Problem 44 - Chemistry The Central Science In Si Units

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 \(5.0 \mathrm{~min} ?(\mathbf{b})\) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol} ?\) (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) 0.00612 mol, (b) 2.68 minutes, (c) 1.69 minutes.

Step 1: Determine the initial concentration

Start by calculating the initial concentration of \( \mathrm{N}_2 \mathrm{O}_5 \) using the initial moles and volume of the container.\[\text{Initial concentration } [\mathrm{N}_2 \mathrm{O}_5]_0 = \frac{0.0250 \text{ mol}}{2.0 \text{ L}} = 0.0125 \text{ M}\]

Step 2: Use the first-order decay formula for part (a)

The first-order decay formula is \([A] = [A]_0 e^{-kt}\), where \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time. Plug the known values into this formula to find the concentration after 5 minutes. Convert 5 minutes to seconds: \(t = 5 \times 60 = 300 \text{ s}\). Use:\[[N_2O_5] = 0.0125 \cdot e^{-6.82 \times 10^{-3} \times 300}\]

Step 3: Solve the equation for remaining moles

Calculate the concentration after 5 minutes using the value computed in the previous step and then find the moles remaining by multiplying the concentration by the volume.\[[N_2O_5] = 0.0125 \cdot e^{-2.046} \approx 0.00306 \text{ M}\]\[\text{Remaining moles} = 0.00306 \times 2.0 = 0.00612 \text{ mol}\]

Step 4: Apply the first-order decay equation for part (b)

To find the time for the concentration to decrease to 0.010 mol in 2 L, set up the decay equation. Substitute \([A] = \frac{0.010}{2} = 0.005 \text{ M}\) and solve for \(t\).\[0.005 = 0.0125 \cdot e^{-6.82 \times 10^{-3} \times t}\]Take the natural logarithm on both sides and solve for \(t\):\[\ln{\frac{0.005}{0.0125}} = -6.82 \times 10^{-3} \times t\]

Step 5: Calculate the time for decrease to 0.010 moles

Rearrange and solve the equation to find \(t\):\[t = \frac{\ln{\left(\frac{0.005}{0.0125}\right)}}{-6.82 \times 10^{-3}} \approx \frac{\ln{0.4}}{-6.82 \times 10^{-3}} \approx 160.62 \text{ s}\]Convert seconds back to minutes:\[t \approx \frac{160.62}{60} \approx 2.68 \text{ minutes}\]

Step 6: Calculate the half-life for part (c)

For a first-order reaction, the half-life \(t_{1/2}\) is given by \(t_{1/2} = \frac{0.693}{k}\). Substitute the known rate constant into the formula:\[t_{1/2} = \frac{0.693}{6.82 \times 10^{-3}} \approx 101.6 \text{ s}\]Convert to minutes:\[t_{1/2} \approx \frac{101.6}{60} \approx 1.69 \text{ minutes}\]

Key Concepts

Rate Constant

The rate constant, often represented by the symbol \( k \), is a crucial parameter in determining the speed of a chemical reaction. It is a unique value for every reaction under specific conditions of pressure and temperature. For first-order reactions, this value plays a significant role because the reaction rate is directly proportional to the concentration of one reactant.

In our example, the rate constant for the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \) at \( 70^{\circ} \mathrm{C} \) is \( 6.82 \times 10^{-3} \mathrm{~s}^{-1} \). This means that the reaction's speed is determined by how fast the \( \mathrm{N}_2 \mathrm{O}_5 \) decomposes due to this rate constant.

Understanding this concept helps you see how altering conditions like temperature can change the value of \( k \), and hence the speed of the reaction. Recall that in first-order reactions, the dimensions of the rate constant are \( \mathrm{s}^{-1} \), indicating the reaction's timing element.

Half-Life Calculation

The half-life of a reaction, denoted by \( t_{1/2} \), is the time it takes for half of the reactant to be consumed. For first-order reactions, the half-life is independent of initial reactant concentration. This makes analysis straightforward.

For a first-order reaction, we calculate the half-life using the formula:

• \( t_{1/2} = \frac{0.693}{k} \)

In our exercise, substituting the given rate constant \( k = 6.82 \times 10^{-3} \mathrm{~s}^{-1} \) leads to:

• \( t_{1/2} = \frac{0.693}{6.82 \times 10^{-3}} = 101.6 \text{ seconds} \)
• Convert seconds to minutes: \( t_{1/2} \approx 1.69 \text{ minutes} \)

This consistent half-life is handy because you don't have to worry about the starting amount of material when determining when half of it will have reacted. You can calculate how quickly or slowly a reaction progresses over time.

Concentration Changes

In first-order reactions, concentration changes according to an exponential decay process. This means that the concentration falls by a fixed proportion in equal time intervals. The formula used to describe the concentration at any time \( t \) is:

• \([A] = [A]_0 e^{-kt}\)

Here, \([A]_0\) is the initial concentration, and the exponential factor \( e^{-kt} \) evaluates how much of the reactant is left after a time \( t \).

In our case, after 5 minutes (or 300 seconds), we find:

• \([N_2O_5] = 0.0125 \cdot e^{-6.82 \times 10^{-3} \times 300} \approx 0.00306 \text{ M}\)
• This equates to \(0.00612 \text{ mol} \) remaining in the 2 L volume.

This exponential decrease helps predict how quickly a reaction proceeds and allows you to calculate how much reactant remains at any given time.