Question

Decomposition of H₂O₂ follows a first order reaction. In 50 min, the concentration of H₂O₂ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H₂O₂ reaches 0.05 M, the rate of formation of O₂ will be

(a) 6.93 x10^-4 mol min^-1 (b) 2.66 L min^-1 at STP

(c) 1.34 x10^-2 mol min^-1 (d) 693 x10^-2 mol min^-1

Answer 1

Correct option(a) 6.93 x10^-4 mol min^-1

Explanation:

Comments

Excellent

Answer 2

Correct option is (a) 6.93 x 10^-4 mol min^-1

\(2H_2O_2 \to 2H_2O + O_2\)

In 50 minutes, the concentration is reduced to one fourth

\((\because \frac{0.5}{0.125} = 4)\)

Hence 2 half life periods corresponds to 50 minutes.

t_1/2 = 25 min

The rate constant \(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{25}\)

Rate of decomposition of

\(H_2O_2 = k[H_2O_2] = \frac{0.693}{25} \times 0.05 = 1.39\times 10^{-3}\) mol/min

Rate of formation of oxygen is one half the rate of decomposition of H₂O₂.

It is \(\frac 12 \times 1.39 \times 10^{-3} = 6.93 \times 10^{-4}\) mol/min.