Problem 33 - Chemistry and Chemical Reactivity

Question

Gaseous NO, decomposes at \(573 \mathrm{K}.\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ The concentration of \(\mathrm{NO}_{2}\) was measured as a function of time. A graph of \(1 /\left[\mathrm{NO}_{2}\right]\) versus time gives a straight line with a slope of \(1.1 \mathrm{L} / \mathrm{mol} \cdot\) s. What is the rate law for this reaction? What is the rate constant?

Step-by-Step Solution

Answer

Rate law: Rate = k [NO2]^2; Rate constant k = 1.1 L/mol·s.

Step 1: Identify the Order of Reaction

Since a graph of \( \frac{1}{[\mathrm{NO}_2]} \) versus time gives a straight line, this indicates that the reaction follows a second-order rate law with respect to \( \mathrm{NO}_2 \). For second-order reactions, the integrated rate law is linear for the plot \( \frac{1}{[A]} \) versus time.

Step 2: Write the General Rate Law Formula

The general rate law for a second-order reaction can be expressed as \( \text{Rate} = k [\mathrm{NO}_2]^2 \). This equation shows that the rate of reaction is proportional to the square of the concentration of \( \mathrm{NO}_2 \).

Step 3: Determine the Rate Constant

The slope of the line in the plot \( \frac{1}{[\mathrm{NO}_2]} \) versus time is equal to the rate constant \( k \) for a second-order reaction. Therefore, the rate constant \( k \) is \( 1.1 \, \mathrm{L/mol \, s} \).

Key Concepts

Second-order reactions

Understanding second-order reactions involves recognizing that the rate of these reactions is proportional to the square of the concentration of one reactant. In simpler terms, as the concentration of the reactant decreases, the reaction slows down significantly because of this squared relationship. Second-order reactions often appear in systems where two molecules of the same kind collide and react.
In the case of the decomposition of NO₂, the reaction follows a second-order rate law. This is because when plotting the reciprocal of the concentration of NO₂ over time, a straight line results. This linear relationship is a hallmark of second-order reactions and helps chemists identify the reaction order from experimental data.

Integrated rate law

The integrated rate law is a powerful tool used to describe how the concentration of reactants changes over time in a chemical reaction. For second-order reactions, the integrated rate law is specifically: \[ \frac{1}{[A]} = kt + \frac{1}{[A_0]} \] where

• \([A]\) is the concentration of the reactant at time \(t\).
• \([A_0]\) is the initial concentration of the reactant.
• \(k\) is the rate constant.
• \(t\) is the time elapsed.

This formula reveals that plotting \( \frac{1}{[A]} \) against time yields a straight line for second-order reactions, making it easier to determine both the reaction order and the rate constant from experimental data. The equation helps in converting complex laboratory data into comprehensible quantitative relations, allowing calculations and predictions about reaction progress.

Rate constant

The rate constant, often denoted as \(k\), is a crucial parameter in chemical kinetics, as it quantifies the speed of a reaction. For second-order reactions, the rate constant has units of \(\text{L/mol \; s}\). This is an important distinction because it differs from the units used in first-order (\(\text{s}^{-1}\)) reactions.
In the process of determining \(k\), experimental data is plotted. For second-order reactions specifically, the plot of \( \frac{1}{[\mathrm{NO}_2]} \) versus time will yield a line whose slope is equal to the rate constant. In the given reaction, the slope of this line is \(1.1 \, \text{L/mol \; s}\), directly indicating the value of \(k\). This allows scientists and students alike to understand how quickly or slowly a reaction proceeds under particular conditions, offering insights into reaction mechanisms and rates.