How do you determine rate law and rate constant with experimental data?

Test Preparation AP Test Preparation
1

<p>The reaction</p>

<p>2I- (aq) + S2O8 2- (aq) -----> I2 (aq) + 2SO4 2- (aq)</p>

<p>was studied at 25 degsC. The following results were obtained where</p>

<p>Rate = -delta[S2O8 2-]/delta[T]</p>

<p>.[I-]o.................[S2O8]o....................Initial Rate
(mol/L)...............(mol/L)......................(mol/L*s)
0.080..................0.040.......................12.5x10^-6
0.040..................0.040.......................6.25x10^-6
0.080..................0.020.......................6.25x10^-6
0.032..................0.040.......................5.00x10^-6
0.060..................0.030.......................7.00x10^-6</p>

<p>A. Determine the rate law.
B. Calculate a value for the rate constant for each experiment and an average value for the rate constant.</p>

<p>Can someone PLEASE explain, simply, how to do this type of problem? I don't want you to do it for me. I just need simplified steps, because my textbook is not very clear. Thanks in advance!</p>

2

<p>This is quite complicated to do. You need to find out, first, the exponents for the different components in the rate law. To do that for, say S2O8, you would find two sets of data where the I- concentrations were the same. Then you'd compare the concentration of S2O8 and the initial rate. So, here it would be the two experiments with I- concentration at 0.080. The S2O8 concentrations are 0.04 and 0.02. So divide 0.02 by 0.04 and you get 1/2. You also have to compare the rates in the same way. So, 6.25x10^-6/12.5x10^-6 is still 1/2. Then you setup an exponential equation like so: (concentration ratio)^X=(rate ratio). So, here it would be (1/2)^X=1/2 where X = 1. The exponent on S2O8 is X in the equation so it is 1. Now, do the same for I-. The exponent is 1 for I- as well.</p>

<p>Then setup the rate law for the reaction. It would be Rate = [I-][S2O8]/[I2][2SO4]. Then you can punch in the concentrations to find the rate. However, there is a problem. Are you sure this is everything? The data isn't enough to come up with a rate law.</p>

3

<p>Aight man this is how you do the problem...</p>

<p>First, you write down a GENERALIZED rate law (in this case you have two reactants 2I^- and S2O9^2- ... so the GENERAL rate law (not the actual rate law) is ... rate= K[I-]^n[S2O82-]^m
- You want to find n and m using the method of intial rates (which i will show u in a second), then when u find n and m you plug them in and that is your ACTUAL rate law.</p>

<p>So when starting the problem you want to look for two rows of data in which one of the columns of data is the same.. (that probably doesnt make sense but when i do the math you will see it)</p>

<p>Rate 2 = 6.25x10^-6 = k[.04]^n[.04]^m </p>

<hr>

<p>Rate 1 = 12.5x10^-6 = k[.08]^n[.04]^m</p>

<p>Then before you compute this you can see that the pair of [.04]^m concentrations and the pair of k's cancel leaving you with one variable (n).</p>

<ul>
<li><p>Now when computing this you get .5= (.04/.08)^n ----> .5=.5^n ---> thus, n = 1</p></li>
<li><p>Now since you have found n, you can find m by using a new set of data in which you can cancel the concentrations raised to the nth power, leaving you with one variable (m). See below.</p></li>
</ul>

<p>Rate 3 = 6.25x10^-6 = k[.08]^n[.02]^m</p>

<hr>

<p>Rate 1 = 12.5x10^-6 = k[.08]^n[.04]^m</p>

<ul>
<li>the left side of the equation yields .5</li>
<li>and now you can see that the pair of [.08]^n and the pair of k cancel out, leaving you with (.02/.04)^m.</li>
</ul>

<p>Thus..</p>

<p>.5= (.02/.04)^m</p>

<p>isnt (.02/.04) = .5</p>

<p>so, .5= .5^m , and thus m = 1</p>

<ul>
<li>Now since you have computed numerical values for n and m which in this case are both you you can write an ACTUAL rate law... </li>
</ul>

<p>rate= k[I-]^1[S208]^1 or you can just leave it as rate= k[I-][S208] because when you dont put a raised to the first power (^1) its implied.</p>

<p>Now you have your rate law...but what about the rate constant ( and if you did not know, the rate constant is k)? </p>

<p>Well choose any set of data..in this case i will choose the first set... and plug it in literally and solve for k.</p>

<p>rate= k[I-]^1[S208]^1</p>

<p>12.5x10^-6= k[.08]^1[.04]^1</p>

<p>12.5x10^-6
--------------- = .0039 <em>significant digits</em>
[.08]^1[.04]^1</p>

<ul>
<li>WAIT!, your not done, you have to remember with rate laws k HAS units, unlike in equilibrium and where K is just a value with no digits.</li>
</ul>

<p>So what are the units..well jus plug the units into the math you did..see below</p>

<h2>mol </h2>

<h2>LxS </h2>

<h2>(mol) (mol)</h2>

<p>(L)(L)</p>

<p>which becomes...</p>

<p>mol (L)(L)</p>

<hr>

<p>LxS (mol)(mol)</p>

<p>which becomes...</p>

<h2>(L)</h2>

<p>(mol x S)</p>

<ul>
<li>So the units are (L/mol x s)</li>
</ul>

<p>So the answers are:</p>

<p>a.) Rate= k[I-]^1[S208]^1
b.) .0039 (L/mol x S)</p>

4

<p>The same would work if you had three reactants, correct?</p>

<p>And with the initial rates… I did a lab with this, so do I use the Relative Rate of the reaction (1000/t) as the initial rates?</p>