Problem 89 - 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 reac- tion 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 at room temperature is calculated using the formula \(t_{1/2} = \frac{0.693}{k}\), resulting in a half-life of 990 seconds. (b) To calculate the rate constant at 415°C, first convert the given half-life of 56.3 minutes to seconds (3378 seconds). Then, use the same half-life formula to solve for the rate constant, resulting in a rate constant of \(2.05 \times 10^{-4} s^{-1}\).

Step 1: Use the half-life equation for first-order reactions

Use the formula \(t_{1/2} = \frac{0.693}{k}\), where k is the rate constant. In this case, the rate constant k is given as \(7.0 \times 10^{-4} s^{-1}\).

Step 2: Calculate the half-life

Now, plug the value of k into the formula and solve for the half-life as follows: \(t_{1/2} = \frac{0.693}{7.0 \times 10^{-4} s^{-1}} = 990~s\) So, the half-life at this temperature is 990 seconds. (b) Calculate the rate constant at 415°C

Step 1: Convert the half-life to seconds

Given the half-life is 56.3 minutes, we must first convert this to seconds. Recall that there are 60 seconds in a minute: \(56.3~minutes \times \frac{60~s}{1~minute} = 3378~s\)

Step 2: Use the half-life equation for first-order reactions

Use the formula \(t_{1/2} = \frac{0.693}{k}\), where this time, we are solving for the rate constant k, and we have the half-life \(t_{1/2} = 3378~s\).

Step 3: Calculate the rate constant

Now, rearrange the formula and solve for k: \(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3378~s} = 2.05 \times 10^{-4} s^{-1}\) So, the rate constant at 415°C is \(2.05 \times 10^{-4} s^{-1}\).

Key Concepts

Understanding Half-Life Calculation for First-Order Reactions

In the context of chemical reactions, the half-life is an important measure. It tells us how long it takes for half of the reactant to convert into product. This concept is crucial for first-order reactions, where the rate at which the reaction occurs is directly proportional to the concentration of the reactant.

The formula to calculate the half-life of a first-order reaction is given by the equation:

• \(t_{1/2} = \frac{0.693}{k}\)

Here, \(t_{1/2}\) represents the half-life, and \(k\) is the rate constant, specific to the reaction conditions.

This equation shows that the half-life of a reaction is inversely proportional to the rate constant. The larger the rate constant, the shorter the half-life. For instance, in the exercise, a rate constant of \(7.0 \times 10^{-4} \mathrm{s}^{-1}\) results in a half-life of 990 seconds. This means, over 990 seconds, half of the reactant decomposes, making it a handy measure to predict the speed of a first-order reaction.

The Importance of the Rate Constant

The rate constant \(k\) is a proportionality factor that plays a significant role in the rate law of a chemical reaction. In first-order reactions, it can be a direct indicator of how fast a reaction will proceed under constant conditions.

The rate constant's units vary with the order of reaction. For first-order reactions, the unit is \(\mathrm{s}^{-1}\), exemplified in the rate constant calculation of the reaction provided in the exercise. Here, given a half-life of 56.3 minutes (or 3378 seconds) for the reaction occurring at 415°C, we calculate \(k\) using the rearranged half-life formula:

• \(k = \frac{0.693}{t_{1/2}}\)

Substituting the half-life into the formula gives a rate constant of \(2.05 \times 10^{-4} \mathrm{s}^{-1}\).

This indicates that, at the given temperature, the reaction is relatively slower than the first example, as this \(k\) value is lower, showing that rate constants are key to comparing reaction rates at different conditions or for different reactions.

An Overview of Chemical Kinetics

Chemical kinetics is a branch of chemistry that studies the speed or rate at which chemical reactions occur and the factors affecting them. Understanding this helps predict how altering conditions like temperature or concentration influences reaction speed.

Central to chemical kinetics are several concepts:

• Reaction Rate: How quickly reactants turn into products. For first-order reactions, this depends on the concentration of a single reactant.
• Rate Law: An equation expressing the relationship between the rate of reaction and the concentrations of reactants. For first-order reactions, it is represented as \(\text{rate} = k[A]\), where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.
• Half-Life: The time required for the concentration of a reactant to reduce by half, as calculated using the rate constant.

In the given exercise, both parts are excellent examples of how chemical kinetics is applied to calculate the half-life and to determine the rate constant, underlining the predictability of first-order reactions.

Overall, mastering chemical kinetics provides valuable insights into reaction mechanisms and the optimization of industrial chemical processes, making it an invaluable area of study for chemists.