Problem 97 - Chemistry: The Central Science in SI Units, Global Edition

Question

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is \(3.05 \times 10^{4} \mathrm{~s},\) what is the rate constant \(k\) at this temperature? What percentage of A will not have reacted after one day?

Step-by-Step Solution

Answer

The rate constant \(k\) at 25°C is \(2.27 \times 10^{-5} \text{ s}^{-1}\), and after one day, 46.7% of A will not have reacted.

Step 1: Calculate the Rate Constant k

Using the half-life formula for a first-order reaction, we can solve for the rate constant k: \(t_{1/2} = \frac{0.693}{k}\), Rearranging the equation to solve for k: \(k = \frac{0.693}{t_{1/2}}\), Now, substitute the given half-life value of \(3.05 \times 10^4\) s: \(k = \frac{0.693}{3.05 \times 10^4 \text{ s}}\). Calculating k: \(k = 2.27 \times 10^{-5} \text{ s}^{-1}\).

Step 2: Calculate the Percentage of A Remaining after One Day

To find the remaining percentage of A after one day, we can use the integrated rate law for a first-order reaction: \[A_t = A_0 e^{-kt}\]. We are interested in finding the ratio between the concentration of A after one day to the initial concentration: \(\frac{A_t}{A_0} = e^{-kt}\). First, let's find the value of t in seconds after one day: 1 day = 24 hours × 60 minutes/hour × 60 seconds/minute = 86400 seconds. Now, substitute the value of k and t into the equation: \(\frac{A_t}{A_0} = e^{-(2.27 \times 10^{-5} \text{ s}^{-1})(86400 \text{ s})}\), Calculating the ratio: \(\frac{A_t}{A_0} = 0.467\).

Step 3: Calculate the Remaining Percentage of A

To find the remaining percentage of A, multiply the ratio by 100: Remaining percentage = \(\frac{A_t}{A_0} \times 100\% = 0.467 \times 100\% = 46.7\%\). Therefore, 46.7% of A will not have reacted after one day.

Key Concepts

Rate Constant

In the study of first-order reactions, the rate constant, often represented by the symbol \(k\), plays a crucial role in determining how fast a reaction proceeds. For a first-order reaction, the rate of reaction depends linearly on the concentration of the reactant. This makes the concept of the rate constant integral in predicting and calculating reaction dynamics.

The rate constant can be estimated using the half-life (\(t_{1/2}\)) of a reaction, which is the time required for the concentration of a reactant to decrease to half of its initial value. For first-order reactions, there is a specific relationship:

• \(t_{1/2} = \frac{0.693}{k}\)

This formula shows that the half-life is inversely proportional to the rate constant, meaning a higher rate constant indicates a faster reaction with a shorter half-life.

Half-life

Half-life is the period it takes for half of the reactant to be consumed in a chemical reaction. It is a unique characteristic of each type of reaction, particularly useful in first-order kinetics where it remains constant throughout the reaction.

For first-order reactions, the relationship between half-life and the rate constant provides insights into the speed of the reaction. The formula used is:

• \(t_{1/2} = \frac{0.693}{k}\)

This equation highlights the relationship between the half-life and the rate of reaction, where a smaller half-life indicates a larger rate constant and vice versa. In practical applications, knowing the half-life aids in predicting how quickly reactants are consumed and products are formed.

Integrated Rate Law

The integrated rate law for first-order reactions is a powerful tool to model the concentration of reactants over time. This law can be expressed mathematically as:

• \(A_t = A_0 e^{-kt}\)

Where:- \(A_t\) is the concentration of reactant \(A\) at time \(t\)- \(A_0\) is the initial concentration- \(k\) is the rate constant
Using this equation, we can derive how the concentration changes, understanding the behavior at any given moment. Furthermore, it helps in calculating the concentration ratio \(\frac{A_t}{A_0}\) after a specific period, demonstrating how much reactant remains after time \(t\). In practical scenarios, this allows us to predict the progress of reactions over time.

Percentage Remaining

When exploring reaction kinetics, determining the percentage of the reactant remaining after a certain time is a key inquiry. Using the integrated rate law, we can find the fraction of the reactant that remains unreacted. For first-order kinetics:

• \(\frac{A_t}{A_0} = e^{-kt}\)

To find what percentage of the reactant remains, we multiply this fraction by 100. This was illustrated in the original exercise where after computing \(\frac{A_t}{A_0} = 0.467\), it was found that approximately 46.7% of the reactant remained after one day.

Understanding this percentage is crucial in many fields including pharmaceutical and chemical manufacturing where it can dictate dosage timings and reaction completion.