Question

A first order reaction has a rate constant of $6.93\times10^{-4}\ \text{s}^{-1}$ at $300\ \text{K}$. What is the half-life period of the reaction in seconds at the same temperature?

• A. 693
• B. 6930
• C. 10000
• D. 1000
• E. 500

Hint:

$0.693$ is approximately the natural log of 2 ($\ln 2$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem relates the rate constant of a first-order reaction to its half-life. For a first-order reaction, the half-life is independent of the initial concentration.
Step 2: Key Formula or Approach:
The half-life (\( t_{1/2} \)) for a first-order reaction is given by the formula: \[ t_{1/2} = \frac{\ln(2)}{k} \] where k is the rate constant. We can use the approximation \( \ln(2) \approx 0.693 \). So, the formula becomes: \[ t_{1/2} = \frac{0.693}{k} \] Step 3: Detailed Explanation:
We are given: - The reaction is first-order. - The rate constant, \( k = 6.93 \times 10^{-4} \text{ s}^{-1} \). The temperature (300 K) is extra information, as the half-life is to be calculated at the same temperature. Substitute the value of k into the half-life formula: \[ t_{1/2} = \frac{0.693}{6.93 \times 10^{-4}} \] We can write \( 0.693 \) as \( 6.93 \times 10^{-1} \). \[ t_{1/2} = \frac{6.93 \times 10^{-1}}{6.93 \times 10^{-4}} \] The \( 6.93 \) terms cancel out: \[ t_{1/2} = \frac{10^{-1}}{10^{-4}} = 10^{-1 - (-4)} = 10^{-1 + 4} = 10^3 \] \[ t_{1/2} = 1000 \text{ seconds} \] Step 4: Final Answer:
The half-life period of the reaction is 1000 seconds.