Problem 104 - 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 min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

Step-by-Step Solution

Answer

(a) The half-life of the reaction at room temperature is \(t_{1/2} = 990\) seconds. (b) The rate constant for the reaction at 415°C is \(k = 2.05 × 10^{-4} s^{-1}\).

Step 1: Identify the given information

We are given that the reaction is a first-order reaction and the rate constant \(k = 7.0 × 10^{-4} s^{-1}\).

Step 2: Apply the half-life formula

Now, we can plug the rate constant into the half-life formula for a first-order reaction: \[t_{1/2} = \frac{0.693}{7.0 × 10^{-4} s^{-1}}\]

Step 3: Calculate the result

Dividing 0.693 by the rate constant, we get: \[t_{1/2} = 990 s\] So, the half-life of the reaction at room temperature is 990 seconds. (b) Calculate the rate constant at 415°C.

Step 4: Identify the given information

We are given that the reaction is a first-order reaction with a half-life of \(t_{1/2} = 56.3\) minutes at 415°C.

Step 5: Convert the half-life to seconds

Since the rate constant will be given in \(s^{-1}\), we need to convert the half-life to seconds: 56.3 minutes × 60 s/minute = 3378 s So, \(t_{1/2} = 3378 s\).

Step 6: Apply the half-life formula

Now, we can use the half-life formula for a first-order reaction to find the rate constant: \[k = \frac{0.693}{t_{1/2}}\]

Step 7: Calculate the result

Plugging in the half-life, we get: \[k = \frac{0.693}{3378 s}\] \[k = 2.05 × 10^{-4} s^{-1}\] So, the rate constant for the reaction at 415°C is \(2.05 × 10^{-4} s^{-1}\).

Key Concepts

Half-life Calculation

Half-life, commonly represented by the symbol \(t_{1/2}\), is a critical concept in chemical kinetics that signifies the time required for the concentration of a reactant to decrease by half. In a first-order reaction, the half-life is independent of the initial concentration and can be calculated using the formula:

\[t_{1/2} = \frac{0.693}{k}\]
Here, \(k\) is the rate constant of the reaction. The constant 0.693 is derived from the natural logarithm of 2, due to the relationship between half-life and exponential decay in first-order kinetics. The simplicity of this formula is especially useful, as it provides a quick means of understanding the reaction's behavior over time.
For instance, in the given exercise, by substituting the rate constant value \(7.0 \times 10^{-4} s^{-1}\), we calculated the half-life of the decomposition of \(\mathrm{H}_2\mathrm{O}_2\) at room temperature. This highlights half-life as a convenient and intuitive characteristic for comparing the speed of different first-order reactions.

Rate Constant Determination

Determining the rate constant, typically denoted as \(k\), is crucial for quantifying the speed of a chemical reaction. It is an intrinsic part of the rate law, which for a first-order reaction, takes the form:

\[\text{rate} = k[\text{A}]\]
In this expression, \([\text{A}]\) represents the concentration of the reactant. The rate constant can be deduced from experimental data, such as concentration measurements over time. Alternatively, if the half-life of the reaction is known, as in the second part of the exercise dealing with the decomposition of \(\mathrm{(CH_2)_2 O}\), the rate constant can be calculated using the correlation:

\[k = \frac{0.693}{t_{1/2}}\]
By converting the half-life into seconds and applying the formula, we determined the rate constant at 415°C. The calculation of \(k\) enables us to predict the rate of the reaction at different concentrations and to compare the reactivity under various conditions.

Chemical Kinetics

Chemical kinetics is the area of chemistry that deals with the speed or rate of a chemical reaction and the mechanism by which the reaction occurs. A fundamental understanding of kinetics enables chemists to manipulate conditions to control the rate and predict the outcome of reactions.

The study of kinetics often involves various order reactions with first-order kinetics being one of the most common. A first-order reaction is characterized by its rate being directly proportional to the concentration of a single reactant. This means that as the reactant is consumed, the rate of the reaction decreases linearly with its concentration. The mathematical representation of a first-order rate law is:

\[ \text{rate} = k[\text{A}]^1 \]
Determining reaction order is vital as it influences how the rate reacts to changes in concentration and how the rate constant \(k\) is used to infer various dynamic aspects of the reaction. Kinetics also informs the stability and shelf-life of substances, crucial in disciplines like pharmaceuticals, food chemistry, and environmental science. In educational settings and in the field, understanding kinetics is pivotal for predicting and controlling the complex nature of chemical reactions.