Question

If the rate constant for a first order reaction is k, the time(t) required for the completion of 99% of the reaction is given by

• A. \(t=\frac{4.606}{k}\)
• B. \(t=\frac{2.303}{k}\)
• C. \(t=\frac{0.693}{k}\)
• D. \(t=\frac{6.909}{k}\)

Answer 1

The Correct Option is A

For a first-order reaction, the time required for the completion of a certain percentage of the reaction can be calculated using the following equation: \[ t = \frac{2.303}{k} \log \left(\frac{1}{1 - \text{fraction of reaction completed}}\right) \] For the completion of 99\%, the fraction of reaction completed is 0.99, so the equation becomes: \[ t = \frac{2.303}{k} \log \left(\frac{1}{1 - 0.99}\right) = \frac{2.303}{k} \log (100) \] Since \( \log(100) = 2 \), the equation simplifies to: \[ t = \frac{2.303}{k} \times 2 = \frac{4.606}{k} \]

So, the correct answer is (A): \(t=\frac{4.606}{k}\)