Decomposition of h2o2 follows a first order reaction

Asked by admin @ 29/08/2021 in Chemistry viewed by 350 People

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 :

Answered by admin @ 29/08/2021

Hey Dear,

◆ 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.

Hope this is helpful...

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