1. Thermodynamics and what it means to you

2.  Energy: Ability to do work or produce heat  First Law of Thermodynamics: Law of conservation of energy  Heat: an extensive property – total kinetic energy of a sample  Temperature: an intensive property – average kinetic energy of a sample

3.  Heat units ◦ Calorie - heat required to raise temperature of 1 g water 1ºC ◦ One dietary calorie (Calorie) = 1000 calories (1 kcal). ◦ SI unit – joule  1 joule = 1 kgm 2 /s 2, or  1 joule = 1 newton-meter  1 calorie = 4.184 joules

4.  An intensive property  Equals the heat necessary to raise the temperature of one gram of a substance one degree Celsius.  Unit of specific heat: J/gºC  Higher value means substance is harder to heat – specific heat is inversely proportional to heating rate.

5. q = cm  T where q = heat transferred, c = specific heat, m = mass, and  T = change in temperature.  Example: Calculate the heat necessary to raise the temperature of 55 g water from 32ºC to 78ºC.  Solution:  q = 4.184j/gºC(55g)(78-32ºC) = 11kj

6.  Example: Calculate the heat necessary to raise the temperature of a 5.0g iron nail (specific heat 0.448 J/gºC) from 22ºC to 150ºC.  Solution: q = 0.448j/gºC(5.0g)(150-22ºC) = 290j

7.  Example. You drop the hot nail into 25 g cold (18ºC) water. What is the final temperature of the water?  Solution: q w = q n 4.184j/gºC(25g)(T-18) =0.448j/gºC(5.0g)(150-T) 106.84T = 2218.8 T = 21ºC

8. 445 g of an unknown metal is heated to 99ºC. The hot metal is dropped into 97g cold water (16ºC). The temperature of the water increases to 23ºC. What is the heat capacity of the metal? SSolution: q m = q w 45(c)(99-23) = 97(4.184)(23-16) c = 97(4.184)(23-16)/[45(99-23)] = 0.83J/gºC

9.  What are exothermic/endothermic processes?  Melting is an endothermic process.  Heat of fusion (  H f ): The heat absorbed by one gram of a substance at its melting point as it melts. Units: J/g  Temperature of a melting substance stays constant.  The heat of fusion is also the heat released to the surroundings during freezing (exothermic).

10.  Heat of vaporization is the heat necessary to vaporize one gram of a substance at its boiling point is the heat of vaporization (  H v ).  This is numerically equal to the heat released to the surroundings during the condensation of one gram of the substance.

11.  Heat (q)= mass x heat of phase change  Calculate the heat necessary to completely boil away one cup of water (250. g).  H v for water = 2260 J/g  q = 250g(2260J/g) = 565KJ  Calculate the heat given off when one gallon (3.79L) of water freezes.  H f for water is 334 J/g.  q = 3790g(334j/g) = 1266KJ

12.  Total heat = heat needed to change temperature + heat needed for phase changes.  Find the heat released when 15 g water cools from 135ºC to 100ºC, condenses, and further cools to 21ºC. Note that the specific heat of steam (2.0J/gºC) is not the same as liquid water.  q = 15(2.0)(35) + 15(2260) + 15(4.184)(79) = 40.KJ

14.  Chemical changes always involve energy changes and ∴ changes in heat.  Exothermic changes – heat moves from system to surroundings. Since the heat content of the system is lowered,  H is negative  In endothermic changes, heat moves from surroundings to system. Since heat content of system is rising,  H is positive.

15.  Generally, exothermic changes are more favored to proceed spontaneously than endothermic changes.  In chemical reactions, the magnitude of  H depends on the difference of the energy in the bonds of the reactants and products. A-B + C-D  A-D + C-B   H = H products - H reactants

16.  Enthalpy Change – heat absorbed or released in a reaction, given for the number of moles represented by the coefficients (extensive property)  2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O   H = -2855.4 kJ (-1427.7kJ/mol ethane)  C (s) + O 2(g)  CO 2(g)   H f = - 393.5 kJ

17.  A way to calculate the enthalpy change of a reaction  The enthalpy of a reaction is the same whether the reaction happens in one step or several.  The total enthalpy is the sum of the enthalpies for any set of steps that leads to the product (enthalpy is a state function)  Using formation of each reactant and product from their elements is most convenient

18.   H =  [  H f(products) -  H f(reactants) ]  Example: Find the heat of combustion of ethane (C 2 H 6 ) using Hess’ Law. 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O  2(C 2 H 6  2C + 3H 2 )   H = 2(84.7kJ) = 169.4 kJ  4(C (s) + O 2(g)  CO 2(g) )   H = 4(-393.5kJ) = -1574 kJ  6(H 2 + ½O 2  H 2 O)   H = 6(-241.8kJ)=-1450.8kJ

19.  Overall enthalpy is the sum of the enthalpies of the steps.   H = 169.4 kJ + -1574 kJ + -1450.8kJ = -2855.4 kJ

20.  Heat is measured in a calorimeter

21.  Heat is measured by the change in temperature of the water  Some heat is lost to the calorimeter  Calorimeter constant – empirical measure of how much heat is absorbed by the calorimeter per degree of temperature change  Units are J/ºC – similar to specific heat, but mass is absent because it is constant.  Heat evolved: q =  T(calorimeter constant)

22.  Example: 0.345g diethyl ether (C 4 H 10 O) is ignited in a calorimeter having a calorimeter constant of 1.24KJ/ºC. The measured temperature change is 10.24ºC. Find the molar heat of combustion of diethyl ether.

23. SSolution: Write a balanced chemical equation. C 4 H 10 O + 6O 2  4CO 2 + 5H 2 O FFind the heat evolved. q = 10.24ºC(1.24J/ºC) = 12.7KJ SScale result to one mole. 0.345g/74.1g/mol = 4.66x10 -3 mol ether 12.7KJ/(4.66x10 -3 mol) = 2730KJ/mol

24. RReactions that happen on their own are said to be spontaneous. SSpontaneity depends on enthalpy and entropy. EEntropy is the measure of the disorder of a system. MMore disordered = higher entropy

25.  Changes that result in higher entropy: ◦ Increase in number of particles ◦ Increase in volume of a gas ◦ State changes: solid to liquid, liquid to gas ◦ Dissolving a solute in a solvent ◦ Raising temperature  A closed system will always tend toward greater entropy (second law of thermodynamics).

26.  An open system will also tend toward greater disorder unless energy in put in and/or some ordering principle is present.  Calculating entropy  Entropy changes are designated  S.  S = S final - S initial or  S = S products – S reactants  Entropy is expressed in units of J/K

27.  Why does increasing entropy drive reactions? ◦ Microscopic reversibility – any reaction can run in reverse, and will to some extent ◦ Competing rates – a spontaneous reaction runs faster in the forward direction than the reverse  Probability – it is more probable that few particles will get together to react than many, ∴ the reaction that produces more particles is unlikely to be reversed 2KClO 3  2KCl +3O 2

28.  Measure of energy available to do work after effects of entropy are removed – gives the limit of the efficiency of a system  G =  H – T  S where T is in Kelvins.  Negative  G means a spontaneous reaction  Positive  S contributes to negative  G, as does high temperature.

29.  Spontaneous reaction conditions  H  S Result -+-+ +-+- +- S at all T NS at all T S at high T, NS at low T NS at high T, S at low T