The half‑life of a first‑order reaction is measured to be $10 min$ . (a) Calculate the rate constant $k$ in $s^{- 1}$ . (b) How much time (in minutes) will it take for the concentration of the reactant to fall to $25%$ of its initial value?

easy 2 marks Class 12 · Chemistry · Chemical Kinetics

Answer

k = 1.16 \times 1 0^{- 3} s^{- 1}

and

t = 20 min

The correct answer is $k = 1.16 \times 1 0^{- 3} s^{- 1}$ and $t = 20 min$ — Both required values are obtained.

Step-by-step solution

For a first‑order reaction $t_{1/2} = \frac{l n 2}{k}$ , therefore $k = \frac{l n 2}{t _{1/2}}$ .
Insert $t_{1/2} = 10 min = 600 s$ : $k = \frac{0.693}{600} = 1.155 \times 1 0^{- 3} s^{- 1}$ .
To reach $25%$ of the initial concentration two half‑lives are required because each half‑life halves the concentration: $C = \frac{1}{2 ^{2}} C_{0}$ .
Hence $t = 2 \times t_{1/2} = 2 \times 10 min = 20 min$ .
Both required values are obtained.

Common mistake: Applying the zero‑order half‑life expression

t_{1/2} = \frac{[ A ] _{0}}{2 k}

instead of the first‑order formula.

Concept tested: First‑order half‑life and integrated rate law