1 / 25

Ch 17: Kinetics Pt 1

Ch 17: Kinetics Pt 1. Dr. Harris Lecture 18 HW: Ch 17: 5, 11, 18, 23, 41 , 50. Reactions Rates. Chemical kinetics is the area of chemistry that investigates how fast reactions occur Different reactions proceed with different rates

xarles
Télécharger la présentation

Ch 17: Kinetics Pt 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ch 17: Kinetics Pt 1 Dr. Harris Lecture 18 HW: Ch 17: 5, 11, 18, 23, 41, 50

  2. Reactions Rates • Chemical kinetics is the area of chemistry that investigates how fast reactions occur • Different reactions proceed with different rates • The rate of a reaction depends on several factors, including: • reactant concentration • temperature • catalysts • surface area • Today, we will focus exclusively on the relationship between reaction rates and reactant concentration

  3. Intro • Lets take the reaction: A ---> B. • This reaction tells us that as A is consumed, B is formed at an equal rate. We can express this mathematically in terms of changing concentrations by: • Imagine we have 10 moles of A in 1 L of solution. If we can freeze time for an instant, such that the reaction has not yet begin (t=0), the concentration of A is 10M.

  4. [A] = 7 M [B] = 3 M A B [A] = 10 M After 10 seconds, 3 moles of B have formed. t = 0 t = 10 = 1 mol of A 10 more seconds = 1 mol of B t = 40 40 more seconds 20 more seconds t = 20 t = 80 [A] = 3 M [B] = 7 M [A] = 4 M [B] = 6 M [A] = 5 M [B] = 5 M

  5. Plotting the data from previous slide

  6. Reactions Follow a Rate Law • The graph in the previous slide shows that the disappearance of A (formation of B) is not linear. • As [A] decreases, the reaction slows down. This means that reaction rates depend on reactant concentration. • This dependence of rate on concentration suggests that reaction rates follow a rate law, a mathematical expression that ties concentration and rate together

  7. Instantaneous Rates • Although the rate of the reaction is constantly changing with reactant concentration, we can determine the instantaneous rate(reaction rate at a specific time and concentration) • Instantaneous rate at t=0 is theinitial rate • We can determine the instantaneous rate by taking the slope of the tangent at the point of interest • Note: a tangent line is linear and ONLY touches the point in question. It does NOT cross the curve Tangent at t = 20s, [A] = 5M Instantaneous rate of disappearance of A at t=20 sec

  8. Rates and Stochiometry • In the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any given time, the rate of disappearance of A equals the rate of formation of B. If the stoichiometry is NOT 1:1, we have a much different situation, as shown below: • As you can see, 2 moles of HI are consumed for every 1 mole of H2 and 1 mole of I2 formed. Thus, the disappearance of HI is twice as fast as the appearance of the products.

  9. Example: N2O5(g) ----> 2NO2(g) + ½ O2(g) Looking at averagerates average rate of disappearance after 10 minutes average rate of disappearance after 100 minutes

  10. N2O5(g) ----> 2NO2(g) + ½ O2(g) slow fast

  11. Rate Laws • We see that reducing reactant concentration lowers the reaction rate, but to what extent? What is the mathematical correlation? • The equation that relates the concentration of the reactants to the rate of reaction is called the rate lawof the reaction. • We can derive the rate law of a reaction by seeing HOW THE REACTION RATE CHANGES WITH REACTANT CONCENTRATION. • For any reaction aA + bB ----> cC + dD • In this expression, k is the rate constant, m and n are reaction orders.

  12. Reaction Orders and the Method of Initial Rates N2O5(g) 2NO2(g) + ½ O2(g) • Lets go back to the previous reaction: • Below is a table of data, showing the initial reaction rate as a function of the starting concentration of N2O5 (g). We perform multiple experiments to collect enough data to determine our rate law. • We see that when we double [N2O5]o, the rate also doubles. When we quadruple [N2O5]o, the rate quadruples. Thus, the rate is directly proportional to [N2O5]oby the rate constant, k. • This means that the reaction is FIRST ORDER WITH RESPECT TO [N2O5] (m=1). We can write the rate law as:

  13. Reaction Orders • The overall reaction order is the sum of the individual reaction orders. In our previous example, there was only one reactant, so the overall order is 1 (1st order reaction). • We can easily solve for k by plugging in any corresponding rate and concentration. Lets plug in the values from run # 1

  14. Rate Laws/Reaction Orders • Reaction orders must be determined experimentally. You can not assume based on the stoichiometry. • When you have multiple reactants, you must determine the reaction order of each one. To do this, you must vary the concentration of only one reactant at a time while holding the others fixed. • Let’s attempt to determine the rate law for the reaction below: 2NO(g) + O2(g) ---> 2NO2

  15. Example: 2NO(g) + O2(g) ---> 2NO2 • Using the data below, determine the rate law of this reaction in the form: • This time, we have two reactants. Lets start by determining the value of ‘m’. To do so, we hold [O2]o fixed and vary [NO]o. This will show how the rate depends on [NO]o. • In experiments #1 and #3, [O2]o is fixed, so we will use these to find ‘m’.

  16. Remember, rate is proportional to [NO] by the power m. The factor of change in the rate is equal to the factor of change of [NO] to the mth power: order factor of rate change factor of change in [NO] m = 1 • The reaction is 1st order with respect to[NO]

  17. Now we can find ‘n’ by varying [O2]oand holding [NO]o fixed. We can use experiments #2 and #3 for this. This will show how the rate depends on [O2]o. order factor of rate change factor of change in [O2] n = 1 The reaction is 1st order with respect to [O2] and 2nd order overall.

  18. Pay Attention to the Units of k, As They Change with Overall Reaction Order • The rate constant, k, is the constant of proportionality between rate and concentration. • Higher values of k = faster reactions • It is important to note that the units of k depend on the overall reaction order. • Ex: • Rate is always in molarity per unit time (sec, hr, etc). Concentration is always M (mol/L). Thus, we have: • Recall for a 1st order reaction: Units of k for a 2nd order reaction Units of k for a 1st order reaction

  19. Example • Determine the relative (m & n) and overall (m+n) reaction order of the reaction below. Then, derive the rate law and determine the value of k (with correct units) • Tripling [NO2] causes the rate to increase nine-fold. This means that the rate is proportional to the square of [NO2], so the reaction is second order with respect to NO2(n=2). Doubling [CO] does nothing. Thus, the rate does not depend on [CO], and is zero order with respect to CO (m=0). Overall 2nd order. k = 1.11 x 108 M-1s-1

  20. Example: 2NO(g) + Br2(L) ---> 2NOBr (g) • Using the information below, determine the rate law of this reaction in the form: m = 2 1.) Find m. We can use runs 1 & 2: 2.) Find n. We can use runs 1 & 3 n = 1 • The reaction is 2nd order with respect to [NO], 1st order with respect to Br2, and the reaction is overall 3rd order.

  21. Determining the Overall Rate Order of A Reaction Graphically • As we have shown, a first-order reaction depends on the concentration of a single reactant to the 1st power. For the reaction: A----> products • Using calculus, we can convert this to: • This equation is in y = mx + b form. Therefore, for any 1st order reaction, the plot of the natural logof [A]tvs time will be linear. The slope of the line will be –k. natural log of starting concentration natural log of concentration at time t rate constant time x axis b m (slope) y axis

  22. Plotting 1stOrder Reactions b slope = -k units: s-1 time values on x-axis natural log of [A]t on y-axis

  23. Determining the Overall Rate Order of A Reaction graphically • A second-order reaction depends on the concentration of [A] to the 2nd power. For the reaction: A ----> B • Therefore, for any 2nd order reaction, the plot of the inverse of [A]tvs time will be linear. The slope of the line will be k. y b m•x

  24. Plotting a 2nd Order Reaction b slope = k units = M-1 s-1 time values on x-axis 1/[A]t on y-axis

  25. Determining Overall Rate Order From Plotting Time-Dependent Data linear! 2nd order not linear: NOT 1st order • We can determine if a process is first or second order by plotting the data against both equations. Which ever fitting method yields a linear plot gives the overall order.

More Related