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Entry Task: Nov 30 th Friday

Entry Task: Nov 30 th Friday. Question: Name three variables that can affect the rate of a chemical reaction? You have 5 minutes!. Agenda:. Sign off and discuss Ch. 14 section 1 HW: Reaction rates ws . I can…. Write rate expressions and calculate the average rate of reaction.

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Entry Task: Nov 30 th Friday

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  1. Entry Task: Nov 30th Friday Question: Name three variables that can affect the rate of a chemical reaction? You have 5 minutes!

  2. Agenda: • Sign off and discuss Ch. 14 section 1 • HW: Reaction rates ws.

  3. I can… • Write rate expressions and calculate the average rate of reaction.

  4. Chapter 14Chemical Kineticssections 1

  5. Define Chemical Kinetics • Studies the rate at which a chemical process occurs. • Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

  6. Factors That Affect Reaction Rates 1. Concentration of Reactants • As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

  7. Factors That Affect Reaction Rates 2. Temperature • At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

  8. Factors That Affect Reaction Rates 3. Presence of a Catalyst • Catalysts speed up reactions by changing the mechanism of the reaction. • Catalysts are not consumed during the course of the reaction.

  9. Factors That Affect Reaction Rates 4. The surface area of solid or liquid reactants • Reaction time increases when the solids’ surface area increases.

  10. 14.1- Reaction Rates- What is the relationship between speed and a chemical reaction? It’s the time it takes for the change of reactants to become products is the “speed” of the reaction

  11. What is the reaction rate measuring? Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

  12. So what does that mean? We expect- reactant concentration will go down as the concentration of product goes up!! During a reaction, what is happening to the reactants- what do we expect?

  13. Provide the mathematical expression: Concentration at end – Concentration at start Average rate = change in the number of moles of B change in time Time ended – time started

  14. Look at Figure 14.4, explain what is happening to the reactants and to the products and the trend for each. As the concentration of reactants decreases the concentration of products increases.

  15. Why is the delta (moles of B) a positive number? The reaction: AB, the concentration of B is increasing therefore its positive.

  16. Why is the delta (moles of A) a negative number? The reaction: AB, the concentration of A is decreasing therefore its negative.

  17. Rates in Terms of ConcentrationsHow do measure (units) for concentration rates? The units of reaction rates are molarity per second (M/s)

  18. Reaction Rates C4H9Cl(aq)+ H2O(l) C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times. The concentration is going down!

  19. Average rate = [C4H9Cl] t Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time:

  20. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

  21. Provide the equation for average rate of concentration change: Average rate = -∆ [C4H9Cl] ∆ t Average rate = - [C4H9Cl] finial time - [C4H9Cl] initial time (final time – initial time) What do the brackets mean? What about the minus? Brackets means concentration of substance (in this case its reactant) and the minus is the rate of its disappearance.

  22. How are instantaneous rates determined- provide a detailed step-by step? C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • A plot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

  23. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • All reactions slow down over time. • Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

  24. Go through the Sample Exercise 14.1, then belowSHOW how to do the practice exercise:

  25. Sample Exercise 14.1Calculating an Instantaneous Rate of Reaction Using Figure 14.2, calculate the instantaneous rate of disappearance of C4H9Cl over the time interval from 50.0 to 150.0 s (b) Using Figure 14.5, estimate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate). Solution Analyze- Look up the concentrations on table. Planlook up the time 150.0s and the concentration for that time and like wise for the 50.0 s interval and concentration. SolvePlace the information in the equation. Average rate = - (0.0741– 0.905) M (150.0 – 50.0) s = 1.64 x 10-4 M/s

  26. Sample Exercise 14.1Calculating an Instantaneous Rate of Reaction Using Figure 14.2, calculate the instantaneous rate of disappearance of C4H9Cl over the time interval from 50.0 to 150.0 s (b) Using Figure 14.5, estimate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the initial rate). Solution Analyze We are asked to determine an instantaneous rate from a graph of reactant concentration versus time. Plan To obtain the instantaneous rate at t = 0s, we must determine the slope of the curve at t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (that is, change in molarity over change in time). Solve The tangent line falls from [C4H9Cl] = 0.100 M to 0.060 M in the time change from 0 s to 210 s. Thus, the initial rate is

  27. Sample Exercise 14.1Calculating an Instantaneous Rate of Reaction Continued Practice Exercise Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s. Answer: 1.5  104 M/s Average rate = ­ –­(0.0549– 0.1000) M (300 – 0) s = 1.5 x 10-4 M/s

  28. Reaction Rates and Stoichiometry How does stoichiometry help determine that rates of reactions between two substances? C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) As one mole of a substance disappears (C4H9Cl), one mole of another substance appears (C4H9OH).

  29. -[C4H9Cl] t Rate = = [C4H9OH] t Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.

  30. Provide the equation for the decomposition of hydrogen iodide: 2 HI (g) H2 (g) + I2 (g) NOTICE: The stoich relationship is NOT the same. It is 2HI to every 1 H2 and I2.

  31. 1 2 [HI] t Rate = − = [I2] t Explain WHY in the equation that HI is divided into two (or ½ in the equation). 2 HI (g) H2 (g) + I2 (g) • Therefore, As two moles of HI disappears, one mole of I2or H2 substance appears, so HI disappears at twice the rate of the products- so ½ the rate of the appearance of the products.

  32. aA + bB cC + dD 1 b 1 d 1 a 1 c [D] t [C] t [A] t [B] t Reaction Rates and Stoichiometry • To generalize, then, for the reaction = = Rate = − =− The coefficients (lower case) is in the denominator which is the rate for the (capital letter) divided by the change of time.

  33. Sample Exercise 14.2Relating Rates at Which Products Appear and Reactants Disappear (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O3(g) 3 O2(g)?(b) If the rate at which O2 appears, [O2]/t, is 6.0  10–5M/s at a particular instant, at what rate is O3 disappearing at this same time,  [O3]/t? Plan We can use the coefficients in the chemical equation to express the relative rates of reactions. Solution Analyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. Solve(a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have: (b) Solving the equation from part (a) for the rate at which O3 disappears, [O3]/t, we have: 2 O3(g) 3 O2(g)

  34. 4 2 1 2 1 4 1 1 1 4 1 2 [N2O5] t [N2O5] t [NO2] t [NO2] t [O2] t Sample Exercise 14.2Relating Rates at Which Products Appear and Reactants Disappear Practice Exercise If the rate of decomposition of N2O5 in the reaction 2 N2O5(g) 4 NO2(g) + O2(g) at a particular instant is 4.2  107M/s, what is the rate of appearance of (a) NO2 and (b) O2 at that instant? = = Rate = − 4.2  107M/s is how long it took for N2O5 to disappear This is our “X” what we are looking for, so get 4 out of there = 4.2  107M/s = Rate of NO2 8.4 x 10-7 M/s

  35. 1 2 1 2 1 4 1 1 1 1 1 2 [N2O5] t [N2O5] t [NO2] t [O2] t [O2] t Sample Exercise 14.2Relating Rates at Which Products Appear and Reactants Disappear Practice Exercise If the rate of decomposition of N2O5 in the reaction 2 N2O5(g) 4 NO2(g) + O2(g) at a particular instant is 4.2  107M/s, what is the rate of appearance of (a) NO2 and (b) O2 at that instant? = = Rate = − 4.2  107M/s is how long it took for N2O5 to disappear This is our “X” what we are looking for, so get 1 out of there = 4.2  107M/s = Rate of O2 2.1 x 10-7 M/s

  36. 14.5 14.5. The isomerization of methyl isonitrile, CH3NC, to acetonitrile, CH3CN, was studied in the gas phase at 215°C, and the following data were obtained: Calculate the average rate of reaction M/s for the time interval between each measurement. 2.7 x 10-6 M/s 0.0110 M – 0.0165 M 2000 s – 0 s 0.0055 2000 = 2.7 x 10-6 M/s

  37. 14.5 14.5. The isomerization of methyl isonitrile, CH3NC, to acetonitrile, CH3CN, was studied in the gas phase at 215°C, and the following data were obtained: Calculate the average rate of reaction M/s for the time interval between each measurement. 2.7 x 10-6 M/s 1.7 x 10-6 M/s 9 x 10-7 M/s 4 x 10-7 M/s 2 x 10-7 M/s

  38. 1 1 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 3 [H2O2] t [N2O2] t [NH3] t [O2] t [O2] t [N2] t [H2] t [N2] t [H2] t 14.9 14.9 For each of the following gas-phase reactions, write the rate of reaction expression. a) H2O2 (g)  H2(g) + O2 (g) b) 2N2O (g)  2N2(g) + O2 (g) c) N2(g) + 3H2 (g)  2NH3 (g) = = Rate = − = = Rate = − = Rate = − =−

  39. 1 2 1 1 1 2 1 2 1 1 1 1 1 2 1 1 [C4H8] t [C2H4] t [SO2] t [SO3] t [HBr] t [Br2] t [O2] t [H2] t 14.10 14.10 For each of the following gas-phase reactions, write the rate of reaction expression. a) 2HBr (g)  H2(g) + Br2 (g) b) 2SO2 (g) + O2 (g)  2SO3(g) c) C4H8(g)  2C2H4 (g) = = Rate = − = Rate = − =− Rate = − =−

  40. 1 2 1 2 1 1 2 2 1 2 [H2O] t [H2O] t [O2] t [H2] t 14.11 a) Consider the combustion of H2 (g): 2H2 (g) + O2 (g)  2H2O(g) If hydrogen is burning at the rate of , what is the rate of consumption of oxygen? What is the rate of formation of water vapor? = Rate = − =− 4.6 mol/s This is our “X” what we are looking for, so get 2 out of there 4.6 mol/s

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