The half life of a first order reaction is 2000 years. If the concentration after 8000 years is 0.02 M, then the initial concentration was :

Question

Accepted Answer

C. 1. For a **first-order reaction**, the concentration is reduced to half of its previous value after each **half-life period**, regardless of the starting concentration.

2. **Finding the number of half-lives** that have passed in the total time given:

$$
n = \frac{	ext{Total time}}{	ext{Half-life}} = \frac{8000}{2000} = 4
$$

3. Since the concentration halves with each half-life, after $n$ half-lives, the remaining concentration relates to the initial concentration $[A]_0$ by:

$$
[A]_t = \frac{[A]_0}{2^n}
$$

4. **Given** that the concentration after 4 half-lives is $0.02\ M$, we substitute the known values to solve for the initial concentration.

$$
0.02 = \frac{[A]_0}{2^4}
$$

5. **Therefore**, solving for $[A]_0$:

$$
[A]_0 = 0.02 	imes 2^4 = 0.02 	imes 16 = 0.32\ M
$$

6. **Hence**, the initial concentration was **0.32 M**, so the answer is **option C**.