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Chapter 2 Elements of Chemical Thermodynamics Silberberg: Ch 6 & 20Atkins: Ch6 & 7.

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Presentation on theme: "Chapter 2 Elements of Chemical Thermodynamics Silberberg: Ch 6 & 20Atkins: Ch6 & 7."— Presentation transcript:

1 Chapter 2 Elements of Chemical Thermodynamics Silberberg: Ch 6 & 20Atkins: Ch6 & 7

2 Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. Thermodynamics is the study of heat and its transformations. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings. The system is composed of particles with their own internal energies (U or E). Therefore the system has an internal energy. When a change occurs, the internal energy changes.

3 A chemical system and its surroundings. the system the surroundings

4 System & Surroundings System - the specific part of the universe that is of interest in the study. often the reacting vessel in the chemistry lab –A beaker of NaOH and HCl Surroundings - the rest of the universe –When heat is given off from the beaker, the energy is transferred out of the system

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6 open mass & energyExchange: closed energy isolated nothing

7 Exothermic & Endothermic Reactions Exothermic - transfer of heat from the system to the surroundings 2H 2 (g) + O 2 (g)  2H 2 O(l) + energy Endothermic - the transfer of heat from the surroundings to the system energy + 2HgO (s)  2Hg (l) + O 2(g)

8 Endothermic vs. Exothermic

9 Copyright McGraw-Hill 2009 Internal Energy Internal energy: 2 components Kinetic energy - molecular motion Potential energy - attractive/repulsive interactions

10 Copyright McGraw-Hill 2009 Internal Energy We cannot calculate the changes in internal energy with any certainty We can calculate the changes in energy of the system experimentally  U =  U (products) -  U (reactants)‏

11  U = U final - U initial = U products - U reactants Energy diagrams for the transfer of internal energy (U) between a system and its surroundings.

12  U universe =  U system +  U surroundings Units of Energy Joule (J) Calorie (cal) British Thermal Unit 1 cal = 4.18J 1 J = 1 kg*m 2 /s 2 1 Btu = 1055 J

13 Copyright McGraw-Hill 2009 First Law of Thermodynamics Energy can be converted from one form to another but cannot be created nor destroyed –(the Law of Energy Conservation)‏

14 Copyright McGraw-Hill 2009 Energy and the Universe  U system +  U surroundings = 0 When a system releases heat, some of the chemical energy is released as thermal energy to the surroundings but this does not change the total energy of the universe.

15 Copyright McGraw-Hill 2009 Energy and the Universe  U system = -  U surroundings When a system undergoes a change in energy, the surroundings must undergo a change in energy of equal magnitude but opposite in sign.

16 Copyright McGraw-Hill 2009 Work and Heat  U sys = q + w “q” = heat “w” = work

17 Pressure-volume work.

18 The Sign Conventions* for q, w and  U qw += UU depends on sizes of q and w * For q : + means system gains heat; - means system loses heat. * For w : + means work done on system; - means work done by system.

19 Copyright McGraw-Hill 2009 Work and Heat Calculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the system. “q” = + (heat absorbed) “w” = - (work done)  U sys = q + w = (+125 J) + (-141 J)‏ = -16 J

20 State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature  U = U final - U initial  P = P final - P initial  V = V final - V initial  T = T final - T initial

21 Copyright McGraw-Hill 2009 Constant Volume Constant volume: (  V = 0)  U = q - P  V q v =  U (not easy to achieve in the chem lab)

22 Copyright McGraw-Hill 2009 Constant Pressure Constant pressure: (common in the chem lab)‏  U = q + w  U = q p - P  V q p =  U + P  V

23 Copyright McGraw-Hill 2009 Enthalpy and Enthalpy Changes Enthalpy (H) is a state function Enthalpy measures heat content Constant pressure process: (most are)‏ q p =  H Enthalpy of reaction:  H = H(products) - H(reactants)  H = - (exothermic)  H = + (endothermic)

24 Copyright McGraw-Hill 2009 Thermochemical Equations Equations that represent mass and enthalpy relationships H 2 O (s)  H 2 O (l)  r H = kJ/mol This is an endothermic process. It requires 6.01 kJ to melt one mole of ice. (the “per mole” means per mole of reaction as written - respect stoichiometry)

25 Copyright McGraw-Hill 2009 Thermochemical Equation Guidelines Always specify state of reactants and products. When multiplying an equation by a factor (n), multiply the  H value by same factor. Reversing an equation changes the sign but not the magnitude of the  H.

26 Enthalpy and the First Law of Thermodynamics  U = q + w  U =  H - P  V  H =  U + P  V q =  H and w = -P  V At constant pressure:

27 The Meaning of Enthalpy w = - P  V  H =  U + P  V q p =  U + P  V =  H  H ≈  U in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> P  V. H = U + PV where H is enthalpy

28 Enthalpy diagrams for exothermic and endothermic processes. Enthalpy, H CH 4 + 2O 2 CO 2 + 2H 2 O H initial H final H 2 O( l ) H 2 O( g ) heat outheat in  H < 0  H > 0 A Exothermic processB Endothermic process CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O( g )H 2 O( l ) H 2 O( g )

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30 The standard enthalpy of reaction (  r H 0 ) is the enthalpy of a reaction carried out at 1 atm. aA + bB cC + dD rH0rH0 d  f H 0 (D) c  f H 0 (C) = [+] - b  f H 0 (B) a  f H 0 (A) [+] rH0rH0 n  f H 0 (products) =  m  f H 0 (reactants)  - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

31 Sample Problem 2.1 Using Hess’s Law to Calculate an Unknown  H SOLUTION: PLAN: PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO( g ) + NO( g ) CO 2 ( g ) + 1/2 N 2 ( g )  H = ? Given the following information, calculate the unknown  H: Equation A: CO( g ) + 1/2 O 2 ( g ) CO 2 ( g )  H A = kJ/mol Equation B: N 2 ( g ) + O 2 ( g ) 2NO( g )  H B = kJ/mol Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation. Multiply Equation B by 1/2 and reverse it.  H B = kJ/mol CO( g ) + 1/2 O 2 ( g ) CO 2 ( g )  H A = kJ/mol NO( g ) 1/2 N 2 ( g ) + 1/2 O 2 ( g )  H rxn = kJ/mol CO( g ) + NO( g ) CO 2 ( g ) + 1/2 N 2 ( g )

32 Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  f H 0 ) as a reference point for all enthalpy expressions. Standard enthalpy of formation (  f H  m ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero.  f H  m (O 2 ) = 0  f H 0 (O 3 ) = 142 kJ/mol  f H 0 (C, graphite) = 0  f H 0 (C, diamond) = 1.90 kJ/mol

33 Selected Standard Heats of Formation at 25 0 C(298K) Formula  f H  m (kJ/mol) calcium Ca( s ) CaO( s ) CaCO 3 ( s ) carbon C(graphite) C(diamond) CO( g ) CO 2 ( g ) CH 4 ( g ) CH 3 OH( l ) HCN( g ) CS s ( l ) chlorine Cl( g ) hydrogen nitrogen oxygen Formula  f H  m (kJ/mol) H( g ) H2(g)H2(g) N2(g)N2(g) NH 3 ( g ) NO( g ) O2(g)O2(g) O3(g)O3(g) H 2 O( g ) H 2 O( l ) Cl 2 ( g ) HCl( g ) Formula  f H  m (kJ/mol) silver Ag( s ) AgCl( s ) sodium Na( s ) Na( g ) NaCl( s ) sulfur S 8 (rhombic) S 8 (monoclinic) SO 2 ( g ) SO 3 ( g )

34 The general process for determining  r H 0 from  f H 0 values. Enthalpy, H Elements Reactants Products  r H 0 =  m  f H 0 (products) -  n  f H 0 (reactants) decomposition -fH0-fH0 fH0fH0 formation rH0rH0 H initial H final

35 Sample Problem 2.2Calculating the Heat of Reaction from Heats of Formation SOLUTION: PLAN: PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: Calculate  H 0 rxn from  H 0 f values. 4NH 3 ( g ) + 5O 2 ( g ) 4NO( g ) + 6H 2 O( g ) Look up the  H 0 f values and use Hess’s Law to find  H rxn.  H rxn =  m  f H 0 (products) -  n  f H 0 (reactants)  r H = [4(  f H 0 NO( g ) + 6(  f H 0 H 2 O( g )]- [4(  f H 0 NH 3 ( g ) + 5(  f H 0 O 2 ( g )] = (4)(90.3 kJ/mol) + (6)( kJ/mol) - [(4)(-45.9 kJ/mol) + (5)(0 kJ/mol)]  r H = -906 kJ/mol

36 Some Important Types of Enthalpy Change heat of combustion (  comb H) heat of formation (  f H) heat of fusion (  fus H) heat of vaporization (  vap H) C 4 H 10 ( l ) + 13/2 O 2 ( g ) 4CO 2 ( g ) + 5H 2 O( g ) K( s ) + 1/2 Br 2 ( l ) KBr( s ) NaCl( s ) NaCl( l ) C 6 H 6 ( l ) C 6 H 6 ( g )

37 Limitations of the First Law of Thermodynamics  U = q + w U universe = U system + U surroundings  U system = -  U surroundings The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.  U system +  U surroundings = 0 =  U universe


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