Question

What is the time required for 99% completion of a first order reaction if rate constant is 23.03 min$^{-1}$?

• A. 0.2 minute
• B. 0.4 minute
• C. 6.2 minute
• D. 8.1 minute

Hint:

For first order kinetics, memorize these shortcut relationships to save time on exams:
$t_{99%} = 2 \times t_{90%}$
$t_{99.9%} = 3 \times t_{90%}$
$t_{75%} = 2 \times t_{50%}$ (two half-lives)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are asked to calculate the exact duration required for a specific chemical reaction to reach 99% completion, given that it follows first-order kinetics and we are provided its specific rate constant ($k$).

Step 2: Key Formula or Approach:
The integrated rate law equation governing first-order reactions is:
$$t = \frac{2.303}{k} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$$
Where:
$t$ = time required
$k$ = rate constant ($23.03 \text{ min}^{-1}$)
$[A]_0$ = initial concentration of reactant (assumed to be 100%)
$[A]_t$ = remaining concentration of reactant at time $t$

Step 3: Detailed Explanation:
If the reaction is strictly 99% complete, it implies that 99% of the initial reactant has been consumed.
Therefore, the amount of unreacted material remaining is:
$[A]_t = 100% - 99% = 1%$
Now, substitute the known values directly into our rate law formula:
$$t = \frac{2.303}{23.03} \log_{10} \left( \frac{100}{1} \right)$$
Simplify the numerical fraction:
$$\frac{2.303}{23.03} = 0.1 \text{ min}$$
Simplify the logarithmic term:
$$\log_{10}(100) = \log_{10}(10^2) = 2$$
Multiply them together to find the time:
$$t = 0.1 \times 2 = 0.2 \text{ minute}$$

Step 4: Final Answer:
The time required is 0.2 minute, matching option (a).