Problem 45 - Chemistry: The Central Science

Question

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\), is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{ll} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the value of the rate constant?

Step-by-Step Solution

Answer

The reaction is second-order with respect to the concentration of NO₂. The rate constant, k, for the gas-phase decomposition of NO₂ is approximately 0.0225 M⁻¹s⁻¹.

Step 1: Determine the order of the reaction with respect to NO₂

First, let's consider that the reaction has a first-order rate with respect to NO₂: Rate = k[NO₂]^n, where n can be 1 (first-order) or 2 (second-order). If the reaction is first-order, we would expect the graph of ln[NO₂] vs. time to be a straight line. If second-order, we would expect the graph of 1/[NO₂] vs. time to be a straight line. Using the given data, we can calculate ln[NO₂] and 1/[NO₂] for each time point: $$ \begin{array}{c|c|c|c} \hline \text { Time (s) } & [\mathrm{NO}_{2}] (M) & \ln [\mathrm{NO}_{2}] & \frac{1}{[\mathrm{NO}_{2}]} \\ \hline 0.0 & 0.100 & \ln(0.100) & \frac{1}{0.100} \\ 5.0 & 0.017 & \ln(0.017) & \frac{1}{0.017} \\ 10.0 & 0.0090 & \ln(0.0090) & \frac{1}{0.0090} \\ 15.0 & 0.0062 & \ln(0.0062) & \frac{1}{0.0062} \\ 20.0 & 0.0047 & \ln(0.0047) & \frac{1}{0.0047} \\ \hline \end{array} $$ Now, we can check which relationship is linear by making the respective graphs. After plotting the graphs, it becomes clear that 1/[NO₂] vs. time gives a straight line. This indicates that the reaction is second-order with respect to the concentration of NO₂.

Step 2: Calculate the rate constant

Now that we know the reaction is second-order, we will use the second-order rate equation to find the rate constant: Rate = k[NO₂]^2 Since the reaction is second-order, the graph of 1/[NO₂] vs. time should yield a straight line with a slope equal to the rate constant k. The equation of the straight line for a second-order reaction is: \( \frac{1}{[\mathrm{NO}_{2}]} = k\cdot t + \frac{1}{[\mathrm{NO}_{2}]_0} \) We will use the first two data points (time = 0.0 s and 5.0 s) to calculate the slope of the line, which represents the rate constant. \( k = \frac{\frac{1}{[\mathrm{NO}_{2}(t=5.0\:s)]} - \frac{1}{[\mathrm{NO}_{2}(t=0)]}}{5.0\:s - 0} = \frac{\frac{1}{0.017\:M} - \frac{1}{0.100\:M}}{5.0\:s} \) Upon calculating the value of k, we get: k ≈ 0.0225 M⁻¹s⁻¹ The value of the rate constant, k, for the gas-phase decomposition of NO₂ is approximately 0.0225 M⁻¹s⁻¹.

Key Concepts

Second-order reaction

In chemical kinetics, the concept of a second-order reaction is crucial when understanding how reactions proceed over time. A second-order reaction signifies that the rate of the reaction depends on the concentration of one reactant squared or on the product of the concentrations of two different reactants. In simple terms, this means that the speed at which the reaction happens is directly influenced by the reactant concentration in a non-linear manner.

This type of reaction is primarily identified through experimentation and plotting data. For the example given, the decomposition of \(\mathrm{NO}_2\), the concentration data over time can be plotted in two ways:

• ln[Reactant] vs. time for first-order
• 1/[Reactant] vs. time for second-order

From the provided data, if the latter results in a straight line, the reaction is confirmed to be second-order. This indicates that as time progresses, the inverse of the reactant's concentration decreases linearly. Understanding this concept helps in predicting the behavior of reactions under different conditions.

Rate constant

The rate constant, often denoted as \(k\), is a fundamental part of rate equations, in this case, for a second-order reaction. The value of \(k\) provides an insight into how fast a reaction proceeds. For a given reaction at a specific temperature, \(k\) remains constant. It incorporates all the variables of the reaction environment that affect speed except the concentration of the reactants.

Calculating \(k\) for second-order reactions uses the linear relationship between the inverse of the reactant's concentration and time, as shown in the equation:
\[ \frac{1}{[\mathrm{NO}_{2}]} = k\cdot t + \frac{1}{[\mathrm{NO}_{2}]_0} \]This linear equation shows that by finding the slope of the plot of 1/[Reactant] vs. time, we retrieve the value of \(k\). This slope directly gives us the rate constant, and in this case, it equates to approximately 0.0225 \(M^{-1}s^{-1}\), indicating how fast the decomposition reaction occurs at 383°C.

Decomposition reaction

Decomposition reactions involve a single compound breaking down into two or more elements or new compounds. They are commonly observed in everyday life and are vital in various industrial and scientific processes. In the decomposition of \(\mathrm{NO}_{2}\) into \(\mathrm{NO}\) and \(\mathrm{O}_{2}\), the compound loses its integrity and forms products that are simpler structures.

These reactions often require an energy input to break the chemical bonds holding the compound together, which, in turn, releases new products. This is particularly important in this reaction, as the decomposition is being studied at a relatively high temperature of 383°C. The energy at this temperature allows the bonds in \(\mathrm{NO}_{2}\) to break, facilitating the formation of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\). This basic understanding of decomposition helps provide a comprehensive view of the types of reactions evident in both natural and industrial chemistries.