Question

The rate constant for the second-order reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.65 M\) to \(0.18 M ?\)

Short answer

It takes approximately 7.44 seconds.

Step-by-step solution

Understand the Second-Order Reaction

In a second-order reaction involving one reactant \[ 2A \rightarrow Products \]the integrated rate law is given by:\[ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \]where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), and \(k\) is the rate constant.

Identify Given Values

From the problem, we know:- The initial concentration, \([A]_0\), is 0.65 M.- The final concentration, \([A]_t\), is 0.18 M.- The rate constant, \(k\), is 0.54 \(M^{-1} s^{-1}\).

Plug Values into Integrated Rate Law

Substitute the given values into the integrated rate law formula:\[ \frac{1}{0.18} - \frac{1}{0.65} = 0.54t \]

Calculate Left Side of the Equation

Calculate the values individually first:\(\frac{1}{0.18} \approx 5.56\)\(\frac{1}{0.65} \approx 1.54\)Subtract these to find:\[ 5.56 - 1.54 = 4.02 \]

Solve for Time \(t\)

Now that we have plugged \(4.02\) into the equation:\[ 4.02 = 0.54t \]We solve for \(t\) by dividing both sides by 0.54:\[ t = \frac{4.02}{0.54} \approx 7.44 \]

Convert Calculation Into Contextual Answer

The time taken for the concentration to drop from 0.65 M to 0.18 M is approximately 7.44 seconds.

Key concepts

Understanding Second-Order Reactions

In chemical kinetics, reactions are often classified by their order, which gives insight into how reaction rates depend on reactant concentration. A second-order reaction means that the rate depends on the concentration of one reactant raised to the second power, or on the product of the concentrations of two different reactants. For the reaction involving nitrogen dioxide, the chemical equation can be simplified as:

• 2 NO₂ (g) → 2 NO (g) + O₂ (g)

When we refer to it being a second-order reaction, it specifically involves a single reactant, NO₂, leading to products. This implies that if you were to double the concentration of NO₂, the reaction rate would increase fourfold due to the \([NO_2]^2\) dependency. This fundamental understanding helps when calculating how the concentration changes over time using kinetic equations.

Role and Understanding of Rate Constants

The rate constant (\(k\)) is a crucial factor in chemical kinetics, as it provides the link between the concentration of reactants and the rate of reaction. For second-order reactions, the units of the rate constant differ from those in reactions of other orders:

• For a second-order reaction, the rate constant is typically expressed in units of \(M^{-1} s^{-1}\).

In the context of the given reaction, the rate constant is specified as 0.54 \(M^{-1} s^{-1}\). Therefore, this value allows us to substitute into the integrated rate law to determine changes over time. Larger values of \(k\):tend to indicate faster reactions, while smaller values suggest slower rates.

Integrated Rate Law for Second-Order Reactions

The integrated rate law for a second-order reaction provides a straightforward means to connect concentrations with time. This law is formulated as:\[ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt \]where \([A]_0\) represents the initial concentration of the reactant, and \([A]_t\) is the concentration at a specific time \(t\).

• This equation offers the method to calculate how long it takes to reach a certain concentration if the initial concentration and rate constant are known.
• It also provides a way to predict the concentration of a reactant at any given time if the rate constant and initial concentration are available.

Thus, it becomes particularly helpful in laboratory settings where determining the time required for certain transformations is essential.

Calculating Reaction Time

Once we have the formula from the integrated rate law, calculating the reaction time between two concentrations is simplified into sequential plug-and-play steps. For the given reaction, we first place the values of initial concentration (\([A]_0 = 0.65 M\)), final concentration (\([A]_t = 0.18 M\)), and the rate constant (\(k = 0.54 M^{-1} s^{-1}\)) into the rate law.

• Compute \(\frac{1}{0.18}\) and get approximately 5.56
• Compute \(\frac{1}{0.65}\) and get approximately 1.54
• The difference is 4.02
• Substitute back to find \(t\) using 4.02 = 0.54 \(t\)
• Divide and simplify to get \(t\approx 7.44 \, \text{seconds}\)

So, this calculation gives the estimated time for the nitrogen dioxide concentration to decrease to the desired level. This approach not only helps in practice but strengthens the understanding of how integrated rate laws are used to determine reaction dynamics.