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 formation of O₂ will be ______.

विकल्प

• 6.93 × 10⁻² mol min⁻¹

• 6.93 × 10⁻⁴ mol min⁻¹

• 2.66 L min⁻¹ at STP

• 1.34 × 10⁻² mol min⁻¹

Answer 1

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 formation of O₂ will be 6.93 × 10⁻⁴ mol min⁻¹.

Explanation:

Given: The decomposition of H₂O₂:

\[\ce{2H2O2 -> 2H2O + O2}\] 

Initial concentration of H₂O₂ (A₀) = 0.5 M, 

Final concentration after 50 minutes (A) = 0.125 M

Time (t) = 50 min

For first order reaction, `k = 2.303/t log_10 ([A]_0/[A])`

∴ `k = 2.303/50 log_10  0.2/0.125`

= `2.303/50 log_10 (4)`

= `(2.303 xx 0.6021)/50`

= `1.387/50`

= 0.02774 min⁻¹

Rate of decomposition of H₂O₂ when [H₂O₂] is 0.05:

Use the formula

`"Rate"_("H"_2"O"_2)` = k[H₂O₂]

= 0.02774 × 0.05

= 1.387 × 10⁻³ mol L⁻¹ min⁻¹

From reaction

\[\ce{2H2O2 -> 2H2O + O2}\] 

So,

Rate of O₂ = `1/2` × Rate of H₂O₂

= `1/2 xx 1.387 xx 10^-3`

= 6.935 × 10⁻⁴ mol min⁻¹

= 6.93 × 10⁻⁴ mol min⁻¹