91Ó°ÊÓ

The rate law of a chemical reaction describes how the rate depends on the concentration of the reactants. In the context of our given problem, the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), we see a typical rate law expression. The rate law can be written as follows:
\[ \text{Rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}] \]Here, **\(k\)** represents the rate constant, and **\([\mathrm{N}_{2} \mathrm{O}_{5}]\)** is the concentration of the reactant. In this equation:

• The rate constant \(k\) is a proportionality factor that is specific to a given reaction at a particular temperature, which, in this case, is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\) at \(64^{\circ} \mathrm{C}\).
• The order of the reaction tells us how the concentration of a reactant affects the rate. Since our reaction is first-order in \(\mathrm{N}_{2} \mathrm{O}_{5}\), any change in its concentration results in an equivalent change in rate.

A first-order reaction is one where the rate is directly proportional to the concentration of a single reactant. This means that if you double the concentration of the reactant, the rate of reaction will also double. In the given problem, since the reaction is first-order concerning \(\mathrm{N}_{2} \mathrm{O}_{5}\), the general form of the rate law is:
\[ \text{Rate} = k [\mathrm{N}_{2} \mathrm{O}_{5}]^1 \]Characteristics of a first-order reaction include:

• The rate depends linearly on only one reactant’s concentration.
• Doubling the concentration doubles the reaction rate. In our problem, changing the concentration from \(0.0240\) M to \(0.0480\) M exactly doubled the calculated rate.

This direct relationship is useful for predicting the behavior of such reactions when the concentrations change.

To calculate the rate of a reaction, you apply the rate law, using the provided rate constant and reactant concentration values. In our exercise:
1. For an initial concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(0.0240 \mathrm{M}\): The rate is calculated as:\[ \text{Rate} = (4.82 \times 10^{-3} \mathrm{~s}^{-1}) \times (0.0240 \mathrm{M}) = 1.156 \times 10^{-4} \mathrm{~M \cdot s^{-1}} \]2. When the concentration is increased to \(0.0480 \mathrm{M}\):The calculation becomes:\[ \text{Rate} = (4.82 \times 10^{-3} \mathrm{~s}^{-1}) \times (0.0480 \mathrm{M}) = 2.314 \times 10^{-4} \mathrm{~M \cdot s^{-1}} \]These calculations show how changes in the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) impact the rate of reaction due to its first-order nature. This method of calculation is simple and reliable for assessing how different concentrations affect the speed of reaction.