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 = 4.606/k
• B. t = 2.303/k
• C. t = 0.693/k
• D. t = 6.909/k

Answer 1

The Correct Option is A

 To solve this problem, we need to use the formula for the time required to achieve a certain completion percentage in a first-order reaction. The time \( t \) for completion of a particular fraction of a first-order reaction can be determined using the first-order kinetics equation:

\(t = \frac{\ln\left(\frac{[A]_0}{[A]}\right)}{k}\)

Where \([A]_0\) is the initial concentration and \([A]\) is the final concentration of the reactant at time \( t \), and \( k \) is the rate constant.

When we want to calculate the time required for the reaction to reach 99% completion, the concentration of the reactant left is 1% of the initial concentration. Therefore, we set \([A] = 0.01[A]_0\).

Substituting these values into the equation gives us:

\(\ln\left(\frac{[A]_0}{0.01[A]_0}\right) = \ln(100)\)

The natural logarithm of 100 is:

\(\ln(100) = 4.605\)

Thus, the time \( t \) to reach 99% completion is given by the equation:

\(t = \frac{4.605}{k}\)

Therefore, the correct answer is: \(t = \frac{4.605}{k}\), which corresponds to the option

t = 4.606/k

.

 

Let's now rule out the other options:

t = 2.303/k

• is the time for a different percentage completion, specifically about 90% of the first-order reaction completion.

t = 0.693/k

• is the half-life for a first-order reaction, not the time for 99% completion.

t = 6.909/k

• does not correspond to any standard milestone in first-order kinetics related to 99% completion.

Conclusively, for a first-order reaction to reach 99% completion, the formula confirms that the correct relation is \(t = \frac{4.605}{k}\).