Chepter 15 Chemical Kinetics.pptx

Chepter 15 Chemical Kinetics.pptx - Chapter 14 Chemical...

This preview shows page 1 - 9 out of 52 pages.

Chapter 14 Chemical Kinetics
Image of page 1

Subscribe to view the full document.

Thermodynamics verses Kinetics Thermodynamics tells us which way a chemical reaction will go. Kinetics tells us how fast a chemical reaction will go as well has how to control the rate. Fast Slow
Image of page 2
Reaction Rates Chemical rates are changes in concentration over a time interval . Rates can be either an (1) average or (2) instantaneous . The advantage of an (1) average rate is they are easy to calculate . The disadvantage is they tend to be very general, and not exact . The advantage of an (2) instantaneous rate is it give very specific, giving exact information . The disadvantage is the time it takes to setup and make the calculation.
Image of page 3

Subscribe to view the full document.

Average Reaction Rates Average rates change in molar concentration of reactants , R Δ[R] = [R] t2 - [R] t1 divided by the time interval Δt = t 2 - t 1 : Average rate of consumption of R = - Note on the “-” meaning reactants disappear. For products P = Remember that [ ] stands for concentration or mol·L -1
Image of page 4
Example 1: Suppose we are studying the reaction 2 HI(g) H 2 (g) + I 2 (g) and find that, in an interval of 100. s, the concentration of HI decreases from 4.00 mmol·L -1 to 3.50 mmol·L -1 . What is the average reaction rate ? Since we are watching reactants go away, we expect to see a negative slope. R = - = - = - = 5.0 × 10 -3 mmol HI·L -1 ·s -1 We just calculated the blue line. Note: 10 -3 mmol = μmol, so in this example -5.0 × 10 -3 mmol HI·L -1 ·s -1 = -5.0 μmol HI·L -1 ·s -1
Image of page 5

Subscribe to view the full document.

To avoid confusion when reporting rates of a reaction, rates are always reported as stoichiometric coefficients relationships. This way the rate is the same for that reaction , no matter which species the rate is reported for. For example, 2 HI(g) H 2 (g) + I 2 (g) R = - = = Average Reaction Rates based on Stoichiometry (a) What is the average rate of formation of H 2 in the same reaction? R = (5.0 × 10 -3 mmol HI·L -1 ·s -1 ) = 2.5 × 10 -3 mmol H 2 ·L -1 ·s -1 (b) What is the unique average rate , both over the same period? R = - = = (5.0 × 10 -3 mmol HI·L -1 ·s -1 ) = 2.5 × 10 -3 mmol H 2 ·L -1 ·s -1
Image of page 6
Monitoring rates of chemical reactions In a stopped-flow experiment , the driving syringes on the left push reactant solutions into the mixing chamber, and the stopping syringe on the right stops the flow. The progress of the reaction is then monitored spectroscopically as a function of time.
Image of page 7

Subscribe to view the full document.

Instantaneous Rate of Reaction Reactions slow down as reactants are used up. An “instant” rate is found by measuring two concentration, as close together in time as possible. A better method is to find the slope of the tangent that provides the instantaneous rate . Though a computer can calculate tangent lines, we need to devise methods to determine the equations for these lines , which are called rate laws .
Image of page 8
There are two parts to every rate law. (1) how fast the line changes, or speed (initial rate) and (2) the shape of the line, called the rate law.
Image of page 9
You've reached the end of this preview.
  • Fall '16
  • Dr. Behrang Madani
  • Kinetics

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern