Problem 112 - Chemistry The Molecular Science

Question

Let \(x\) represent the number of half-lives that have elapsed during the course of a first-order reaction. That is, the elapsed time \(t\) is \(x\) times the half-life \(t_{1 / 2}: t=x t_{1 / 2}\). (a) Show that at time \(t\) the concentration \([\mathrm{A}]_{t}\) of reactant is related to the initial concentration \([\mathrm{A}]_{0}\) by $$ [\mathrm{A}]_{t}=[\mathrm{A}]_{0}\left(\frac{1}{2}\right)^{x} $$ (b) Use your result in part (a) to show that $$ \log \left(\frac{[\mathrm{A}]_{t}}{[\mathrm{~A}]_{0}}\right)=\frac{t}{t_{1 / 2}}(-\log 2)=\frac{t}{t_{1 / 2}}(-0.301) $$ (c) Use your result in part (b) to show that for any firstorder reaction a plot of \(\log [\mathrm{A}]_{t},\) versus \(t\) will be a straight line and that half-life can be obtained from the slope of the line.

Step-by-Step Solution

Answer

(a) Concentration reduces by half for each half-life. (b) Log relation reflects number of half-lives. (c) Linear plot shows reaction's kinetics.

Step 1: Understand the problem

We need to show the relationship between the concentration of a reactant after a time period and its initial concentration based on its half-life. Then connect that to the logarithmic expression, and finally establish why a plot of \( \log [A]_t \) versus \( t \) results in a straight line.

Step 2: Express concentration at time \( t \)

Given that \( x \) represents the number of half-lives that have passed, the concentration at time \( t \) can be expressed as after each half-life, the concentration halves. Thus, \([A]_t = [A]_0 \left(\frac{1}{2}\right)^x\), proving part (a).

Step 3: Derive the logarithmic relationship

Take the logarithm of the concentration at time \( t \) over the initial concentration:\[\log \left(\frac{[A]_t}{[A]_0}\right) = \log \left(\left(\frac{1}{2}\right)^x\right)\]This simplifies to:\[x \log \left(\frac{1}{2}\right) = -x \log 2\]Since \( x = \frac{t}{t_{1/2}} \), this becomes:\[\frac{t}{t_{1/2}}(-\log 2)\]For \( \log 2 \approx 0.301 \), it simplifies further:\[\frac{t}{t_{1/2}}(-0.301)\]This proves part (b).

Step 4: Prove linearity of the log plot

By expressing \( \log [A]_t \) as follows:\[\log [A]_t = \log [A]_0 + \frac{t}{t_{1/2}}(-\log 2)\]This is in the form of a linear equation, \( y = mx + b \), where:- \( y = \log [A]_t \)- \( x = t \)- \( m = \frac{-\log 2}{t_{1/2}} \), hence the slope is proportional to \( \frac{-\log 2}{t_{1/2}} \)Thus, plotting \( \log [A]_t \) against \( t \) will result in a straight line where the slope provides the half-life, verifying part (c).

Key Concepts

Understanding Half-life in First-order Reactions

In first-order reactions, the concept of half-life plays a vital role. The half-life \(t_{1/2}\) is the time it takes for half of the reactant to be consumed or decay.
This time period is consistent throughout the reaction, regardless of how much reactant is left.

Why is half-life constant in first-order reactions?

• Because the reaction rate is directly proportional to the concentration.
• Every half-life reduces the concentration by half, maintaining a consistent rate of decay.

Consider a reaction where \(x\) number of half-lives have elapsed. We can express the time elapsed as \(t = x t_{1/2}\).
At time \(t\), the concentration of the reactant \([A]_t\) becomes:\[ [A]_t = [A]_0 \left(\frac{1}{2}\right)^x \]This relationship demonstrates how each half-life consistently halves the concentration.

Exploring Reaction Rate through Logarithms

Logarithms help us understand how concentration changes over time in a first-order reaction. They provide a mathematical way to express exponential decay.

Taking the logarithm of the concentration ratio:\[ \log \left(\frac{[A]_t}{[A]_0}\right) = \log \left(\left(\frac{1}{2}\right)^x\right) \]The expression simplifies to:\[ x \log \left(\frac{1}{2}\right) = -x \log 2 \]Substituting \(x = \frac{t}{t_{1/2}}\), this becomes:\[ \frac{t}{t_{1/2}}(-\log 2) \]Since \(\log 2 \approx 0.301\),\[ \frac{t}{t_{1/2}}(-0.301) \]This formula shows how logarithms translate the exponential processes into linear relationships, making them easier to analyze.

Graphing Reaction Rate for Clarity

In graphing first-order reactions, the use of logarithms transforms data into a straight line. This makes trends clear and quantifiable.

The equation for the straight line is:\[ \log [A]_t = \log [A]_0 + \frac{t}{t_{1/2}}(-\log 2) \]This matches the linear form \(y = mx + b\), where:

• \(y = \log [A]_t\)
• \(x = t\)
• \(m = \frac{-\log 2}{t_{1/2}}\) is the slope
• The slope gives useful data like the half-life \(t_{1/2}\).

Plotting \(\log [A]_t\) against \(t\) helps visually illustrate the reaction rate, confirming both the nature and kinetics of the reaction.