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 step by step, we will use the concepts of first-order kinetics and the half-life of a reaction.

### Step 1: Understand the Given Information
We are given:
- The half-life (t₁/₂) of a first-order reaction is 10 minutes.
- We need to find the time required for the concentration of A to be reduced to 10% of its original concentration.

### Step 2: Calculate the Rate Constant (k)
For a first-order reaction, the relationship between the half-life and the rate constant (k) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

Rearranging the formula to find k:
\[ k = \frac{0.693}{t_{1/2}} \]

Substituting the half-life:
\[ k = \frac{0.693}{10 \text{ minutes}} \]
\[ k = 0.0693 \text{ min}^{-1} \]

### Step 3: Set Up the Equation for Concentration
We need to find the time (t) when the concentration of A is reduced to 10% of its original concentration. If we assume the initial concentration of A is 100 units, then:
- Initial concentration (A₀) = 100
- Final concentration (A) = 10 (which is 10% of 100)

Using the first-order kinetics equation:
\[ \ln\left(\frac{A_0}{A}\right) = kt \]

Substituting the values:
\[ \ln\left(\frac{100}{10}\right) = 0.0693 \cdot t \]
\[ \ln(10) = 0.0693 \cdot t \]

### Step 4: Calculate ln(10)
The natural logarithm of 10 is approximately:
\[ \ln(10) \approx 2.303 \]

### Step 5: Solve for Time (t)
Now substituting back into the equation:
\[ 2.303 = 0.0693 \cdot t \]

Rearranging to find t:
\[ t = \frac{2.303}{0.0693} \]
\[ t \approx 33.22 \text{ minutes} \]

### Step 6: Conclusion
The time required for the concentration of A to be reduced to 10% of its original concentration is approximately 33.22 minutes.

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