[Skip]

So I have no idea how to do this...help me please.
Using a first order plot, determine the rate constant, K_1, for the oxidation of benzyl alcohol with potassium permanganate at the temperature T_1 = 20.4 Celsius, from the information in the table below. (Molar absorptivity, E = 2350, path length = 1cm)

Time (min)..Absorbance
15....0.794
30....0.604
45....0.431

[ARGOS++]

Dear Skip;

With the help of the Beer-Lambert Law and your Epsilon (= 2350 L mol^-1 cm^-1) you are able to translate the absorptions into concentrations as the first step.

For the second step use the following Link as guidance/help:

http://dbhs.wvusd.k12.ca.us/webdocs/Kinetics/First%20Order%20Kinetics.html

Good Luck!
                    ARGOS⁺⁺

[Skip]

Okay, I have all my molarities, but the equation in the lab book (to plot a straight line) is

ln[MnO4] = ln[MnO4]₀ - kt

To solve for k, would I not need the original molarity (ln[MnO4]₀)

How do I find that?

[ARGOS++]

Dear Skip;

Read the last line of the help in the given link (y-intercept of the graph!)!

Good Luck!
                    ARGOS⁺⁺

[Skip]

Yeah, I understand that it's the y-intercept of the graph, but is the graph not ln[MnO4] vs time? To get the ln[MnO4] values, I thought I would need the value of ln[MnO4]₀ already?

By the way, thanks for the continuous help, Argos.

[ARGOS++]

Dear Skip;

Please read the line correctly!, -  It is the ln()!

Good Luck!
                    ARGOS⁺⁺

[Skip]

I guess I'm NOT understanding...
"and the y-intercept is ln [A]o" so, how would I get that number without knowing all other variables? I only know ln[A], and t, which leaves k and ln [A]o to be solved for.

[ARGOS++]

[Skip]

Oh my goodness, how silly of me. It's a straight line graph.. I would just have to plot three point and join them, then continue the line.

[ARGOS++]

Dear Skip;

Correct!

Good Luck!
                    ARGOS⁺⁺