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 halflife of only 30 min at room temperature: $$\mathrm{HOF}(\mathrm{g}) \longrightarrow \mathrm{HF}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If the partial pressure of HOF in a \(1.00-\mathrm{L}\). flask is initially \(1.00 \times 10^{2} \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what is the total pressure in the flask and the partial pressure of HOF after exactly 30 min? After 45 min?

Step-by-Step Solution

Answer

After 30 min, HOF is 50 mmHg, total 125 mmHg; after 45 min, HOF is 35.36 mmHg, total 132.32 mmHg.

Step 1: Understanding the reaction

The decomposition of hypofluorous acid (HOF) is a first-order reaction. This means the rate of reaction depends linearly on the concentration of HOF. The products are hydrofluoric acid (HF) and oxygen gas (O₂). Given the half-life of 30 minutes, every 30 minutes, the amount of HOF left is halved.

Step 2: First-Order Reaction Formula

For a first-order reaction, the concentration as a function of time is given by:\[[A]_t = [A]_0 \cdot e^{-kt}\]where \([A]_t\) is the concentration at time \( t \),\([A]_0\) is the initial concentration,\(k\) is the rate constant, and\(e\) is Euler's number.

Step 3: Calculate the Rate Constant

Use the half-life to find the rate constant \(k\) using the formula:\[k = \frac{\ln 2}{t_{1/2}}\]where \(t_{1/2} = 30 \text{ min} = 30 \cdot 60 \text{ s} = 1800 \text{ s}\).Thus,\[k = \frac{\ln 2}{1800} \approx 3.85 \times 10^{-4} \text{ s}^{-1}\].

Step 4: Calculate Partial Pressure after 30 min

The pressure of HOF after 30 minutes can be calculated using:\[P_t = P_0 \cdot e^{-kt}\]where the initial pressure \(P_0 = 100 \text{ mmHg}\).Substitute \(t = 1800 \text{ s}\):\[P_{30} = 100 \cdot e^{-3.85 \times 10^{-4} \times 1800}\]\[\approx 50 \text{ mmHg}\].

Step 5: Calculate Total Pressure after 30 min

After 30 minutes, half of the HOF decomposes to form HF and O₂. For every mole of HOF, one mole of HF and 0.5 mole of O₂ are produced. Therefore, the pressure from the products is 50 mmHg (from HF) + 25 mmHg (from O₂) = 75 mmHg. Thus, total pressure = 50 mmHg (HOF) + 75 mmHg = 125 mmHg.

Step 6: Calculate Partial Pressure after 45 min

Now calculate the pressure of HOF after 45 min (2700 s):\[P_{45} = 100 \cdot e^{-3.85 \times 10^{-4} \times 2700}\]\[\approx 35.36 \text{ mmHg}\].

Step 7: Calculate Total Pressure after 45 min

After 45 minutes, 64.64 mmHg (from the initial 100 mmHg) has decomposed into products, which results in 64.64 mmHg pressure of HF and half of that from O₂, 32.32 mmHg. Total pressure = 35.36 mmHg (HOF) + 64.64 mmHg (HF) + 32.32 mmHg (O₂) = 132.32 mmHg.

Key Concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. In a closed container, like a flask, the total pressure is the sum of the partial pressures of all gases present. For a first-order reaction, partial pressures allow us to track the concentration of gases involved.
In the exercise, hypofluorous acid (HOF) decomposes into hydrofluoric acid (HF) and oxygen gas ( O₂ ). Initially, the partial pressure of HOF is 100 mmHg. As the reaction progresses, this pressure declines as more HOF decomposes into HF and O₂ .

• After 30 minutes, the partial pressure of HOF decreases to approximately 50 mmHg due to the halved concentration.
• After 45 minutes, it further reduces to about 35.36 mmHg as more decomposition occurs.

This change in partial pressure helps to determine how much gas remains and how much has turned into the products at any given time.

Half-life

In the context of chemistry, the half-life is the time required for half of the reactant to decompose in a first-order reaction. It's a useful measure because it gives a quick way to estimate how fast a reaction proceeds.
For HOF, the half-life is given as 30 minutes. This means that every 30 minutes, the concentration of HOF will reduce by half. Knowing the half-life allows us to predict concentrations at various times without needing the precise rate constant calculation each time.

• At 0 minutes, the concentration is at 100% or 100 mmHg partial pressure.
• At 30 minutes, it reduces to 50% (or 50 mmHg).
• By 45 minutes, it decreases to less than half again, reflecting an exponential process.

Understanding half-life puts reaction kinetics in a simpler perspective, highlighting the exponential decay nature of first-order reactions.

Reaction Kinetics

Reaction kinetics studies the rates of chemical processes and how different conditions affect these rates. In a first-order reaction like the decomposition of HOF, the reaction rate depends on the concentration of just one reactant.
The formula for a first-order reaction is given by:\[[A]_t = [A]_0 \cdot e^{-kt}\]where:- [A]_t is the concentration at time t,- [A]_0 is the initial concentration,- k is the rate constant, and- e is Euler's number.
The rate constant k can be determined using the reaction's half-life. For HOF with a half-life of 30 minutes, k is calculated as approximately 3.85 \times 10^{-4} s^{-1}. This rate constant tells us how quickly the reaction proceeds and helps model how the concentration changes over time.

Chemical Decomposition

Chemical decomposition is the process where a compound breaks down into simpler products. In the case of HOF, it decomposes into HF and O₂ through a first-order reaction.
The decomposition process can be monitored by observing the change in partial pressures:

• The decomposition of HOF results in gas products, which means the total pressure in the flask increases even as the HOF pressure diminishes.
• After 30 minutes, half of the initial HOF has decomposed, creating 75 mmHg of pressure from HF and O₂ combined.
• After 45 minutes, more pressure is from the decomposed products compared to the remaining HOF, showing how the substance transitions to its products and affects total pressure.

The chemical decomposition not only shows change in specific chemical composition but also illustrates how energy and matter conservation principles account for changes in measurable quantities, like pressure in a confined space.