A. Dependence of Reaction Rate on Concentration Mixture xTime (t) for Relative r

Question

A. Dependence of Reaction Rate on Concentration Mixture xTime (t) for Relative rate Reactant Concentration in Mixture Temperature, "C color change of reaction (sec) (1000/t) [BrO,1 [H'] 1T4 9 2 3t.tt_10.00 5 I. Show what Equation 3 (shown below) would look like for Mixture #1 if you substitute in the in concentrations and the rate from the table above. 2. Show what Equation 3 would look like, if you substitute in values for Mixture #2. Divide the rate equation for Reaction Mixture 2 by the rate equation for Reaction Mixture 1. The resultant equation should have the ratio of Rates on the left side (Rate/Rate) and a ratio of [I] raised to the power m on the right. Note that you just have iodine on the right side of the equation as the concentrations of hydrogen and bromate were unchanged and cancelled each other out. Write the resulting relationship below: 3. 2-a 4.a4 a- Take the log of both sides of your equation and determine the power m round to the nearest whole number) 4.

Explanation / Answer

I'll guide you with example so you can do it. This really it's not difficult, and calculations are easy to make.

To answer question 7, all you need to do is apply equation from question 1 table 1 which is:

rate = k [I]m [BrO3]n[H]p

You already calculate m, n and p in the previous question, so all you need to do is substitute the values of mixture 1, 2, 3 and 4 in this equation with the value of m, n and p, and solve for k. After you do that, you can actually calculate the average. You can do it like the next way:

Mixture 1:

k' = rate / [I]m [BrO3]n[H]p

k'1 = 5.68 / (0.002)1 * (0.008)1 * (0.02)2 = 8.875x108

Doing the same thing with the other 3 mixture and you'll get the following k' values:

k'2 = 1.1164x109;

k'3 = 1.28x1010

k'4 = 1.35x109

kaverage = 4.04x109

k' is nearly equivant basically because concentrations of every compound there are equivalents too. the concentration of H is almost the same in all the mixture, and concentrations of BrO3 in one mixture, can be the concentration of I of another mixture. So, all of these concentrations are nearly equivalent, therefore, k is equivalent too.

Finally for question 9, use the value of k and concentrations of reaction 5, so you can solve for rate and then time:

rate = 4.04x109 * (0.0016) * (0.004) * (0.03)2 = 23.2704

and rate = 1000/t ----> t = 1000 / rate

t = 1000/23.2704 = 42.97 s

Observed time = 83 s

% error = (42.97 / 83) * 100 = 51.77%

Hope this helps.

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