Animated Solution for Chemistry - Chemical Kinetics: $t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $3/4$ of its initial value. If the rate constant for a first order reaction is $k$, the $t_{1/4}$ can be written as

Decoding Fractional Lifetimes in First-Order Kinetics

When we study chemical kinetics, we often obsess over the half-life ($t_{1/2}$) of a reaction. It's the famous benchmark—the time it takes for exactly $50\%$ of our reactant to vanish. But what happens when we want to track the reaction at a different milestone? What if we want to know the time it takes for the concentration to drop to exactly $3/4$ of its initial value?

This is where the concept of fractional lifetimes, like $t_{1/4}$, comes into play. Let's break down the mathematics and the physical intuition behind this specific time marker.

The language of the question is our first critical clue. It asks for the time taken for the concentration to "drop to $3/4$ of its initial value."

Imagine you have a bucket of water representing your initial concentration, $[A{]}_0=a$. If the water level drops to $3/4$ of its original height, the remaining amount of water is simply $\frac{3}{4}a$.

It is crucial not to confuse "drop to" with "drop by". If the concentration had dropped by $3/4$, the remaining amount would be $1/4$. Always read the prepositions carefully!

So, at time $t=t_{1/4}$, our final concentration is:

For any first-order reaction, the relationship between time, rate constant ($k$), and concentration is governed by the integrated rate law:

This equation is our master key. We want to find the specific time $t_{1/4}$, so let's rearrange the formula to solve for time:

Notice how beautifully the initial concentration $a$ cancels out. This is a hallmark of first-order kinetics: the time it takes to reach any fractional completion is completely independent of the starting amount!

After canceling $a$, we are left with a simple logarithmic term:

To solve this without a calculator, we rely on the fundamental properties of logarithms. The log of a quotient is the difference of their logs:

If you are preparing for competitive exams, memorizing basic log values is non-negotiable. We know that $\log 4\approx 0.6020$ and $\log 3\approx 0.4771$.

Subtracting these gives:

Now, we simply plug this value back into our time equation:

Rounding to two decimal places, we get our final, elegant expression:

Before we move on, let's compare this result to the famous half-life equation, $t_{1/2}=\frac{0.693}{k}$.

Our calculated $t_{1/4}$ is $\frac{0.29}{k}$. Notice that $0.29$ is less than half of $0.693$. Why is that?

In a first-order reaction, the rate of decay is directly proportional to the concentration. At the very beginning of the reaction, the concentration is at its highest, meaning the reaction is proceeding at its absolute fastest. Therefore, it takes comparatively less time to burn through the first $25\%$ of the reactant than it does to burn through the next $25\%$. The mathematics perfectly mirrors the physical reality of exponential decay!