Question

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, \quad 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) 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 \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C} ?\)

Short answer

(a) After 5.0 min, there will be approximately 0.0168 mol of N2O5 remaining. (b) It will take approximately 5.81 minutes for the quantity of N2O5 to drop to 0.010 mol. (c) The half-life of N2O5 at 70°C is approximately 1.70 minutes.

Step-by-step solution

Part A: Calculate moles of N2O5 remaining after 5.0 min

Given: Initial moles (A0) = 0.0250 mol Time (t) = 5.0 min = 300 s (converted to seconds) Rate constant (k) = 6.82 × 10^{-3} s^{-1} Use the formula: \[ ln(\frac{[A]_0}{[A]}) = kt \] Rearrange the formula to find the remaining moles (A): \[ [A] = \frac{[A]_0}{e^{kt}} \] Substitute the values: \[ [A] = \frac{0.0250}{e^{(6.82 × 10^{-3})(300)}} \] Calculate the result: \[ [A] = 0.0168 \; \text{mol} \] After 5.0 min, there will be approximately 0.0168 mol of N2O5 remaining.

Part B: Calculate time for the quantity of N2O5 to drop to 0.010 mol

Given: Initial moles (A0) = 0.0250 mol Final moles (A) = 0.010 mol Use the formula: \[ ln(\frac{[A]_0}{[A]}) = kt \] Rearrange the formula to find the time (t): \[ t = \frac{ln(\frac{[A]_0}{[A]})}{k} \] Substitute the values: \[ t = \frac{ln(\frac{0.0250}{0.010})}{6.82 × 10^{-3}} \] Calculate the result: \[ t ≈ 348.88 \; \text{s} \] Convert to minutes: \[ time \approx 5.81 \; \text{min} \] It will take approximately 5.81 minutes for the quantity of N2O5 to drop to 0.010 mol.

Part C: Calculate the half-life of N2O5 at 70°C

To find the half-life (t1/2) of a first-order reaction, we use the formula: \[ t_{1/2} = \frac{ln(2)}{k} \] Given: Rate constant (k) = 6.82 × 10^{-3} s^{-1} Substitute the value: \[ t_{1/2} = \frac{ln(2)}{6.82 × 10^{-3}} \] Calculate the result: \[ t_{1/2} ≈ 101.73 \; \text{s} \] Convert to minutes: \[ half-life ≈ 1.70 \; \text{min} \] The half-life of N2O5 at 70°C is approximately 1.70 minutes.

Key concepts

First-Order Reactions

When it comes to chemical kinetics, understanding reaction orders is crucial. A first-order reaction is a type where the reaction rate depends linearly on only one reactant's concentration. This means, as time progresses, the rate of reaction decreases at a rate proportionate to the concentration of one reactant. It's like having a single-track train line where the speed is controlled solely by the number of passengers or goods on board.

In mathematical terms, this is expressed with the rate equation:

• Rate = k[A],

where \( k \) is the rate constant and \([A]\) is the concentration of the reactant. Often, with first-order reactions, logarithmic functions are used to express changes concerning time, leading to an exponential decay of the reactant's concentration. This characteristic makes it easier to predict and analyze how reactants behave over time, allowing chemists to make accurate predictions.

Rate Constant Calculation

The rate constant \( k \) of a reaction is fundamental to understanding the speed of reactions in chemical kinetics. For first-order reactions, \( k \) can be found by examining how the concentration of a reactant changes over time. This value essentially provides insight into the efficiency or speed at which a reaction progresses under certain conditions.

To calculate the rate constant, the equation:

• \( ln\left(\frac{[A]_0}{[A]}\right) = kt \),

is used, where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \( t \). By measuring the concentration at various intervals, one can determine \( k \) by rearranging and solving the equation accordingly.

The unit of the rate constant for first-order reactions is \( s^{-1} \), reflecting how the concentration changes per second. This small yet potent value relays how fast a reaction can proceed, making it a critical factor in the world of chemical kinetics.

Half-Life Determination

Half-life is a particularly intriguing concept in kinetics, especially for understanding the longevity and decay of substances. In a first-order reaction, the half-life is the time required for half of the reactant to be consumed or transformed. This means that in every half-life period, the concentration of the reactant decreases by 50%.

For first-order reactions, the half-life \( t_{1/2} \) is calculated using the formula:

• \( t_{1/2} = \frac{ln(2)}{k} \),

where \( k \) is the rate constant. A unique feature of first-order reactions is that the half-life remains constant, regardless of the reactant's initial concentration.

This stability in half-life makes first-order reactions particularly useful in real-world applications such as radioactive decay, pharmacokinetics, and the breakdown of unstable compounds, providing a predictable timeline for when a substance will reach certain levels of concentration.