# Rate of a chemical reaction

**URL:** https://boards.straightdope.com/t/rate-of-a-chemical-reaction/683291
**Category:** Factual Questions
**Created:** [March 9, 2014, 8:03pm UTC](https://boards.straightdope.com/t/rate-of-a-chemical-reaction/683291 "2014-03-09T20:03:48Z")
**Posts on this page:** 1
**Page:** 1

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### Author: ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)
#### Post date: [March 9, 2014, 8:03pm UTC](https://boards.straightdope.com/t/rate-of-a-chemical-reaction/683291/1 "2014-03-09T20:03:48Z")

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> [@](#):
>
> Please do not ask other members to do your homework…

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