The half life period of a first order reaction, A `to` Product is 10 minutes. In how much time is the concentration of A reduced to 10% of its original concentration?

Options

A. 26 min
B. 33 min
C. 71 min
D. 90 min

Answer

To solve the problem, we need to determine the time it takes for the concentration of reactant A to be reduced to 10% of its original concentration in a first-order reaction, given that the half-life (t₁/₂) is 10 minutes.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We know the half-life (t₁/₂) of the reaction is 10 minutes.
- We want to find the time (t) when the concentration of A is reduced to 10% of its initial concentration (A₀).

2. **Set Up the Equation**:
- For a first-order reaction, the relationship between the concentration at time t (A_t) and the initial concentration (A₀) can be expressed as:
\[ A_t = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
- In our case, we want A_t to be 10% of A₀:
\[ A_t = 0.1 A_0 \]

3. **Substitute Values**:
- Substitute A_t into the equation:
\[ 0.1 A_0 = A_0 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]
- We can cancel A₀ from both sides (assuming A₀ ≠ 0):
\[ 0.1 = \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

4. **Take Logarithms**:
- Taking the logarithm (base 10) of both sides:
\[ \log(0.1) = \log\left(\left( \frac{1}{2} \right)^{\frac{t}{10}}\right) \]
- Using the property of logarithms:
\[ \log(0.1) = \frac{t}{10} \log\left(\frac{1}{2}\right) \]

5. **Calculate Logarithms**:
- We know that:
\[ \log(0.1) = -1 \]
- And:
\[ \log\left(\frac{1}{2}\right) = -\log(2) \approx -0.3010 \]
- Substituting these values into the equation:
\[ -1 = \frac{t}{10} \cdot (-0.3010) \]

6. **Solve for t**:
- Rearranging gives:
\[ t = 10 \cdot \frac{-1}{-0.3010} \approx \frac{10}{0.3010} \approx 33.22 \text{ minutes} \]

7. **Conclusion**:
- Therefore, the time required for the concentration of A to be reduced to 10% of its original concentration is approximately 33 minutes.

### Final Answer:
The time required is approximately **33 minutes**.