Question

A first-order reaction is 25% complete in 30 minutes. What is its half-life?

• A. 72.33
• B. 70
• C. 75.33
• D. 77.66

Hint:

In first-order reactions, use $\displaystyle t_{1/2} = \frac{0.693}{k}$ and apply logarithmic equations carefully.

Answer 1

The Correct Option is A

For a first-order reaction: \[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] 25% complete means 75% of reactant remains. So, \[ \frac{[A]_0}{[A]} = \frac{100}{75} = \frac{4}{3}, \quad t = 30 \text{ min} \] \[ k = \frac{2.303}{30} \log \left( \frac{4}{3} \right) = \frac{2.303}{30} \times 0.1249 \approx 0.00958 \text{ min}^{-1} \] Now, use half-life formula for first-order reaction: \[ t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00958} \approx 72.33 \ \text{min} \]

Answer 2

To find the half-life of a first-order reaction that is 25% complete in 30 minutes, we can use the first-order integrated rate law equation:

Integrated Rate Law (First-Order):
\[\ln\left(\frac{[A]_0}{[A]}\right) = kt\]

Where:

• \([A]_0\) = initial concentration
• \([A]\) = concentration at time \(t\)
• \(k\) = rate constant
• \(t\) = time

Given that the reaction is 25% complete, this means 75% of the reactant remains.

So, \(\frac{[A]}{[A]_0} = 0.75\).

Substitute the values into the equation:

\[\ln\left(\frac{1}{0.75}\right) = kt\]

\[t = 30 \text{ minutes}\]

\[k = \frac{ln(\frac{1}{0.75})}{30}\]

Next, calculate \(k\):

\[k = \frac{\ln(1.333)}{30} = \frac{0.2877}{30} \approx 0.00959 \text{ min}^{-1}\]

The half-life (\(t_{1/2}\)) for a first-order reaction is given by:

\[t_{1/2} = \frac{0.693}{k}\]

Using the calculated \(k\):

\[t_{1/2} = \frac{0.693}{0.00959} \approx 72.33 \text{ minutes}\]