Question

\(A \rightarrow\) products, is a first order reaction. The time required to decompose \(A\) to half its initial amount is \(60\) minutes. The rate constant of the reaction \((in\ s^{-1})\) is

• A. \(1.05\times10^{-2}\)
• B. \(1.15\times10^{-2}\)
• C. \(1.25\times10^{-4}\)
• D. \(1.92\times10^{-4}\)

Hint:

For first order reactions: \[ t_{1/2}=\frac{0.693}{k} \] The half-life is independent of initial concentration.

Answer 1

The Correct Option is D

Step 1: Recall the half-life formula for first order reactions.
For a first order reaction: \[ t_{1/2} = \frac{0.693}{k} \] This applies regardless of initial concentration.
Step 2: Convert half-life to seconds.
Given \( t_{1/2} = 60 \) minutes. Converting: \[ t_{1/2} = 60 \times 60 = 3600 \text{ s} \]
Step 3: Solve for k.
Rearranging the formula: \[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3600} \]
Step 4: Calculate the value.
\[ k = \frac{0.693}{3600} \approx 1.925 \times 10^{-4} \ s^{-1} \]
Step 5: Check units.
First order rate constants have units of \( s^{-1} \). Our result is in \( s^{-1} \), which is correct.
Step 6: Match with options.
\( 1.92 \times 10^{-4} \ s^{-1} \) matches option 4. \[ \boxed{1.92 \times 10^{-4} \ s^{-1}} \]