Problem 98 - Chemistry: The Central Science

Question

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1}\). Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g)\). If the reaction is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

Step-by-Step Solution

Answer

For part (a), the half-life of the reaction at room temperature is approximately 990 seconds. For part (b), the rate constant of the reaction at 415°C is approximately \(2.05 \times 10^{-4} s^{-1}\).

Step 1: Gather information

We are given the rate constant (k) of the reaction at room temperature, which is \(7.0 \times 10^{-4} s^{-1}\). Our task is to determine the half-life (t₅₀) of this reaction.

Step 2: Calculate the half-life using the first-order relationship

Use the relationship between the half-life and the rate constant for a first-order reaction: \(t_{\frac{1}{2}} = \frac{0.693}{k}\) Plug in the given rate constant: \(t_{\frac{1}{2}} = \frac{0.693}{7.0 \times 10^{-4} s^{-1}}\)

Step 3: Solve for t₅₀

Calculate the value of t₅₀: \(t_{\frac{1}{2}} \approx 990s\) For part (a), the half-life of the reaction at room temperature is approximately 990 seconds. For part (b):

Step 1: Gather information

We are given the half-life (t₅₀) of the reaction at 415°C, which is 56.3 minutes. We need to calculate the rate constant (k) of this reaction in \(s^{-1}\) .

Step 2: Convert the half-life to seconds

To calculate k in \(s^{-1}\), we need the half-life in seconds: \(56.3 \text{ min} \times \frac{60 \text{ s}}{1\text{ min}} \approx 3378\text{ s}\) The half-life in seconds is 3378s.

Step 3: Calculate the rate constant using the first-order relationship

Use the relationship between the half-life and the rate constant for a first-order reaction: \(t_{\frac{1}{2}} = \frac{0.693}{k}\) Now, plug in the half-life in seconds: \(3378\text{ s} = \frac{0.693}{k}\)

Step 4: Solve for k

Calculate the value of k: \(k \approx 2.05 \times 10^{-4} s^{-1}\) For part (b), the rate constant of the reaction at 415°C is approximately \(2.05 \times 10^{-4} s^{-1}\).

Key Concepts

First-Order Reaction

When we talk about a first-order reaction, we are referring to a reaction where the rate depends on the concentration of one reactant raised to the first power. In simpler terms, this means that if you double the concentration of the reactant, the reaction rate also doubles. This is represented in the rate law as:

• Rate = k [A]
• Where "Rate" is the speed of the reaction, "k" is the rate constant, and "[A]" is the concentration of the reactant.

For a reaction to be classified as first-order, it must have this straightforward relationship between concentration and rate. This is an essential concept because it helps us predict how the reaction will behave as conditions change.
The given reactions in the problem

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Rate Constant Calculation

The rate constant is a crucial factor in understanding reaction kinetics. It provides insights into the speed of a reaction, how fast or slow it might proceed under certain conditions. Calculating the rate constant, "k", involves understanding the relationship between the reaction rate and the reactant concentrations.
In a first-order reaction, this calculation takes advantage of the straightforward relationship between the reaction rate and the concentration of the reactant.
For instance, in part (b) of the exercise, we can determine the rate constant from the given half-life using the first-order equation for half-life:

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This equation directly relates half-life with the rate constant, which underscores how critical the rate constant is in reaction kinetics.

Half-Life Determination

The term half-life refers to the time it takes for the concentration of a reactant to decrease to half its initial value during a reaction. This concept is crucial in understanding how a reaction progresses over time.
For a first-order reaction, the half-life is constant and does not depend on the concentration of the reactant. It is calculated with the formula:

• \(t_{\frac{1}{2}} = \frac{0.693}{k}\)
• When we know the rate constant "k", we can directly calculate the half-life.

This means that for any given rate constant, the half-life will always be the same, no matter the initial concentration.
In part (a) of the problem, we find that the half-life of the reaction at room temperature is approximately 990 seconds, highlighting the beauty and simplicity of first-order reactions. Understanding and calculating half-life allows chemists to predict how long a reaction will take, which is invaluable in both research and industry settings.