Problem 81 - Chemistry and Chemical Reactivity

Question

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2},\) with a half-life of \(30 .\) minutes at room temperature: $$ \mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ If the partial pressure of HOF in a 1.00-L flask is initially \(1.00 \times 10^{2} \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what are the total pressure in the flask and the partial pressure of HOF after exactly 30 minutes? After 45 minutes?

Step-by-Step Solution

Answer

After 30 minutes: HOF = 50 mm Hg, Total Pressure = 125 mm Hg. After 45 minutes: HOF ≈ 37.68 mm Hg, Total Pressure ≈ 131.16 mm Hg.

Step 1: Understand the First-order Reaction

The decomposition of hypofluorous acid follows a first-order kinetic reaction. This means the rate at which HOF decomposes depends solely on its concentration. Equations related to first-order kinetics will be used to calculate changes in concentration or partial pressure over time.

Step 2: Calculate the Rate Constant (k)

For a first-order reaction, the relationship between half-life and the rate constant ( k ) is given by: \[ t_{1/2} = \frac{0.693}{k} \] With a half-life \( t_{1/2} = 30 \) minutes, solve for \( k \): \[ k = \frac{0.693}{30} = 0.0231 \text{ min}^{-1} \]

Step 3: Calculate Partial Pressure After 30 Minutes

The expression for first-order kinetics is: \[ [ ext{HOF}]_t = [ ext{HOF}]_0 e^{-kt} \] After 30 minutes (which is one half-life period), the partial pressure of HOF is halved from its initial value: \[ [ ext{HOF}]_{30} = 0.5 imes 100 \text{ mm Hg} = 50 \text{ mm Hg} \]

Step 4: Calculate Partial Pressure After 45 Minutes

Using the formula \([\text{HOF}]_t = [\text{HOF}]_0 e^{-kt}\), substitute t = 45 and calculate: \[ [\text{HOF}]_{45} = 100 \, \text{mm Hg} \, e^{-0.0231 \times 45} \] Solve for the remaining pressure: \[ [\text{HOF}]_{45} \approx 37.68 \, \text{mm Hg} \]

Step 5: Calculate Total Pressure after 30 Minutes

After 30 minutes, since HOF has decomposed into HF and \( \text{O}_2 \), the lost partial pressure of HOF translates to equivalent contributions from the products HF and \( \text{O}_2 \). For every \( \text{HOF} \rightarrow \text{HF} + \frac{1}{2} \text{O}_2 \), the pressure contribution is equal, but for \( \text{O}_2 \) it's half: \[ \Delta P = 50 \text{ mm Hg} \text{ (HOF lost)} \] Thus, \( \text{HF} = 50 \text{ mm Hg} \) and \( \text{O}_2 = \frac{1}{2} \times 50 = 25 \text{ mm Hg} \). Total pressure = remaining \( \text{HOF} \) + \( \text{HF} \) + \( \text{O}_2 \): \[ P_{\text{total}} = 50 + 50 + 25 = 125 \text{ mm Hg} \]

Step 6: Calculate Total Pressure after 45 Minutes

After 45 minutes, calculate the reacted pressure: \( 100 - 37.68 = 62.32 \text{ mm Hg} \). For \( \text{HF} \), the pressure is the same as that lost by HOF, so \( 62.32 \text{ mm Hg} \). For \( \text{O}_2 \), the lost pressure is divided by two: \( 31.16 \text{ mm Hg} \). Total pressure = remaining \( \text{HOF} \) + \( \text{HF} \) + \( \text{O}_2 \): \[ P_{\text{total}} = 37.68 + 62.32 + 31.16 \approx 131.16 \text{ mm Hg} \]

Key Concepts

Kinetics

Understanding kinetics is key to dominating chemical reactions, particularly first-order reactions. In kinetics, we study the rates at which chemical processes occur and how different conditions affect these rates. In a first-order reaction like the decomposition of hypofluorous acid (HOF), the reaction proceeds at a rate that depends solely on the concentration of one reactant—in this case, HOF.
The equation for a first-order reaction can be expressed as:
\[ [A]_t = [A]_0 e^{-kt} \]
Here is a rundown of this equation:

• \([A]_t\) is the concentration of A at time \(t\).
• \([A]_0\) is the initial concentration of A.
• \(k\) is the rate constant.
• \(t\) is the time elapsed.

In our exercise, we interpret this equation to calculate how much HOF is left at different time points, leading to insights into the kinetic behavior of the reaction as time progresses.

Partial pressure

Partial pressure is crucial when dealing with reactions involving gases. It refers to the pressure exerted by a particular gas in a mixture of gases. For our reaction, we need to determine how the partial pressure of HOF changes as it decomposes into HF and \(\mathrm{O}_2\).
Consider all gases as ideal to simplify calculations. The partial pressure of each product formed can be deduced using stoichiometric relationships. With the decomposition of HOF, the lost pressure (from HOF) distributes between HF and \(\mathrm{O}_2\) based on the stoichiometry of the reaction:
\[\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g})\]
For every mole of HOF that disappears, one mole of HF and 0.5 moles of \(\mathrm{O}_2\) are formed. Understanding how these pressures add up gives us an accurate picture of the total pressure in the system at any time.

Rate constant

In first-order reactions, the rate constant \(k\) plays a major role. It's a proportionality factor that links the rate of reaction to the concentration of reactants. For any first-order reaction, the half-life \(t_{1/2}\) is a constant and is calculated using:
\[ t_{1/2} = \frac{0.693}{k} \]
From this equation, we can isolate \(k\) and determine its value using given half-life data. This is fundamental for predicting how fast a reaction proceeds. In the step-by-step solution, we see that with a half-life of 30 minutes, the rate constant \(k\) is computed as:
\[ k = \frac{0.693}{30} = 0.0231 \, \text{min}^{-1} \]
This rate constant helps predict how quickly HOF concentration decreases over time and is essential when projecting the reaction profile across different time frames. A larger \(k\) would imply a faster reaction under the same conditions.