Question

♣For a reaction 2H2O2→2H2O + O2 if rate of formation of O2 is 3.2×10^-3 kg/min then what will be rate of disappearance of H2O2 1)in mol/min
2)in g/sec 
3)in kg/min​

Answer 1

Answer:

1) 1.88 × 10⁻¹ mol/min

2) 1.07 × 10⁻¹ g/sec

3) 6.4 × 10⁻³ kg/min

Explanation:

Hi dear!

The given reaction is,

$2H_2O_2 \longrightarrow 2H_2O + O_2$

For this reaction , we know that

$\boxed {-\dfrac{1}{2} \bigg(\dfrac{d[H_2O_2]}{dt} \bigg) = + \dfrac{1}{2} \bigg(\dfrac{d[H_2O]}{dt} \bigg) = + \bigg(\dfrac{d[O_2]}{dt} \bigg)}$

[-ve and +ve signs are only to describe the completion and formation]

$\text{Given that, rate of formation of O_2 is 3.2 \times 10^{-3} that is } \\\\ \bigg(\dfrac{d[O_2]}{dt} \bigg) = 3.2 \times 10^{-3} \ kg/min$

So, the rate of disappearance of H₂O₂ is  $\bigg(\dfrac{d[H_2O_2]}{dt} \bigg)$

So, from the above relation,

$\boxed{\bigg(\dfrac{d[H_2O_2]}{dt} \bigg) = 2\bigg(\dfrac{d[O_2]}{dt} \bigg)}$

$: \longmapsto \bigg(\dfrac{d[H_2O_2]}{dt} \bigg) = 2 \times 3.2 \times 10^{-3} \text{ kg/min}$

$: \longmapsto \bigg(\dfrac{d[H_2O_2]}{dt} \bigg) = \underline{ 6.4 \times 10^{-3} \text{ kg/min}}$

This is the 3rd answer.

----------------------------------------------------

Now,

$So, \ 6.4 \times 10^{-3}\ kg/min = 6.4 \times 10^{-3} \times \dfrac{ 10 ^ 3g}{60 sec}$  

$: \longmapsto 6.4 \times 10^{-3} \times \dfrac{10^3 g}{ 60 sec} = 1.07 \times 10^{-1} \ g/sec$

This is the 2nd answer

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Now,

$Mole = \dfrac{\text{Weight of substance in gms}}{\text{Molecular weight in gms}}$

$6.4 \times 10^{-3} \ kg/min = 6.4 \times 10^{-3} \times \dfrac{10^3 g}{34g . min} \\\\: \longmapsto 1.88 \times 10^{-1} \ mol/min$

[Molecular weight of H₂O₂ is 34]

this is the 1st answer

Hope it helps!!