Chapter 5.2 notes: Hess’ Law chapter 5.3 notes: bond enthalpies 5.2: In chemical transformations energy can neither be created nor destroyed (the first law of thermodynamics) 5.3: energy is absorbed when bonds are broken and is released when bonds are formed

Important terms for these sections Chapter 5.2 Hess’ Law Energy cycle Σ (sum) Standard enthalpy of formation Chapter 5.3 Bond enthalpy Average bond enthalpies Free radicals Steady state

Enthalpy cycles Sometimes it’s hard to determine the enthalpy change directly, so an indirect route is necessary This is shown in the example cycles

Enthalpy cycles This is shown for the reaction in the cycle below Experimental values shown: Law of conservation of energy: enthalpy change in complete cycle = 0 = Hess’ Law

Hess’ Law The enthalpy change for any chemical reaction is independent of the route, provided the starting conditions and final conditions, and reactants and products are the same It’s like a logic and math problem put together! A little puzzle! Note: some steps in the enthalpy cycle may be hypothetical! Requirement is that the reactions must balance for the whole cycle

Standard enthalpy changes of reaction & formation ΔHѳf std enthalpy change of formation Enthalpy change that occurs when 1 mole of a substance is formed from its elements in their std states 298K (25 °C) 1.00 x 105 Pa These give a measure of the stability of a substance relative to its elements Can be used to calculate the enthalpy changes of all reactions, either hypothetical or real General cycle and expression: Σ means ‘sum of’ ΔHѳf for any element in its standard state is zero

Calculating Std enthalpy change of reaction Problem 1 Given these enthalpy changes: 2C(s) + O2(g)  2CO(g) ΔHѳ = -222 kJ mol-1 C(s) + O2(g)  CO2(g) ΔHѳ = -394 kJ mol-1 Calculate the enthalpy change for the reaction 2CO(g) + O2(g)  2CO2(g) Remember ΔHѳr is std enthalpy change of reaction There are 2 ways to solve this problem Using a cycle Manipulating the equations

Problem 1 Method 1: using a cycle Given these enthalpy changes: 2C(s) + O2(g)  2CO(g) ΔHѳ = -222 kJ mol-1 ΔH1 REACTION 1 C(s) + O2(g)  CO2(g) ΔHѳ = -394 kJ mol-1 ΔH2 REACTION 2 Calculate the enthalpy change for the reaction 2CO(g) + O2(g)  2CO2(g) ΔHѳ 2CO(g) + O2(g) 2CO2(g) ΔH1 2C(s) + O2(g) + O2(g) ΔH1 2C(s) + O2(g) + O2(g) 2ΔH2

Problem 1 Method 1: using a cycle Calculate the enthalpy change for the reaction 2CO(g) + O2(g)  2CO2(g) ΔHѳ 2CO(g) + O2(g) 2CO2(g) ΔH1 2C(s) + O2(g) + O2(g) ΔH1 2C(s) + O2(g) + O2(g) 2ΔH2 ΔHѳr = - ΔH1 + 2ΔH2 = - (-222) + (2x -394) = -566 kJ mol-1

Problem 1 Method 2: manipulating the equations Given these enthalpy changes: 2C(s) + O2(g)  2CO(g) ΔHѳ = -222 kJ mol-1 C(s) + O2(g)  CO2(g) ΔHѳ = -394 kJ mol-1 Calculate the enthalpy change for the reaction 2CO(g) + O2(g)  2CO2(g) This is like a math problem! need CO on the left side: 2CO(g)  2C(s) + O2(g) ΔHѳ = +222 kJ mol-1 need to multiply the C by 2: 2C(s) + 2O2(g)  2CO2(g) ΔHѳ = -788 kJ mol-1 2CO(g)  2C(s) + O2(g) ΔHѳ = +222 kJ mol-1 2C(s) + 2O2(g)  2CO2(g) ΔHѳ = -788 kJ mol-1 2CO(g) + O2(g)  2CO2(g) ΔHѳ = -566 kJ mol-1

Problem 2: Std. Enthalpy change of formation Given these standard enthalpy change values (kJ mol-1): ΔHѳf [NH3(g)] = -46 kJ mol-1 ΔHѳf [NO(g)] = +90 kJ mol-1 ΔHѳf [H2O(g)] = -242 kJ mol-1 Calculate the enthalpy change for the reaction, ΔHѳr : 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) ΣΔHf (products) = (4 x 90) + [6 x (-242)] = -1092 kJ mol-1 ΣΔHf (reactants) = 4 x (-46) = -184 kJ mol-1 (remember that O2 is zero = element at std state) ΔHѳr = -1092 – (-184) = -908 kJ mol-1

Chapter 5.3: Bond enthalpies Covalent bond: electrostatic attraction between shared pair of electrons and the positive nuclei of the bonded atoms. Bond enthalpy is the enthalpy change when 1 mole of covalent bonds, in a gaseous molecule, is broken under standard conditions Bond enthalpies are used for reactions occurring in the gaseous state It takes energy to BREAK BONDS!!! endothermic Forming bonds is typically favorable (releases energy)

Bond enthalpies to calculate ΔHr Example: combustion of methane (here we assume all gas states) Use the bond enthalpies to calculate:

Ozone The ozone is important, m’kay! O2 and O3 play important roles in making sure the ozone layer is protecting the earth. Remember E=hv OR c = λν where E = photon emission (EM spectrum) h = Planck’s constant v = frequency λ = wavelength c = speed of light (3.00x 108 ms-1) Looking at the structures, O2 vs O3, Draw the bonds below: O O vs O O O Why does it take lower/less energy to break ozone (330nm) than oxygen (242nm)?

Ozone