Problem 69 - Chemistry: An Atoms-Focused Approach

Question

In addition to being studied in the gas phase, the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) has been evaluated in solution. In carbon tetrachloride (CC1,) at \(45^{\circ} \mathrm{C}\), $$2 \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$$ is a first-order reaction and \(k=6.32 \times 10^{-4} \mathrm{s}^{-1} .\) How much \(\mathrm{N}_{2} \mathrm{O}_{5}\) remains in solution after \(1 \mathrm{h}\) if the initial concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) was \(0.50 \mathrm{mol} / \mathrm{L} ?\) What percent of the \(\mathrm{N}_{2} \mathrm{O}_{5}\) has reacted at that point?

Step-by-Step Solution

Answer

Answer: The final concentration of N₂O₅ after 1 hour is approximately 0.0513 mol/L, and about 89.74% of N₂O₅ has reacted.

Step 1: Convert the time units

First, we need to convert the given time of 1 hour into seconds, since the rate constant is given in seconds: $$1\,\text{h} \times \frac{60\,\text{min}}{1\,\text{h}} \times \frac{60\,\text{s}}{1\,\text{min}} = 3600\,\text{s}$$

Step 2: Use the first-order reaction formula to find the final concentration

Now, we can plug in our values into the first-order reaction formula to find the concentration of \(\mathrm{N}_{2}\mathrm{O}_{5}\) after 1 hour (3600 seconds): $$[A]_t = [A]_0 e^{-kt} = (0.50\,\text{mol/L}) e^{-(6.32\times 10^{-4}\,\text{s}^{-1})(3600\,\text{s})} = 0.50\,\text{mol/L}\cdot e^{-2.2752}$$ Now calculate the concentration: $$[A]_t \approx 0.50\,\text{mol/L}\cdot 0.1025 \approx 0.0513\,\text{mol/L}$$

Step 3: Calculate the percent of \(\mathrm{N}_{2}\mathrm{O}_{5}\) that has reacted

To find the percent of \(\mathrm{N}_{2}\mathrm{O}_{5}\) that has reacted, we need to compare the initial amount with the final amount of \(\mathrm{N}_{2}\mathrm{O}_{5}\): $$\text{Percentage reacted} = \frac{[A]_0 - [A]_t}{[A]_0} \times 100$$ Using our calculated values, we have: $$\text{Percentage reacted} = \frac{(0.50-0.0513)\,\text{mol/L}}{0.50\,\text{mol/L}} \times 100 \approx \frac{0.4487\,\text{mol/L}}{0.50\,\text{mol/L}} \times 100 \approx 89.74\%$$ Therefore, approximately 89.74% of the \(\mathrm{N}_{2}\mathrm{O}_{5}\) has reacted after one hour.

Key Concepts

First-order reaction

A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one reactant. This means that the reaction rate decreases as the concentration of the reactant decreases. The mathematical expression that describes a first-order reaction is given as: \[[A]_t = [A]_0 e^{-kt}\]- \([A]_t\) is the concentration of the reactant at time \(t\)- \([A]_0\) is the initial concentration of the reactant.- \(k\) is the rate constant.- \(t\) is the time.First-order reactions are unique because they have a constant half-life regardless of the concentration. This characteristic makes them relatively easy to analyze and predict over time.
In the context of our exercise, the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) is a first-order reaction, which means it can be calculated using the mentioned formula. Understanding these principles is crucial for determining how much of a reactant remains after a given period.

Rate constant

The rate constant, often designated as \(k\), is a crucial component in the study of chemical kinetics. It provides valuable insight into the speed of the reaction equation and is influenced by factors like temperature and the solvent. The unit of \(k\) can vary depending on the reaction order, but for first-order reactions, it typically has units of \(s^{-1}\).
The rate constant is integral to the first-order reaction formula: \[[A]_t = [A]_0 e^{-kt}\]In our problem on \(\mathrm{N}_2\mathrm{O}_5\) decomposition, \(k\) is given as \(6.32 \times 10^{-4} \, \text{s}^{-1}\), underlining the role of time in this chemical process. As temperature increases, the rate constant, \(k\), often increases, reflecting a faster reaction. Accurate determination of \(k\) is vital for predicting how reactions progress over time.

Reaction rate

The reaction rate is a measure of how quickly a chemical reaction occurs. It's determined by how fast the concentration of the reactants decreases or the products increase in a given period. For a first-order reaction, the rate is proportional to the concentration of the reactant, meaning that as \(\mathrm{N}_2\mathrm{O}_5\) decreases, so does the rate.
In the decomposition reaction of \(\mathrm{N}_2\mathrm{O}_5\), the reaction rate can be described by the equation \[\text{rate} = k[A]\]- \(k\): rate constant- \[A\]: concentration of reactantThis proportionality means that first-order reactions have a decreasing reaction speed as the reactant concentration decreases over time, an important consideration for practical applications, like determining how long it will take for a reactant to reach a certain concentration.

Decomposition reaction

A decomposition reaction involves breaking down a compound into simpler substances. In chemistry, these reactions are important because they show how complex molecules can be dissociated into more basic elements or molecules.
The chemical equation for the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) into nitrogen dioxide (\(\mathrm{NO}_2\)) and oxygen (\(\mathrm{O}_2\)) is:\[2 \mathrm{N}_2\mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2 + \mathrm{O}_2\]This indicates that each mole of \(\mathrm{N}_2\mathrm{O}_5\) generates several moles of product gases.c Decomposition reactions can be driven by energy in the form of heat, light, or catalysis.

• In the exercise, the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) occurs in a solution of carbon tetrachloride at \(45^{\circ} \mathrm{C}\).
• This is a temperature-sensitive process where increased heat often speeds up the decomposition.

Understanding these kinds of reactions is fundamental for analyzing how substances transform under specific conditions.