THERMOCHEMISTRY Thermodynamics To play the movies and simulations included, view the presentation in Slide Show Mode.

ENTHALPY CHANGES be able to define enthalpy and enthalpy change (ΔH) be able to define standard enthalpy chemical reactions involve the making and breaking of bonds be able to determine bond enthalpies from data use mean bond enthalpies to estimate ΔH for reactions be able to calculate enthalpy change from experimental data use Hess’s law to calculate enthalpy changes in reactions

Internal Energy and Enthalpy e.g. Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) Heat change at constant pressure = Change in internal energy + Work done on the surroundings (Heat change at constant volume) Enthalpy change

THERMODYNAMICS First Law Energy can be neither created nor destroyed but It can be converted from one form to another Second Law Energy flows from hot to cold Third Law A perfectly ordered crystal has zero entropy. Chemical reactions either absorb or give off energy Exothermic Energy is given out Endothermic Energy is absorbed Every Day Examples Exothermic combustion of fuels respiration (oxidation of carbohydrates) Endothermic photosynthesis thermal decomposition of calcium carbonate

Exothermic and Endothermic Reactions An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)

Exothermic and Endothermic Reactions An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)

Bond breaking - endothermic Energy is always required to be inputted to break a bond. Bond breaking is always endothermic.

Bond making - exothermic Energy is always released when a bond is formed. Bond making is always exothermic.

Enthalpy IN to Breaking Bonds e.g. CH4 + 2O2 CO2 + 2H2O In an exothermic reaction, the energy required in breaking the bonds in the reactants is less than the energy released in forming the bonds in the products (products contain stronger bonds).

Energy out, to make…. Bonds In an endothermic reaction, the energy in breaking the bonds is more than the energy out making the bonds

Single bond < Double bond < Triple bond The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H2 (g) H (g) + DH0 = 436.4 kJ Cl2 (g) Cl (g) + DH0 = 242.7 kJ HCl (g) H (g) + Cl (g) DH0 = 431.9 kJ O2 (g) O (g) + DH0 = 498.7 kJ N2 (g) N (g) + DH0 = 941.4 kJ O Bond Energies Single bond < Double bond < Triple bond N Notes • strength of bonds also depends on environment; MEAN values quoted • making bonds is exothermic as it is the opposite of breaking a bond

Calculation of enthalpy of reaction from bond energies Only the net number and types of bonds broken (red bonds above) and formed (blue bonds above) are included in the calculation. 12

Combustion of methane CH4(g) + 2O2(g) 2H2O(l) + CO2(g) + ENERGY

Burning Methane CH4 + 2O2 2H2O + CO2 Methane Oxygen Water Carbon dioxide

Bond energies C-H = 413KJ O=O = 498 KJ Total for breaking bonds = 4x413 + 2x497 = 2658 KJ/mol H-O = 464 Kj C=O = 805 KJ Total for making bonds = 2x805 + 4x464 = 3466 KJ/mol Total energy change = 2658-3466 = -808 KJ/mol

Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bond Bond energy KJ/mol H-H 436 Cl-Cl 242 H-Cl 431 C-H 413 C-C 347 C-O 335 O=O 498 Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bonds broken = 4 x (C-H) + 2 x (O=O) = 4 x 413 + 2 x 498 = 1662 + 996 = 2658 KJ/mol

ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bonds made = 4 x (O-H) + 2 x (C=O) Bond energy KJ/mol H-H 436 Cl-Cl 242 H-Cl 431 C-H 413 C-C 347 C-O 335 O=O 498 ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bonds broken = 4 x (C-H) + 2 x (O=O) (Previous page) = 4 x 413 + 2 x 498 = 1662 + 996 = 2658 KJ/mol Bonds made = 4 x (O-H) + 2 x (C=O) = 4 x -464 + 2 x -805 = -1856 + -1610 = -3466 KJ/mol

ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bond Bond energy KJ/mol H-H 436 Cl-Cl 242 H-Cl 431 C-H 413 C-C 347 C-O 335 O=O 498 CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bonds broken = 4 x (C-H) + 2 x (O=O) = 4 x 413 + 2 x 498 = 1662 + 996 = 2658 KJ/mol Bonds made = 4 x (O-H) + 2 x (C=O) = 4 x -464 + 2 x -805 = -1856 + -1610 = -3466 KJ/mol Overall Energy change = 2658 + -3466 = -808 KJ/mol (Exothermic)

Standard Enthalpy Changes CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -808 kJ mol-1 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1 As enthalpy changes depend on temperature and pressure, chemists find it convenient to report enthalpy changes based on an internationally agreed set standard conditions: 1. elements or compounds in their normal physical states; 2. a pressure of 1 atm (101.325 KPa); and 3. a temperature of 250C (298 K) Enthalpy change under standard conditions denoted by symbol: H ø

Standard Enthalpy Changes of Reactions Standard Enthalpy Change of Formation The standard enthalpy change of formation (Hf) is the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions. ø e.g. 2Na(s) + Cl2(g)  2NaCl(s) H = -822 kJ mol-1 ø Standard enthalpy change of formation of NaCl is -411 kJ mol-1. Na(s) + ½Cl2(g)  NaCl(s) Hf = -411 kJ mol-1 1 mole ø

STANDARD ENTHALPY OF FORMATION Definition The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DH°f This little f is VERY important Example(s) C(graphite) + O2(g) ———> CO2(g) H2(g) + ½O2(g) ———> H2O(l) 2C(graphite) + ½O2(g) + 3H2(g) ———> C2H5OH(l) Notes Only ONE MOLE of product on the RHS of the equation Elements In their standard states have zero enthalpy of formation.

Enthalpies of Formation CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Using Enthalpies of Formation, Calculate Enthalpies of Reaction For a reaction (Method #2) Note: n & m are stoichiometric coefficients. Calculate heat of reaction for the combustion of methane gas giving carbon dioxide and water. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆Hof (kJ/mol): Try this

One more time!!!! C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) + ???

Hess’s Law Second Method reactants products Hreaction elements  Hf [reactants] Methane and oxygen destroyed ø Hf [products] CO2 and water formed ø Hreaction =  Hf [products] -  Hf [reactants] ø

Standard Enthalpy Changes of Reactions Standard Enthalpy Change of Combustion e.g. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H1 = -2220 kJ 2C3H8(g) + 10O2(g) 6CO2(g) + 8H2O(l) H2 = ? H2 = -4440 kJ  It is more convenient to report enthalpy changes per mole of the main reactant reacted/product formed. ANY IDEAS WHAT IT WOULD BE FOR THE FORMATION OF 18.02 g OF WATER?

3rd Method For Heats of Rx Really this is NOT another law, it is still Hess’s Law BUT It looks different, yet we get to the same place , the same answer. There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.

Hess’s Law Flip Method https://www.youtube.com/watch?v=9oaNmEqY-ss

4C(s) + 5H2(g) → C4H10(g) Example 1 Use these equations to calculate the molar enthalpy change to make butane gas. C4H10(g) + 6½O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol C(s) + O2(g) → CO2(g) ∆H= -393.5kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

4C(s) + 5H2(g) → C4H10(g) 4C(s) + 5H2(g) → C4H10(g) ∆H = -125.6kJ/mol Example 1 4C(s) + 5H2(g) → C4H10(g) 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) 5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol) _____________________________________________________ ∆H = -125.6kJ/mol 4C(s) + 5H2(g) → C4H10(g)

Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse & x 2 Found in more than one place, SKIP IT (its hard). O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ

4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x2 Found in more than one place, SKIP IT. O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ Cancel terms and take sum. + 5O2 + 6H2O H = -906.3 kJ 4NH3  4NO Is the reaction endothermic or exothermic?

Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

Importance of Mathematical Hess’s Law The enthalpy change of some chemical reactions cannot be determined directly because: the reactions cannot be performed in the laboratory the reaction rates are too slow the reactions may involve the formation of side products