TODAY 0. Why H = q (if p ext =p=constant and no useful work) 1. Constant Pressure Heat Capacity (what we usually use)

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Cuthbert Miles
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1 361 Lec 7 Fri 9sep15 TODAY 0. Why H = q (if p ext =p=constant and no useful work) 1. Constant Pressure Heat Capacity (what we usually use) 2. Heats of Chemical Reactions: r H (mechanics of obtaining from the table of standard heats of formation ) 3. I can get r H from table, BUT, how to get r U??? 4. What exactly is f H 0? 5. Hesse s Law (adding/subtracting chemical reactions): an example of exploiting the fact that changes in state functions are independent of path. 1

2 Enthalpy: (defined because of the small work against atmosphere) H = U + pv by definition (always true) dh =? = du + d(pv) = H = U + (pv) = (not U + p V + V p) H = q (IF p ext = p = constant, and only p atmosphere V work done) because U = q + w = q - p atmosphere V + w useful and pv = p atmosphere V The annoying -p atmosphere V term cancels! du + pdv + Vdp U + p 2 V 2 p 1 V 1 H for an ideal gas? H = U + (pv) by definition pv = nrt (ideal gas) (pv) = (nrt) = nr T We know U = 0 if T = 0, so H also = 0 for ideal gas if isothermal process 2

Heat Capacity of Gases at Constant Pressure: C p 1 atm p = 1 atm What happens if we heat the gas while keeping p = p ext = constant = 1 bar? But, expansion does some work, which has a cooling effect.

3 Heat Capacity of Gases at Constant Pressure: C p 1 atm p = 1 atm What happens if we heat the gas while keeping p = p ext = constant = 1 bar? But, expansion does some work, which has a cooling effect. T is smaller C = q/ T, so C is larger. q Heating make the molecules move faster, making the pressure increase, which causes expansion. pv= nrt p V= nr T More heat will be needed to raise the temperature than if volume is kept constant. C p > C v, always 3

4 How much more heat is required? H = U + pv what is pv? pv = nrt if ideal gas but H = q (if p= p ext = constant and only p atmosphere V work possible) At constant Volume U = q (if no useful work done) q = C V T = U At constant Pressure H = q (if no useful work done) q = C P T = H= U + (pv) = U + (nrt) C P T = C V T + nr T C P = C V + nr 4

5 361 Lec 7 Fri, 6sep13 Very very high pressure applied to a solid will often turn it into another crystal form that is more dense, e.g., graphite into diamond. 5

6 Heats of CHEMICAL REACTIONS ( r H) (and Phase Changes treated the same way) State 1 State 2 Reactants Products Tabulated at 1 bar and some T (usually 298 K) (note that this is constant T and p) so, q = q p = H In general: if weak bonds STRONG BONDS then the reaction is very EXOTHERMIC == chemical energy r H = Σ H i (products) - Σ H i (reactants) (this is an abstract useless statement) In practice, we use a Table of standard heats of formation r H 0 = Σ f H (products) - Σ f H (reactants) 6

7 From a table of f H for a few dozen reactions we can know the r H 0 for thousands of reactions that may have never been measured. 7

8 Consider a generic chemical reaction aa + bb cc + dd where the a,b,c,d = the stoichiometric numbers and A,B,C,D are chemicals r H 0 = c f H 0 (C) + d f H 0 (D) a f H 0 (A) b f H 0 (B) Example: C 6 H 12 O 6 (s) + 6O 2 ---> 6CO 2 (g) + 6H 2 O(g) f H 0 kj mol-1 r H 0 = 6 ( 393.5) + 6 ( 241.8) ( ( ( 0 ) ) = kj/mol as written you MUST always associate the r H 0 with a balanced reaction 8

9 What EXACTLY is f H 0 298??? f H = Standard Heat of Formation (at 25 o C) The superscript 0 means that all reactants and products are in their standard state, which means: Gases: 1 bar and ideal Liquids and solids: 1 bar applied and pure Solutes: 1 molar (usually) and ideal (no solute-solute interaction) Note that temperature is NOT part of the definition. (There is a different table for every temperature.) f means formation of 1 mole from the most stable form of the elements at the given temperature (298 in this case) Quiz: What chemical reaction has: H 2 (g) + 1/2O 2 (g) H 2 O(g) H 2 (s) + 1/2O 2 (s) H 2 O(g) 1. H 298 = f H for H 2 O (g) 2. H 5 = f H 0 5 for H 2 O (g) 3. Why is f H = 0 for H 2 (g), O 2 (g), N 2 (g)...? H 2 (g) H 2 (g) 9

10 We will constantly be using: Powerful Exploitation of State Function Concept in Thermodynamics State 1 H unknown or unmeaureable State 3 known H 1 known H 2 State 2 can be ANYTHING Path does not matter: H is a STATE FUNCTION H unknown = H 1 + H 2 When applied to chemical reactions this trick is known as Hesses Law 10

11 Hesse s Law is a: Powerful Exploitation of State Function Concept in Thermodynamics Reactants known H f 0 (1) H unknown or unmeaureable Products known + H f 0 (3) Elements in most stable form (solid, liquid or gas) Path does not matter: H is a STATE FUNCTION H 0 unknown =+ H f 0 (prod) - H f 0 (react) 11

12 Bond Enthalpies Another Exploitation of State Function Concept in Thermodynamics Reactants sum of reactant bond enthalpies H unknown or unmeaureable Products sum of reactant bond enthalpies Gaseous ATOMS Path does not matter: H is a STATE FUNCTION H 0 unknown =+ H f 0 (reactants) - H f 0 (products) Why the sign change??? Bond Enthalpy DEFINED as Enthalpy of bond breaking (not making) 12

13 Bond Energies (Enthalpies) 13

14 14

Enthalpy and Adiabatic Changes

Enthalpy and Adiabatic Changes Chapter 2 of Atkins: The First Law: Concepts Sections 2.5-2.6 of Atkins (7th & 8th editions) Enthalpy Definition of Enthalpy Measurement of Enthalpy Variation of Enthalpy