Question

Decomposition of H2O2 follows a first order reaction. In fifty minutes 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:

give reason also......................​

Answer 1

Answer:

d[O2]/dt = 6.93×10^-4 mol/min

Explanation:

Concentration of [H2O2] decreases from 0.5 m to 0.125 m i.e. 1/4th in 50 min.

Thus, half life of [H2O2] is

t½ = 50/2

t½ = 25 min.

Rate constant for first order reaction,

k = 0.693/t½

k = 0.693/25

Decomposition of H2O2 takes place as follows -

2H2O2 --> 2H2O + O2

From this, we can write -

d[O2]/dt = 1/2 d[H2O2]/dt

d[O2]/dt = 1/2 × 0.693/25 × 0.05

d[O2]/dt = 1/2 × k × [H2O2]

d[O2]/dt = 6.93×10^-4 mol/min

Hence, the rate of formation of O2 will be 6.93×10^-4 mol/min.

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Answer 2

Please refer to the attachment.

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