Decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ follows a first order reaction. In fifty minutes the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ reaches 0.05 M , the rate of formation of $\mathrm{O}_{2}$ will be :

(A) $6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}$

(B) $2.66 \mathrm{~L} \mathrm{~min}^{-1}$ at STP

(C) $1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}$

(D) $6.93 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}$

JEE Main · 2016 · 03 Apr 2016 · Offline · Chemical Kinetics

Decomposition of

H_{2} O_{2}

follows a first order reaction. In fifty minutes the concentration of

H_{2} O_{2}

decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of

H_{2} O_{2}

reaches 0.05 M , the rate of formation of

O_{2}

will be :

(A) $6.93 \times 1 0^{- 4} mol min^{- 1}$ ✓ Correct
(B) $2.66 L min^{- 1}$ at STP
(C) $1.34 \times 1 0^{- 2} mol min^{- 1}$
(D) $6.93 \times 1 0^{- 2} mol min^{- 1}$

Answer: (A)

Solution

The decomposition is first order, and the fall from 0.5 M to 0.125 M in 50 minutes is two successive halvings, from 0.5 to 0.25 to 0.125, so the half-life is 25 minutes and

k = \frac{0.693}{25} min^{- 1}

. The rate of loss of

H X_{2} O X_{2}

k [H X_{2} O X_{2}]

, and since

H X_{2} O X_{2} H X_{2} O + \frac{1}{2} O X_{2}

makes half a mole of oxygen per mole decomposed, the rate of oxygen formation is

\frac{1}{2} k [H X_{2} O X_{2}] = \frac{1}{2} \times \frac{0.693}{25} \times 0.05 = 6.93 \times 1 0^{- 4} mol min^{- 1}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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JEE Main · 2016 · 03 Apr 2016 · Offline · Chemical Kinetics

Solution

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