Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry

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Ginger Greene
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Recall the equation. w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 10 2 J -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate. 8.3145 J/mol K 0.

1 Recall the equation. w = -PΔV = -(1.20 atm)(1.02 L)( = J -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate J/mol K L atm/mol K J/L atm ) Winter 2009 Page 23 The First Law of Thermodynamics Internal Energy U. Total energy (potential and kinetic) in a system. Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons. Winter 2009 Page 24 1

2 A system contains only internal energy. A system does not contain heat or work. These only occur during a change in the system. ΔU = q + w Law of Conservation of Energy The energy of an isolated system is constant Winter 2009 Page 25 Winter 2009 Page 26 2

Any property that has a unique value for a specified state of a system is said to be a State Function. Water at 293.15 K and 1.00 atm is in a specified state. d = 0.

3 Any property that has a unique value for a specified state of a system is said to be a State Function. Water at K and 1.00 atm is in a specified state. d = g/ml This density is a unique function of the state. It does not matter how the state was established. Winter 2009 Page 27 U is a function of state. Not easily measured. ΔU has a unique value between two states. Is easily measured. Winter 2009 Page 28 3

4 Changes in heat and work are not functions of state. a one step expansion of gas could occur in multiple steps Winter 2009 Page 29 Heats of Reaction: ΔU and ΔH Reactants Products U i U f ΔU = U f - U i Δ = q rxn + w In a system at constant volume: ΔU = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v? Winter 2009 Page 30 4

5 Winter 2009 Page 31 q V = q P + w We know that w = - PΔV and ΔU = q P therefore: ΔU = q P - PΔV q P = ΔU + PΔV These are all state functions so define a new function. Let Then H = U + PV ΔH = H f H i = ΔU + PΔV If we work at constant pressure and temperature: ΔH = ΔU + PΔV = q P Winter 2009 Page 32 5

5 kj/mol = q V Winter 2009 Page 33 Molar enthalpy of vaporization: H 2

6 q P = -566 kj/mol = ΔH PΔV = P(V f V i ) = RT(n f n i ) = -2.5 kj ΔU = ΔH - PΔV = kj/mol = q V Winter 2009 Page 33 Molar enthalpy of vaporization: H 2 O (l) H 2 O(g) ΔH = 44.0 kj at 298 K Molar enthalpy of fusion: H 2 O (s) H 2 O(l) ΔH = 6.01 kj at K Winter 2009 Page 34 6

7 Enthalpy Changes Accompanying Changes in States of Matter. Calculate ΔH for the process in which 50.0 g of water is converted from liquid at 10.0 C to vapor at 25.0 C. Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step. Set up the equation and calculate: q P = mc H 2O ΔT + nδh vap = (50.0 g)(4.184 J/g C)( ) C + = 3.14 kj kj = 125 kj 50.0 g 18.0 g/mol 44.0 kj/mol Winter 2009 Page 35 Standard States and Standard Enthalpy Changes Define a particular state as a standard state. Standard enthalpy of reaction ΔH The enthalpy change of a reaction in which all reactants and products are in their standard states. Standard State The pure element or compound at a pressure of 1 bar and at the temperature of interest. Winter 2009 Page 36 7

8 Winter 2009 Page 37 Indirect Determination of ΔH: Hess s Law ΔH is an extensive property. Enthalpy change is directly proportional to the amount of substance in a system. N 2 (g) + O 2 (g) 2 NO(g) ΔH = kj ½N 2 (g) + ½O 2 (g) NO(g) ΔH = kj ΔH changes sign when a process is reversed Winter 2009 NO(g) ½N 2 (g) + ½O 2 (g) ΔH = kj Page 38 8

9 Hess s Law Topic 2.1: Hess s Law Hess s law of constant heat summation If a process occurs in stages or steps (even hypothetically) the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2 (g) + ½O 2 (g) NO(g) NO(g) + ½O 2 (g) NO 2 (g) ΔH = kj ΔH = kj ½N 2 (g) + O 2 (g) NO 2 (g) ΔH = kj Winter 2009 Page 39 Topic 2.1: Hess s Law Winter 2009 Page 40 9

10 Standard Enthalpies of Formation ΔH f Topic 2.1: ΔH formation The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. The standard enthalpy of formation of a pure element in its reference state is 0. Winter 2009 Page 41 Example: Topic 2.1: ΔH formation Given the Reactions: CH 2 =CHCH 3 (g) + H 2 (g) CH 3 CH 2 CH 3 (g) Δ rx H = -124 KJ mol-1 CH 3 CH 2 CH 3 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l) Δ rx H = KJ mol-1 Calculate the standard enthalpy of propane. CH 2 =CHCH 3 (g) + 9/ 2 O 2 (g) 3CO 2 (g) + 3H 2 O (l) Δ rx H =? You will also need: H 2 O (l) H 2 (g) + ½ O 2 (g) Δ rx H = +286 KJ mol-1 Winter 2009 Page 42 10

CH 3 (g) + 5O 2 (g) 3CO 2 (g) + 34 H 2 O (l) Δ rx H = -2220 KJ mol-1 H 2 O (l) H 2 (g) + ½ O 2

11 Result: Topic 2.1: ΔH formation CH 2 =CHCH 3 (g) + H 2 (g) CH 3 CH 2 CH 3 (g) Δ rx H = -124 KJ mol-1 CH 3 CH 2 CH 3 (g) + 5O 2 (g) 3CO 2 (g) + 34 H 2 O (l) Δ rx H = KJ mol-1 H 2 O (l) H 2 (g) + ½ O 2 (g) Δ rx H = +286 KJ mol-1 CH 2 =CHCH 3 (g) + 9/ 2 O 2 (g) 3CO 2 (g) + 3H 2 O (l) Δ rx H = KJ mol-1 Winter 2009 Page 43 Topic 2.1: ΔH formation Winter 2009 Page 44 11

12 Topic 2.1: ΔH formation ΔH rxn = ΣΔH f products - ΣΔH f reactants Winter 2009 Page 45 Topic 2.1: ΔH formation Winter 2009 Page 46 12

13 Example: Determine the Δ rx H for the following : Topic 2.1: ΔH formation 2HN 3 (l) + 2NO (g) H 2 O 2 (l) + 4N 2 (g) Δ rx H = [Δ f H (H 2 O 2 (l) ) + 4Δ f H (N 2 (g) )] - [2Δ f H (HN 3 (l) ) + 2Δ f H (NO (g) )] = [ (0) KJ mol-1] - [2(264.0) + 2 (90.25) KJ mol-1] = KJ mol-1 Winter 2009 Page 47 Topic 2: Global warming Winter 2009 Page 48 13

14 Winter 2009 Page 49 Topic 2: Global warming Winter 2009 Page 50 14

15 Topic 2: Global warming total petroleum coal natural gas Winter 2009 Page 51 Topic 2: Global warming Winter 2009 Page 52 15

Spontaneous Change A process that occurs in a system left to itself. Topic 2:Entropy Once started no external actions is necessary to make the process continue.

16 Spontaneous Change A process that occurs in a system left to itself. Topic 2:Entropy Once started no external actions is necessary to make the process continue. A non-spontaneous process will not occur without external action continuously applied. 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) H 2 O(s) H 2 O(l) Potential energy decreases. For chemical systems the internal energy U is equivalent to potential energy. Winter 2009 Page 53 Entropy (S) ΔU = ΔH = 0 Topic 2:Entropy The greater the number of configurations of the microscopic particles among the energy levels in a particular system the greater the entropy of the system. ΔS > 0 spontaneous Degree of RANDOMNESS Winter 2009 Page 54 16

The Boltzmann Equation for Entropy S = k lnw Where: Entropy = S Topic 2:Entropy States. The microscopic energy levels available in a system. Microstates W.

17 The Boltzmann Equation for Entropy S = k lnw Where: Entropy = S Topic 2:Entropy States. The microscopic energy levels available in a system. Microstates W. The particular way in which particles are distributed amongst the states. Number of microstates = W. The Boltzmann constant k. Effectively the gas constant per molecule = R/N A. Winter 2009 Page 55 Boltzmann Distribution Topic 2:Entropy Winter 2009 Page 56 17

18 Entropy Change Topic 2:Entropy ΔS = q rev T For changes occurring at constant temperature Winter 2009 Page 57 Evaluating Entropy and Entropy Changes Topic 2:Entropy Phase transitions. Exchange of heat can be carried out reversibly. ΔS = ΔH T tr H 2 O(s 1 atm) H 2 O(l 1 atm) ΔH fus = 6.02 kj at K ΔS fus = ΔH fus 6.02 kj mol-1 = T tr K = kj mol -1 K -1 = 22.0 J mol -1 K -1 Winter 2009 Page 58 18

Evaluating Entropy and Entropy Changes Topic 2:Entropy Phase transitions. Exchange of heat can be carried out reversibly. ΔS = ΔH H 2 O(l 1 atm) H 2 O(g 1 atm) ΔH vap = 40.

19 Evaluating Entropy and Entropy Changes Topic 2:Entropy Phase transitions. Exchange of heat can be carried out reversibly. ΔS = ΔH H 2 O(l 1 atm) H 2 O(g 1 atm) ΔH vap = 40.7 kj at 373 K T tr ΔS vap = ΔH vap 40.7 kj mol-1 = = kj mol 373 K -1 K -1 T tr = 109 J mol -1 K -1 Winter 2009 Page 59 Trouton s Rule Topic 2:Entropy ΔS = ΔH vap T bp 87 J mol-1 K -1 Winter 2009 Page 60 19

20 Topic 2:Entropy The Second Law of Thermodynamics. Spontaneous Change: ΔS total = ΔS universe = ΔS system + ΔS surroundings The Second Law of Thermodynamics: ΔS universe = ΔS system + ΔS surroundings > 0 All spontaneous processes produce an increase in the entropy of the universe. Winter 2009 Page 61 Absolute Entropies Topic 2:Entropy Third law of thermodynamics. The entropy of a pure perfect crystal at 0 K is zero. Standard molar entropy. Tabulated in Appendix D. ΔS = [ Σν p S (products) - Σν r S (reactants)] Winter 2009 Page 62 20

21 ΔS = [ Σν p S (products) - Σν r S (reactants)] Topic 2:Entropy Example: Determine the ΔS for the following : 2NO (g) + O 2 (g) 2NO 2 (g) ΔS = [2ΔS (NO 2 (g) )] - [2ΔS (NO (g) ) + ΔS (O 2 (g) )] = [2 x J K -1 ] - [2 x J K -1 ] = J K -1 Winter 2009 Page 63 Entropy as a Function of Molecular Complexity Topic 2:Entropy ΔS NO = J K -1 ΔS NO 2 = J K -1 #vib modes = 3N-5 linear molecule #vib modes = 3N-6 non-linear molecule Winter 2009 Page 64 21

22 Entropy as a Function of Temperature Topic 2:Entropy Winter 2009 Page 65 Free Energy and Free Energy Change Hypothetical process: only pressure-volume work at constant T and P. Topic 2:Free Energy q surroundings = -q p = -ΔH sys Make the enthalpy change reversible. large surroundings infinitesimal change in temperature. Under these conditions we can calculate entropy. ΔS univ = ΔH sys T TΔS univ = ΔH sys Winter 2009 Page 66 22

23 Free Energy and Free Energy Change For the universe: TΔS univ. = TΔS sys ΔH sys = -(ΔH sys TΔS sys ) Topic 2:Free Energy -TΔS univ. = ΔH sys TΔS sys For the system: G = H - TS ΔG = ΔH - TΔS ΔG sys = - TΔS universe Winter 2009 Page 67 Criteria for Spontaneous Change Topic 2:Free Energy ΔG sys < 0 (negative) ΔG sys = 0 (zero) ΔG sys > 0 (positive) the process is spontaneous. the process is at equilibrium. the process is non-spontaneous. J. Willard Gibbs Winter 2009 Page 68 23

The change in free energy for a reaction in which a substance in its standard state is formed from

24 Topic 2:Free Energy Winter 2009 Page 69 Standard Free Energy Change ΔG Topic 2:Free Energy The standard free energy of formation ΔG f. The change in free energy for a reaction in which a substance in its standard state is formed from its elements in reference forms in their standard states. The standard free energy of reaction ΔG. ΔG = [ Σν p ΔG f (products) - Σν r ΔG f (reactants)] Winter 2009 Page 70 24

25 Example: Topic 2:Free Energy ΔG = [ Σν p ΔG f (products) - Σν r ΔG f (reactants)] Determine the ΔG for the following : 2NO (g) + O 2 (g) 2NO 2 (g) ΔG = [2ΔG (NO 2 (g) )] - [2ΔG (NO (g) ) + ΔG (O 2 (g) )] = [2 x KJ mol -1 ] - [2 x KJ mol -1 ] = KJ mol -1 Winter 2009 Page 71 ΔG = ΔH - TΔS Topic 2:Free Energy Example: Determine the ΔG for the following : 2NO (g) + O 2 (g) 2NO K We previously determined ΔH = KJ ; ΔS = J K -1 ΔG = ΔH - TΔS = KJ ( K x KJ K -1 ) = KJ Winter 2009 Page 72 25

26 Free Energy Change and Equilibrium Topic 2:Free Energy Winter 2009 Page 73 Topic 2:Free Energy Relationship of ΔG to ΔG - Non-standard Conditions 2 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔG = ΔH - TΔS ΔG = ΔH - TΔS For ideal gases ΔH = ΔH ΔG = ΔH - TΔS Winter 2009 Page 74 26

27 Relationship Between S and S Topic 2:Free Energy V i q rev = -w = RT ln Vf ΔS = q rev T = R ln V f V i ΔS = S f S i = R ln V f V i = R ln P i P f = -R ln P f P i S = S - R ln P P = S - R ln P 1 = S - R ln P Winter 2009 Page 75 27

Ch. 19 Entropy and Free Energy: Spontaneous Change

Ch. 19 Entropy and Free Energy: Spontaneous Change 19-1 Spontaneity: The Meaning of Spontaneous Change 19-2 The Concept of Entropy 19-3 Evaluating Entropy and Entropy Changes 19-4 Criteria for Spontaneous