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

Options

A. `138.6 sec`
B. `346.5 sec`
C. `693.0 sec`
D. `238.6 sec`

Explanation

To solve the problem, we need to determine how much time it will take for a first-order reaction to reduce the quantity of a reactant from 20 g to 5 g, given that the specific reaction rate (k) is \(10^{-2} \, \text{sec}^{-1}\).

### Step-by-Step Solution:

1. **Identify the Initial and Final Amounts of the Reactant:**
- Initial amount (A₀) = 20 g
- Final amount (A) = 5 g

2. **Calculate the Change in Amount:**
- The change in amount is from 20 g to 5 g, which means the reactant decreases by 15 g.

3. **Use the First-Order Reaction Formula:**
The integrated rate law for a first-order reaction is given by:
\[ \ln\left(\frac{A_0}{A}\right) = kt \]
Where:
- \(A_0\) = initial concentration (20 g)
- \(A\) = final concentration (5 g)
- \(k\) = specific reaction rate (\(10^{-2} \, \text{sec}^{-1}\))
- \(t\) = time taken

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

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

6. **Set Up the Equation:**
\[ 1.386 = (10^{-2})t \]

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

### Final Answer:
It will take approximately **138.6 seconds** for the amount of the reactant to reduce from 20 g to 5 g.

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