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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

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Ch 17: Kinetics Pt 1

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  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.

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