In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values)

In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values) 2. Energies required to break bonds vary from molecule to molecule

In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values) 2. Energies required to break bonds vary from molecule to molecule 3. Bond energies are average bond breaking energies (all positive numbers)

In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values) 2. Energies required to break bonds vary from molecule to molecule 3. Bond energies are average bond breaking energies (all positive numbers) 4. Enthalpy is a state function – the path doesn’t matter, only the net change

In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values) 2. Bond energies are average bond breaking energies (all positive numbers) 3. Enthalpy is a state function – the path doesn’t matter, only the net change

In Summary from Yesterday: 1. Breaking bonds requires E (endothermic, + values) and forming bonds releases E (exothermic, - values) 2. Energies required to break bonds vary from molecule to molecule 3. Bond energies are average bond breaking energies (all positive numbers) 4. Enthalpy is a state function – the path doesn’t matter, only the net change

Unit 10: Thermodynamics

First Law of Thermodynamics Law of Conservation of Energy – Energy can not be created or destroyed, only change formge forms The energy change for a closed system is the sum of energy transferred as heat and the energy transferred as work. Formula:

First Law of Thermodynamics Formula: ∆U = q + w How can the Energy from a chemical reaction “do work”? Dry ice in a plastic bag …

So then, what is Enthalpy exactly? Formula: ∆U = q + w Since most times in chemistry we aren’t worried about the work that a reaction does, we measure only the heat it exchanges with the surroundings Enthalpy, is the measure of this heat, ignoring the work ∆H = qp

Enthalpy: ∆H are stoichiometric quantities and are paired with physical or chemical changes on a per mole-rxn basis. Units: ∆H = 𝑘𝐽 𝑚𝑜𝑙−𝑟𝑥𝑛 Read as: “kilojoules per mole of reaction” Sign conventions: -∆H = exothermic = heat is released +∆H = endothermic = heat is absorbed

Enthalpy: ∆H are stoichiometric quantities and are paired with physical or chemical changes on a per mole-rxn basis. Example 1: How much energy is evolved or absorbed when 15 g of ethane is burned? 2 C2H6 + 7 O2  4 CO2 + 6 H2O ∆H = -2857.3 kJ/molrxn

1. Bond dissociation energies Enthalpy: Methods of Calculation 1. Using bond dissociation energies

1. Bond dissociation energies Enthalpy: Methods of Calculation 1. Using bond dissociation energies Example 2: Determine the ∆H for the reaction of hydrogen gas with oxygen gas to produce water.

1. Bond dissociation energies 2. Hess’ Law Enthalpy: Methods of Calculation 2. Using Hess’ Law Enthalpy is a state function. It doesn’t matter how you get there, the net change in H is the same!!

1. Bond dissociation energies 2. Hess’ Law Enthalpy: Methods of Calculation 2. Using Hess’ Law Enthalpy is a state function. It doesn’t matter how you get there, the net change in H is the same!!

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation 3. Enthalpies of Formation

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation 3. Enthalpies of Formation

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation 4. Lab Method - Calorimetry 4. Calorimetry For our next lab day …

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation Try #1, 4 and 6 NOW! 1. How much heat will be transferred when 5.81g of graphite reacts with excess H2 according to the following equation? Is the reaction endothermic or exothermic?   6C+ 3H2 → C6H6 ∆Ho = 49.03 kJ/molrxn

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation Try #1, 4 and 6 NOW!

1. Bond dissociation energies 3. Enthalpies of formation 2. Hess’ Law Enthalpy: Methods of Calculation Try #1, 4 and 6 NOW!