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): \(t=\frac{4.606}{k}\)

Answer 2

1. Recall the integrated rate law for a first-order reaction:
The integrated rate law for a first-order reaction is:

$$ \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt $$

where:
- $[A]_t$ is the concentration of reactant A at time $t$
- $[A]_0$ is the initial concentration of reactant A
- $k$ is the rate constant
- $t$ is the time

2. Convert to base-10 logarithm:
We can also express this equation using the base-10 logarithm:

$$ 2.303 \log\left(\frac{[A]_t}{[A]_0}\right) = -kt $$

3. Determine the remaining concentration after 99% completion:
If the reaction is 99% complete, then 1% of the reactant remains. So:
$$ [A]_t = 0.01[A]_0 $$

4. Substitute the values into the integrated rate law:
$$ 2.303 \log\left(\frac{0.01[A]_0}{[A]_0}\right) = -kt $$
$$ 2.303 \log(0.01) = -kt $$
$$ 2.303 \log(10^{-2}) = -kt $$
$$ 2.303(-2) = -kt $$
$$ -4.606 = -kt $$

5. Solve for $t$:
$$ t = \frac{4.606}{k} $$

Final Answer:
(A) $t = \frac{4.606}{k}$