Lecture 304/18/07

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 340 mL of water to cool it from 20.5°C to 0°C?

A rainstorm deposits 2.5 x kg of rain. Calculate the quantity of thermal energy in joules transferred when this much rain forms. (∆H cond = KJ/mol) Exothermic or endothermic?

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 (aq)  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 (coffee cup calorimetry) heat lost = heat gained Measure change in temperature of water

Constant pressure calorimetry (coffee 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 coffee 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 combustion chamber

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 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 Sounds great BUT.....

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 universe = ∆S system + ∆S surroundings 2 nd Law of Thermodynamics Entropy of the universe increases with spontaneous reactions Reversible reactions vs. Irreversible reaction

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 Key points

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

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)

Example Sulfur (2.56 g) was burned in a bomb calorimeter with excess O 2. The temperature increased from ºC to ºC. The bomb had a heat capacity of 923 J/ºC and the calorimeter contained 815 g of water. Calculate the heat evolved per mole of SO 2 formed. S(s) + O 2 (g)  SO 2 (g)

Standard heat of reaction (∆H° rxn ) Same standard conditions as before:

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

Degrees of freedom translational motion molecules in gas > liquid > solid vibrational motion movement of a atom inside a molecule rotational motion rotation of a molecule

Entropy trends Entropy increases: with more complex molecules with dissolution of pure gases/liquids/solids with increasing temperature with increasing volume with increasing # moles of gases

Which has higher entropy? dry ice orCO 2 liquid water at 25°Corliquid water at 50°C pure Al 2 O 3 (s)orAl 2 O 3 with some Al 2+ replaced with Cr 3+ 1 mole of N 2 at 1 atmor1 mol of N 2 at 10 atm CH 3 CH 2 CH 2 CH 3 (g)or CH 3 CH 3 (g)

Is the reaction spontaneous?

Gibbs Free Energy ( ∆G) ∆G° = ∆H° - T∆S° ∆G = ∆H - T∆S ∆G° = Σn∆G f ° (products) - Σn∆G f ° (reactants)

Gibbs Free Energy ∆G = ∆H - T∆S ∆H∆S-T∆S∆G spontaneous? example -+ 2O 3 (g)  3O 2 (g) +- 3O 2 (g)  2O 3 (g) -- H 2 O (l)  H 2 O (s) ++ H 2 O (s)  H 2 O (l)

Gibbs Free Energy (∆G) and equilibrium R = J/mol-K

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?

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?