Q.A first order reaction is 50% completed in $1.26 \times 10^{14}$ s. How much time would it take for 100% completion?

Concept: Average Rate Of Reaction — for a first order reaction, the half-life is constant and independent of initial concentration.

Reasoning:

1. For a first order reaction, $t_{1/2} = \frac{0.693}{k}$. Given $t_{1/2} = 1.26 \times 10^{14}$ s.
2. 100% completion would mean the reactant concentration becomes zero.
3. From the integrated rate law $[A] = [A]_0 e^{-kt}$, $[A] = 0$ only when $t \to \infty$ (exponential never reaches zero in finite time). …

Why this question is a classic trap

Many students see “50% completed in $1.26 \times 10^{14}$ s” and instinctively think: If half the reaction takes that long, then the whole reaction should take twice as long — $2.52 \times 10^{14}$ s. That would be true only for a zero‑order reaction, where the rate is constant and the concentration decreases linearly. But this is a first‑order reaction, and the behaviour is fundamentally different.

In a first‑order reaction, the rate at any moment is proportional to the concentration remaining. As the reactant gets used up, the reaction slows down. It never actually stops — it just keeps getting slower and slower, approaching completion asymptotically. That is why the time for “100% completion” is not a finite number; it is infinite.

Step‑by‑step reasoning

1. Recall the integrated rate law for a first‑order reaction

For a reaction $A \rightarrow \text{products}$ that is first order in $A$:

where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant.

2. Relate half‑life to the rate constant

The half‑life $t_{1/2}$ is the time when $[A]_t = \frac{1}{2}[A]_0$. Substituting into the rate law:

So:

Given $t_{1/2} = 1.26 \times 10^{14}$ s, we can find $k$:

But we don’t actually need the numerical value of $k$ to answer the question — the key insight is conceptual.

3. What does “100% completion” mean?

“100% completion” means $[A]_t = 0$. But look at the integrated rate law:

If $[A]_t = 0$, then $\frac{[A]_0}{[A]_t} \rightarrow \infty$, and $\ln(\infty) \rightarrow \infty$. That would require $t \rightarrow \infty$.

In other words, no finite value of $t$ can make $[A]_t$ exactly zero. The concentration decays exponentially:

$[A]_t = [A]_0 e^{-kt}$ …

Method: First-Order Kinetics — Half-Life Concept

For a first-order reaction, the time required for completion does not depend on the initial concentration in a simple multiple-of-half-life way. The key insight is:

A first-order reaction never reaches 100% completion in finite time — it approaches completion asymptotically.

Steps:

1. Recall the half-life formula for a first-order reaction For a first-order reaction:

Here, $t_{1/2} = 1.26 \times 10^{14} \text{ s}$.

2. Understand what 100% completion means For a first-order reaction, the concentration $[A]$ at time $t$ is:

For 100% completion, $[A] = 0$. This requires $e^{-kt} = 0$, which only happens as $t \to \infty$.

3. Conclusion …

Common Mistakes Students Make on This Question

Mistake 1: Assuming 100% Completion is Finite

The error: Students see "50% completed in $1.26 \times 10^{14}$ s" and think "100% will take double the time" → picking option (ii) $2.52 \times 10^{14}$ s.

Why it's wrong: For a first order reaction, the time for completion is infinite because the rate depends on the concentration of the reactant. As concentration approaches zero, the rate slows down asymptotically — it never truly reaches zero.

How to avoid: Remember the key property:

• For zero order: finite time for 100% completion
• For first order: infinite time for 100% completion

The integrated rate law for first order:

When $[A]_t = 0$, $\log \frac{[A]_0}{0} \to \infty$, so $t \to \infty$.

Mistake 2: Using Half-Life Formula Incorrectly

The error: Students calculate $k$ from half-life:

Then try to find time for 100% by plugging $[A]_t = 0$ into:

This gives a division by zero error, but some students force a finite answer.

How to avoid: Recognize that $\log(0)$ is undefined. The moment you see "100% completion" for a first order reaction, the answer must be infinite — no calculation needed.

Mistake 3: Misinterpreting "100% Completion"

The error: Students think "100% completion" means the reaction is completely over in a measurable time, like a physical process.

Why it's wrong: In chemical kinetics, "100% completion" for first order means all reactant is consumed. Mathematically, this requires infinite time because the rate law is exponential decay.

How to avoid: Visualize the concentration vs. time graph for first order — it approaches zero asymptotically but never reaches it. Compare with zero order where the graph is a straight line hitting zero at finite time.

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