FLOW OF ENERGY Heat, Enthalpy, & Thermochemical Equations THERMOCHEMISTRY FLOW OF ENERGY Heat, Enthalpy, & Thermochemical Equations

THERMOCHEMICAL EQUATIONS Although it is NOT form of matter, in a chemical equation enthalpy change (DH) for the reaction can be written as either a reactant or a product. RECALL:

THERMOCHEMICAL EQUATIONS & HEATS OF REACTION EXOTHERMIC REACTION CaO(s) +H2O(l) Ca(OH)2 + 65.2 kJ A chemical equation that includes the enthalpy change is known as a thermochemical equation Heat of Reaction is the enthalpy change for the chemical equation exactly as written

THERMOCHEMICAL EQUATIONS & HEATS OF REACTION EXOTHERMIC REACTION CaO(s) +H2O(l) Ca(OH)2 + 65.2 kJ The above equation can be written as: CaO(s) +H2O(l) Ca(OH)2 DH = -65.2 kJ Where the DH is written on the side. (Recall it is negative because it is exothermic.)

THERMOCHEMICAL EQUATIONS & HEATS OF REACTION ENDOTHERMIC REACTION 2NaHCO3(s) + 129 kJ Na2CO3(s) + H2O(g) + CO2(g) OR 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) DH =+129 kJ DH IS POSITIVE BECAUSE ENDOTHERMIC

Enthalpy diagrams ENTHALPY DIAGRAMS are graphic representations of energy changes in a system

ENTHALPY DIAGRAMS

ENTHALPY DIAGRAM EXAMPLE ENDOTHERMIC

ENTHALPY DIAGRAM EXAMPLE EXOTHERMIC

Laws of Thermochemistry In order to make effective use of thermochemical equations, certain basic laws can be applied: Here come the LAWS….

23.5 kcal of heat is produced when 2 moles decompose 1. ∆H is directly proportional to the amount of substance produced or reacting in a reactions Consider hydrogen peroxide decomposing: 2H2O2 (l) 2 H2O (g) + O2 (g) + 23.5 kJ 23.5 kcal of heat is produced when 2 moles decompose

What if 1 gram of H2O2 Decompose? 2H2O2 (l) 2 H2O (g) + O2 (g) + 23.5 kJ 2 moles = 2(34 g H2O2)= 68 g if- 68 g H2O2 produce 23.5 kJ then- 1 g produces X kJ X kJ = .35 kJ

2. ∆H for a reaction is equal in magnitude but opposite in sign from the reverse reaction if- 2H2O2 (l) 2 H2O (g) + O2 (g) ∆H= -23.5 kJ then- 2 H2O (g) + O2 (g) 2H2O2 (l) ∆H=+23.5 kJ

Law of Hess or Hess’s Law 3. If a reaction can be regarded as the sum of 2 or more other reactions, ∆H for the overall reaction must be the sum of the enthalpies of the component reactions

Consider: Sn(s) +Cl2(g) SnCl2 (s) ∆H= -83.6 kJ SnCl2(s) +Cl2(g) SnCl4(l) ∆H = -46.7 kJ then Sn(s) +2l2(g) SnCl4(l) ∆H =H1 + H2 ∆H = -130.3 kJ Another example

A fairly easy one! Calculate the heat evolved in the formation of 1 mole of PbSO4(s) from its elements, given the following: Pb(s) + S(s) → PbS(s) + 94 kJ/mol PbS(s) + 2O2(g) → PbSO4(s) + 824 kJ/mol a) ∆H = 730 kJ/mol b) ∆H = -730 kJ/mol c) ∆H = 918 kJ/mol d) ∆H = -918 kJ/mol  -918 kJ/mol 

Here are three more- a tad more challenging 1. Calculate the standard enthalpy change, ΔHo, for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements. -1220 kJ

2. The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is +15.3 kJ

This one is REALLY a test of your knowledge!! 3. One reaction involved in the conversion of iron ore to the metal is -11.0 kJ

Let’s go on-- ANOTHER METHOD of Finding ∆H HEATS OF FORMATION 1

As a concise way of recording thermochemical data, tables listing the quantities of heat accompanying many chemical changes have been compiled, in units of both joules and calories.

HeATs OF FORMATION The Heat of Formation (∆Hf) is the amount of heat released when one mole of a substance is formed from its elements at 25oC and 1 atm pressure. Here are some Heats of Formation- Write the reactions they represent(1st one is given): C(s) + ½ O2(g) –> CO(g)

HEATS OF FORMATION Here’s what a real table of Standard Heats of Formation (DHºf) looks like:

ΔHreaction =S∆Hfproducts – S∆Hfreactants These values allow us to calculate ΔH for a chemical reaction indirectly If we know the ∆Hf of the reactants and the ∆Hf of the products (these will be given), we can use the following equation: ΔHreaction =S∆Hfproducts – S∆Hfreactants This states that the enthalpy change of a reaction is equal to the heats of formation of the products minus the heats of formation of the reactants. S MEANS SUMMATION!

Heats of formation EXAMPLE CH4 + 2O2  2H2O + CO2 ΔHf of CH4 = -75 kJ/mol ΔHf of O2 = 0 kJ/mol ΔHf of H2O = -242 kJ/mol ΔHf of CO2 = -394 kJ/mol ΔHreaction = [product energy – reactant energy] = [2 (-242) + (-394)] – [-75 + 2 (0)] = -803 kJ GIVEN in TABLE

Calculate ΔH for the following reaction: Your turn Calculate ΔH for the following reaction: 8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s) ΔH = Σ ΔHf products - Σ ΔHf reactants ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s) ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ) ΔH = -3316.5 kJ Heats of formation Table

A Question from 2000 AP Chem Exam

ANOTHER METHOD of Finding ∆H Bond Energies 1

The strength of a chemical bond is measured by the energy required to break the bond, i.e. to separate the atoms and leave them as distinct isolated gaseous atoms. Bond Energies are ALWAYS +

Bond Energies for commonly occurring diatomic molecules range from: N2 946 kJ to 150 kJ I2 HF 569 kJ to 295 kJ HI

Molecules with 3 or more atoms have 2 or more bonds The ∆Hf for such molecules are equal to the sum of the Bond Energies

∆Hf for NH3 Equals 3 B.E. N-H

∆Hf for CH3OH Equals 3B.E.’s C-H + BE C-0 + BE 0-H

B.E.’s can be used to calculate ∆H values for reactions – IF reaction is viewed as the forming and breaking of bonds

NH3 + H2O ---> NH4OH H [H-N-H+ ] +[ O-H]- H ∆H = BE H-O + (-BE N-H) H-N-H + H-O-H --> H H [H-N-H+ ] +[ O-H]- H ∆H = BE H-O + (-BE N-H) Note: BE’s are + when bonds broken BE’s are - when bonds formed Bond Formed Bond Broken

C(s) + 2H2(g) --> CH4(g) ∆H = 2 (+BE H-H) + 4(-BE C-H) 2nd Example: C(s) + 2H2(g) --> CH4(g) ∆H = 2 (+BE H-H) + 4(-BE C-H) = 2(432 kJ) + 4(-413k j) ∆H = -788 KJ

BREAK EVERYTHING IN THE REACTANTS--- FORM EVERYTHING IN THE PRODUCTS!! Generally- For those who don’t want to think about what bonds are broken or formed- BREAK EVERYTHING IN THE REACTANTS--- FORM EVERYTHING IN THE PRODUCTS!!

Use the formula- ∆H= S BE’s bonds broken - S BE’s bonds formed

A Table of Bond Energies looks like: H C N O F S Cl Br I 436 415 390 464 569 340 432 370 295 H 345 290 350 439 340 260 330 275 C 160 200 270 --- 200 245 --- N 140 185 --- 205 --- 200 O 285 255 195 --- F REMEMBER: ∆H= S BE’s bonds broken - S BE’sbonds formed

We could use this table to find ∆H for H2 + Cl2 --> 2HCl ∆H = BE H-H + + BE Cl-Cl - 2BE H-Cl From Table ∆H = 436 kj + 243 kj - 2(432 kJ) ∆H = 186 kj/mol

SUMMARY

∆H= S BE’s bonds broken - S BE’sbonds formed Source The net energy change during a reaction is the sum of the energy required to break the reactant bonds and the energy released in forming the product bonds. The net energy change may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released. Any bond that can be formed can be broken. These processes are in opposition. (their enthalpy changes are equal in magnitude, opposite sign) ΔH bonds breaking  ENDOTHERMIC (+) ΔH bonds forming  EXOTHERMIC (-) To find ΔHrxn, apply Hess’s Law: ΔHrxn = ΣΔH bonds breaking (+) + Σ ΔH bonds forming (-) To calculate or estimate ΔHrxn from Bond Energy: Draw the Lewis Structure. Don’t forget about double and triple bonds! Add up ΔH bonds breaking. It’s + (kJ) Add up ΔH bonds forming. It’s - (kJ). Add the two terms. Units are kJ/mol rxn. . Table of Bond Energies H C N O F S Cl Br I 436 415 390 464 569 340 432 370 295 H 345 290 350 439 340 260 330 275 C 160 200 270 --- 200 245 --- N 140 185 --- 205 --- 200 O 285 255 195 --- F ∆H= S BE’s bonds broken - S BE’sbonds formed

Video REVIEW

∆H= S BE’s bonds broken - S BE’sbonds formed Source The net energy change during a reaction is the sum of the energy required to break the reactant bonds and the energy released in forming the product bonds. The net energy change may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released. Any bond that can be formed can be broken. These processes are in opposition. (their enthalpy changes are equal in magnitude, opposite sign) ΔH bonds breaking  ENDOTHERMIC (+) ΔH bonds forming  EXOTHERMIC (-) To find ΔHrxn, apply Hess’s Law: ΔHrxn = ΣΔH bonds breaking (+) + Σ ΔH bonds forming (-) To calculate or estimate ΔHrxn from Bond Energy: Draw the Lewis Structure. Don’t forget about double and triple bonds! Add up ΔH bonds breaking. It’s + (kJ) Add up ΔH bonds forming. It’s - (kJ). Add the two terms. Units are kJ/mol rxn. . Table of Bond Energies H C N O F S Cl Br I 436 415 390 464 569 340 432 370 295 H 345 290 350 439 340 260 330 275 C 160 200 270 --- 200 245 --- N 140 185 --- 205 --- 200 O 285 255 195 --- F ∆H= S BE’s bonds broken - S BE’sbonds formed Video

ΔH= Bonds Broken – Bonds Formed = [H-H + Cl-Cl]-[2(H-Cl)] = [436 kJ/mol+242 kJ/mol] – [2(431)] = -184 kJ/mol   = [4 H-C + C=C + F-F] – [C-C + 4 C-H + 2 C-F] = [4(413) + 614 + 155]-[348 + 4(413) + 2(485)] = 2421 – 2970 kJ/mol = -549 kJ/mol = [6 H-C + 2 C-O + 2 O-H] – [6 H-C + 2 C-O + 2 O-H] = [6(413) + 2(358) + 2(463)]-[ 6(413) + 2(358) + 2(463)] = 4120 – 4120 kJ/mol = 0 kJ/mol = [H-C + 3 C-Cl] – [2 C-Cl + C=O + H-Cl] = [413 + 3(328)]- [2(328) + 799 + 431] = 1397 – 1886 kJ/mol = -489 kJ/mol = [4 H-C + 2 O=O] – [2 C=O + 4 H-O] = [4(413) + 2(495)]- [2(799) + 4(463)] = 2642 – 3450 kJ/mol = -808 kJ/mol