Question

The first order rate constant for the decomposition of \( \text{N}_2\text{O}_5 \) is \( 6.93 \times 10^{-4}\, \text{sec}^{-1} \). Its half-life period is

• A. \( 1000\,s \)
• B. \( 100\,s \)
• C. \( 10\,s \)
• D. \( 1\,s \)
• E. \( 10000\,s \)

Hint:

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks for the half-life (\(t_{1/2}\)) of a first-order reaction, given the rate constant (\(k\)). The half-life of a first-order reaction is constant and depends only on the rate constant.
Step 2: Key Formula or Approach:
The relationship between the half-life (\(t_{1/2}\)) and the rate constant (\(k\)) for a first-order reaction is:
\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \] Step 3: Detailed Explanation:
We are given the rate constant:
\[ k = 6.93 \times 10^{-4} \text{ sec}^{-1} \] Now, substitute this value into the half-life formula:
\[ t_{1/2} = \frac{0.693}{6.93 \times 10^{-4} \text{ sec}^{-1}} \] To simplify the calculation, notice that 6.93 is 10 times 0.693.
\[ t_{1/2} = \frac{0.693}{0.693 \times 10 \times 10^{-4}} \text{ s} \] \[ t_{1/2} = \frac{1}{10 \times 10^{-4}} \text{ s} \] \[ t_{1/2} = \frac{1}{10^{-3}} \text{ s} \] \[ t_{1/2} = 10^3 \text{ s} = 1000 \text{ s} \] Step 4: Final Answer:
The half-life period for the decomposition of N\(_2\)O\(_5\) is 1000 s. This corresponds to option (A).