A first order reaction has a specific reaction rate of `10^(-2) s^(-1)`. How much time will it take for 20 g of the reactant to reduce to 5 g ?

Options

A. 2.38.6 second
B. 138.6 second
C. 346.5 second
D. 693.0 second

Answer

To solve the problem, we will use the integrated rate law for a first-order reaction. The equation we will use is:

\[ \ln \left( \frac{[A_0]}{[A_t]} \right) = k t \]

Where:
- \( [A_0] \) is the initial concentration (or mass) of the reactant,
- \( [A_t] \) is the concentration (or mass) of the reactant at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial mass of the reactant, \( A_0 = 20 \, \text{g} \)
- Final mass of the reactant, \( A_t = 5 \, \text{g} \)
- Rate constant, \( k = 10^{-2} \, \text{s}^{-1} \)

2. **Set Up the Integrated Rate Law:**
Since this is a first-order reaction, we can use the formula:
\[ \ln \left( \frac{A_0}{A_t} \right) = k t \]

3. **Substitute the Values into the Equation:**
\[ \ln \left( \frac{20}{5} \right) = 10^{-2} t \]

4. **Calculate the Ratio:**
\[ \frac{20}{5} = 4 \]

5. **Calculate the Natural Logarithm:**
\[ \ln(4) \approx 1.386 \]

6. **Substitute Back into the Equation:**
\[ 1.386 = 10^{-2} t \]

7. **Rearrange to Solve for Time \( t \):**
\[ t = \frac{1.386}{10^{-2}} = 138.6 \, \text{s} \]

### Final Answer:
The time required for the reactant to reduce from 20 g to 5 g is approximately **138.6 seconds**.

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