Problem 19 - Chemistry and Chemical Reactivity

Question

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}),\) decomposes to \(\mathrm{H}_{2} \mathrm{O}(\ell)\) and \(\mathrm{O}_{2}(\mathrm{g})\) in a reaction that is first order in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and has a rate constant \(k=1.06 \times 10^{-3} \mathrm{min}^{-1}\) at a given temperature. (a) How long will it take for \(15 \%\) of a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to decompose? (b) How long will it take for \(85 \%\) of the sample to decompose?

Step-by-Step Solution

Answer

(a) 152.75 min, (b) 1871.57 min.

Step 1: Identify the First-Order Reaction Formula

For a first-order reaction, the relationship between concentration and time is given by the formula: \[ ext{ln}[ ext{A}] = ext{ln}[ ext{A}_0] - kt \] where \([ ext{A}]\) is the concentration at time \(t\), \([ ext{A}_0]\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.

Step 2: Determine the Fraction of \\([\text{H}_2\text{O}_2]\\) Remaining

For part (a), since 15% of the sample decomposes, 85% remains. Thus, \( \frac{[ ext{A}]}{[ ext{A}_0]} = 0.85 \).For part (b), 85% decomposes, so only 15% remains. Therefore, \( \frac{[ ext{A}]}{[ ext{A}_0]} = 0.15 \).

Step 3: Substitute Values into the First-Order Formula

Use the fraction values determined in Step 2 and substitute them into the formula from Step 1. The equation becomes:For part (a):\[ \frac{[ ext{A}]}{[ ext{A}_0]} = 0.85 = e^{-kt} \]For part (b):\[ \frac{[ ext{A}]}{[ ext{A}_0]} = 0.15 = e^{-kt} \]

Step 4: Solve for Time \(t\) in Each Scenario

Rearrange each equation to solve for \(t\):For part (a): solve for \(t\) using \[ t = \frac{- ext{ln}(0.85)}{k} = \frac{- ext{ln}(0.85)}{1.06 \times 10^{-3} \, ext{min}^{-1}} \]For part (b):\[ t = \frac{- ext{ln}(0.15)}{k} = \frac{- ext{ln}(0.15)}{1.06 \times 10^{-3} \, ext{min}^{-1}} \]

Step 5: Calculate the Time \(t\) from the Equations

For part (a):\[ t = \frac{- ext{ln}(0.85)}{1.06 \times 10^{-3}} \approx 152.75 \, ext{minutes} \]For part (b):\[ t = \frac{- ext{ln}(0.15)}{1.06 \times 10^{-3}} \approx 1871.57 \, ext{minutes} \]

Key Concepts

First-Order Reaction

In chemical kinetics, a first-order reaction is characterized by its rate being directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases over time, so does the rate of the reaction.

• The mathematical expression for a first-order reaction can be written as: \[\text{Rate} = k [A]\]where \(k\) is the rate constant, and \([A]\) is the concentration of the reactant.
• First-order reactions are common in processes like radioactive decay and the decomposition of hydrogen peroxide.

Understanding first-order reactions is crucial because it helps in determining how long a reaction will take to reach a certain extent of completion. By knowing just a few variables, we can predict the progress of the reaction over time.

Rate Constant

The rate constant, often denoted as \(k\), is a crucial factor in the rate equation of a reaction. It gives us an idea about how fast or slow a reaction is proceeding.

• In a first-order reaction, the units of the rate constant \(k\) are typically inverse time, such as \(\mathrm{min}^{-1}\), indicating that the concentration changes exponentially over time.
• The value of \(k\) is influenced by factors such as temperature and the presence of a catalyst.

The rate constant in a first-order reaction remains constant at a given temperature, allowing us to simplify the relationship between the concentration of reactants and time. In our hydrogen peroxide decomposition example, the rate constant is \(1.06 \times 10^{-3} \mathrm{min}^{-1}\). This constant allows us to calculate how long it takes for a specified fraction of the reactant to decompose.

Reaction Time Calculation

To calculate the time it takes for a given percentage of a reaction to occur, particularly in first-order reactions, you can use the first-order kinetics formula: \[\text{ln}[A] = \text{ln}[A_0] - kt\]This formula helps in determining the decay rate of reactants over time. Here's how you can apply it:

• First, determine the fraction of the reactant that remains after a certain percentage has decomposed. For instance, if 15% decomposes, then 85% remains, giving you \(\frac{[A]}{[A_0]} = 0.85\).
• Substitute this fraction into the equation, along with the rate constant \(k\), to solve for time \(t\).

For example, for hydrogen peroxide, with a 15% decomposition, the formula becomes:\[0.85 = e^{-kt}\]Solving for \(t\) gives us how long it will take for these reactions to proceed to the given extent. The computations involve taking the natural logarithm of the fraction remaining and dividing by the rate constant. This technique is invaluable for estimating reaction timelines under known conditions.