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

Chemical Kinetics Agenda: Sign off and discuss Ch. 14 section 1 HW: Reaction rates ws.

Chemical Kinetics I can… Write rate expressions and calculate the average rate of reaction.

Chemical Kinetics Chapter 14 Chemical Kinetics sections 1

Chemical Kinetics 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).

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

Chemical Kinetics 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.

Chemical Kinetics 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.

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

Chemical Kinetics 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

Chemical Kinetics 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.

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

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

Chemical Kinetics 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.

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

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

Chemical Kinetics Rates in Terms of Concentrations How do measure (units) for concentration rates? The units of reaction rates are molarity per second (M/s)

Chemical Kinetics Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. The concentration is going down!

Chemical Kinetics Reaction Rates Average rate =  [C 4 H 9 Cl]  t C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

Chemical Kinetics Reaction Rates 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. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

Chemical Kinetics Provide the equation for average rate of concentration change: Average rate = -∆ [C 4 H 9 Cl] ∆ t Average rate = - [C 4 H 9 Cl] finial time - [C 4 H 9 Cl] 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.

Chemical Kinetics How are instantaneous rates determined- provide a detailed step-by step? 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. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

Chemical Kinetics Reaction Rates All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

Chemical Kinetics Go through the Sample Exercise 14.1, then below SHOW how to do the practice exercise:

Chemical Kinetics Sample Exercise 14.1 Calculating an Instantaneous Rate of Reaction Using Figure 14.2, calculate the instantaneous rate of disappearance of C 4 H 9 Cl over the time interval from 50.0 to s (b) Using Figure 14.5, estimate the instantaneous rate of disappearance of C 4 H 9 Cl at t = 0 s (the initial rate). Solution Analyze- Look up the concentrations on table. Plan look up the time 150.0s and the concentration for that time and like wise for the 50.0 s interval and concentration. Solve Place the information in the equation. Average rate = - ( – 0.905) M (150.0 – 50.0) s = 1.64 x M/s

Chemical Kinetics Sample Exercise 14.1 Calculating an Instantaneous Rate of Reaction Using Figure 14.2, calculate the instantaneous rate of disappearance of C 4 H 9 Cl over the time interval from 50.0 to s (b) Using Figure 14.5, estimate the instantaneous rate of disappearance of C 4 H 9 Cl 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 [C 4 H 9 Cl] = M to M in the time change from 0 s to 210 s. Thus, the initial rate is

Chemical Kinetics Sample Exercise 14.1 Calculating an Instantaneous Rate of Reaction Practice Exercise Using Figure 14.4, determine the instantaneous rate of disappearance of C 4 H 9 Cl at t = 300 s. Answer: 1.5  10  4 M/s Continued Average rate = ­ – ­( – ) M (300 – 0) s = 1.5 x M/s

Chemical Kinetics How does stoichiometry help determine that rates of reactions between two substances? As one mole of a substance disappears (C 4 H 9 Cl), one mole of another substance appears (C 4 H 9 OH). Reaction Rates and Stoichiometry C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

Chemical Kinetics Reaction Rates and Stoichiometry Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

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

Chemical Kinetics Explain WHY in the equation that HI is divided into two (or ½ in the equation). 2 HI (g)  H 2 (g) + I 2 (g) Therefore, Rate = − 1212  [HI]  t =  [I 2 ]  t As two moles of HI disappears, one mole of I 2 or H 2 substance appears, so HI disappears at twice the rate of the products- so ½ the rate of the appearance of the products.

Chemical Kinetics Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = −= − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t = The coefficients (lower case) is in the denominator which is the rate for the (capital letter) divided by the change of time.

Chemical Kinetics Sample Exercise 14.2 Relating Rates at Which Products Appear and Reactants Disappear 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 O 3 disappears,  [O 3 ]/  t, we have: (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g)  3 O 2 (g)? (b) If the rate at which O 2 appears,  [O 2 ]/  t, is 6.0  10 –5 M/s at a particular instant, at what rate is O 3 disappearing at this same time,  [O 3 ]/  t? Plan We can use the coefficients in the chemical equation to express the relative rates of reactions. 2 O 3 (g)  3 O 2 (g)

Chemical Kinetics Practice Exercise If the rate of decomposition of N 2 O 5 in the reaction 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) at a particular instant is 4.2  10  7 M/s, what is the rate of appearance of (a) NO 2 and (b) O 2 at that instant? Sample Exercise 14.2 Relating Rates at Which Products Appear and Reactants Disappear Rate = − 1212 [N2O5]t[N2O5]t = 1414  [NO 2 ]  t 1111 [O2]t[O2]t = 4.2  10  7 M/s is how long it took for N 2 O 5 to disappear 1212 [N2O5]t[N2O5]t = 1414  [NO 2 ]  t This is our “X” what we are looking for, so get 4 out of there  10  7 M/s = Rate of NO x M/s

Chemical Kinetics Practice Exercise If the rate of decomposition of N 2 O 5 in the reaction 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) at a particular instant is 4.2  10  7 M/s, what is the rate of appearance of (a) NO 2 and (b) O 2 at that instant? Sample Exercise 14.2 Relating Rates at Which Products Appear and Reactants Disappear Rate = − 1212 [N2O5]t[N2O5]t = 1414  [NO 2 ]  t 1111 [O2]t[O2]t = 4.2  10  7 M/s is how long it took for N 2 O 5 to disappear 1212 [N2O5]t[N2O5]t = 1111 [O2]t[O2]t This is our “X” what we are looking for, so get 1 out of there  10  7 M/s = Rate of O x M/s

Chemical Kinetics The isomerization of methyl isonitrile, CH 3 NC, to acetonitrile, CH 3 CN, 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. Time (s)[CH 3 NC] (M)Average rate M/s , , , , , M – M 2000 s – 0 s = 2.7 x M/s 2.7 x M/s

Chemical Kinetics The isomerization of methyl isonitrile, CH 3 NC, to acetonitrile, CH 3 CN, 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. Time (s)[CH 3 NC] (M)Average rate M/s , , , , , x M/s 1.7 x M/s 9 x M/s 4 x M/s 2 x M/s

Chemical Kinetics For each of the following gas-phase reactions, write the rate of reaction expression. a) H 2 O 2 (g)  H 2 (g) + O 2 (g) b) 2N 2 O (g)  2N 2 (g) + O 2 (g) c) N 2 (g) + 3H 2 (g)  2NH 3 (g) Rate = − 1111  [H 2 O 2 ]  t = 1111  [H 2 ]  t 1111  [O 2 ]  t = Rate = − 1212  [N 2 O 2 ]  t = 1212  [N 2 ]  t 1111  [O 2 ]  t = Rate = − 1111  [N 2 ]  t = −= − 1313  [H 2 ]  t = 1212  [NH 3 ]  t

Chemical Kinetics For each of the following gas-phase reactions, write the rate of reaction expression. a) 2HBr (g)  H 2 (g) + Br 2 (g) b) 2SO 2 (g) + O 2 (g)  2SO 3 (g) c) C 4 H 8 (g)  2C 2 H 4 (g) Rate = − 1212  [HBr]  t = 1111  [H 2 ]  t 1111  [Br 2 ]  t = Rate = − 1212  [SO 2 ]  t = −= − 1111  [O 2 ]  t = 1212  [SO 3 ]  t Rate = − 1111  [C 4 H 8 ]  t = −= − 1212  [C 2 H 4 ]  t

Chemical Kinetics a) Consider the combustion of H 2 (g): 2H 2 (g) + O 2 (g)  2H 2 O(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 = − 1212  [H 2 ]  t = −= − 1111  [O 2 ]  t = 1212  [H 2 O]  t 4.6 mol/s  [H 2 O]  t This is our “X” what we are looking for, so get 2 out of there 4.6 mol/s

Chemical Kinetics