Problem 71 - 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 \(1.00-\mathrm{L}\). 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 min: P(HOF) = 50 mm Hg, Total P = 100 mm Hg. After 45 min: P(HOF) = 35.4 mm Hg, Total P = 100 mm Hg.

Step 1: Understanding First-Order Reactions

First-order reactions have a specific characteristic: the rate at which the reactant concentration decreases is proportional to the reaction:\[ \frac{d[A]}{dt} = -k[A] \]where \(k\) is the rate constant. The half-life for a first-order reaction is given by \( t_{1/2} = \frac{0.693}{k} \).

Step 2: Calculate the Rate Constant

Given the half-life \( t_{1/2} = 30 \) minutes, we can find the rate constant \( k \) using the relation:\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} \, \text{min}^{-1} \approx 0.0231 \, \text{min}^{-1} \]

Step 3: Calculate Partial Pressure of HOF after 30 Minutes

Using the first-order decay formula:\[ [A] = [A]_0 e^{-kt} \]where \([A]_0 = 1.00 \times 10^2 \text{mm Hg}\), and \(t = 30 \text{ minutes}\):\[ [A] = (1.00 \times 10^2) \times e^{-0.0231 \times 30} \approx 50 \text{mm Hg}\]Thus, the partial pressure of HOF after 30 minutes is 50 mm Hg.

Step 4: Calculate Total Pressure after 30 Minutes

After 30 minutes, half of the initial pressure of HOF has decomposed. The reaction:\[ \mathrm{HOF(g)} \rightarrow \mathrm{HF(g)} + \frac{1}{2} \mathrm{O}_{2}(g) \]indicates that the decomposition of HOF results in new gases (HF and O2), conserving the number of moles (because 1 mole of HOF produces 1 mole of products). Thus, the total pressure remains the same, 100 mm Hg, as initial.

Step 5: Calculate Partial Pressure of HOF after 45 Minutes

Using the decay formula again for 45 minutes:\[ [A] = [A]_0 e^{-kt} \]\[ [A] = (1.00 \times 10^2) \times e^{-0.0231 \times 45} \approx 35.4 \text{mm Hg} \]The partial pressure of HOF after 45 minutes is approximately 35.4 mm Hg.

Step 6: Calculate Total Pressure after 45 Minutes

Initially, the pressure was 100 mm Hg. After 45 minutes, 64.6% (since 100 mm Hg - 35.4 mm Hg = 64.6 mm Hg) of HOF decomposed producing equal moles of gas. Therefore, the total pressure still remains 100 mm Hg, as the moles of gas do not change.

Key Concepts

Reaction Rate Constant

In chemical kinetics, the reaction rate constant is crucial for understanding how quickly a reaction proceeds. For first-order reactions, this constant (\( k \)) describes the proportionate rate at which the concentration of a reactant decreases over time. Imagine the reactant molecules rushing to react with each other; the rate constant helps us mathematically capture the speed of this process. The formula for the reaction rate in such reactions is:

• \( \frac{d[A]}{dt} = -k[A] \)

Here, \([A]\) is the concentration of the reactant, and \(k\) represents the rate constant. A helpful relation for students to remember is that the half-life equation for a first-order reaction is \( t_{1/2} = \frac{0.693}{k} \). This formula tells us how long it takes for half of the reactant to decompose. At a given temperature, smaller values of \(k\) mean the reaction is slower because it takes longer for half of the substance to react. Conversely, a larger \(k\) indicates a faster reaction.

Half-life Calculation

The term half-life refers to the time required for half of a given amount of a substance to decompose or react. In the context of first-order reactions, calculating the half-life helps predict how quickly a chemical species will diminish to half its original amount. This calculation is essential in reactions involving unstable compounds like hypofluorous acid (HOF). To determine the rate constant, we use the relationship:

• \( k = \frac{0.693}{t_{1/2}} \)

In the exercise given, the half-life of HOF is 30 minutes. Plugging this value into the formula, we calculate the rate constant:

• \( k = \frac{0.693}{30} \, \text{min}^{-1} \approx 0.0231 \, \text{min}^{-1} \)

This value allows us to predict how the concentration of HOF decreases over time, informing other calculations, such as the partial pressure after certain time intervals.

Partial Pressure Calculation

Partial pressure is a measure used to describe the pressure exerted by a single type of gas in a mixture of gases. In our exercise, hypofluorous acid (HOF) decomposes, altering its partial pressure in a sealed flask over time. Understanding how partial pressures change is essential when dealing with reactions that generate or consume gases.

The formula for first-order decay, used in calculating partial pressures, is given by:

• \( [A] = [A]_0 e^{-kt} \)

Where \([A]_0\) is the initial concentration and \(t\) is the time elapsed. By knowing the rate constant \(k\), you can calculate how the partial pressure changes. For example, after 30 minutes in the problem, the partial pressure of HOF changes from 100 mm Hg to approximately 50 mm Hg. Similarly, after 45 minutes, it reduces further to nearly 35.4 mm Hg.

Interestingly, even though individual gas pressures change, the total gas pressure in the flask remains 100 mm Hg. This constancy occurs because any decomposed HOF molecules convert into an equivalent number of HF and O2 molecules, maintaining the total number of moles and thus the total pressure.