Question

The decomposition of dinitrogen pentaoxide $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ has the following rate equation: Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) It has been found experimentally that the decomposition is \(20.5 \%\) complete in 13.0 hours at \(298 \mathrm{K}\). Calculate the rate constant and the half-life at \(298 \mathrm{K}.\)

Short answer

Rate constant \(k\) is approximately 0.0177 hr\(^{-1}\), and half-life \(t_{1/2}\) is approximately 39.2 hours.

Step-by-step solution

Understand the Rate Equation

The rate of the decomposition of dinitrogen pentaoxide is given by the rate law: Rate \( = k \left[ \text{N}_2\text{O}_5 \right] \). This indicates it's a first-order reaction because the rate depends linearly on the concentration of \( \text{N}_2\text{O}_5 \).

Use First-Order Kinetics Formula

For a first-order reaction, the relationship between concentration and time is given by the formula: \[ ext{ln}\left( \frac{[A]_0}{[A]} \right) = kt\]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.

Determine Initial and Remaining Concentrations

Assume the initial concentration of \( \text{N}_2\text{O}_5 \) is \([\text{N}_2\text{O}_5]_0 = 1\). Since the decomposition is 20.5% complete, the concentration left is 79.5% of the initial:\[[\text{N}_2\text{O}_5] = 0.795 \]

Calculate the Rate Constant \(k\)

Substitute the given values of time \( t = 13.0 \, \text{hours} \) and the concentrations into the first-order kinetics formula:\[ ext{ln}\left( \frac{1}{0.795} \right) = k \times 13.0\]Solve for \(k\):\[k = \frac{ ext{ln}(1.258)}{13.0} = \frac{0.230}{13.0} \approx 0.0177 \, \text{hr}^{-1}\]

Calculate the Half-Life \( t_{1/2} \)

For a first-order reaction, the half-life \( t_{1/2} \) is calculated using the formula:\[t_{1/2} = \frac{0.693}{k}\]Substitute the previously calculated value of \(k\):\[t_{1/2} = \frac{0.693}{0.0177} \approx 39.2 \, \text{hours}\]

Key concepts

Decomposition of Dinitrogen Pentaoxide

When studying chemical reactions, it's important to understand the specific processes involved. Dinitrogen pentaoxide (\(\text{N}_2\text{O}_5\)) is a compound that can decompose into nitrogen dioxide (\(\text{NO}_2\)) and oxygen (\(\text{O}_2\)) in a reaction that is frequently used to explain first-order kinetics. This type of reaction is crucial because it involves a simple breakdown of a molecule into smaller components.
In the case of \(\text{N}_2\text{O}_5\), it decomposes according to the equation:\[\text{N}_2 \text{O}_5 \rightarrow 2 \text{NO}_2 + \frac{1}{2} \text{O}_2\]which tells us about the products formed from the reactant. This reaction is significant because it is a first-order reaction, meaning that the rate at which the reaction occurs depends directly on the concentration of \(\text{N}_2\text{O}_5\).
First-order reactions are characterized by a linear relationship between the natural logarithm of the concentration of the reactant and time. Understanding this linear relationship is key to analyzing how quickly the reaction proceeds and calculating essential parameters like the rate constant and half-life.

Rate Constant Calculation

In physics and chemistry, the rate constant (\(k\)) plays a central role in defining the speed of a chemical reaction. Measuring how rapidly reactions occur allows scientists to predict reaction specifics under various conditions. For first-order reactions like our decomposition of dinitrogen pentaoxide, the rate constant can be calculated using the formula:\[\ln\left(\frac{[A]_0}{[A]}\right) = kt\]where \([A]_0\) is the initial concentration of the reactant, \([A]\) is the remaining concentration, and \(t\) is time.
To illustrate, if 20.5% of \(\text{N}_2\text{O}_5\) reacts in 13 hours, this means 79.5% remains. Assigning a convenient initial concentration \([\text{N}_2\text{O}_5]_0 = 1\), the concentration after 13 hours is 0.795. Simply plug these values into the formula:\[\ln\left(\frac{1}{0.795}\right) = k \times 13.0\]Solving for \(k\) gives us the rate constant, determining how quickly our reaction advances under given conditions. \(k\) is calculated to be approximately 0.0177 \(\text{hr}^{-1}\). Calculating \(k\) is vital because it is intrinsic to the reaction, unaffected by the concentration change.

Half-Life Determination

Another concept crucial to our understanding of reaction kinetics is half-life, denoted \(t_{1/2}\). For first-order reactions, this is the time required for precisely half of the reactant to decompose. Determining half-life provides insight into the stability and duration of a substance's persistence.
The half-life for a first-order reaction can be easily calculated using the rate constant:\[t_{1/2} = \frac{0.693}{k}\]and substituting our specific rate constant of 0.0177 \(\text{hr}^{-1}\) produces:\[t_{1/2} = \frac{0.693}{0.0177} \approx 39.2 \text{ hours}\]This simple formula shows that, independent of the initial concentration, the time it takes for half of the reactant to decompose is constant. This property is unique to first-order reactions. Understanding half-life is fundamental in fields like pharmacology and environmental science, where determining how long a compound remains active is pivotal.