Problem 40 - Chemistry: The Central Science

Question

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1} .\) Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .\) (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) After 5 minutes, there are approximately 0.0129 moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) remaining. (b) It will take approximately 6.8 minutes for the quantity of \(\mathrm{N}_{2}\mathrm{O}_{5}\) to drop to 0.010 mol. (c) The half-life of \(\mathrm{N}_{2}\mathrm{O}_{5}\) at \(70^{\circ}\mathrm{C}\) is approximately 1.7 minutes.

Step 1: (a) Calculate the remaining moles of N₂O₅ after 5 minutes

To find the remaining moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) after 5 minutes, we can use the first-order integrated rate law: \[N_t = N_0e^{-kt}\] Where \(N_t\) is the moles at time \(t\), \(N_0\) is the initial moles, \(k\) is the rate constant, and \(t\) is the time in seconds. First, convert 5 minutes to seconds: \(5.0 \, \text{min} \times \frac{60 \, \text{s}}{1 \, \text{min}} = 300 \, \text{s}\) Next, plug the given values into the equation: \[N_t = (0.0250 \, \text{mol})e^{-(6.82 \times 10^{-3} \, \text{s}^{-1})(300 \, \text{s})}\] Calculate the remaining moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\): \[N_t = 0.0250 \, \text{mol} \times e^{-2.046} \approx 0.0129 \, \text{mol}\] So after 5 minutes, there are approximately 0.0129 moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) remaining.

Step 2: (b) Calculate the time for N₂O₅ to drop to 0.010 mol

Rearrange the first-order integrated rate law to solve for \(t\): \[t = -\frac{\ln{\frac{N_t}{N_0}}}{k}\] Plug in the given values and the desired final quantity of 0.010 mol: \[t = -\frac{\ln{\frac{0.010 \, \text{mol}}{0.0250 \, \text{mol}}}}{6.82 \times 10^{-3} \, \text{s}^{-1}}\] Calculate the time in seconds, then convert to minutes: \[t \approx 406.9 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \approx 6.8 \, \text{min}\] So it will take approximately 6.8 minutes for the quantity of \(\mathrm{N}_{2}\mathrm{O}_{5}\) to drop to 0.010 mol.

Step 3: (c) Find the half-life of N₂O₅ at 70°C

The half-life for a first-order reaction is given by: \[t_{1/2} = \frac{\ln{2}}{k}\] Plug in the given rate constant: \[t_{1/2} = \frac{\ln{2}}{6.82 \times 10^{-3} \, \text{s}^{-1}}\] Calculate the half-life in seconds, then convert to minutes: \[t_{1/2} \approx 101.7 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \approx 1.7 \, \text{min}\] The half-life of \(\mathrm{N}_{2}\mathrm{O}_{5}\) at \(70^{\circ}\mathrm{C}\) is approximately 1.7 minutes.

Key Concepts

First-order Reaction

In chemical kinetics, reactions are classified based on their order, which refers to how the rate of reaction is dependent on the concentration of reactants. A **first-order reaction** is one where the rate is directly proportional to the concentration of a single reactant. This means that if the concentration of the reactant is doubled, the reaction rate also doubles. In mathematical terms, the rate law for a first-order reaction can be expressed as: - \[ ext{Rate} = k[N_2O_5]\] where \(k\) is the rate constant, and \[N_2O_5\] is the concentration of reactant.
Such reactions often involve a single molecule breaking down into smaller molecules or atoms. For example, the decomposition of \(N_2O_5\) to \(NO_2\) and \(O_2\) as given in the exercise is a perfect example of a first-order reaction.

Rate Constant

The **rate constant** (**k**) is a crucial factor in the rate equation for any chemical reaction. For first-order reactions, it provides insights into how fast the reaction occurs independent of the reactant's initial concentration. The units of the rate constant for a first-order reaction are \( ext{s}^{-1}\)\.
💡 Key roles of the rate constant:

• Indicates reaction speed: A larger value of \(k\) means the reaction proceeds more rapidly. A small \(k\) suggests a slower reaction process.
• Unique for every reaction: It varies depending on the reaction and its conditions, such as temperature.
• Essential for calculating other parameters: It's used in the integrated rate law to determine the concentration of reactants at any time.

In our exercise, the rate constant is \(6.82 imes 10^{-3} \, ext{s}^{-1}\), which helps in calculating both the remaining concentration of \(N_2O_5\) at any given time and its half-life.

Half-life

The **half-life** of a reaction is the time required for the concentration of a reactant to reduce to half its initial value. It is a useful measure because it describes how quickly a reaction progresses. For first-order reactions, the half-life is constant and can be calculated using the following formula: - \[ t_{1/2} = \frac{\ln{2}}{k} \] where \(\ln{2}\) is approximately equal to 0.693.
Features of half-life in first-order reactions:

• Independent of initial concentration: This is a unique characteristic of first-order reactions. The half-life remains constant regardless of how much reactant you start with.
• Can be used to predict how long it will take for the reactant to decrease to a certain level.

In our example, calculating the half-life of \({N_2O_5}\) using \(t_{1/2} = \frac{0.693}{6.82 \times 10^{-3} \, ext{s}^{-1}}\) gives approximately 1.7 minutes. This indicates that every 1.7 minutes, the amount of \(N_2O_5\) is halved.

Integrated Rate Law

The **integrated rate law** for first-order reactions is pivotal for understanding how the concentration of reactants decreases over time. It is derived from the basic rate law and provides a relation between the concentration of reactants and time. The formula can be expressed as: - \[[ ext{N}_t] = [ ext{N}_0] e^{-kt} \] where \[N_t\] is the concentration at time \(t\), \[N_0\] is the starting concentration, \(k\) is the rate constant, and \(e\) is the base of the natural logarithm.
Here’s why the integrated rate law is crucial:

• Allows calculation of concentration: It helps us determine how much of a reactant is left after a certain period.
• Time prediction for desired concentration: We can rearrange the formula to find out how long it takes for a reactant to reach a specific concentration.
• Useful in experimental data analysis: Scientists can use it to plot and examine the progress of a reaction over time.

In the exercise, we use the integrated rate law to find how many moles of \(N_2O_5\) remain, or to determine how long it takes to reach a certain molarity after starting with \(0.0250 \, ext{mol}\).