Problem 35 - Chemistry and Chemical Reactivity

Question

Data for the decomposition of dinitrogen oxide $$2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{N}_{2} (\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ on a gold surface at \(900^{\circ} \mathrm{C}\) are given below. Verify that the reaction is first order by preparing a graph of \(\ln \left[\mathrm{N}_{2} \mathrm{O}\right]\) versus time. Derive the rate constant from the slope of the line in this graph. Using the rate law and value of \(k\), determine the decomposition rate at \(900^{\circ} \mathrm{C}\) when \(\left[\mathrm{N}_{2} \mathrm{O}\right]=\) \(0.035 \mathrm{mol} / \mathrm{L}.\) $$\begin{array}{cc}\hline \begin{array}{c}\text { Time } \\\\\text { (min) }\end{array} & \begin{array}{c}{\left[\mathrm{N}_{2} 0\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 15.0 & 0.0835 \\\30.0 & 0.0680 \\\80.0 & 0.0350 \\\120.0 & 0.0220 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Graph straightens for first-order; \(k\) from slope. Rate = \(k \times 0.035\).

Step 1: Understand the Task

We need to verify whether the decomposition of \(\text{N}_2\text{O}\) is a first-order reaction. To do this, we prepare a graph of \(\ln[\text{N}_2\text{O}]\) versus time. If the graph is a straight line, the reaction is first-order, and we can determine the rate constant \(k\) from the slope.

Step 2: Calculate Natural Logarithms

Calculate the natural logarithm \(\ln[\text{N}_2\text{O}]\) for each concentration value given in the data table at different times. This will help create the graph to verify the order of the reaction.

Step 3: Prepare the Graph

Plot the calculated \(\ln[\text{N}_2\text{O}]\) values on the y-axis against time on the x-axis. If the plot is approximately a straight line, this verifies that the reaction is first-order.

Step 4: Determine the Rate Constant

The slope of the straight line from the graph in Step 3 is the negative of the rate constant \(k\) for a first-order reaction. Measure the slope to find the value of \(k\).

Step 5: Verify the Reaction Order

Since the graph is a straight line, the reaction is confirmed to be first-order, as first-order reactions show a linear relationship between \(\ln[\text{N}_2\text{O}]\) and time.

Step 6: Calculate the Rate of Decomposition

Using the rate law for a first-order reaction \(\text{Rate} = k [\text{N}_2\text{O}]\), substitute \([\text{N}_2\text{O}] = 0.035\, \text{mol/L}\) and the calculated \(k\) to find the decomposition rate at \(900^{\circ} \text{C}\).

Key Concepts

Reaction Order

In chemical kinetics, the reaction order is a crucial concept that helps to describe the rate at which a reaction occurs. When we talk about the reaction order, we refer to the power to which the concentration of a reactant is raised in the rate law. This could be zero, first, second order, or even fractional, each indicating different influences of reactant concentration on the rate.

- **Zero-Order Reaction**: The rate is constant and does not depend on concentration.
- **First-Order Reaction**: The rate is directly proportional to the reactant's concentration.
- **Second-Order Reaction**: The rate is proportional to the square of the reactant concentration.

Understanding the order of a reaction is essential for predicting how concentration changes affect reaction rates. Importantly, reaction order is determined experimentally, not just from the stoichiometry of the reaction.

First-Order Reaction

First-order reactions are quite common, especially in processes like radioactive decay and certain chemical decompositions. For these reactions, the rate of reaction is directly proportional to the concentration of one reactant.

This means that if you double the concentration of the reactant, the rate of reaction will also double. The mathematical expression for a first-order reaction is given by the equation: \[ ext{Rate} = k[ ext{A}] \] where \(k\) is the rate constant and \([ ext{A}]\) is the concentration of the reactant.

First-order kinetics are characterized by a straight-line graph when plotting \( ext{ln}[ ext{A}]\) versus time, indicating that the natural logarithm of the concentration decreases linearly with time. This linearity is a hallmark of first-order reactions and provides a simple way to determine the reaction order through visual or computational analysis.

Rate Constant

The rate constant, denoted as \(k\), is a critical parameter in the rate equation of a chemical reaction. It provides insights into the speed of the reaction under specified conditions such as temperature.

For a first-order reaction, the relation between the reaction rate and the reactant concentration is linear, allowing us to extract \(k\) by analyzing the slope of the line on a graph of \( ext{ln}[ ext{A}]\) versus time. The rate constant has specific units depending on the overall order of the reaction, often expressed in \( ext{s}^{-1}\) for first-order reactions.

Understanding the rate constant is essential for calculating how quickly a reaction occurs and how it might be influenced by factors such as temperature or the presence of a catalyst. Moreover, accurate determination of \(k\) helps in predicting how a reaction progresses over time, facilitating better control in industrial or laboratory settings.

Decomposition Reaction

A decomposition reaction involves the breakdown of a compound into two or more simpler substances. They are a subset of chemical reactions that are quite common both in nature and in industrial processes.

A classic example is the decomposition of dinitrogen oxide \(( ext{N}_2 ext{O})\), the process analyzed in the given exercise. These reactions often require an input of energy, such as heat, light, or electricity, to proceed. In this case, heating to \(900^{ ext{o}} ext{C}\) on a gold surface catalyzes the reaction, causing \( ext{N}_2 ext{O}\) to break down into nitrogen \(( ext{N}_2)\) and oxygen \(( ext{O}_2)\).

Decomposition reactions are essential in various domains, from biochemical pathways to industrial chemical production. They help release or consume energy, making them integral to processes like metabolism and material recovery.