NO2 Dissociation

Dissociation is a process in which a compound separates into particles such as atoms, ions, or molecules. When nitrogen dioxide (\text{NO}_{2}) gas dissociates, it participates in a reversible reaction where it breaks down into nitrogen monoxide (\text{NO}) and oxygen (\text{O}_{2}). This can be represented by the equation:
\[2 \text{NO}_{2}(g) \rightleftharpoons 2 \text{NO}(g) + \text{O}_{2}(g)\]
Understanding this process is crucial for grasping chemical equilibrium. It's important to note that, in dynamic equilibrium, the rate of the forward reaction (dissociation in this case) is equal to the rate of the reverse reaction, leading to constant concentrations of reactants and products over time, even though actual reactions still occur.

Equilibrium Concentrations Calculation

Calculating equilibrium concentrations involves a few key steps. Initially, you determine the starting concentration of reactants and products. In absence of products, their concentrations start as zero. From there, changes in concentration as the reaction reaches equilibrium are determined. Here's a simplified example: If \text{NO}(g) has an equilibrium concentration of 2.0 M and originally there was none, then \text{NO}_{2}(g) must have decreased by 1.0 M since it dissociates in a 1:1 molar ratio. Using stoichiometry and initial conditions, you deduce the final concentrations of all species at equilibrium.
\[ [ \text{NO}_{2} ]_{eq} = [ \text{NO}_{2} ]_{0} + \text{Change in} [\text{NO}_{2}] \]
Through similar logic, you can calculate for other products and reactants. This detailed look into the shifts within the chemical reaction allows for a more comprehensive understanding of the equilibrium state.

Law of Mass Action

The Law of Mass Action expresses the relationship between the concentrations of reactants and products present in a chemical equilibrium. According to this law, the rate of the reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. This concept allows us to define an equilibrium constant (\text{K}), which is a quantitative measure of the propensity of a reaction mix to reach equilibrium given specific concentrations. For the dissociation of \text{NO}_{2}, the equilibrium constant is expressed as:
\[ K = \frac{[ \text{NO} ]_{eq}^{2} [ \text{O}_{2} ]_{eq}}{[ \text{NO}_{2} ]_{eq}^{2}} \]
The equilibrium constant varies with temperature, but it is independent of the initial concentrations of the reactants and products. It is critical for students to grasp that the value of \text{K} offers insight into the extent of a reaction: a larger \text{K} value implies a greater proportion of products at equilibrium, while a smaller value indicates a reaction mixture richer in reactants.