Welcome to Thermochemistry!

Energy in Chemistry Energy in Chemistry (11:23)  Energy is the ability to do work or produce heat. The sum of the potential energy and kinetic energy is equal to the internal energy of the system.  Potential energy is the energy stored in chemical bonds. To break a bond, energy must be added to the system and so the P.E. goes up. (ie: changing liquid water to water vapor, you must add heat or lower pressure). The attraction between the nucleus of an atom and the shared electrons is called the coulombic attraction which is the electrostatic attraction between two charged particles.

Coulombic attractions  Electrostatic attraction (coulombic forces) are the strongest of all intermolecular forces. They cause opposite charges to attract and like charges to repel. When they act between separate charged particles (ion-ion interactions) they are especially strong.  The energy it takes to pull a Na + and a Cl- apart in a sodium chloride crystal is much greater than that needed to pull apart a hydrogen molecule.

 When atoms bond, the coulombic attraction results in energy being released and a subsequent lower potential energy.  Kinetic energy is the energy of motion. KE = ½ mv 2  First Law of Thermodynamics is the Law of Conservation of Energy.  Heat vs. Temperature: Heat is the total amount of kinetic energy in a system and is represented by the letter q. Temperature is the average kinetic energy in a system and is measured in o C or K.

Enthalpy (H): The flow of energy (heat exchange) at constant pressure when two systems are in contact.  Enthalpy of reaction: ΔH rxn is the amount of heat released in an exothermic reaction (negative values) or amount of heat absorbed in an endothermic reaction (positive values). kJ/mol rxn  Enthalpy of combustion: ΔH comb is the heat absorbed or released by burning something usually with oxygen. kJ/mol rxn  Enthalpy of formation: ΔH f is the heat released or absorbed when ONE mole of a compound is formed from elements in their standard state. kJ/mol rxn  Enthalpy of fusion: ΔH fus is the heat absorbed to melt (overcome IMFs) of 1 mole of a solid to a liquid at its melting point. kJ/mol rxn  Enthalpy of vaporization: Δh vap is the heat absorbed to vaporize 1 mole of a liquid to vapor at its boiling point. kJ/mol rxn

Other definitions  Endothermic: Energy absorbed by the system. + ΔH  Exothermic: Energy released by the system. - ΔH  Entropy: The measure of the dispersal of matter and energy; increase dispersal +ΔS; decrease dispersal –ΔS  Gibbs Free Energy (G): criteria for determining thermodynamic favorability and calculating the theoretical amount of energy to do work.  Standard Conditions: 1 atm, 25 o C, and 1.0 M solutions. Standard conditions are indicated by adding the symbol o to G, S, or H. So if you see ΔH o, ΔS o, ΔG o then you know you’re dealing with STP.  ΔE = q(heat) + w (work)  + w if work is done on the system (compression of gases)  -w if work is done by the system (expansion of gases)  Work = -PΔV

Gibbs Free EnergyGibbs Free Energy (13:00) It is the energy associated with a chemical reaction that can be used to do work and is the sum of its enthalpy (H) plus the product of the temperature and the entropy (S) of the system. This quantity can be defined as: G=H−TS or more completely as G=U+PV−TS where U = internal energy (SI unit: joule) P = pressure (SI unit: pascal) V = volume (SI unit: m 3 ) T = temperature (SI unit: kelvin) S = entropy (SI unit: joule/kelvin) H = enthalpy (SI unit: joule)

Enthalpy ΔH  Measure only the change in enthalpy (the difference between the potential energies of the products and reactants.)  Enthalpy can be calculated from several sources including:  Stoichiometry  Calorimetry  Tables of standard values  Hess’s Law  Bond energies

Potential energy diagrams ΔH = PE products – PE reactants E A is always +

Potential Energy Diagram with Catalyst

CalorimetryCalorimetry (12:00) q = m· C · ΔTorJ = g · ΔT · SH q = heat energy m = mass of water C = specific heat capacity ΔT = the change in temperature SH of water = J/g o C Calculate the amount of energy it takes to heat g of water from 0 o C to boiling at sea level.

Hess’s Law  2C(s) + H 2 (g) ---> C 2 H 2 (g)ΔH° = ??? kJ  Given the following thermochemical equations:  C 2 H 2 (g) + (5/2)O 2 (g) ---> 2CO 2 (g) + H 2 O(l)ΔH° = kJ  C(s) + O 2 (g) ---> CO 2 (g)ΔH° = kJ  H 2 (g) + (1/2)O 2 (g) ---> H 2 O(l)ΔH° = kJ  a) first eq: flip it so as to put C 2 H 2 on the product side  b) second eq: multiply it by two to get 2C  c) third eq: do nothing. We need one H 2 on the reactant side and that's what we have.

 2CO 2 (g) + H 2 O(l) ---> C 2 H 2 (g) + (5/2)O 2 (g)ΔH° = kJ  2C(s) + 2O 2 (g) ---> 2CO 2 (g)ΔH° = -787 kJ  H 2 (g) + (1/2)O 2 (g) ---> H 2 O(l)ΔH° = kJ  Examine what cancels:  2CO 2 ⇒ first & second equation  H 2 O ⇒ first & third equation  (5/2)O 2 ⇒ first & sum of second and third equation  4) Add up ΔH values for our answer:  kJ + (-787 kJ) + ( kJ) = kJ

 Problem #2: The standard molar enthalpy of formation, ΔH f °, of diborane cannot be determined directly because the compound cannot be prepared by reaction of boron and hydrogen. However, the value can be calculated. Calculate the standard enthalpy of formation of gaseous diborane (B 2 H 6 ) using the following thermochemical information:  a) 4 B(s) + 3 O 2 (g) ---> 2 B 2 O 3 (s)ΔH° = kJ  b) 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(l)ΔH° = kJ  c) B 2 H 6 (g) + 3 O 2 (g) ----> B 2 O 3 (s) + 3 H 2 O(l)ΔH° = kJ  1) An important key is to know what equation we are aiming for. The answer is in the word 'formation:‘  2B + 3H 2 ---> B 2 H 6  Remember that formation means forming one mole of our target substance. This means that a one MUST be in front of the B 2 H 6

 2) In order to get to our formation reaction, the following must happen to equations (a), (b) and (c):  equation (a) - divide through by two equation (b) - multiply through by 3/2 equations (c) – flip  a) 2 B(s) + (3/2) O 2 (g) ---> B 2 O 3 (s)ΔH° = kJ  b) 3 H 2 (g) + (3/2) O 2 (g) ---> 3 H 2 O(l)ΔH° = kJ  c) B 2 O 3 (s) + 3 H 2 O(l) ---> B 2 H 6 (g) + 3 O 2 (g) ΔH° = kJ  2B + 3H 2 ---> B 2 H 6 ΔH° = kJ

Using bond enthalpies and Hess’s Law: ΔH = Σ E reactant bonds broken minus Σ E product bonds broken  Problem #1: Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane:  H−C≡C−H(g) + 2H 2 (g) ---> H 3 C−CH 3 (g)  Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)  1) You have to put energy into a bond (any bond) to break it. Bond breaking is endothermic. Let's break all the bonds of the reactants:  one C≡C ⇒ +839 kJ  two C−H ⇒ 413 x 2 = +826 kJ  two H−H ⇒ 432 x 2 = +864 kJ  The sum is kJ

 Note there are two C−H bonds in one molecule of C 2 H 2 and there is one H−H bond in each of two H 2 molecules. Two different types of reasons for multiplying by two.  2) You get energy out when a bond (any bond) forms. Bond making is exothermic. Let's make all the bonds of the one product:  one C−C ⇒ −347 kJ  six C−H ⇒ −413 x 6 = −2478  The sum is −2826 kJ  3) ΔH = the energies required to break bonds (positive sign) plus the energies required to make bonds (negative sign):  (−2825) = −296 kJ/mol

Phase Changes  For water:  H vap = 2330 J/g  H fus = 335 J/g  SH solid = 2.06 J/g o C  SH liquid = J/g o C  SH vap = 1.87 J/g o C