Two substances A `(t_(1//2)=5 min)` and B`(t_(t//2)=10min)` are taken in such a way that initially [A]- 4[B]. The time after which both the concentrations will be equal is : (Assume that reaction is first order)

Options

A. 5 min
B. 15 min
C. 20 min
D. concentration can never be equal

Answer

To solve the problem, we need to find the time after which the concentrations of substances A and B will be equal, given their initial concentrations and half-lives.

### Step-by-Step Solution:

1. **Define Initial Concentrations**:
- Let the concentration of B be \( [B] = x \).
- According to the problem, the concentration of A is \( [A] = 4[B] = 4x \).

2. **Understand Half-Life**:
- The half-life of substance A is \( t_{1/2} = 5 \) minutes.
- The half-life of substance B is \( t_{1/2} = 10 \) minutes.

3. **Calculate Concentration of A Over Time**:
- After 5 minutes (1 half-life), concentration of A:
\[ [A] = \frac{4x}{2} = 2x \]
- After another 5 minutes (10 minutes total, 2 half-lives), concentration of A:
\[ [A] = \frac{2x}{2} = x \]
- After another 5 minutes (15 minutes total, 3 half-lives), concentration of A:
\[ [A] = \frac{x}{2} = \frac{x}{2} \]
- After another 5 minutes (20 minutes total, 4 half-lives), concentration of A:
\[ [A] = \frac{x/2}{2} = \frac{x}{4} \]

4. **Calculate Concentration of B Over Time**:
- After 10 minutes (1 half-life), concentration of B:
\[ [B] = \frac{x}{2} \]
- After another 10 minutes (20 minutes total, 2 half-lives), concentration of B:
\[ [B] = \frac{x/2}{2} = \frac{x}{4} \]

5. **Set Concentrations Equal**:
- At 20 minutes, we have:
\[ [A] = \frac{x}{4} \quad \text{and} \quad [B] = \frac{x}{4} \]
- Therefore, the concentrations of A and B are equal after 20 minutes.

### Final Answer:
The time after which both concentrations will be equal is **20 minutes**.