Final answer:
The average rate of the reaction in a 10-second interval where I⁻ concentration fell from 1.000 M to 0.868 M is calculated to be 0.00440 M/s for I₃⁻ formation.
Step-by-step explanation:
The student has asked to calculate the average rate of the reaction over a 10-second interval, given the decrease in concentration of I⁻ from 1.000 M to 0.868 M. To find this rate, you subtract the final concentration of I⁻ from the initial concentration, then divide by the time interval. The reaction shows that 3 moles of I⁻ are consumed for every mole of I₃⁻ produced. Therefore, the change in concentration of I⁻ (Δ[I⁻]) is (1.000 - 0.868) M = 0.132 M over 10 seconds.
The average rate of disappearance of I⁻ is 0.132 M divided by 10 s, which gives 0.0132 M/s. However, as per the stoichiometry of the balanced reaction, the rate of consumption of I⁻ is three times the rate of formation of I₃⁻. Therefore, the average rate of formation of I₃⁻ would be one-third of 0.0132 M/s, which equals 0.00440 M/s.