Rate of a chemical reaction

I’m not asking anyone to do my homework; just asking if I’m going about it correctly.

In the Iodine Clock reaction, we are given that 6.0 x 10[sup]-4[/sup] M of I[sub]2[/sub] must be consumed before the colour changes for all tests.

For one run, which took 39.56 seconds, I[sup]-[/sup] = 0.02 moles/L x 0.008 L = M[sub]2[/sub] x 0.02 L = 0.08 moles. That tells me how many moles of I[sup]-[/sup] I start with. If I divide the number of moles I start with (0.08) by ∆[I[sub]2[/sub]] (6.0 x 10[sup]-4[/sup]) and then divide that by 39.56 seconds, I get 3.37 seconds. Since there is half the amount of I[sub]2[/sub] product compared to I[sup]-[/sup] reactant, I divide by 2 and come up with 1.69 seconds. In other words:

Rate = (½ x 0.08 / 6.0 x 10[sup]-4[/sup]) / 39.56 = 1.69 M/s

So is this equation correct, to find the rates of reaction for the remaining tests?

Rate = (½ [I[sup]-[/sup]] / ∆[I[sub]2[/sub]]) / ∆t