The hydrolysis of tert‑butyl chloride ((CH3)3CCl) in excess aqueous NaOH at 30 °C follows first‑order kinetics with a half‑life of 2.5×10³ s. Calculate the rate constant k in s⁻¹.
Answer
2.8×10⁻⁴ s⁻¹
The correct answer is 2.8×10⁻⁴ s⁻¹ — Step 5: This is the rate constant for the SN1 hydrolysis of tert‑butyl chloride.
Step-by-step solution
- Step 1: For a first‑order reaction, the half‑life t½ is related to the rate constant by t½ = ln 2 / k.
- Step 2: Rearrange to k = ln 2 / t½.
- Step 3: Substitute the given values: k = 0.693 / (2.5×10³ s) = 2.772×10⁻⁴ s⁻¹.
- Step 4: Round to one significant figure in the exponent: k ≈ 2.8×10⁻⁴ s⁻¹.
- Step 5: This is the rate constant for the SN1 hydrolysis of tert‑butyl chloride.
Common mistake: Using the second‑order half‑life relation 1/(k t½) instead of the first‑order one
Concept tested: First‑order rate law and half‑life
Published · How this solution was prepared and checked
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