If time taken for a first-order reaction to get 90% complete is 24 min, its t_99.9% will be:

Question

If time taken for a first-order reaction to get 90% complete is 24 min, its t99.9% will be:

Prepp Team · Accepted Answer

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:

Percentage Completion	Remaining Concentration ($[A]_t$)	Integrated Rate Law Expression
90%	$0.1 [A]_0$	$kt_{90\%} = \ln(10)$
99%	$0.01 [A]_0$	$kt_{99\%} = \ln(100) = 2 \ln(10)$
99.9%	$0.001 [A]_0$	$kt_{99.9\%} = \ln(1000) = 3 \ln(10)$
99.99%	$0.0001 [A]_0$	$kt_{99.99\%} = \ln(10000) = 4 \ln(10)$

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

Concept	Description	Formula/Example
Rate Law	Describes how the reaction rate depends on reactant concentrations. For A → Products, Rate = $k[A]^1$.	Rate $\propto [A]$
Rate Constant ($k$)	Proportionality constant in the rate law. Specific for a reaction at a given temperature.	Units: time$^{-1}$ (e.g., s$^{-1}$, min$^{-1}$)
Integrated Rate Law	Relates concentration of reactants to time. Useful for finding concentration at a given time or time for a given concentration change.	$kt = \ln\left(\frac{[A]_0}{[A]_t}\right)$
Half-Life ($t_{1/2}$)	Time taken for the concentration of a reactant to become half of its initial value. For first-order, it's independent of initial concentration.	$t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{k}$
Percentage Completion	Fraction or percentage of initial reactant consumed over a certain time.	e.g., 90% complete means $[A]_t = 0.1[A]_0$

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.