Chapters 14 and 15

Chapter 14 Reaction Rates

Factors That Affect Reaction Rates Physical state Concentration Temperature Presence of a catalyst Reaction Rates are expressed as change in concentration per unit time. 1.00 mol A, 0.00 mol B 0.54 mol A, 0.46 mol B 0.30 mol A, 0.70 mol B Rate can be expressed as the rate of disappearance of A or the rate of appearance of B Calculate rates from t = 0 to t = 20 and t = 20 to t = 40 it is typical for rates to decrease over time.

Instantaneous Rate

Reaction Rates and Stoichiometry 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)? If the rate at which O2 appears is 6.0 x 10-5 M/s, at a particular instant, at what rate is O3 disappearing at this time? 2 HI  H2 + I2 general reaction: aA + bB  cC + dD Homework: 14.2, 14.6, 14.20, 14.22

The Effect of Concentration on Rate NH4+(aq) + NO2-(aq)  N2(g) + 2 H2O(l) Experiment Number Initial [NH4+] Initial [NO2-] Observed Rate 1 0.010 0.20 5.4 x 10-7 2 0.020 10.8 x 10-7 3 0.040 21.5 x 10-7 4 0.0202 5 0.0404 21.6 x 10-7 6 0.0808 43.3 x 10-7 Rate = k[NH4+][NO2-] aA + bB  cC + dD Rate = k[A]m[B]n Reaction order Units of the rate constant “k”

Using Initial Rates to Determine Rate Laws Experiment Number [A] (M) [B] (M) Initial Rate (M/s) 1 0.100 4.0 x 10-5 2 0.200 3 16.0 x 10-5 Sample exercise 14.6 p 584 Homework: 14.26, 14.30, 14.32

First Order Reactions The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr-1 at 12oC. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the average temperature of the lake is 12oC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to decrease to 3.0 x 10-7 g/cm3? First order reactions are ones whose rate law depends on the concentration of a single reactant (THIS DOES NOT MEAN THERE IS ONLY ONE REACTANT!) Differential rate law Integrated Rate Law: ln[A]t – ln[A]0 = -kt (rearrange)

Second Order Reactions 2nd order reactions depend on a single reactant raised to the second power or on two 1st order reactants. Integrated rate law p. 588. How to determine if a reaction is 1st or second order by graphing.

2nd Order Reactions vs 1st Order

Half Life Half Life (t1/2) is the time required to reach ½[A]0 equation for half life of first order reaction on page 589 and info sheet equation for half life of second order reaction on page 591 Homework: 14.38, 14.40, 14.42

Temperature and Rate For most chemical reactions increasing the temperature increases the rate. Collision model Orientation Factor Acitivation Energy: Activated complex/Transition state Lower Activation Energy means faster rate of reaction: Sample exercise 14.10 on page 595 Arrhenius Equation: Natural log of both sides results in linear equation, that plots 1/T(x) vs lnk (y) and the slope is –Ea/R

Determination of Activation Energy Temp (K) k(s-1) 1/T lnk 462.9 2.52 x 10-5 2.16 x 10-3 -10.589 472.1 5.25 x 10-5 2.11 x 10-3 -9.855 503.5 6.30 x 10-4 1.986 x 10-3 -7.370 524.4 3.16 x 10-3 1.907 x 10-3 -5.757 Catalysts lower Ea and speed up reactions.

Reaction Mechanism Reaction Mechanism: The process by which a reaction occurs. Elementary Reaction: Happens in one step. Unimolecular Bimolecular Termolecular Rate laws of elementary reactions: Table 14.3 on page 600 Multistep mechanisms: determine overall reaction and intermediates Rate determining step: Slow first step: Example on page 601 Fast initial step: Example on page 603 Homework: 14.52, 14.58, 14.68, 14.100

Example Problems The reaction between nitrogen monoxide and oxygen is shown below. 2 NO(g) + O2(g)  2 NO2(g) One proposed mechanism is the following: Step I: NO(g) + O2(g)  NO2(g) + O(g) (Slow) Step II: NO(g) + O(g)  NO2(g) (Fast) Which of the following rate expressions agrees best with this possible mechanism? Rate = k[NO][O] Rate = k[NO][O2] Rate = k[NO]2[O2] Rate = k[NO]/[O2] Rate = k[NO]/[O] Answer: B

C6H6(l) + 15/2 O2(g)  6 CO2(g) + 3 H2O(l) The above reaction represents the combustion of benzene. If a sample of benzene is burning at 0.25 mol/L*s, which of the following is the rate at which CO2(g) is being produced. 0.25 mol/L*s 0.75 mol/L*s 1.5 mol/L*s 2.5 mol/L*s 3.0 mol/L*s Answer: C

Which of the following correctly describes the reaction? 2 N2O5(g)  4 NO2(g) + O2(g) A sample of N2O5 was placed in an evacuated container, and the reaction represented above occurred. The partial pressure of N2O5(g), was measured during the reaction and recorded in the table. Which of the following correctly describes the reaction? Zero-order reaction 1st order reaction 2nd order reaction 3rd order reaction T (min) PN2O5 ln(PN2O5) 1/PN2O5 150 5.0 0.0067 100 75 4.3 0.013 200 38 3.6 0.027 300 19 2.9 0.053 Answer B

Chapter 15 Equilibrium

Equilibrium N2O4  2 NO2 Rate laws are equal, rearrange to solve for rate constants. At equilibrium concentrations do not change but are not necessarily equal For equilibrium to occur neither reactants nor products can be allowed to escape.

N2 + 3 H2  2 NH3 k = [NH3]2/[N2][H2]3 aA + bB -> cC + dD

N2O4(g)  2 NO2 Experiment Initial [N2O4] Initial [NO2] Equilibrium [N2O4] Equilibrium [NO2] kc 1 0.0 0.02 0.0014 0.0172 0.211 2 0.03 0.0028 0.0243 3 0.04 0.00452 0.031 4 All kc values calculate to 0.211

Equilibrium Constants and Pressure The synthesis of ammonia from hydrogen and nitrogen has a kc of 9.6 at 300oC. Calculate kp for this reaction at this temperature Equilibrium constants of reactions where all components are in the gaseous state can be calculated in terms of the partial pressures of each gas. Kc is not always equal to Kp Problem answer: 4.34 x 10-3

Using Equilibrium Constants CO(g) + Cl2(g)  COCl2(g) kc = 4.56 x 109 How does the numerical value of kc influence the equilibrium concentrations? The kc for a reverse reaction is the inverse of the kc for the forward reaction. Hess’s law applications on page 638.

Heterogeneous Equalibria Equilibria where the components are NOT all in the same physical state. Solids and pure liquids are left out of kc expressions. PbCl2(s)  Pb2+(aq) + Cl-(aq) CaCO3(s)  CaO(s) + CO2(g) Homework; 15.2, 15.12, 15.14, 15.16, 15.22

Calculating the Value of kc When all equilibrium concentrations are known. Sample exercise 15.8 on page 642 When all initial concentrations are known but not all equilibrium concentrations are known. Sample exercise 15.9 on page 643.

Applications of Equilibrium Constants Example on page 644. Reaction Quotient is used to predict the direction the reaction will go to reach equilibrium. Sample Exercise 15.10 on page 645 Calculating equilibrium concentrations sample exercise 15.11 p. 646 Calculating equilibrium concentrations from initial concentrations only. sample exercise 15.12 on page 647

Le Chatelier’s Principle When an equilibrium system is disturbed it will move in the direction necessary to reestablish equilibrium. Removing or adding a component causes the equilibrium to shift to compensate.

Effects of Volume and Pressure Changes Reducing the volume of a gaseous system at equilibrium will cause a shift in the direction that reduces the total number of moles of particles. Example: N2O4(g)  2 NO2 Example problem on page 651.

Effect of Temperature Change In an endothermic reaction we treat heat as a reactant, thus adding heat (increasing temp) results in a shift towards products. In an exothermic reaction we treat heat as a product, thus adding heat (increasing temp) results in a shift towards reactants. Show Co(H2O)62+ pink/blue equilibrium demo. Sample exercises 15.13 and 15.14 on page 653 Homework: 15.28, 15.32, 15.38, 15.42, 15.52

Example Questions Iodine and bromine are combined at high pressure and high temperature and react as follows: I2 + Br2  2 IBr The equilibrium constant for the above reaction is 280 at 250oC. If the equilibrium concentrations of IBr and Br2 are 1.3 M and 0.23 M respectively, what is the equilibrium concentration of I2 at 250oC? 1.3 x 10-3 M 8.4 x 10-3 M 1.2 x 10-2 M 2.6 x 10-2 M 6.7 x 10-2 M Answer: D

More H2(g) and I2(g) will form More HI(g) will form H2(g) + I2(g)  2 HI At 450oC, 2.0 moles each of H2(g), I2(g) and HI(g) are combined in a 1.0 L rigid container. The value of Kc at 450oC is 50. Which of the following will occur as the system moves toward equilibrium? More H2(g) and I2(g) will form More HI(g) will form The total pressure will decrease No net reaction will occur, because the number of molecules is the same on both sides of the equation. Answer: B

The following components are at equilibrium in a 1 L flask at 25oC: P4(g) + 5 O2(g)  P4O10 If 0.25 mol of gaseous chlorine is added to the flask, which of the following is true? [P4] will increase [P4O10] will increase [O2] will decrease I only II only I and II only II and III only I, II and III answer: D

COCl2(g)  CO(g) + Cl2(g) COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at 690 K, the pressure in the flask is initially 1.0 atm. After the reaction reaches equilibrium at 690 K, the total pressure in the flask is 1.2 atm. What is the value of Kp for the reaction at 690 K? 0.040 0.050 0.80 1.0 Answer: B

SO2Cl2(g)  SO2(g) + Cl2(g) A 4.32 g sample of liquid SO2Cl2 is placed in a ridged, evacuated 1.50 L reaction vessel. As the container is heated to 400 K, the sample vaporizes completely and starts decomposing according to the equation below. The decomposition reaction is endothermic. SO2Cl2(g)  SO2(g) + Cl2(g) If no decomposition occurred, what would be the pressure, in atm, of the SO2Cl2 in the vessel at 400 K. When the system has reached equilibrium at 400 K, the total pressure in the container is 1.26 atm. Calculate the partial pressures, in atm, of SO2Cl(g), SO2(g) and Cl2(g) in the container at 400 K. For the decomposition reaction at 400 K, Write the equilibrium expression for Kp for the reaciont and Calculate the value of the equilibrium constant Kp The temperature of the equilibrium mixture is increased to 425 K. Will the value of Kp increase, decrease or remain the same? Justify your prediction. In another experiment, the original partial pressuresof SO2Cl2(g), SO2(g), and Cl2(g) are 1.0 atm each at 400 K. Predict whether the amount of SO2Cl2(g) in the container will increase, decrease or remain the same. Justify your prediction.

Step 2: NO + NOBr2  2 NOBr Fast 2 NO(g) + Br2(g)  2 NOBr(g) NO reacts with Br2 according to the equation above. An experiment was performed to study the rate of the reaction at 546 K. Data from three trials are shown in the table below. Using the data above determine the order of the reaction with respect to both NO(g) and Br2(g). Write the rate law for the reaction. Determine the value of the rate constant, k, for the reaction, include its units. At a later time during trial 2, the concentration of Br2(g) is determined to be 0.16 M. Calculate the concentration of NO(g) at that time. Calculate the rate of consumption of Br2 at that time. A two step mechanism is proposed for this reaction. Step 1: NO + Br2  NOBr2 Slow Step 2: NO + NOBr2  2 NOBr Fast is the proposed mechanism consistent with the rate law determined in part (b)? Justify your answer