Q.The rate constant for a first order reaction is $60\ \text{s}^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

The key idea is First Order Kinetics, where the integrated rate law relates concentration to time:

$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$.

Step 1: The concentration is reduced to $\frac{1}{16}$ of its initial value, so

$\frac{[A]_0}{[A]} = 16$.

Step 2: Substitute $k = 60\ \text{s}^{-1}$ into the formula:

$t = \frac{2.303}{60} \log 16$.

Step 3: $\log 16 = \log 2^4 = 4 \log 2 \approx 4 \times 0.3010 = 1.204$. …

First-order kinetics is one of the cleanest models in chemical kinetics because the rate depends only on the concentration of one reactant. The key insight: the time to go from any concentration to a fraction of it is constant — that’s the half-life property. For a first-order reaction, each half-life reduces the concentration by half. So if you want to go from $[A]_0$ to $[A]_0/16$, you’re asking: how many half-lives does it take to drop to $1/16$th?

$1/16 = (1/2)^4$, so it takes exactly 4 half-lives. That’s the conceptual shortcut. But let’s verify it formally using the integrated rate law, because exams often test both the formula and the reasoning.

1. Write the integrated rate law for a first-order reaction. The standard form is:

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

2. Plug in the given fraction. We want $[A] = \frac{[A]_0}{16}$. So:

3. Simplify $\ln 16$. $16 = 2^4$, so $\ln 16 = 4 \ln 2$. Thus:

4. Solve for $t$.

5. Substitute $k = 60\ \text{s}^{-1}$. Using $\ln 2 \approx 0.693$: …

Method: Integrated Rate Law for First-Order Reactions

This is the standard approach for solving time–concentration problems in first-order kinetics.

Steps

1. Recall the integrated rate law For a first-order reaction:

where

• $k$ = rate constant ($60\ \text{s}^{-1}$)
• $[A]_0$ = initial concentration
• $[A]_t$ = concentration at time $t$

2. Identify the given ratio The concentration is reduced to $\frac{1}{16}$ of its initial value:

Therefore:

3. Substitute into the equation

4. Simplify the logarithm

Using $\log 2 \approx 0.3010$:

5. Solve for $t$

Quick Check (Alternative Method) …

Here are the most common mistakes students make when solving this first-order kinetics problem, along with how to avoid each.

Mistake 1: Using the wrong formula for the number of half-lives

Students often try to use the half-life formula directly without checking if the fraction $1/16$ is a simple power of $1/2$.

• The error: They might calculate $t_{1/2} = \frac{0.693}{k}$ and then multiply by 4 (since $1/16 = (1/2)^4$). This is actually correct for this specific fraction, but the mistake is doing this blindly without verifying the fraction is a power of $1/2$.
• How to avoid: Always check: $1/16 = (1/2)^4$ → 4 half-lives. Then $t = 4 \times t_{1/2} = 4 \times \frac{0.693}{60\ \text{s}^{-1}}$.
• Key insight: This shortcut only works when the fraction is exactly $(1/2)^n$. For fractions like $1/10$ or $1/3$, you must use the integrated rate law.

Mistake 2: Forgetting to convert units or misplacing the rate constant

The rate constant is given as $60\ \text{s}^{-1}$. Some students treat it as if it were in minutes or hours, or they incorrectly invert it.

• The error: Writing $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$ but then plugging $k = 60$ without checking units, or writing $t = \frac{0.693}{60}$ and forgetting the unit is seconds.
• How to avoid:
  • Always write the unit alongside the number: $k = 60\ \text{s}^{-1}$.
  • The answer will be in seconds (since $k$ is in $\text{s}^{-1}$).
  • If the question expects an answer in minutes, convert at the end: divide by 60.

Mistake 3: Using the wrong logarithm base in the integrated rate law

The first-order integrated rate law is:

Some students use natural log ($\ln$) but forget the factor $2.303$, or they use $\log_{10}$ but omit the $2.303$.

• The error: Writing $t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$ (correct) but then using $\log_{10}$ tables without converting, or writing $t = \frac{1}{k} \log_{10} \frac{[A]_0}{[A]}$ (wrong).
• How to avoid:
  • Stick to one form and be consistent.
  • If using $\ln$: $t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$
  • If using $\log_{10}$: $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]}$
  • For this problem, since $[A]_0/[A] = 16$, $\log_{10} 16 = 1.2041$ and $\ln 16 = 2.7726$. Both give the same $t$ if you use the correct factor.

Mistake 4: Misidentifying the ratio $[A]_0/[A]$

Students sometimes invert the fraction.

• The error: They set $\frac{[A]_0}{[A]} = \frac{1}{16}$ instead of $16$.
• How to avoid: The question says "reduce to its $1/16$th value". So $[A] = \frac{1}{16}[A]_0$. Therefore $\frac{[A]_0}{[A]} = \frac{[A]_0}{\frac{1}{16}[A]_0} = 16$. Always write it out: final concentration = initial concentration divided by 16.

Mistake 5: Arithmetic errors in the final calculation …