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Consider the following reaction:
2H2O2(aq)→2H2O(l)+O2(g)
The graph (Figure 1) shows the concentration of H2O2 as a function of time.
If the instantaneous rate of formation of O2 is 3.3*(10^-3) moles/(liters*seconds), then...
If the initial volume of the H2O2 solution is 1.5 L, what total amount of O2 (in moles) is formed in the first 50 s of reaction?
Express your answer using two significant figures.

Consider the following reaction 2H2O2aq2H2OlO2g The graph Figure 1 shows the concentration of H2O2 as a function of time If the instantaneous rate of formation class=

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Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

[tex]\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }[/tex]

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

[tex]\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }[/tex]

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

[tex]\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}[/tex]

adefioyelaoye
  • If the instantaneous rate of formation of O₂ is 3.3*(10^-3) moles/(liters*seconds), then d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

  • The total amount of O₂ (in moles) is formed in the first 50s of reaction is 0.58 mol

Parameters and Calculation

2H₂O₂     ⟶      2H₂O     +     O₂

We were given ΔO₂/Δt

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt

= -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =

-6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt =  2d[O₂]/dt

=  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹

=  6.6 × 10⁻³mol·L⁻¹s⁻¹.

  • Moles of O₂

       Mole= (1.5L × 1mol)/ 1L  = 1.5mol

  • Final moles of H₂O₂

       Moles =  (1.5L × 0.22mol)/ 1L = 0.33mol

  • Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

  • Moles of O₂ formed = 1.33mol H₂O₂ ₓ ( 1mol O₂ / 2mol  H₂O₂)

                                 = 0.58 mol  H₂O₂

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