A first order reaction has specific 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. 238.6 second
B. 138.6 second
C. 346.5 second
D. 693.0 second

Explanation

To solve the problem of how much time it will take for a first-order reaction to reduce 20 g of a reactant to 5 g, we can follow these steps:

### Step 1: Identify the given data
- The specific rate constant (k) for the reaction is \(10^{-2} \, s^{-1}\).
- The initial amount of the reactant (\(A_0\)) is 20 g.
- The final amount of the reactant (\(A\)) is 5 g.

### Step 2: Calculate the half-life (\(t_{1/2}\)) of the reaction
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \(k\):
\[ t_{1/2} = \frac{0.693}{10^{-2}} = 0.693 \times 10^{2} = 69.3 \, seconds \]

### Step 3: Determine how many half-lives are required
To find out how many half-lives are needed to reduce the amount from 20 g to 5 g:
- After 1 half-life, the amount will be:
\[ A_0 \times \frac{1}{2} = 20 \, g \times \frac{1}{2} = 10 \, g \]
- After 2 half-lives, the amount will be:
\[ 10 \, g \times \frac{1}{2} = 5 \, g \]

Thus, it takes 2 half-lives to reduce the amount from 20 g to 5 g.

### Step 4: Calculate the total time required
The total time (\(t\)) required is given by:
\[ t = n \times t_{1/2} \]
where \(n\) is the number of half-lives. Here, \(n = 2\):
\[ t = 2 \times 69.3 \, seconds = 138.6 \, seconds \]

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

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