The rate of a first-order reaction is followed by spectroscopy,
m...

Question

The rate of a first-order reaction is followed by spectroscopy,
monitoring the absorbance of a colored reactant at 520 nm.
The reaction occurs in a 1.00-cm sample cell, and the only
colored species in the reaction has an extinction coefficient
of 5.60 * 103 M-1 cm-1 at 520 nm. (d) How long does it take for the absorbance to fall to 0.100?

Accepted Answer

Hey everyone today, we're given a first order reaction that is monitored by spectra for tom a tree. Were given the absorptive Itty at a specific wavelength as well as the length of the cell sample. And the initial and final absorbency is that we're looking for given a specific rate constant. There's a lot of information to take in. But first thing we need to look at is that we're dealing with a first order reaction. So since the reaction is first order, that means we have an integrated rate law that looks like so natural log of the final concentration of the reactant is equal to negative K. Times T. Plus the natural log of the initial reacting concentration. We'll come back to this soon. But we need in order to utilize this. We need to actually use beer lambert's law to find the initial and final concentrations of the reacted. Now, Beer lambert's law states that absorbent excuse me, is equal to the absorptive. Itty times the length of the sample times its concentration. Therefore the concentration is equal to the observed activity. Divided by or sorry, the absorbent divided by the absorptive Itty times the sample length. So the initial substrate or concentration reacting concentration which is just hmm, is equal to 0.245 Hoops to 4 5 divided by 6.4. 2 times 10 to the second. Multi negative one times two negative one, Multiplied by one cm. Let's give this a final value of 3. Times 10 to the -4 Moller. We can use the same process for the final concentration. So that will be zero divided by The same thing. 6.42 times 10 to the second molinari, T. To the native one times centimeters to the negative one multiplied by the length which is one centimeter. And you'll get a value of. And let's write this in red, 2.57 times 10 to the negative 4th Waller. So with this in mind we can go ahead and plug everything into the integrated rate law. So we can say that the natural log of the final concentration Just 2. times 10 to the -4 Moller, oops is equal to negative K. The rate constant which is 3.48 times 10 to the negative two seconds. The negative one multiplied by T. Plus the natural log of The initial reacting concentration which is 3. times 10 to the Native four Mueller. Simplifying and equating to T. We get that T. Is finally equal to negative 0. divided by oops negative 3.48 times to the negative second seconds to the negative one Whiskers. As a final value of 11.4 seconds. Therefore the time that absorbent will fall to 0.165. If the rate constant is 3.48 attempts 10 to the negative two seconds to the negative one. At 300 kelvin is 11.4 seconds. I hope this helps, and I look forward to seeing you all in the next one.

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