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

Concept:
For a first-order reaction: \[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] where:

• \([A]_0\) = initial concentration
• \([A]\) = concentration after time \(t\)

Step 1: Substitute given values. \[ [A]_0 = 20,\quad [A] = 8,\quad t = 40 \text{ min} \] \[ k = \frac{2.303}{40} \log \frac{20}{8} \]
Step 2: Simplify ratio. \[ \frac{20}{8} = 2.5 \] \[ \log 2.5 \approx 0.398 \]
Step 3: Calculate k. \[ k = \frac{2.303}{40} \times 0.398 \] \[ k \approx \frac{0.916}{40} \approx 0.023 \]
Step 4: Conclusion. \[ k = 0.023 \text{ minute}^{-1} \]