The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mole/litre and concentration at some instant ‘t’ is 0.01 mol/litre, then the value of ‘t’ is :

Options

A. 10 min.
B. 30 min.
C. 20 min.
D. 40 min.

Explanation

To solve the problem, we need to determine the time 't' it takes for the concentration of a reactant in a first-order reaction to decrease from 0.08 mol/L to 0.01 mol/L, given that the half-life of the reaction is 10 minutes.

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life**:
The half-life (t₁/₂) of a first-order reaction is the time required for the concentration of the reactant to decrease to half of its initial value. For this reaction, the half-life is given as 10 minutes.

2. **Calculate the Number of Half-Lives**:
- Initial concentration (C₀) = 0.08 mol/L
- Final concentration (C) = 0.01 mol/L
- We need to find how many half-lives it takes to go from 0.08 mol/L to 0.01 mol/L.

- The concentration decreases as follows:
- After 1 half-life (10 minutes): C = 0.08 mol/L / 2 = 0.04 mol/L
- After 2 half-lives (20 minutes): C = 0.04 mol/L / 2 = 0.02 mol/L
- After 3 half-lives (30 minutes): C = 0.02 mol/L / 2 = 0.01 mol/L

Thus, it takes 3 half-lives to go from 0.08 mol/L to 0.01 mol/L.

3. **Calculate the Total Time**:
Since each half-life is 10 minutes:
- Total time (t) = Number of half-lives × Half-life duration
- Total time (t) = 3 × 10 minutes = 30 minutes

4. **Conclusion**:
The value of 't' is 30 minutes.

### Final Answer:
The value of 't' is 30 minutes.

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