Understanding First-Order Reaction Kinetics

A first-order reaction is a reaction whose rate depends on the concentration of only one reactant raised to the power of one. The rate constant ($k$) for a first-order reaction has units of time$^{-1}$.

The integrated rate law for a first-order reaction can be expressed in various forms. A common form relating the initial concentration $[A]_0$ at time $t=0$ and the concentration $[A]_t$ at time $t$ is:

\begin{equation*} kt = \ln\left(\frac{[A]_0}{[A]_t}\right) \end{equation*}

We can use this equation to solve the problem.

Calculating the Rate Constant ($k$) from 90% Completion Data

We are given that the reaction takes 24 minutes to get 90% complete. This means that if we start with an initial concentration $[A]_0$, after 24 minutes, 90% of $[A]_0$ has reacted. The remaining concentration $[A]_t$ will be the initial concentration minus the reacted amount:

\begin{equation*} [A]_t = [A]_0 - 0.90 [A]_0 = 0.10 [A]_0 \end{equation*}

Now we can plug these values into the integrated rate law with $t = 24$ min:

\begin{equation*} k \times 24 \text{ min} = \ln\left(\frac{[A]_0}{0.10 [A]_0}\right) \end{equation*}

The $[A]_0$ terms cancel out:

\begin{equation*} 24k = \ln\left(\frac{1}{0.10}\right) \end{equation*}

\begin{equation*} 24k = \ln(10) \end{equation*}

So, the rate constant $k$ is:

\begin{equation*} k = \frac{\ln(10)}{24} \text{ min}^{-1} \end{equation*}

Calculating Time for 99.9% Completion

Now we need to find the time ($t_{99.9\%}$) when the reaction is 99.9% complete. This means that 99.9% of the initial concentration $[A]_0$ has reacted. The remaining concentration $[A]_t$ at time $t_{99.9\%}$ will be:

\begin{equation*} [A]_t = [A]_0 - 0.999 [A]_0 = 0.001 [A]_0 \end{equation*}

Using the same integrated rate law with the calculated value of $k$ and the concentration at 99.9% completion:

\begin{equation*} k \times t_{99.9\%} = \ln\left(\frac{[A]_0}{0.001 [A]_0}\right) \end{equation*}

Again, the $[A]_0$ terms cancel out:

\begin{equation*} k \times t_{99.9\%} = \ln\left(\frac{1}{0.001}\right) \end{equation*}

\begin{equation*} k \times t_{99.9\%} = \ln(1000) \end{equation*}

We know that $\ln(1000) = \ln(10^3) = 3 \ln(10)$. So,

\begin{equation*} k \times t_{99.9\%} = 3 \ln(10) \end{equation*}

Now, substitute the value of $k = \frac{\ln(10)}{24}$ into this equation:

\begin{equation*} \left(\frac{\ln(10)}{24}\right) \times t_{99.9\%} = 3 \ln(10) \end{equation*}

To solve for $t_{99.9\%}$, divide both sides by $\frac{\ln(10)}{24}$:

\begin{equation*} t_{99.9\%} = \frac{3 \ln(10)}{\frac{\ln(10)}{24}} \end{equation*}

The $\ln(10)$ terms cancel out:

\begin{equation*} t_{99.9\%} = 3 \times 24 \text{ min} \end{equation*}

\begin{equation*} t_{99.9\%} = 72 \text{ min} \end{equation*}

Thus, the time taken for the first-order reaction to get 99.9% complete is 72 minutes.

Summary of First-Order Reaction Calculations

Here is a summary of the percentage completion and corresponding time expressions for a first-order reaction:

From the table, we can see the relationship:

• $t_{99\%} = 2 \times t_{90\%}$ (since $k t_{99\%} = 2 \ln(10)$ and $k t_{90\%} = \ln(10)$)
• $t_{99.9\%} = 3 \times t_{90\%}$ (since $k t_{99.9\%} = 3 \ln(10)$ and $k t_{90\%} = \ln(10)$)

Given $t_{90\%} = 24$ min, $t_{99.9\%} = 3 \times 24$ min = 72 min.

Revision Table: First-Order Kinetics Concepts

Additional Information: Properties of First-Order Reactions

• The rate of a first-order reaction slows down as the concentration of the reactant decreases.
• The half-life of a first-order reaction is constant. This is a key characteristic. It means it takes the same amount of time for the concentration to drop from 1M to 0.5M as it does from 0.5M to 0.25M, and so on.
• Radioactive decay is a classic example of a first-order process.
• Plotting $\ln[A]_t$ versus time ($t$) for a first-order reaction yields a straight line with a slope equal to $-k$. This is often used experimentally to determine the rate constant.

Understanding the relationship between the rate constant, time, and percentage completion is crucial for solving problems involving first-order reactions.