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For the reaction, N₂ + 3H₂ → 2NH₃, -d[NH₃]/dt = 2 × 10⁻⁴ mol L⁻¹ s⁻¹, the value of -d[H₂]/dt would be:

1. 4 × 10⁻⁴ mol L⁻¹ s⁻¹
2. 6 × 10⁻⁴ mol L⁻¹ s⁻¹
3. 1 × 10⁻⁴ mol L⁻¹ s⁻¹
4. 3 × 10⁻⁴ mol L⁻¹ s⁻¹

Correct answer: 3 × 10⁻⁴ mol L⁻¹ s⁻¹

Solution

The stoichiometry of the reaction shows that 3 moles of H₂ are consumed for every 2 moles of NH₃ formed. Using the rate relationship, -d[H₂]/dt = (3/2) × (-d[NH₃]/dt) = (3/2) × 2 × 10⁻⁴ = 3 × 10⁻⁴ mol L⁻¹ s⁻¹.

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