Animated Solution for Chemistry - Chemical Kinetics: The half-life period of a first order chemical reaction is $6.93\text{ min}$. The time required for the completion of $99\%$ of the chemical reaction will be ($\log 2=0.301$)

Welcome to the fascinating world of Chemical Kinetics! Today, we are tackling a classic problem involving a first-order reaction. Imagine a radioactive sample or a drug degrading in your bloodstream; these processes often follow first-order kinetics. The defining characteristic of such reactions is that their half-life ($t_{1/2}$) is completely independent of the initial concentration.

In our problem, we are given a half-life of $6.93\text{ min}$. Our mission is to find out how long it takes for the reaction to reach $99\%$ completion. Let's break this down step-by-step.

The rate constant is the heartbeat of any chemical reaction. For a first-order reaction, the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is beautifully simple:

Why $0.693$? It's actually an approximation of the natural logarithm of 2 ($\ln 2$). By substituting our given half-life into this formula, we get:

Notice how the numbers were perfectly crafted to give us a clean, easy-to-work-with rate constant. This is a common theme in competitive exams!

Now that we have our rate constant, we can predict the future of this reaction using the integrated rate law for first-order kinetics:

Here, $[A_0]$ is the initial concentration, and $[A]$ is the concentration remaining at time $t$. The question asks for the time when the reaction is $99\%$ complete. This is where many students make a silly mistake. If $99\%$ of the reactant has been consumed, how much is left? Exactly $1\%$!

To make the math intuitive, let's assume our initial concentration $[A_0]$ is $100$. Therefore, the remaining concentration $[A]$ will be $100-99=1$. Let's plug these values into our master equation:

We know that $\log (100)$ is simply $2$ (since $10^2=100$). Substituting this back in:

And there we have it! It takes exactly $46.06\text{ min}$ for the reaction to reach $99\%$ completion.

Before we wrap up, let's look at a powerful shortcut. If you compare the time for $99\%$ completion ($46.06\text{ min}$) with the half-life ($6.93\text{ min}$), you'll notice that $t_{99\%}\approx 6.64\times t_{1/2}$.

Even more impressively, the time for $99.9\%$ completion is exactly $10\times t_{1/2}$. Memorizing these standard ratios can save you precious minutes during high-pressure exams like JEE and NEET. Keep practicing, and let the kinetics flow!