Chapter 6 Principles of Reactivity: Energy and Chemical Reactions - PowerPoint PPT Presentation

chapter 6 principles of reactivity energy and chemical reactions n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 6 Principles of Reactivity: Energy and Chemical Reactions PowerPoint Presentation
Download Presentation
Chapter 6 Principles of Reactivity: Energy and Chemical Reactions

play fullscreen
1 / 22
Chapter 6 Principles of Reactivity: Energy and Chemical Reactions
178 Views
Download Presentation
keilah
Download Presentation

Chapter 6 Principles of Reactivity: Energy and Chemical Reactions

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chapter 6Principles of Reactivity: Energy and Chemical Reactions Read/Study:Chapter 6 in e-Textbook! Learn Key Definitions:Class Lecture Notes OWL Assignments: Chapter 6 OWL Quiz for Chapter 6: NONE!

  2. 2 1. INTRODUCTION CHEMISTRY -The study of the properties, composition, and structure of matter, the physical and chemical changes it undergoes, and theENERGYliberated or absorbed during those changes. THERMODYNAMICS -Derived from the Greek words for heat and power; it is the study of all forms of energy and the interconversions among the different forms. THERMOCHEMISTRY -The study of the energy liberated (released) or absorbed by chemical or physical changes of matter.

  3. SYSTEM AND SURROUNDINGS - 1)System -The part of the universe a scientist is interested in. 2)Surroundings -Everything else in the universe that is outside of the system. 2 • 3)Boundary -A real or imaginary barrier between • the system and its surroundings through which • THERMAL ENERGYmay flow, work may • appear or disappear, and matter may or may • not be exchanged. • 4)Closed System -A system that does not • exchange matter with its surroundings.

  4. 5)Open System -A system that mayexchangeTHERMAL ENERGYandMATTER with its surroundings. 6)Isolated System -A system that does not exchangeMATTERorTHERMAL ENERGYwith its surroundings. • 2. CHANGE - WHY DOES IT HAPPEN? • A. SPONTANEOUS CHANGE -A change that • takes place by itself. • B. NON-SPONTANEOUS CHANGE -The • opposite of a spontaneous change.

  5. C. Factors Affecting Change - • 1)Energy -The ability to do work. • a) Thermal • b) Electrical • c) Radiant • d) Chemical • e) Mechanical • f) Nuclear • g) Kinetic • h) Potential • 2)Entropy -A measure ofdisorder. • 3)The Rate of Change -(Kinetics)

  6. D. Predicting Change -Requires a knowledge of • how energy and entropy of a system will change • as a result of the change and whether the change • will take place at a practical speed. • 3. Temperature, Thermal Energy, Heat • A. Temperature -An indirect measure of the • average kinetic energy of the molecules, atoms, • or ions in the material; A measure of the “hot- • ness” or “coldness” of a material; anINTENSIVE • property of matter. • B. Thermal Energy -The energy that is transferred • from hotter objects to colder objects due to the • kinetic energy of the molecules, atoms, or ions; • anEXTENSIVEproperty of matter.

  7. C. Heat -The transfer of thermal energy that • results from a difference in temperature; the • process of transferring thermal energy from • hotter objects to colder objects. • D. Potential Energy - Stored • energy that is a related to an • objects relative position. • E. Kinetic Energy -The energy of • motion. • K.E. = ½ mv2 K.E. = ½(2 kg)(1 m/s)2 = 1kg-m2/s2 = 1 Joule

  8. 4. The First Law of Thermodynamics • Energy is neither created nor destroyed. • Energy given up bythe systemmust be absorbed • bythe surroundings. This type of change is • EXOTHERMIC. • Energy absorbed bythe systemmust come from • the surroundings. This type of change isENDO- • THERMIC. • A. Internal Energy, E -The total energy of a • system. D E = q + w = Efinal - Einitial

  9. Constant Volume Calorimetry • E = q + w (w = 0 at constant V)

  10. B. Enthalpy, H -The Thermal Energy gained • or lost by a system when the system under- • goes a change under constant pressure. DH = Hfinal - Hinitial We can only measure The Change in enthalpy, not the absolute enthalpy. Enthalpy is a state function. Exothermic Change - Changes during which the system gives off thermal energy and DH< 0 (negative). Endothermic Change - Changes during which the system absorbs thermal energy and DH> 0 (positive).

  11. D H Constant Pressure Calorimetry

  12. Examples ofExothermicChanges - H2O (l) H2O (s) DH < 0 H2O (l) This is an exothermic process because the finalenergy state is lower than the initial energy state. DH = - H2O (s) 2H2 (g) + O2 (g) 2 H2O (l) DH < 0 2 H2 (g) + O2 (g) Water is lower in enthalpy than hydrogen and oxygen are. Exothermic DH = - 2 H2O (l)

  13. Examples ofEndothermicChanges - H2O (l) H2O (g) DH > 0 H2O (g) This is Endothermicbecause the final state is higher in enthalpy than the initial state. DH = + H2O (l) CaCO3 (s)  CaO (s) + CO2 (g) DH > 0 CaO (s) + CO2 (g) CaO (s) and CO2 (g) are higher in enthalpy than CaCO3 (s). DH = + CaCO3 (s)

  14. Class Exercises - Endothermic or Exothermic ? 3 H2 (g) + N2 (g) 2 NH3 (g) DH = - 46.1 kJ/mol N2 (g) + 2 O2 (g) 2 NO2 (g) DH = + 33.2 kJ/mol C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Sign of DH ?? Negative NH4Cl dissolves in water with a decrease in temperature. Sign of DH ?? Positive

  15. Thermochemical Equations - A chemical equa- tion that includes an enthalpy change explicitly. 2 H2 (g, 1 atm) + O2 (g, 1 atm) 2 H2O (l) H298 K = - 571.7 kJ • A. H subscript indicates temperature of rxn. • B. H represents thermal energy evolved when • 2 moles of H2O (l) are formed! H2 (g, 1 atm) + 1/2 O2 (g, 1 atm) H2O (l) H298 K = - 285.8 kJ • C. The sign of H is reversed when the equation • is reversed!

  16. H2O (l)  H2 (g, 1 atm) + 1/2 O2 (g, 1 atm) H298 K = + 285.8 kJ D. The enthalpy change during a reaction can be considered as a reactant (endothermic) or as a product (exothermic). H2 (g, 1 atm) + 1/2 O2 (g, 1 atm)  H2O (l) + 285.8 kJ 285.8 kJ + H2O (l)  H2 (g, 1 atm) + 1/2 O2 (g, 1 atm)

  17. Practice Problem:When 2.0 moles of isooctane are burned, 10 930.9 kJ of thermal energy are liberated under constant temperature conditions.How many kJ will be liberated when 369 g of isooctane are burned? Problem: 369 g Equation: 2 C8H18 (l) + 25 O2 (g)  16 CO2 (g) + 18 H2O (l) D H = -10 930.9 kJ Molar Masses: 114.231 ? kJ “When in doubt, calculate MOLES!” (369 g C8H18)(1 mol C8H18/114.231 g C8H18) = 3.230 mol C8H18

  18. (3.230 mol C8H18)(-10 930.9 kJ/2 mol C8H18) = - 1.77 x 104 kJ Practice Problem:The DHrxn for the burning of H2 (g) to form H2O (l) is –285.83 kJ/mol H2O. How many grams of H2 (g) are needed to produce 539.63 kJ of thermal energy? Problem: ? g - 539.63 kJ Equation: H2 (g) + ½ O2 (g)  H2O (l) DH = -285.83 kJ Molar Masses: 2.015 88 u

  19. (- 539.63 kJ)(1 mol H2/-285.83 kJ) = 1.887 94 mol H2 (1.887 94 mol H2)(2. 015 88 g H2/mol H2) = 3.8059 g H2 Molar Mass Avogadro’s Number MASS MOLES PARTICLES Molarity PV = nRT Molar Heat of Reaction VOLUME P, V, T Thermal Energy

  20. Standard Enthalpy Changes - A standard enthalpy change, DH0, is the enthalpy change for a reaction in which each reactant and each product is in its “Standard State”. • Solids - Pure solid at 1 atm • Liquids – Pure liquid at 1 atm • Gas – Ideal gas at 1 atm partial pressure • Solute – Ideal solution at 1 M conc. H2 (g, 0.5 atm) + Cl2 (g, 0.5 atm)  2 HCl (g, 1 atm) D Hrxn, NOT DHorxn

  21. Standard Enthalpy of Formation - DHf The enthalpy change accompanying the formation of one mole of a substance in its standard state from its elements, each in their standard states and most stable form. Hg (l) + Cl2 (g)  HgCl2 (s) DHfo = -221.3 kJ S (s) + O2 (g)  SO2 (g) DHfo = -296.8 kJ H2 (g)  H2 (g) DHfo = 0 kJ Elements in their standard states have zero enthalpies of formation!

  22. Standard Enthalpy of Combustion - DHcombThe enthalpy change accompanying the combustion of one mole of a substance in O2 in its standard state from its elements, each in their standard states and most stable form. H2 (g, 1 atm) + ½ O2 (g, 1 atm)  H2O (l, 1 atm) DHcombo = -285.8 kJ C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (l) DHcombo = -5455.6 kJ C2H5OH (l) + 7/2 O2 (g)  2 CO2 (g) = 3 H2O (l) DHcombo = -1366.8 kJ