For the decomposition of hydrogen iodide on a gold surface at 150 °C 2 HI(g)H2(g) + I2(g) the average rate of disappearance of HI over the time period from t = 0 s to t = 3646 s is found to be 1.20×10-4 M s-1.\

The average rate of formation of I₂(g) is 6.00x10⁻⁵Ms⁻¹

Explanation:

The question is:

The average rate of formation of I2 over the same time period isM 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)

put the answers   and i could answer.

let that person have a turn to speak : )