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

664.64 s

The correct answer is option 1. 664.64 s.

To determine the time for 99% of the reaction to occur in a first-order reaction, we can use the formula:

\(t_{\text{99%}} = \frac{-\ln(1 - 0.99)}{k} \)

Given that half of the reaction is completed in 100 seconds, we can use this information to find the rate constant (\( k \)). Since it is a first-order reaction, the half-life (\( t_{1/2} \)) is related to the rate constant by:

\( t_{1/2} = \frac{0.693}{k} \)

Substituting the given half-life of 100 seconds, we can solve for the rate constant

\(k = \frac{0.693}{100} = 0.00693 \)

Now we can substitute this value of \( k \) into the equation for \( t_{\text{99%}} \):

\(t_{\text{99%}} = \frac{-\ln(1 - 0.99)}{0.00693} \)

Evaluating this expression, we find:

\( t_{\text{99%}} \approx 664.64 \) seconds

Therefore, the correct answer is (1) 664.64 seconds.