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For the reaction [N₂O₅(g) → 2NO₂(g) + 1/2 O₂(g)] the value of rate of disappearance of N₂O₅ is given as 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rate of formation of NO₂ and O₂ is given respectively as :

1. 6.25 × 10⁻³ mol L⁻¹ s⁻¹ and 6.25 × 10⁻³ mol L⁻¹ s⁻¹
2. 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹
3. 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 6.25 × 10⁻³ mol L⁻¹ s⁻¹
4. 6.25 × 10⁻³ mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Correct answer: 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Solution

The stoichiometry of the reaction shows that 1 mole of N₂O₅ produces 2 moles of NO₂ and 0.5 moles of O₂. Thus, the rate of formation of NO₂ is twice the rate of disappearance of N₂O₅, and the rate of formation of O₂ is half the rate of disappearance of N₂O₅. Therefore, the rates are 1.25 × 10⁻² mol L⁻¹ s⁻¹ for NO₂ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹ for O₂.

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