Question

In a first order reaction concentration of reactant decreases from \(20\text{ milli mol dm}^{-3}\) to \(8\text{ milli mol dm}^{-3}\) in \(40\text{ minutes}\), find rate constant of reaction?

• A. \(0.011\text{ minute}^{-1}\)
• B. \(0.023\text{ minute}^{-1}\)
• C. \(0.032\text{ minute}^{-1}\)
• D. \(0.041\text{ minute}^{-1}\)

Hint:

In first-order reactions, rate constant is independent of initial concentration—use log formula directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first-order reaction, the integrated rate law connects the rate constant (\(k\)), time (\(t\)), initial concentration (\([A]_0\)), and the concentration at time \(t\) (\([A]_t\)).
Step 2: Key Formula or Approach:
The formula is: \[ k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right) \] Step 3: Detailed Explanation:
Given data:
Initial concentration, \([A]_0 = 20\text{ milli mol dm}^{-3}\)
Final concentration, \([A]_t = 8\text{ milli mol dm}^{-3}\)
Time interval, \(t = 40\text{ minutes}\)
Substitute the values into the formula: \[ k = \frac{2.303}{40} \log_{10} \left( \frac{20}{8} \right) \] \[ k = \frac{2.303}{40} \log_{10} (2.5) \] The value of \(\log_{10}(2.5)\) can be found using log properties. We know \(\log_{10}(10/4) = 1 - 2\log_{10}(2) = 1 - 2(0.3010) = 1 - 0.6020 = 0.3980\).
\[ k = \frac{2.303 \times 0.3980}{40} \] \[ k \approx \frac{0.9166}{40} \] \[ k \approx 0.022915\text{ min}^{-1} \] Rounding to three decimal places, we get \(0.023\text{ min}^{-1}\).
Step 4: Final Answer:
This matches option (B).