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 :

Answer 1

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...

Answer 2

Decomposition reaction of H2O2 is:

\( \ce{H2O2 -> H2O + 1/2O2} \)

2t½ = 50 mins

t½ = 25 mins

For first order reaction,

t½ = 0.693/k

k = 0.693/25

$\sf\dfrac{d[H_2O_2]}{dt} = \dfrac{d[O_2]}{dt} \times 2 \\ \\ \dfrac{d[O_2]}{dt} = \: \sf \red{\dfrac{1}{2} \times \dfrac{0.693}{25} \times 0.05 }$

= 6.93 × 10–4

The correct option is A.