00:01
So we have the data varying the concentrations of three reactants and looking at the effects on the initial rate.
00:15
And if we look at the first two experiments, the hydronium ion h -plus is at 2 times 10 to the minus 2, and the iodide is at 2 times 10 to the minus 2.
00:30
So those two are held constant in the first two experiments.
00:33
And we can look at the effect of doubling the h2 -s -e -o -3, i guess that's selenic acid.
00:41
And so i've set it up to compare the ratio of the rates in the first two experiments to that concentration in the first two experiments.
00:57
And so i plugged in the rates for the first two experiments here.
01:03
And notice that when we determined, the rate law, we're working on the value of that exponent, which is the exponent is on the concentration.
01:13
So you want to be sure to show that on the concentration in each case.
01:16
Basically, the rate doubled when the concentration doubled.
01:19
And so the exponent on the h2 seo3 is a 1.
01:25
Now, and you can see also when you go from the first to the third, that again, the h plus and the i minus remain constant.
01:35
And there was a tripling of the rate.
01:38
And so that one is worked out.
01:40
Now, looking at the last four experiments, the h -2 -s -e -o -3 was constant, and the last and not the one above it, but the second one above it, also holds the h -plus constant.
01:56
So we can look at the iodide there.
01:58
Let's see, that's two, four, six.
02:01
Look at the seventh experiment over the fifth experiment.
02:12
You have to be careful that everything remains constant except the one that you're changing.
02:18
So again, the h2 -sco -3 is 1 times 10 to the minus 4 in both of those experiments.
02:25
The h -plus is 1 times 10 to the minus 2 in both of those experiments.
02:29
And in the seventh one, okay, so this is going to be related to the iodide in trial number 7 and the iodide in trial number 5.
02:43
And so the rate in experiment 7 was 3 .36 times 10 to the minus 7...