Decomposition of \(H_2O_2\) follows a first order reaction. In 50 min, the concentration of \(H_2O_2\) decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of \(H_2O_2\) reaches 0.05 M, the rate of formation of \(O_2\) will be (Given log 2 = 0.3010)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(6.93×10^{-4}\) mol/min
From 0.5 M to 0.125 M the concentration has been halved twice (0.5 → 0.25 → 0.125), so 50 min = 2 half-lives. For a first order reaction the half-life is independent of concentration:
t1/2 = 50/2 = 25 min
k = 0.693/t1/2 = 0.693/25 = 2.772 × 10−2 min−1
(The same k follows from k = (2.303/50) log(0.5/0.125) = (2.303/50) × 2 × 0.3010.)
At \( \ce{[H2O2]} \) = 0.05 M:
Rate of decomposition = k[H2O2] = 2.772 × 10−2 × 0.05 = 1.386 × 10−3 mol L−1 min−1
From \( \ce{2H2O2} \) → \( \ce{2H2O} \) + \( \ce{O2} \), one \( \ce{O2} \) forms for two \( \ce{H2O2} \) used:
Rate of formation of \( \ce{O2} \) = ½ × 1.386 × 10−3 = 6.93 × 10−4 mol min−1 (per litre)
NCERT (2026-27 edition):
Half-Life of a Reaction – “… Half-life for a first order reaction is t1/2 = 0.693/k”
Half-Life of a Reaction – “… For zero order reaction t1/2 ∝ [R]0. For first order reaction t1/2 is independent of [R]0.”