In the first order reaction, half of the reaction is completed in 100 seconds. The time for `99%` of the reaction to occur will be

Options

A. 664.64s
B. 646.6s
C. 660.9s
D. 654.5 s

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Identify the given data
We know that the half-life (T_half) of the reaction is given as 100 seconds.

### Step 2: Calculate the rate constant (k)
For a first-order reaction, the rate constant (k) can be calculated using the formula:
\[ k = \frac{0.693}{T_{1/2}} \]
Substituting the value of T_half:
\[ k = \frac{0.693}{100 \text{ s}} = 0.00693 \text{ s}^{-1} \]

### Step 3: Use the first-order kinetics formula to find time for 99% completion
The formula to calculate the time (T) for a first-order reaction is:
\[ T = \frac{2.303}{k} \log \left( \frac{A}{A - X} \right) \]
Where:
- A = initial amount of reactant
- X = amount of reactant that has reacted

In this case, if 99% of the reaction has occurred, then:
- A = 100 (initial amount)
- X = 99 (amount reacted)
- A - X = 1 (amount remaining)

### Step 4: Substitute values into the formula
Now substituting the values into the equation:
\[ T = \frac{2.303}{0.00693} \log \left( \frac{100}{1} \right) \]
\[ T = \frac{2.303}{0.00693} \log(100) \]

### Step 5: Calculate log(100)
We know that:
\[ \log(100) = 2 \]
So we can substitute this value:
\[ T = \frac{2.303}{0.00693} \times 2 \]

### Step 6: Calculate T
Now, calculating the value:
\[ T = \frac{2.303 \times 2}{0.00693} \]
\[ T = \frac{4.606}{0.00693} \approx 664.64 \text{ seconds} \]

### Step 7: Round off the answer
Rounding off, we find:
\[ T \approx 666.64 \text{ seconds} \]

### Final Answer
The time for 99% of the reaction to occur is approximately **666.64 seconds**.

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