[vanalm]

In lab, there were 2 parts. Part A,we did 3 experiments, each with variations in concentrations of reactants, and using that information we used the initial rate law to determine the reaction order. It was a first oder reaction. In part B, we used the same concentrations as in experiment 1 from part A, but timed the reaction at various temperatures. We tested at 5 different temperatures, and ran 3 trials at each temp. For a total of 15 pieces of data. Here is where I get confused. We are to:

calculate the rate constants for all experiments measured at temperatures other than room temp. Prepare a table for these values which contains the headings: reaction temp, 1/temp, rate constant, ln k. Make a plot of ln k verses 1/temp and determine the slope of the line. Using the slope of the line, calculate the activation energy.
I used the same formula I used to find k for the room temp experiments in part A of the lab: (rate= -1/2 change in concentration of S₂O₃^2-/change in temperature) and found 15 k's.  My class was split into groups, each group got a different temp., I don't know how accurate the results are, but I am getting some crazy activation energies. I don't know if maybe I am putting something in my calculator wrong, or if I am doing this wrong, or if the data is just bad.

This is what I did. I took each of my k values, and hit "ln" button on my calculator, then typed in the k value. Which gave me 15 different ln k values. Then I plotted ln k verses 1/temp, and found the slope using m=(y₂-y₁)/(x₂-x₁). I don't know if it matters, but my plots were not in a straight line, I drew a line through the middle of the plots because they were all were in groups... (ex: the 40 degree were kind of all together in the same general area, but not in a line since the 3 trials at 30 degrees were not exactly 30 degrees, we had a 30.2, 29.9, & 30.0)... So I then took two plot marks that were connected by my line and plugged them into my slope formula. Doing so, I got a slope of 8362. And I plugged that into ln k=(E_a/R)(1/T) +ln A. and I got an E_a of 2,368,167. I just don't see how that can be right.

The data points I used to find my slope are: (3.531 *10^-4, -6.54),(3.299*10^-3,-4.60) where x= 1/T (in K), and y= ln k. Am I doing my math wrong???  or using the wrong formula?    Thanks.

[enahs]

When you solve your equation, the slope is just -E_a/R, and the Y-Intercept is ln(A).

[vanalm]

I understand that the slope is E_a/R.And we are to find the activation energy, which is just E_a. The formula in our book is ln k=(Ea/R)(1/T) +ln A. So to get E_a by itself, I would have to multiply both sides by R and T, right?? when I do this, I get the really big number. Or is this wrong?

[enahs]

No, you do not have to multiply by T.

Also, either you are typing the equation wrong, or reading it wrong; as there is a minus sign missing.