Lecture 314/10/06

Thermodynamics: study of energy and transformations Energy Kinetic energy Potential Energy

Units 1 calorie (cal) 1 calorie (cal) = Joule (J) Calorie (Cal)

1 st Law of Thermodynamics Law of conservation of energy Energy in the universe is conserved System vs. surroundings vs. universe Internal energy

Specific heat capacity (C) Quantity of energy to increase the temperature of 1 gram of a substance by 1 degree C (liquid water)= J/g·K C (ice)= 2.06 J/g·K C (steam)= 1.84 J/g·K C (aluminum)= J/g·K Molar heat capacity Quantity of energy that must be transferred to increase the temperature of 1 mole of a substance by 1 °C

Specific heat capacity (C) q = Cm∆T

Heat transfer q gained + q lost = 0 q gained = - q lost 55.0 g of iron at 99.8°C is plunged into 225 g of water at 21°C. What is the final temperature? C (iron) = J/g-K

Example 59.8 J are required to change the temperature of 25.0 g of ethylene glycol by 1 K. What is the specific heat capacity of ethylene glycol?

Does drinking ice water cause you to lose weight? Does drinking ice cold coke?

Changes in state Temperature stays the same during changes of state Gas/Vapor Liquid Solid ENERGY q = mass x constant q = moles x constant

Change of state constant??? Depends on two things: Identity of substance Which states are changing

Solid/Liquid Heat of fusion Solid  Liquid Endothermic ice  Water (333 J/g or 6 KJ/mol) Heat of crystallization Liquid  Solid Exothermic Water  ice (- 333 J/g or - 6 KJ/mol)

Liquid/Gas Heat of vaporization Liquid  Gas Endothermic water  water vapor (40.7 KJ/mol) Heat of condensation Gas  Liquid Exothermic vapor  Water ( KJ/mol)

Solid/Gas Heat of sublimation Solid  Gas Endothermic Heat of deposition Gas  Solid Exothermic

What is the minimum amount of ice at 0 °C that must be added to a 340 mL of water to cool it from 20.5°C to 0°C? q water + q ice = 0 C water m water ∆T water + m ice ∆H fus = 0 (4.184 J/K-g)(340 g)(0°C °C) + (333 J/g)m ice = 0 m ice = 87.6 g

A rainstorm deposits 2.5 x kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms. (∆H vap = - 44 KJ/mol) Exothermic or endothermic? q = 2.5 x Kg x (10 3 g/kg) x (1 mol/18 g) x -44 KJ/mol q = -6.1 x KJ Exothermic

1 st Law of Thermodynamics revisited ∆E = q + w Change in Energy content heat work

work (w) = - F x d w = - (P x A) x d w = - P∆V if ∆V = 0, then no work

State function property of a system whose value depends on the final and initial states, but not the path driving to Mt Washington route taken vs. altitude change ∆E is a state function q and w are not

Change in Enthalpy (∆H or q p ) equals the heat gained or lost at constant pressure ∆E = q p + w ∆E = ∆H + (-P∆V) ∆H = ∆E + P∆V

∆E vs. ∆H Reactions that don’t involve gases 2KOH (aq) + H 2 SO 4 (aw)  K 2 SO 4 (aq) + 2H 2 O (l) ∆V ≈ 0, so ∆E ≈ ∆H Reactions in which the moles of gas does not change N 2 (g) + O 2 (g)  2NO (g) ∆V = 0, so ∆E = ∆H Reactions in which the moles of gas does change 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆V > 0, but often P∆V << ∆H, thus ∆E ≈ ∆H

Enthalpy is an extensive property Magnitude is proportional to amount of reactants consumed H 2 (g) + ½ O 2 (g)  H 2 O (g) ∆H = KJ 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆H = KJ Enthalpy change for a reaction is equal in magnitude (but opposite in sign) for a reverse reaction H 2 (g) + ½ O 2 (g)  H 2 O (g) ∆H = KJ H 2 O (g)  H 2 (g) + ½ O 2 (g) ∆H = KJ Enthalpy change for a reaction depends on the state of reactants and products H 2 O (l)  H 2 O (g) ∆H = 88 KJ

Constant pressure calorimetry (cofee cup calorimetry) heat lost = heat gained Measure change in temperature of water 10 g of Cu at 188 °C is added to 150 mL of water in a cofee cup calorimeter and the temperature of water changes from 25 °C to 26 °C. Determine the specific heat capacity of copper.

Bomb calorimetry Mainly for combustion experiments ∆V = 0 q rxn + q bomb + q water = 0 Often combine q bomb + q water into 1 calorimeter term with q cal = C cal ∆T combustion chamber

Bomb calorimeter math K & T: q rxn + q bomb + q water = 0 q rxn + C bomb ∆T + C water m water ∆T = 0 In the lab: q rxn + q calorimeter = 0 q calorimeter = q bomb + q water q rxn + C calorimeter ∆T = 0 empirically determined same value On the exam

Bond enthalpies

Enthalpies of formation

Hess’ Law

Example A hot plate is used to heat two 50-mL beakers at the same constant rate. One beaker contains 20.0 grams of graphite (C=0.79 J/g-K) and one contains 10 grams of ethanol (2.46 J/g-K). Which has a higher temperature after 3 minutes of heating?

Standard heat of reaction (∆H° rxn ) Same standard conditions as before: 1 atm for gas 1 M for aqueous solutions 298 K For pure substance – usually the most stable form of the substance at those conditions

Standard heat of formation (∆H° f ) Enthalpy change for the formation of a substance from its elements at standard state Na(s) + ½ Cl 2 (g)  NaCl (s) ∆H° f = kJ Three points An element in its standard state has a ∆H° f = 0 ∆H° f = 0 for Na(s), but ∆H° f = KJ/mol for Na(g) Most compounds have a negative ∆H° f formation reaction is not necessarily the one done in lab

Using ∆H° f to get ∆H° rxn 2 ways to look at the problem Calculate ∆H° rxn for: C 3 H 8 (g) + 5 O 2  3 CO 2 (g) + 4 H 2 O (l) Given: 3 C(s) + 4 H 2 (g)  C 3 H 8 (g) ∆H° f = KJ/mol C(s) + O 2 (g)  CO 2 (g) ∆H° f = KJ/mol O 2 (g) + 2 H 2 (g)  2H 2 O (l) ∆H° f = KJ/mol

Using Hess’s Law and ∆H° f to get ∆H° rxn 1 st way: Hess’s Law C 3 H 8 (g) + 5 O 2  3 CO 2 (g) + 4 H 2 O (l) ∆H° rxn = ∆H 1 + ∆H 2 + ∆H 3 Reverse 1 st equation: C 3 H 8 (g)  3 C(s) + 4 H 2 (g) ∆H 1 = - ∆H° f = KJ Multiply 2 nd equation by 3: 3C(s) + 3O 2 (g)  3CO 2 (g) ∆H 2 = 3x∆H° f = KJ Multiply 3 rd equation by 2: 2O 2 (g) + 4 H 2 (g)  4H 2 O (l) ∆H 2 = 2x∆H° f = KJ ∆H° rxn = ( KJ) + ( KJ) + ( KJ) ∆H° rxn = KJ

Using ∆H° f to get ∆H° rxn 2 nd way C 3 H 8 (g) + 5 O 2  3 CO 2 (g) + 4 H 2 O (l) ∆H° rxn = Σn ∆H° f (products) - Σn ∆H° f (reactants) ∆H° rxn = [3x( KJ/mol) + 2x( KJ/mol)] – [( KJ/mol) + 0] ∆H° rxn = [ KJ] – [ KJ] ∆H° rxn = KJ

Spontaneity Some thought that ∆H could predict spontaneity Exothermic – spontaneous Endothermic – non-spontaneous Sounds great BUT some things spontaneous at ∆H > 0 Melting Dissolution Expansion of a gas into a vacuum Heat transfer Clearly enthalpy not the whole story

Entropy (Measurement of disorder) Related to number of microstates S = klnW ∆S universe = ∆S system + ∆S surroundings 2 nd Law of Thermodynamics Entropy of the universe increases with spontaneous reactions Reversible reactions ∆S universe = ∆S system + ∆S surroundings = 0 Can be restored to the original state by exactly reversing the change Each step is at equilibrium Irreversible reaction ∆S universe = ∆S system + ∆S surroundings > 0 Original state can not be restored by reversing path spontaneous

3 rd Law of thermodynamics S = O at O K S° - entropy gained by converting it from a perfect crystal at 0 K to standard state conditions ∆S° = ΣS°(products) - ΣS°(reactants)