Problem 110 - Chemistry The Molecular Science

Question

Calculate the half-life of a first-order reaction if \(30.5 \mathrm{~s}\) after the reaction starts the concentration of the reactant is \(0.0451 \mathrm{M}\) and \(45.0 \mathrm{~s}\) after the reaction starts it is 0.0321 M. Calculate how many seconds after the start of the reaction it takes for the reactant concentration to decrease to \(0.0100 \mathrm{M}\).

Step-by-Step Solution

Answer

Half-life is 29.5 s; concentration reaches 0.0100 M after 64.1 s.

Step 1: Identify the First-Order Kinetics Formula

For a first-order reaction, the concentration of the reactant at time \( t \) is given by \( [A] = [A]_0 e^{-kt} \), where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), \( k \) is the rate constant, and \( e \) is the base of the natural logarithm.

Step 2: Use Two Points to Find the Rate Constant (k)

We use the given concentrations at two different times to set up the following two equations based on the first-order kinetics formula. \( 0.0451 = [A]_0 e^{-k \times 30.5} \) and \( 0.0321 = [A]_0 e^{-k \times 45.0} \). We can divide these equations to eliminate \( [A]_0 \): \( \frac{0.0321}{0.0451} = e^{-k \times (45.0 - 30.5)} \). Calculating the ratio and taking the natural logarithm allows us to solve for \( k \).

Step 3: Calculate the Rate Constant (k)

Calculate \( \frac{0.0321}{0.0451} \approx 0.7118 \). Take the natural logarithm: \( \ln(0.7118) \approx -0.341 \). Since \( 45.0-30.5=14.5 \), the equation becomes \( -0.341 = -k \times 14.5 \). Solve for \( k \) to get \( k = \frac{0.341}{14.5} \approx 0.0235 \mathrm{~s}^{-1} \).

Step 4: Determine Half-Life using Rate Constant

The half-life \( t_{1/2} \) for a first-order reaction is given by \( t_{1/2} = \frac{0.693}{k} \). Substitute the value of \( k \) to get \( t_{1/2} \approx \frac{0.693}{0.0235} \approx 29.5 \mathrm{~s} \).

Step 5: Calculate Time to Reach 0.0100 M Concentration

Using the rate constant \( k \) and the first-order reaction formula, set \( 0.0100 = [A]_0 e^{-kt} \). Use the initial condition with \( 0.0451 \mathrm{~M} \) at \( 30.5 \mathrm{~s} \) to solve \( 0.0100 = 0.0451 e^{-0.0235t} \). Take the natural logarithm: \( (\frac{0.0100}{0.0451}) = -0.0235t \). Calculate \( (0.2217) \approx -1.507 \). Solve for \( t \) to find \( t \approx \frac{-1.507}{-0.0235} \approx 64.1 \mathrm{~s}\).

Key Concepts

Half-life Calculation

When dealing with first-order reaction kinetics, understanding the half-life of a chemical reaction is crucial. The **half-life** refers to the time it takes for the concentration of a reactant to reduce to half of its initial amount. For first-order reactions, the half-life is independent of the initial concentration of the reactant. This unique characteristic simplifies calculations.

The general formula to calculate the half-life for a first-order reaction is \[ t_{1/2} = \frac{0.693}{k} \]where:

• \( t_{1/2} \) is the half-life, and
• \( k \) is the rate constant.

To find the half-life, once you have determined the rate constant, simply divide 0.693 by the rate constant. As seen from our example, with a rate constant of 0.0235 s^-1, the half-life calculation is straightforward, resulting in approximately 29.5 seconds.

This relation shows the simplicity and beauty of first-order kinetics, where, regardless of the reactant's beginning concentration, the time taken to reach half of that initial amount remains constant.

Rate Constant

The **rate constant** is a fundamental element when exploring reaction kinetics. It provides important insight into the speed at which a reaction proceeds. For first-order reactions, it is denoted by \( k \) and has units of s^-1. Its value is critical for determining other quantities such as the half-life.

To find the rate constant \( k \), we use the first-order rate equation:\[ [A] = [A]_0 e^{-kt}\]Given two sets of concentration and time values, you can solve for the rate constant by leveraging the ratios of the concentrations. This exercise shows how dividing the concentration equations at different times and taking the natural logarithm helps isolate \( k \).

For instance, using the concentration ratio from 30.5s to 45.0s, the calculations yield that \( k \approx 0.0235 \mathrm{~s}^{-1} \). This simplistic approach illustrates that by applying the foundational laws of logarithms and exponential based equations, finding the rate constant becomes a manageable task, driving further understanding of the reaction's behavior.

Natural Logarithm

The **natural logarithm** is an indispensable part of solving problems involving first-order reaction kinetics. Represented by \( \ln \), it helps in expressing the exponential decay of reactant concentrations over time. The natural logarithm has the base \( e \), recognized as an irrational constant approximately equal to 2.718.

In the context of first-order reactions, taking the natural logarithm of both sides of the concentration ratio equation simplifies solving for unknowns like the rate constant or time. For example, simplifying \[ \ln \left( \frac{[A]}{[A]_0} \right) = -kt\]allows solving for \( k \) or \( t \), based on known values.

Understanding how to manipulate logarithms is vital because it transforms non-linear first-order rate equations into linear forms. This transformation simplifies complex calculations, ultimately providing clarity and precision in kinetic problem-solving, as shown in calculating the time for the reactant to reach a specific concentration.