Definitions:

Universe – composed of the system and the surroundings

System – part being studied; that which we are focusing on

Surroundings – everything else in the universe outside the system

Open System – where both energy and matter can be transferred to and from surroundings

Closed System – only energy can be transferred to and from surroundings

Isolated System – one where neither energy nor matter can be transferred to and from surroundings

Exothermic – Energy (as heat) flows out of the system

Endothermic – Energy (as heat) flows into the system

Thermodynamics – the study of energy and its interconversions

Energy – the capacity to produce work or heat

Kinetic Energy – the energy of motion; KE = ½ mass x (velocity)2

Potential Energy – Energy that can be converted to useful work

Heat – involves transfer of energy between two objects (from hot to cold)

Work – Force x distance

State Functions – depend only upon the initial and final state of a substanceExamples: ΔH, ΔS, ΔGA property that is independent of pathway. That is, it does notmatter how you get there, the difference in the value is the same.

For example, you can drive from New York to Los Angeles viamany different routes. No matter which one you take, you are stillgoing from NY to LA. The actual distance between the two citiesis the same.

Enthalpy and CalorimetryΔH = Hproducts – Hreactants

The change in enthalpy (ΔH) of the system is equal to the energy flow asheat at constant pressure.

ΔH = qp

If ΔH>0, the reaction is endothermic (Heat is absorbed by the system)If ΔH<0, the reaction is exothermic (Heat is given off by the system)

Endothermic**Enthalpy of products > enthalpy of reactants** ΔH = positive (energy must be put into reaction to occur)

Exothermic**Enthalpy of products< enthalpy of reactants** ΔH = negative (energy released from reaction as it occurs)

Know the energy diagrams:Endothermic Exothermic

ΔH = +

ΔH = -

**Spontaneous reactions will tend toward conditions of lower enthalpy(more negative ΔH)

Calorimetry-the experimental technique used to determine the heat exchange (q) associated with a reaction.-the amount of heat exchanged in a reaction depends upon:

1. The net temperature change during a reaction2. The amount of substance. The more you have, the more heat can

be exchanged.3. The heat capacity (C) of a substance.

heat absorbed C = increase in temperature = J/°C

-some substances can absorb more heat that others for a given temperature change.

There are three ways of expressing heat capacity:1. Heat capacity (as above) = J /°C2. Specific heat capacity = heat capacity per gram of substance

= J/g °C or J/g K3. Molar heat capacity = heat capacity per mole of substance

= J/ mol °C or J/ mol K

The specific heat of water = 4.18 J/g°C

Specific Heat-the amount of heat energy required to raise one gram of a substance by one degree celsius or kelvin

Cp = specific heat (p = constant pressure)q = heat lost or gainedm = mass in gramsΔT = change in temperature (°C or K)

-Standard Enthalpy of formation (ΔH°f) – the change in enthalpy that accompaniesthe formation of one mole of a compound from its elements with all substances intheir standard states

-the degree symbol indicates that the corresponding process has been carried outunder standard conditions

-For a Compound-standard state for gas is exactly 1 atm.-for a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.-For a substance present in solution, the standard state is a concentration of exactly 1M.

-For an Element-The standard state of an element is the form in which the element exists under conditions of 1 atmosphere and 25° C. (The standard state for oxygen is O2(g) at a pressure of 1 atm; the standard state for sodium is Na(s); the standard state for mercury is Hg(l); and so on.)

Conditions of Enthalpy

For example:

ΔH°f for C2H5OH(l) = -279 kJ/molmeans

2 C(graphite) + 3 H2(g) + ½ O2(g) C2H5OH(l) ΔH°f = -279 kJ/mol

-Standard Enthalpy of Combustion (ΔH°c ) – the enthalpy change when one moleof a substance is completely burned in oxygen under standardconditions.

For example:

ΔH°c for C2H6(g) = -1565 kJ/molmeans

C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g) ΔH°c = -1565 kJ/mol

ΔH°reaction = ΣnpΔH°f (products) - ΣnrΔH°f (reactants)

-When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes-When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer.-Elements in their standard states are not included in the ΔHreaction calculations. That is, H°f for an element in its standard state is zero.

Using standard enthalpies of formation, calculate the standard enthalpy change for thefollowing overall reaction:

4NH3(g) + 7 O2(g) 4NO2(g) + 6H2O (l)

ΔH°f

NH3 (g) = -46 kJ/molNO2 (g) = 34 kJ/molO2 (g) = you figure it outH2O (l) = -286 kJ/mol

Bond Energy & Enthalpy

ΔH = Σ D (bonds broken) - Σ D (bonds formed) Energy required Energy released

the sum of the energies required to break old bonds (positive signs)plus the sum of the energies released in the formation of new bonds(negative signs)

Using bond energies, calculate ΔH for the following reaction:CH4 (g) + 2Cl2 (g) + 2F2 (g) CF2Cl2 (g) + 2HF (g) + 2HCl (g)

Average bond energies (kJ/mol)C-H = 413 Cl-Cl = 239 F-F = 154C-F = 485 C-Cl = 339 H-F = 565H-Cl = 427 C=C = 614 C-C = 347

Calculate the energy required to break the bonds of 3-methyl-4-octene

First law of thermodynamics – The energy of the universe is constant

Second law of thermodynamics – In any spontaneous process there is always an increase in the entropy of the universe (the entropy of the universe is increasing)

ΔSuniv = ΔSsys + ΔSsurr

The sign of ΔSsurr depends on the direction of the heat flowWhat can you tell me about an exothermic reaction?What can you tell me about an endothermic reaction?

The magnitude of ΔSsurr depends on the temperature ΔSsurr = -ΔH

T

ΔG° = ΔH° - T ΔS°

The standard free energy of formation and the standard free energy changethe change in free energy that will occur if the reactants in theirstandard states are converted to the products in their standard states.

ΔG° = ΣnpΔG°f (products) - ΣnrΔG°f (reactants)

Consider the reaction2SO2(g) + O2(g) 2SO3(g)carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°, and ΔG° using the following data:Substance ΔH°f (kJ/mol) S°(J/K·mol) SO2(g) -297 248 SO3(g) -396 257 O2(g) 0 205

The Dependence of Free Energy on Pressure

G = G° + RT ln Pwhere R = 8.3145 J/K·molG° is the free energy of the gas at a pressure of 1 atmG is the free energy of the gas at a pressure P atmT is the Kelvin temperature

can also be written as ΔG = ΔG° + RT ln (Q)

One method for synthesizing methanol involves reacting carbon monoxide andhydrogen gases:

CO(g) + 2H2(g) CH3OH(l)

Calculate ΔG at 25°C for this reaction where carbon monoxide gas at5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.

Standard free energies of formation:CH3OH(l) = -166 kJH2(g) = 0CO(g) = -137 kJ

Free Energy and Equilibriumthe equilibrium position represents the lowest free energy value availableto a particular reaction system

ΔG° = - RT ln (K)

The overall reaction for the corrosion (rusting) of iron by oxygen is4Fe(s) + 3O2(g) 2Fe2O3(s)

Using the following data, calculate the equilibrium constant for thisreaction at 25°C.

Substance ΔH°f (kJ/mol) S° (J/K·mol)Fe2O3(s) -826 90Fe(s) 0 27O2(g) 0 205

Born – Haber Cycle

5 step process1. Sublimation (enthalpy of sublimation)2. Ionization to form ion (1st ionization energy)3. Dissociation of nonmetal (energy to break bond)4. Formation of Anions (electron affinity)5. Form solid compound (lattice energy)

ExampleOverall desired reaction: Li(s) + ½ F2(g) LiF(s)

1. Li(s) Li(g) 161 kJ/mol

2. Li(g) Li+(g) + e- 520 kJ/mol

3. ½ F2(g) F(g) 154/2 = 77 kJ/mol need to form 1 mole of F atoms

4. F(g) + e- F-(g) -328 kJ/mol

5. Li+(g) + F-(g) LiF(s) -1047 kJ/mol

The sum yields the desired overall reaction:161 + 520 + 77 + (-328) + (-1047)= -617 kJ/mol of LiF

Li(s) + ½ F2(g) LiF(s)

Practice ProblemCalculate the Lattice Energy for the followingreaction:

Na(s) + ½ F2(g) NaF(s)

Write out the reaction for each step.Enthalpy of sublimation = 109 kJ/mol1st ionization energy = 494 kJ/molEnergy to break bond = 79 kJ/molElectron affinity = -348 kJ/molLattice energy = ? kJ/molOverall energy for the desired reaction = -569

kJ/mol

Born-Haber Cycle Diagram

Link: http://upload.wikimedia.org/wikipedia/commons/a/af/BornHaberLiF.PNG

Lattice Energy• The principle reason for ionic stability is due to the attraction between oppositely charged ions.• The physical attraction between ions in an ionic compound releases energy as the ions are drawn together.• Once the ions reach the lowest energy possible, distance is minimum given electron repulsion, a crystal lattice is formed.

The strength of the ion attraction is described by the lattice energybetween the ions.

•Lattice Energy (aka lattice enthalpy) = the energy required to separateone mole of a solid ionic compound into its gaseous ions.

= k (Q1Q2) r

k = proportionality constant dependent on structure of solid and on electron configuration of the ions. (8.99 x 109 Jm/C2)

Q1 and Q2 = charges on the ions

r = the shortest distance between the centers of the cation and anion

• Lattice Energy is influenced much more by the product of Q1 and Q2 than by the value of r