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

2 1. INTRODUCTION CHEMISTRY - ENERGY CHEMISTRY - The study of the properties, composition, and structure of matter, the physical and chemical changes it undergoes, and the ENERGY liberated or absorbed during those changes. THERMODYNAMICS - 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 - THERMOCHEMISTRY - The study of the energy liberated (released) or absorbed by chemical or physical changes of matter. 2

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

4 Open System - THERMAL ENERGYMATTER 5) Open System - A system that may exchange THERMAL ENERGY and MATTER with its surroundings. Isolated System - MATTERTHERMAL ENERGY 6) Isolated System - A system that does not exchange MATTER or THERMAL ENERGY with its surroundings. 2. CHANGE - WHY DOES IT HAPPEN? A. SPONTANEOUS CHANGE - A. SPONTANEOUS CHANGE - A change that takes place by itself. B. NON-SPONTANEOUS CHANGE - B. NON-SPONTANEOUS CHANGE - The opposite of a spontaneous change.

5 C. Factors Affecting Change - 1) E nergy - The ability to do work. a) Thermal b) Electrical c) Radiant d) Chemical e) Mechanical f) Nuclear g) Kinetic h) Potential 2) ntropy - A measure of d dd disorder. 3) T he Rate of Change - (Kinetics)

6 D. Predicting Change - 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 - A. Temperature - An indirect measure of the average kinetic energy of the molecules, atoms, or ions in the material; A measure of the “hot- INTENSIVE ness” or “coldness” of a material; an INTENSIVE property of matter. B. Thermal Energy - 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; EXTENSIVE an EXTENSIVE property of matter.

7 C. Heat - 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 - D. Potential Energy - Stored energy that is a related to an objects relative position. E. Kinetic Energy - E. Kinetic Energy - The energy of motion. K.E. = ½ mv 2 K.E. = ½(2 kg)(1 m/s) 2 m/s) 2 = 1kg-m 2 /s 2 = 1 Joule

8 4. The First Law of Thermodynamics Energy is neither created nor destroyed. the system Energy given up by the system must be absorbed the surroundings. by the surroundings. This type of change is EXOTHERMIC. the system Energy absorbed by the system must come from the surroundings. ENDO- the surroundings. This type of change is ENDO- THERMIC. A. Internal Energy, E - A. Internal Energy, E - The total energy of a system. E = q + w= E final E final - E initial

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.   final  final - H initial 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  negative). Endothermic Change - Changes during which the system absorbs thermal energy and  positive).

11 Constant Pressure Calorimetry  H H H H

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

13 Examples of Endothermic Changes - H 2 O (l) H 2 O (g)  H > 0 H 2 O H 2 O (g) (l)  H  H = + This is Endothermic because the final state is higher in enthalpy than the initial state. CaCO 3 CaCO 3 (s) CaO (s) + CO 2 CO 2 (g)  H  H > 0 CaO (s) and CO 2 (g) are higher in enthalpy than CaCO 3 (s). CaO (s) + CO 2 (g) CaCO 3 (s)  H = +

14 Class Exercises - Endothermic or Exothermic ? 3 H 2 (g) + N 2 (g)2 NH 3 (g)  H = kJ/mol N 2 (g) + 2 O 2 (g)2 NO 2 (g)  H = kJ/mol C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 CO 2 (g) +4 H 2 O H 2 O (g) Sign of  H ?? Negative NH 4 Cl dissolves in water with a decrease in temperature. Sign of  H ?? Positive

15 Thermochemical Equations - A chemical equa- tion that includes an enthalpy change explicitly. 2 H 2 (g, 1 atm) + O 2 (g, 1 atm) 2 H 2 O (l)  H 298 K = kJ  H 298 K = kJ A.  H subscript indicates temperature of rxn. B.  H represents thermal energy evolved when 2 moles of H 2 O (l) are formed! H 2 (g, 1 atm) + 1/2 O 2 (g, 1 atm) H 2 O (l)  H 298 K = kJ  H 298 K = kJ C. The sign of  H is reversed when the equation is reversed!

16 H 2 O (l)  H 2 (g, 1 atm) + 1/2 O 2 (g, 1 atm)  H 298 K = kJ  H 298 K = kJ D. The enthalpy change during a reaction can be considered as a reactant (endothermic) or as considered as a reactant (endothermic) or as a product (exothermic). a product (exothermic) kJ + H2O (l)  H2 (g, 1 atm) + 1/2 O2 (g, 1 atm) H2 (g, 1 atm) + 1/2 O2 (g, 1 atm)  H2O (l) kJ

17 Practice Problem: When 2.0 moles of isooctane are burned, 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 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O (l)  H = kJ  H = kJ Molar Masses: ? kJ “When in doubt, calculate MOLES!” (369 g C8H18)(1 mol C8H18/ g C8H18) = mol C8H18

18 ( mol C 8 H 18 )( kJ/2 mol C 8 H 18 ) = x 10 4 kJ Practice Problem: The  H rxn for the burning of H 2 (g) to form H 2 O (l) is – kJ/mol H 2 O. How many grams of H 2 (g) are needed to produce kJ of thermal energy? Problem: ? g kJ Equation: H2 (g) + ½ O2 (g)  H2O (l) H = kJ Molar Masses: u

19 ( kJ)(1 mol H2/ kJ) = mol H2 ( mol H2)( g H2/mol H2) = g H2 MOLESMASSPARTICLES VOLUME P, V, T Molar Mass Avogadro’s Number MolarityPV = nRT Thermal Energy Molar Heat of Reaction

20 Standard Enthalpy Changes - A standard enthalpy change,  H 0, 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. H 2 (g, 0.5 atm) + Cl 2 (g, 0.5 atm)  2 HCl (g, 1 atm)  H rxn, NOT  H o rxn

21 Standard Enthalpy of Formation -  f 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) + Cl 2 (g)  HgCl 2 (s)  f o = kJ S (s) + O 2 (g)  SO 2 (g)  f o = kJ H 2 (g)  H 2 (g)  H f o = 0 kJ Elements in their standard states have zero enthalpies of formation!

22 Standard Enthalpy of Combustion -  comb The enthalpy change accompanying the combustion of one mole of a substance in O 2 in its standard state from its elements, each in their standard states and most stable form. H 2 (g, 1 atm) + ½ O 2 (g, 1 atm)  H 2 O (l, 1 atm)  H comb o = kJ C 8 H 18 (l) + 25/2 O 2 (g)  8 CO 2 (g) + 9 H 2 O (l)  H comb o = kJ C 2 H 5 OH (l) + 7/2 O 2 (g)  2 CO 2 (g) = 3 H 2 O (l)  H comb o = kJ