Answer:
The average rate of formation of I₂(g) is 6.00x10⁻⁵Ms⁻¹
Step-by-step explanation:
The question is:
The average rate of formation of I2 over the same time period is______M s-1.
Based in the reaction:
2 HI(g) ⇄ H₂(g) + I₂(g)
If 2 moles of HI(g) disappears in a rate of 1.20x10⁻⁴Ms⁻¹, 1 mole of I₂(g) will appears at:
1 mole I₂(g) × (1.20x10⁻⁴Ms⁻¹ / 2 mol) = 6.00x10⁻⁵Ms⁻¹ I₂(g)