The rate constant for a zero order reaction was found to be \( 1.0 × 10^{-3}\ mol\, L^{-1}\,s^{-1} \). How long will 1.0 mol of the reactant take to reduce to 0.5 mol in one liter flask?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 500 s
In a zero order reaction the rate does not depend on the concentration of the reactant, so the concentration falls by the same amount in every second. The integrated rate law is a straight line:
\( \displaystyle [\mathrm{R}] = [\mathrm{R}]_0 - kt \quad \Rightarrow \quad t = \frac{[\mathrm{R}]_0 - [\mathrm{R}]}{k} \)
In a one-litre flask, 1.0 mol and 0.5 mol are concentrations of 1.0 and 0.5 mol L−1:
\( \displaystyle t = \frac{1.0 - 0.5}{1.0 \times 10^{-3}}\,\mathrm{s} = \frac{0.5}{1.0 \times 10^{-3}}\,\mathrm{s} = 500\,\mathrm{s} \)
So the reactant takes 500 s to fall to 0.5 mol.
Shortcut: Falling from 1.0 to 0.5 mol is one half-life, and for a zero order reaction \( t_{1/2} = [\mathrm{R}]_0/2k = 1.0/(2 \times 10^{-3}) = 500 \) s.