Question

In a first order reaction the initial concentration of the reactant was \(0.1 \text{ M}\) at \(298 \text{K}\). It is decreased to \(0.01 \text{ M}\) in 8 minutes and 20 seconds. Calculate the rate constant of the reaction at \(298 \text{K}\).

• A. \(2.303 \times 10^{-3} \text{ s}^{-1}\)
• B. \(2.303 \times 10^{-4} \text{ s}^{-1}\)
• C. \(4.606 \times 10^{-4} \text{ s}^{-1}\)
• D. \(2.303 \times 10^{-5} \text{ min}^{-1}\)
• E. \(4.606 \times 10^{-3} \text{ s}^{-1}\)

Hint:

Always check the units of the options. Since the time includes seconds and the options are in \(s^{-1}\), convert the entire time duration to seconds immediately before starting the calculation.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
For a first-order reaction, the rate constant \(k\) is calculated using the integrated rate law, which relates concentration changes over a specific time interval.

Step 2: Key Formula or Approach:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
Where \([A]_0\) is the initial concentration and \([A]\) is the final concentration.

Step 3: Detailed Explanation:
1. Convert time to seconds:
\(t = 8 \text{ minutes } 20 \text{ seconds} = (8 \times 60) + 20 = 480 + 20 = 500 \text{ s}\).
2. Substitute the values into the formula:
\[ k = \frac{2.303}{500 \text{ s}} \log \left( \frac{0.1 \text{ M}}{0.01 \text{ M}} \right) \]
\[ k = \frac{2.303}{500} \log (10) \]
Since \(\log(10) = 1\):
\[ k = \frac{2.303}{500} = 0.004606 \text{ s}^{-1} \]
\[ k = 4.606 \times 10^{-3} \text{ s}^{-1} \]

Step 4: Final Answer:
The rate constant is \(4.606 \times 10^{-3} \text{ s}^{-1}\).