Problem 25 - Chemistry: The Central Science

Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64{ }^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\)

Step-by-Step Solution

Answer

a) The rate law for the reaction is: Rate = k[N2O5]. b) The rate of reaction when [N2O5] = 0.0240 M is \(1.156 \times 10^{-4}\) M·s^{-1}. c) When the concentration of N2O5 is doubled to 0.0480 M, the rate of reaction doubles as well to \(2.314 \times 10^{-4}\) M·s^{-1}.

Step 1: Write the rate law for the reaction.

Since we are given that the reaction is first order in N2O5, we can write the rate law as: Rate = k[N2O5]

Step 2: Calculate the rate of reaction when [N2O5] = 0.0240 M.

We are given the rate constant k = 4.82 x 10^{-3} s^{-1} and the initial concentration of N2O5 as 0.0240 M. We can now use the rate law to calculate the rate of the reaction: Rate = k[N2O5] = (4.82 \times 10^{-3} \mathrm{s}^{-1})(0.0240 \mathrm{M}) = 1.156 \times 10^{-4} \mathrm{M \cdot s^{-1}}

Step 3: Calculate the rate of reaction when [N2O5] is doubled to 0.0480 M.

Now we want to determine what happens to the rate when the concentration of N2O5 is doubled to 0.0480 M. We can again use the rate law: Rate = k[N2O5] = (4.82 \times 10^{-3} \mathrm{s}^{-1})(0.0480 \mathrm{M}) = 2.314 \times 10^{-4} \mathrm{M \cdot s^{-1}} Notice that the rate of the reaction has doubled as well, which is expected for a first-order reaction. So in conclusion: a) The rate law for the reaction is: Rate = k[N2O5]. b) The rate of reaction when [N2O5] = 0.0240 M is 1.156 x 10^{-4} M·s^{-1}. c) When the concentration of N2O5 is doubled to 0.0480 M, the rate of reaction doubles as well to 2.314 x 10^{-4} M·s^{-1}.

Key Concepts

Rate Law

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.

First-Order Reaction

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.

Reaction Rate Calculation

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.