A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Start

Q: Why did we use the $2^n$ formula instead of the standard $e^{-kt}$ formula?

Both formulas are perfectly valid and will give the exact same answer! However, when dealing with ratios that are powers of 2 (like 16, which is $2^4$), using the base-2 half-life formula $[C_t] = \frac{[C_0]}{2^n}$ makes the algebra much faster and avoids the need for natural logarithms.

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

Because the problem states it's an 'equimolar' mixture, both A and B start with the exact same concentration $x$. When we set up the equation $[A_t] = 16 [B_t]$, the $x$ appears on both sides as a multiplier and simply cancels out. The time taken depends only on the ratio, not the absolute starting amount.

Q: Why is the concentration of A 16 times that of B, and not the other way around?

Substance B has a much shorter half-life (18 min) compared to A (54 min). This means B is decaying much faster. Over time, B will disappear quickly while A remains relatively abundant. Therefore, A will always end up having a higher concentration than B.

Q: How did we get the LCM as 54 in the final calculation step?

We had the fractions $\frac{t}{18}$ and $\frac{t}{54}$. Since 54 is a direct multiple of 18 ($18 \times 3 = 54$), the least common multiple (LCM) is simply 54. So, $\frac{t}{18}$ becomes $\frac{3t}{54}$.

Q: Can this method be used if the ratio wasn't a perfect power of 2, like 10 instead of 16?

If the ratio wasn't a perfect power of 2, you could still use this method, but you would eventually have to take the logarithm (either base 10 or natural log) on both sides to solve for $t$. In such cases, starting with the $e^{-kt}$ formula might feel more natural.

Question

A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ......... min.
(Round off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Race of the Decaying Molecules

Imagine you are watching a race, but instead of cars, we have two chemical substances, **A** and **B**. They both start at the exact same starting line—meaning they have the same initial concentration, which we will call $x$. Both of them are decomposing following **first-order kinetics**, but they are running at very different speeds.

Substance A is the slow and steady runner. It has a half-life of $54.0 	ext{ min}$. This means it takes a full 54 minutes for its concentration to drop to half of its initial value. Substance B, on the other hand, is sprinting! Its half-life is only $18.0 	ext{ min}$, meaning it vanishes much faster.

### The Master Equation for Half-Lives

For any first-order reaction, there is a beautiful and simple way to find the concentration left after a certain time $t$. Instead of dealing with messy exponential functions like $e^{-kt}$, we can use the half-life formula directly:

$$[C_t] = \frac{[C_0]}{2^n}$$

Here, $n$ represents the **number of half-lives** that have passed, which is simply the total time $t$ divided by the half-life $t_{1/2}$. So, $n = \frac{t}{t_{1/2}}$.

### Setting Up the Condition

The question asks us to find the exact moment when substance A (the slow decayer) has a concentration that is exactly **16 times** greater than substance B (the fast decayer). Mathematically, we write this as:

$$[A_t] = 16 	imes [B_t]$$

Now, let's substitute our master equation into this condition. Since both started with the same concentration $x$, we get:

$$\frac{x}{2^{t/54}} = 16 	imes \frac{x}{2^{t/18}}$$

### The Final Calculation

The beauty of an equimolar mixture is that the initial concentration $x$ completely cancels out from both sides! We don't even need to know what it was.

Next, we can express 16 as a power of 2, which is $2^4$. This allows us to rewrite the equation entirely in base 2:

$$2^{-t/54} = 2^4 	imes 2^{-t/18}$$

Using the laws of exponents, we combine the terms on the right side:

$$2^{-t/54} = 2^{4 - t/18}$$

Since the bases are identical, their exponents must be equal. Let's equate them:

$$-\frac{t}{54} = 4 - \frac{t}{18}$$

Now, it's just simple algebra. Bring the $t$ terms to one side:

$$\frac{t}{18} - \frac{t}{54} = 4$$

To subtract these fractions, we take the common denominator, which is 54:

$$\frac{3t - t}{54} = 4$$

$$\frac{2t}{54} = 4$$

$$t = \frac{4 	imes 54}{2} = 108 	ext{ min}$$

And there we have it! After exactly **108 minutes**, the fast-decaying substance B will have dwindled so much that substance A will be 16 times more concentrated. A perfect example of how exponential decay creates massive differences over time!

Enter Numerical Value:

Visualized Solution

$[A]_{0} = [B]_{0} = x$

$(t_{1/2})_{A} = 54 min, (t_{1/2})_{B} = 18 min$

$[C_{t}] = \frac{[ C _{0} ]}{2 ^{n}}$

$[A_{t}] = 16 [B_{t}]$

$\frac{x}{2 ^{t /54}} = 16 \times \frac{x}{2 ^{t /18}}$

$2^{- t /54} = 2^{4} \times 2^{- t /18}$

$- \frac{t}{54} = 4 - \frac{t}{18}$

$\frac{t}{18} - \frac{t}{54} = 4$

$t = 108 min$

$Alternative Method$

The Sigma Insight: Order and Molecularity

The Race of the Decaying Molecules

The Master Equation for Half-Lives

Setting Up the Condition

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [A]0​=[B]0​=x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (t1/2​)A​=54 min,(t1/2​)B​=18 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [Ct​]=2n[C0​]​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [At​]=16[Bt​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2t/54x​=16×2t/18x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2−t/54=24×2−t/18

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −54t​=4−18t​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 18t​−54t​=4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=108 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alternative Method