Q.Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life ($t_{1/2}$) of the reaction.

Concept: First Order Kinetics — For a first order reaction, the integrated rate law is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.

Step 1: Half-life $t_{1/2}$ is the time when $[A] = \frac{[A]_0}{2}$.

From the rate law:

$t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0/2} = \frac{2.303}{k} \log 2 = \frac{0.693}{k}$.

Step 2: For 99.9% completion, $[A] = 0.1\%$ of $[A]_0$, i.e., $[A] = \frac{[A]_0}{1000}$. …

To show

For a first-order reaction the integrated rate law is

Half-life. At $t_{1/2}$, $[A] = \dfrac{[A]_0}{2}$:

Time for 99.9% completion. When $99.9\%$ has reacted, $0.1\%$ remains, so $[A] = \dfrac{[A]_0}{1000}$:

Take the ratio. …

Method: Integrated Rate Law Approach for First-Order Kinetics

This method uses the integrated rate equation for a first-order reaction and compares the time for 99.9% completion with the half-life.

Steps

Step 1: Write the integrated rate law for first-order kinetics

For a first-order reaction:

Where:

• $k$ = rate constant
• $t$ = time
• $[A]_0$ = initial concentration
• $[A]$ = concentration at time $t$

Step 2: Express half-life ($t_{1/2}$)

At half-life, $[A] = \frac{[A]_0}{2}$:

Since $\log 2 = 0.3010$:

Step 3: Find time for 99.9% completion ($t_{99.9\%}$)

99.9% completion means only 0.1% remains:

Substitute into the integrated rate law:

Since $\log 1000 = 3$: …

Here are the common mistakes students make when proving that for a first-order reaction, $t_{99.9\%} = 10 \times t_{1/2}$, and how to avoid each.

Mistake 1: Using the wrong formula for $t_{1/2}$

The error:

Students often write $t_{1/2} = \frac{0.693}{k}$ correctly, but then incorrectly use $t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1}$ without checking the remaining fraction.

Why it’s wrong:

For 99.9% completion, the remaining reactant is $0.1\%$, not $0.1$ of the original. The fraction remaining is $\frac{0.1}{100} = 0.001$.

How to avoid:

Always write the integrated rate law as:

For 99.9% completion: $[A]_t = 0.001 [A]_0$, so:

Mistake 2: Incorrectly calculating $\log 1000$

The error:

Some students write $\log 1000 = 3$ but then forget to multiply by $2.303$, or they use $\log_{10}$ and $\ln$ interchangeably without adjusting.

How to avoid:

Remember:

• $\log_{10} 1000 = 3$
• So $t_{99.9\%} = \frac{2.303}{k} \times 3 = \frac{6.909}{k}$

Then compare with $t_{1/2} = \frac{0.693}{k}$:

Key check: Always compute the ratio numerically to confirm it’s exactly 10 (or very close).

Mistake 3: Assuming 99.9% means $[A]_t = 0.1 [A]_0$

The error:

Thinking “99.9% done” means only $0.1$ of the reactant is left — but that’s 90% completion, not 99.9%.

How to avoid:

Use the percentage remaining formula:

For 99.9% completed: fraction remaining $= 1 - 0.999 = 0.001$

Mistake 4: Forgetting to state the final conclusion clearly

The error:

Students derive $t_{99.9\%} = \frac{6.909}{k}$ and $t_{1/2} = \frac{0.693}{k}$, but don’t explicitly write:

How to avoid:

Always end with a clear, boxed statement showing the relationship. In exams, this is often the mark-bearing line.

Mistake 5: Mixing up natural log and log base 10

The error:

Using $\ln$ instead of $\log_{10}$ in the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.

Why it’s wrong:

The integrated first-order rate law is:

$t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}$ …