Problem 25 - Chemistry and Chemical Reactivity

Question

The rate equation for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (giving \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ) is \(^{*}-\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] / \Delta t=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) "The value of \(k\) is \(5.0 \times 10^{-4} \mathrm{s}^{-1}\) for the reaction at a particular temperature. (a) Calculate the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (b) How long does it take for the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration to drop to one tenth of its original value?

Step-by-Step Solution

Answer

Half-life is 1386 seconds. Time to drop to one tenth is 4604 seconds.

Step 1: Understanding the Half-life Formula

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half its initial concentration. For a first-order reaction, the half-life (\(t_{1/2}\)) formula is \(t_{1/2} = \frac{0.693}{k}\).

Step 2: Applying the Half-life Formula

Substitute the given value of the rate constant \(k = 5.0 \times 10^{-4} \mathrm{s}^{-1}\) into the half-life formula:\[t_{1/2} = \frac{0.693}{5.0 \times 10^{-4}}\]Calculate the half-life using this equation.

Step 3: Calculating the Half-life

Perform the division:\[t_{1/2} = \frac{0.693}{5.0 \times 10^{-4}} \approx 1386\]seconds. This is the time it takes for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to reduce to half its initial value.

Step 4: Understanding First-order Decay

For a first-order reaction, the time required for the concentration to drop to a certain fraction of its initial value can be calculated using the formula:\[t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right)\]where \([A]_0\) is the initial concentration and \([A]\) is the final concentration.

Step 5: Calculating Time to Reach One Tenth Original Value

We're given that the concentration drops to one tenth of its initial value, so \([A]_0/[A] = 10\). Substitute \(k = 5.0 \times 10^{-4} \mathrm{s}^{-1}\) and \([A]_0/[A] = 10\) into the equation:\[t = \frac{1}{5.0 \times 10^{-4}} \ln(10)\]

Step 6: Performing the Calculation

First, calculate \(\ln(10)\), which is approximately 2.302. Then solve:\[t = \frac{1}{5.0 \times 10^{-4}} \times 2.302\]\[t \approx 4604\] seconds. Thus, this is the time required for the concentration to drop to 10% of its original value.

Key Concepts

Rate Equation

The rate equation is fundamental in chemical kinetics as it defines the relationship between the reaction rate and the concentration of reactants. It is a mathematical expression that indicates how fast reactants turn into products. For a first-order reaction, such as the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \), the rate equation is expressed as \(-\Delta[\mathrm{N}_2 \mathrm{O}_5]/\Delta t = k [\mathrm{N}_2 \mathrm{O}_5]\).
Here, \( k \) is the rate constant which measures the speed of the reaction. The negative sign indicates that the concentration of \( \mathrm{N}_2 \mathrm{O}_5 \) decreases over time.
This equation shows that the rate of the reaction is directly proportional to the concentration of \( \mathrm{N}_2 \mathrm{O}_5 \). In simple terms, as more \( \mathrm{N}_2 \mathrm{O}_5 \) is present, the faster it decomposes.

• This is characteristic of first-order reactions.
• The unit of the rate constant for a first-order reaction is \( \mathrm{s}^{-1} \).

Half-life Calculation

Half-life is a concept used to describe the time it takes for the concentration of a reactant to reach half its initial value. For first-order reactions, the half-life is constant and does not depend on the initial concentration.
The formula for calculating the half-life \( (t_{1/2}) \) of a first-order reaction is\[ t_{1/2} = \frac{0.693}{k} \]where \( k \) is the rate constant. By plugging in the rate constant \( k = 5.0 \times 10^{-4} \mathrm{s}^{-1} \), we can find the half-life:\[ t_{1/2} = \frac{0.693}{5.0 \times 10^{-4}} \approx 1386 \] seconds.
This means it takes approximately 1386 seconds for half of the \( \mathrm{N}_2 \mathrm{O}_5 \) to decompose.

• The half-life is a valuable tool for predicting how quickly a reactant will decrease in a reaction.
• Understanding half-life helps in assessing the speed of chemical processes in various practical and industrial applications.

First-order Reaction

First-order reactions are those in which the rate depends linearly on the concentration of only one reactant. For the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \), it is a classic first-order reaction.
One of the distinctive features of first-order reactions is that their half-life remains constant regardless of changes in the initial concentration of the reactant. The formula \[ t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right) \] allows us to determine how the concentration of \( \mathrm{N}_2 \mathrm{O}_5 \) changes over time.
In this expression:

• \([A]_0\) is the initial concentration.
• \([A]\) is the final concentration at time \( t \).
• \( \ln\left(\frac{[A]_0}{[A]}\right)\) represents the natural logarithm of the concentration ratio.

These calculations provide valuable insight into how the reaction progress changes with time.

Decomposition Reaction

A decomposition reaction is one where a single compound breaks down into two or more simpler substances. This type of reaction is common in both organic and inorganic chemistry fields.
For example, in the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \), it breaks down to form \( \mathrm{NO}_2 \) and \( \mathrm{O}_2 \):\( \mathrm{N}_2 \mathrm{O}_5 \rightarrow \mathrm{NO}_2 + \mathrm{O}_2 \).
This process involves breaking chemical bonds in the reactant, leading to the formation of new products.

• Decomposition reactions often require energy input, such as heat, light, or electricity, to breakdown the reactant.
• These reactions are important in understanding chemical stability and the conditions needed for a reaction to occur.

In the context of chemical kinetics, understanding decomposition reactions like that of \( \mathrm{N}_2 \mathrm{O}_5 \) can be crucial in designing and controlling chemical processes.