Question

In the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}+(1 / 2) \mathrm{O}_{2}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes by a first-order mechanism. At \(298^{\circ} \mathrm{K}\), the half- life is 340 minutes. Find the value of the reaction rate constant. Calculate the number of minutes required for the reaction to proceed 70 percent towards completion.

Short answer

The reaction rate constant (k) is \(2.04 \times 10^{-3}\ \text{min}^{-1}\). It takes approximately 538.4 minutes for the reaction to reach 70% completion at 298 K.

Step-by-step solution

Step-1: Calculate the reaction rate constant

From the given half-life and the first-order reaction half-life equation, we can find the value of k: \[ k = \frac{0.693}{t_{1/2}} \] Plug in the given half-life of 340 minutes: \[ k = \frac{0.693}{340} \] Now, calculate k: \[ k = 2.04 \times 10^{-3}\ \text{min}^{-1} \] Next, calculate the time for the reaction to reach 70% completion.

Step-2: Find the time for 70% completion of the reaction

Since the reaction needs to be 70% complete, the remaining percentage of the initial concentration will be 100% - 70% = 30%. So we have \( \frac{A}{A_0} = 0.30 \). Now, use the integrated rate law equation for a first-order reaction: \[ \ln\frac{A_0}{A} = kt \] Since we want to find the time (t) when \(\frac{A}{A_0} = 0.30\), we can rewrite the equation as: \[ \ln\frac{A_0}{0.30A_0} = kt \] Simplify and plug in the value of k: \[ \ln3 = (2.04 \times 10^{-3}\ \text{min}^{-1}) \times t \] Now, solve for t: \[ t = \frac{ \ln3 }{ 2.04 \times 10^{-3} \ \text{min}^{-1}} \] Calculate t: \[ t = 538.4\ \text{min} \] So, it takes approximately 538.4 minutes for the reaction to reach 70% completion at 298 K.

Key concepts

Understanding Chemical Kinetics

Chemical kinetics is the study of rates of chemical processes and the factors affecting them. It's essential for predicting how long a reaction will take, which is crucial in all fields of chemistry, from creating new materials to understanding biological systems. Rate laws express the relationship between the reaction rate and the concentrations of reactants. For a first-order reaction, the rate is directly proportional to the concentration of one reactant. An important aspect of chemical kinetics is that it also contemplates the influence of temperature, catalysts, and other conditions on the reaction rate.

When solving problems in chemical kinetics, we often use the rate constant, denoted as 'k', which is unique for each reaction at a given temperature. This constant is a measure of how quickly a reaction proceeds, and calculating it – as in the exercise you've encountered – is fundamental to predicting the behavior of the reaction over time.

Delving Into Reaction Half-Life

The reaction half-life is the time required for half of a reactant to be used up in a reaction. In a first-order reaction, the half-life is constant and does not depend on the initial concentration of the reactant, which is a distinctive property of this type of reaction. Our problem mentions the half-life of a decomposition reaction, which is a classic example of a process where knowing the half-life is beneficial, as it helps in planning and control of the process.

Understanding the half-life provides insight into the stability of a compound and its rate of conversion to products. From a practical standpoint, half-life is a concept often utilized in areas like pharmaceuticals, environmental science, and any field where the breakdown or conversion of substances over time is relevant.

Integrated Rate Law

The integrated rate law for first-order reactions gives us a way to connect the concentration of reactants and time, which can be represented by the equation \( \text{ln}\frac{A_0}{A} = kt \), where \( A_0 \) is the initial concentration, A is the concentration at time t, k is the rate constant, and t is time. When using this equation, we can determine the time it takes for a reaction to progress to a certain extent, as was required in our exercise with the 70% completion.

To utilize this law practically, one usually determines the remaining concentration at a certain time or, inversely, calculates the time needed for the concentration to reach a certain value. This application is particularly useful in predicting the timelines for chemical reactions in industrial processes, biochemical applications, or even the expiration of pharmaceutical drugs.