CHAPTER 16: REACTION ENERGY AND CHAPTER 17: REACTION KINETICS. Honors Chemistry Ms. Agostine

Transcription

1 CHAPTER 16: REACTION ENERGY AND CHAPTER 17: REACTION KINETICS Honors Chemistry Ms. Agostine

2 16.1 Thermochemistry Definition: study of the transfers of energy as heat that accompany chemical reactions and physical changes Measured by a calorimeter

3 Measuring Heat (Review) Units that measure heat/energy: CALORIES! Calorie (cal): the quantity of heat that raises the temperature of 1 g of pure water by 1 o C 1 Calorie = 1000 calories = 1 kilocalorie SI Units: joule (J) Conversion: 1 J = cal or 1 cal = J

4 Specific Heat (c) (Review) How the quantity of heat is actually measured Specific Heat (c) amount of energy required to raise the heat 1 gram of a substance by 1 degree Celsius q = m c T (heat quantity) = (mass)(spec. heat)(temp change) or rearranged to be: c = q m T

5 Specific Heat Water has a high specific heat and metals have low specific heat Therefore it takes less energy to heat up a metal then a nonmetal like water! That is why the beach is cooler in the summer then the city!

6 Heat of Reaction Definition: Quantity of heat released or absorbed during a chemical reaction Enthalpy (H): heat content of a system Enthalpy Change ( H): amount of heat absorbed or lost by a system during a process Can have Exothermic and Endothermic reactions

7 Enthalpy of Reaction Thermochemical Equation : equation that includes the quantity of heat released or absorbed during the reaction as written Depends on the amounts of reactants and products: H 2 (g) + ½ O 2 (g) H 2 O (g) kJ

8 Exothermic Reactions Energy is released to surroundings Products have LESS energy then reactants H is negative 2 H 2 (g) + O 2 (g) 2 H 2 O (g) H = kj

9 Endothermic Reactions Energy is absorbed from surroundings Products have MORE energy then reactants H is positive 2 H 2 O (g) 2 H 2 (g) + O 2 (g) H = kj

10 Standard Molar Heat of Formation ( H fo ) Definition: heat released or absorbed when one mole of a compound is formed by combination of its elements at standard room temperature of 25 o C Appendix Table A-14 p. 862 H 2 (g) + ½ O 2 (g) H 2 O (l) H o f = kj Na (s) + ½ Cl 2 (g) NaCl (s) H o f = kj S (s) + O 2 (g) SO 2 (g) H o f = kj Elements in their standard states have H o f = 0 kj A negative H o f indicates a substance is more stable than the free elements

11 Standard Heats of Formation

12 Molar Heat of Combustion ( H comb ) Definition: heat released when one mole of a compound combusts Appendix Table A-5, p. 856 C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H comb = kj C 8 H 18 (g) O 2 (g) 8 CO 2 (g) + 9 H 2 O (l) H comb = kj

13 Calculating Heats of Reaction Hess s Law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process In order for a reaction to occur: all reactants must have their bonds broken all products must have their bonds formed

14 Calculating Heats of Reactions Hess s Law of Heat Summation: If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction H o = H f o products - H f o reactants ( = sum) This will also show if the net reaction is exothermic (-) or endothermic (+).

15 Sample Problem What is the H f o for the reaction between gaseous carbon monoxide with oxygen to form gaseous carbon dioxide? Balanced equation: 2 CO (g) + O 2 (g) 2 CO 2 (g) LOOK UP H f o CO (g) = kj/mol H f o O 2 (g) = 0 kj/mol (free element) H f o CO 2 (g) = kj/mol

16 2 CO (g) + O 2 (g) 2 CO 2 (g) H o = H f o products - H f o reactants H o = [2 mol x kj/mol] [2 mol x kj/mol] H o = [-791 kj] [-221 kj] H o = kj Therefore, this reaction is an exothermic process!

17 Calculating H Reaction from the Heats of Formation Reactions PbCl 2 + Cl 2 PbCl 4, H o =? Look up: H f o PbCl 2 = kj/mol Rxn of the formation: Pb + Cl 2 PbCl 2 H f o Cl 2 = 0 kj/mol b/c it s an element H f o PbCl 4 = kj/mol Rxn of the formation: Pb + 2 Cl 2 PbCl 4 In the reaction: PbCl 2 has to be broken, so reverse the reaction PbCl 4 has to be made, keep the reaction as written

18 Calculating H Reaction from the Heats of Formation Reactions H f o PbCl 2 = kj/mol FLIP IT! Rxn: PbCl 2 Pb + Cl 2 H f o Cl 2 = 0 kj/mol b/c it s an element H f o PbCl 4 = kj/mol Rxn: Pb + 2 Cl 2 PbCl 4 Complete: PbCl 2 + Pb + 2 Cl 2 Pb + Cl 2 + PbCl 4 Simplify: PbCl 2 + Pb + 2 Cl 2 Pb + Cl 2 + PbCl 4 Net: PbCl 2 + Cl 2 PbCl 4 Add H f o = kj kj = 30.2 kj

19 Calculating H Reaction from the Heats of Formation Reactions Check Answer with: H o = H f o products - H f o reactants H o = [1 mol x kj/mol] [1 mol x kj/mol] H o = [ kj] [ kj] H o = kj Therefore, this reaction is an endothermic process!

20 Calculating H Reaction from the Heats of Formation Reactions Calculate the heat of the reaction for the decomposition of calcium carbonate when heated CaCO 3 (s) CaO (s) + CO(g) H o =?

21 Solution Look up: H f o CaCO 3 = kj/mol Rxn : Ca + C + 3/2 O 2 CaCO 3 H f o CaO = kj/mol Rxn: Ca + ½ O 2 CaO H f o CO 2 = kj/mol Rxn: C + O 2 CO 2 In the reaction: CaCO 3 has to be broken, so reverse the rxn CaO & CO 2 has to be made, keep the rxns as written

22 Answer: Complete equation: CaCO 3 + Ca + C + 3/2 O 2 Ca + C + CaO + 3/2 O 2 + CO 2 x x x x x x Net Eqn: CaCO 3 (s) CaO (s) + CO(g) Add up heats of formation: CaCO 3 is reversed: kj CaO is the same: CO 2 is the same: TOTAL: kj kj kj, endothermic

23 Answer: Check your answer: H = products - reactants = [ kj kj] [-1207 kj] = kj

24 Section 16.2 Driving Force of Reactions What makes a reaction spontaneous? Ex: Ball rolls down a hill obvious, but why? Driving force: Potential energy is too high, so nature wants to lower it. SPONTANEOUS! Nature has a tendency of reactions to occur that lead to a lower energy state Therefore: ball will NOT roll up a hill!

25 Entropy (S) Definition: property that describes the order of a system Nature tends toward disorder Hint: easier to mess up or to clean up a room? We measure S in J/K mol Temperature can effect S

26 Free Energy Depends on two things: Enthalpy: H Entropy: S and the Temperature of the system Gibb s Free Energy: combined enthalpy-entropy of a system G = H - T S If G = (-) = rxn will be favorable/spontaneous If If G = (+) = rxn will be unfavorable/nonspontaneous Note: G o, H o, & S o = at standard temperature

27 Enthalpy, Entropy & Free Energy H S G - Value (exothermic) - Value (exothermic) + Value (endothermic) + Value (endothermic) + Value (more random) - Value (less random) + Value (more random) - Value (less random) Always negative lower temperatures higher temps Never negative

28 Free Energy Sample Problem For the reaction NH 4 Cl (s) NH 3 (g) + HCl (g), at K, H o = 176 kj/mol and S o = kj/(mol K). Calculate G o, and tell whether this reaction is spontaneous in the forward direction at this temperature.

29 Free Energy Sample Problem G = H - T S G = 176 kj/mol [298 K x kj/(mol K)] G = 176 kj/mol 84.9 kj/mol G = 91 kj/mol Non spontaneous because G = positive!

30 Chapter 17: Reaction Kinetics

31 Collision Theory

32 Section 17.1 The Reaction Process Reaction Mechanism: step-by-step sequence of reactions by which the overall chemical change occurs Why does the rate of chemical reactions vary so widely? Collision Theory: Reactant molecules must collide in the proper orientation and with sufficient energy Only a fraction of collisions meet these requirements

33 Collision Theory Consider: Ozone + smog nitrogen dioxide + oxygen

34 Collisions of NO and O3

35 Energy Diagrams Reactants must overcome an energy barrier before they can change to products Activation Energy (E a ): Minimum amount of energy that reactants must have before they can be converted to products

36 Energy Diagrams

37 Energy Diagrams Activated Complex: An unstable, high-energy chemical species that must be formed as reactants convert to products The in-between stage. Designated with a double superscript dagger ( ) such as {O 3 NO}

38 Energy Diagram for NO and O3

39 Exothermic vs. Endothermic

40 Energy Diagrams with Activated Complexes Example: The following is exothermic: 2 HI (g) H 2 (g) + I 2 (g) Draw an energy diagram that shows the relative energies of the reactants, products and the activated complex. Label the diagram with molecular representations of the reactants, products, and a possible structure for the activated complex.

41 Section Reaction Rates What makes milk and other foods spoil more quickly? What happens if you accidentally add a drop of bleach to a dark wash? A cup? What dissolves faster granulated sugar or a sugar cube?

42 Reaction Rates Factors that influence reaction rates 1. Temperature 2. Reaction concentration 3. Surface area 4. Presence of a catalyst Looking at each in the terms of Collision Theory!

43 Concentration Rxns go faster when the concentration of one or more of the reactants increases Increases in concentration increases the number of reactants per unit volume Because molecules are closer together, the number of collisions per unit time increases

44 Surface Area Similar to increase concentration More surface area means more molecules or atoms exposed to the reaction collisions Ex) granulated sugar vs. sugar cubes

45 Temperature The average kinetic energy of a substance increases when the temperature rises Increases the fraction of collisions that are effective Molecules move faster at higher temperatures, so they collide more frequently More molecules attain the required activation energy

46 Temperature Temperature effects on reaction rates are significant For a typical reaction, the rate approximately doubles for every 10 o C increase in temperature

47 Fraction of Collisions vs. Collision Different Temperatures

48 Catalysts Catalyst: A substance that alters the pathway in which a reaction occurs without itself being consumed in the reaction Lower-energy pathway with lower activation energy This way, increases the rate of a reaction Great fraction of reactants can achieve the new minimum energy requirement

49 Catalyst

50 Rate Laws for Reactions Rate Law: an equation that relates reaction rate and concentrations of reactants Specific Rate Laws for the directions of reactions Forward rate law is different then the reverse reaction rate law

51 Rate Laws for Reactions 2 H 2 (g) + 2 NO (g) N 2 (g) + 2 H 2 O (g) Rate Law of the forward reaction: Depends on concentrations of H 2 and NO Depends on temperature R = k [reactants] coeffient R = reaction rate K = rate constant (takes into account temperature) R = k [H 2 ] 2 [NO] 2