Question

Half-life (\( t_{1/2} \)) of a first order reaction is 1386 s. The value of rate constant is:

• A. \(0.5 \times 10^{-4} \, \text{s}^{-1}\)
• B. \(5.0 \times 10^{-4} \, \text{s}^{-1}\)
• C. \(0.5 \times 10^{-5} \, \text{s}^{-1}\)
• D. \(0.5 \times 10^{-3} \, \text{s}^{-1}\)

Hint:

For first order reactions: \[ t_{1/2} = \frac{0.693}{k} \] If half-life is known, rate constant can be found directly without concentration data.

Answer 1

The Correct Option is A

The half-life (\( t_{1/2} \)) of a first-order reaction is given as 1386 seconds. We need to find the rate constant (\( k \)) for this reaction. For a first-order reaction, the relationship between the half-life and the rate constant is defined by the formula:

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

We can rearrange this formula to solve for \( k \):

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

Where \(\ln(2) \approx 0.693\). Now, substituting the given half-life into the equation:

• Given \( t_{1/2} = 1386 \, \text{s} \)
• \( k = \frac{0.693}{1386} \)

Perform the calculation:

• \( k \approx \frac{0.693}{1386} \approx 0.0005 \, \text{s}^{-1} \)

Converting this into scientific notation, we get:

\(k \approx 0.5 \times 10^{-4} \, \text{s}^{-1}\)

Therefore, the correct answer is:

\(<0.5 \times 10^{-4} \, \text{s}^{-1} \)

Here is the reasoning for ruling out other options:

• \(5.0 \times 10^{-4} \, \text{s}^{-1}\): This value is much larger than the calculated rate constant.
• \(0.5 \times 10^{-5} \, \text{s}^{-1}\): This value is smaller than the calculated rate constant.
• \(0.5 \times 10^{-3} \, \text{s}^{-1}\): This value is also larger than the calculated rate constant.

Thus, based on the calculation, the correct option is \( 0.5 \times 10^{-4} \, \text{s}^{-1} \).