A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. T

Q: Why do we use $e^{-Kt}$ instead of $(1/2)^n$?

Both forms are mathematically identical! Using $e^{-Kt}$ is the standard convention in chemical kinetics and makes calculus operations easier. However, as shown in the shortcut method, using base 2 (which is equivalent to $(1/2)^n$) can sometimes bypass logarithms entirely and save you precious time in exams.

Q: What happens to the initial concentration $N_0$? Why didn't we need its value?

Because both compounds started with the exact same initial concentration, and the problem asks for a *ratio* between their final amounts ($[A] = 4[B]$), the $N_0$ term appears on both sides of the equation as a multiplier. Thus, it perfectly cancels out. The time required to reach a specific ratio is independent of the starting amount!

Q: Why is the curve for compound B steeper in the graph?

Compound B has a shorter half-life ($180 \text{ s}$) compared to A ($300 \text{ s}$). A shorter half-life means the substance is more unstable and decays at a faster rate. Therefore, its concentration drops more sharply, resulting in a steeper exponential decay curve.

Q: What if the initial concentrations were not equal? Say, $[A]_0 = 2[B]_0$?

If they weren't equal, they wouldn't cancel out completely. You would substitute them into the equation: $2[B]_0 e^{-K_A t} = 4 [B]_0 e^{-K_B t}$. The $[B]_0$ would cancel, leaving a factor of 2 on the left side. You would then take the natural log of both sides, and the math would proceed similarly, just with an extra $\ln(2)$ term to handle.

Q: I always get confused with the minus signs when moving terms across the equals sign. Any tips?

It's a very common trap! The best way to avoid silly mistakes is to always move the smaller (or more negative) term to the side where it becomes positive. In our equation $-\frac{\ln 2}{300}t = 2\ln 2 - \frac{\ln 2}{180}t$, the term $-\frac{\ln 2}{180}t$ is more negative (since $1/180 > 1/300$). Moving it to the left makes it positive: $\frac{\ln 2}{180}t - \frac{\ln 2}{300}t$, which keeps your overall $t$ coefficient positive and safe!

Question

A flask contains a mixture of compounds $A$ and $B$. Both compounds decompose by first-order kinetics. The half-life for $A$ and $B$ are $300	ext{ s}$ and $180	ext{ s}$, respectively. If the concentrations of $A$ and $B$ are equal initially, the time required for the concentration of $A$ to be four times that of $B$ (in $	ext{s}$) is (Use $\ln 2 = 0.693$)

Options:

(A) 120

(B) 180

(C) 300

(D) 900

Sigma Percentile · Accepted Answer

The problem of decaying molecules is like a race where both runners are constantly slowing down, but one is losing energy much faster than the other. In this classic chemical kinetics problem, we are asked to find the exact moment when the slower-decaying compound becomes exactly four times as abundant as the faster-decaying one.

Let's dive into the beautiful mathematics of exponential decay and see how this unfolds!

### The Setup: A Tale of Two Half-Lives

Imagine you are looking at a flask containing a mixture of two compounds, **A** and **B**. They both start with the exact same initial concentration, which we will call $N_0$.

However, they have different half-lives. Compound **A** has a half-life of $300 	ext{ s}$, meaning it takes 5 minutes for half of it to disappear. Compound **B** is much more unstable; its half-life is only $180 	ext{ s}$, meaning it decays much faster.

Because **B** decays faster, its concentration will drop rapidly compared to **A**. Eventually, the amount of **A** left in the flask will be significantly larger than **B**. We are looking for the specific time $t$ when $[A] = 4[B]$.

### The Master Equation: Exponential Decay

Since both compounds follow **first-order kinetics**, their concentrations at any given time $t$ can be described by the master equation of exponential decay:

$$N(t) = N_0 e^{-Kt}$$

Here, $K$ is the rate constant, which is inversely proportional to the half-life:

$$K = \frac{\ln 2}{t_{1/2}}$$

Let's write down the specific decay equations for both compounds. For compound **A**:

$$[A] = N_0 e^{-\frac{\ln 2}{300}t}$$

And for compound **B**:

$$[B] = N_0 e^{-\frac{\ln 2}{180}t}$$

### The Mathematical Showdown

Now, we apply our core condition. We want the concentration of **A** to be four times that of **B**:

$$[A] = 4[B]$$

Substituting our exponential expressions into this condition, we get:

$$N_0 e^{-\frac{\ln 2}{300}t} = 4 \cdot N_0 e^{-\frac{\ln 2}{180}t}$$

Notice how the initial concentration $N_0$ appears on both sides. Because the compounds started with the same amount, this initial value beautifully cancels out, leaving us with a pure ratio:

$$e^{-\frac{\ln 2}{300}t} = 4 e^{-\frac{\ln 2}{180}t}$$

To solve for $t$, we need to bring the variables down from the exponents. We do this by taking the **natural logarithm** ($\ln$) on both sides. Remember the logarithmic property $\ln(ab) = \ln a + \ln b$, and that $\ln(4) = \ln(2^2) = 2\ln 2$:

$$-\frac{\ln 2}{300}t = 2\ln 2 - \frac{\ln 2}{180}t$$

### The Final Calculation

Let's group all the terms containing $t$ on the left side of the equation:

$$\frac{\ln 2}{180}t - \frac{\ln 2}{300}t = 2\ln 2$$

We can factor out $t \ln 2$ from the left side:

$$t \ln 2 \left( \frac{1}{180} - \frac{1}{300} ight) = 2\ln 2$$

The $\ln 2$ term cancels out perfectly from both sides! This is the elegance of first-order kinetics. Now, we just need to solve the simple fraction inside the bracket:

$$t \left( \frac{300 - 180}{180 	imes 300} ight) = 2$$

$$t \left( \frac{120}{54000} ight) = 2$$

Rearranging to solve for $t$:

$$t = \frac{2 	imes 54000}{120} = 900 	ext{ s}$$

And there we have it! At exactly **900 seconds**, the concentration of compound **A** will be four times the concentration of compound **B**.

### A Brilliant Shortcut (Base 2)

While the exponential method using base $e$ is standard, there is a brilliant shortcut using base $2$. Since the problem deals with half-lives and a ratio of $4$ (which is $2^2$), we can write the decay equation as:

$$N(t) = N_0 2^{-\frac{t}{t_{1/2}}}$$

Applying our condition $[A] = 4[B]$:

$$N_0 2^{-\frac{t}{300}} = 2^2 \cdot N_0 2^{-\frac{t}{180}}$$

Cancel $N_0$ and combine the exponents on the right side:

$$2^{-\frac{t}{300}} = 2^{2 - \frac{t}{180}}$$

Since the bases are the same, we can directly equate the exponents:

$$-\frac{t}{300} = 2 - \frac{t}{180}$$

$$\frac{t}{180} - \frac{t}{300} = 2$$

This leads to the exact same fractional equation we solved earlier, but we bypassed the logarithms entirely! This is a powerful technique to keep in your arsenal for JEE and NEET.

Select Answer:

Visualized Solution

Visualizing the Decay

First-Order Kinetics Formula

Applying the Condition $[A] = 4 [B]$

Substituting the Expressions

Taking the Natural Logarithm

Algebraic Manipulation

Calculating Final Time $t$

Generalizing the Concept

The Sigma Insight: Order and Molecularity

The Setup

The Master Equation

The Mathematical Showdown

The Final Calculation

A Brilliant Shortcut (Base 2)

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