Assign.#6.6 – Enthalpies of Bonds, Phase Changes, and Reactions

Bond Strengths The strength of a covalent bond is determined by the energy required to break it Bond enthalpies is the ΔH for the breaking of a certain bond Bond enthalpies are usually given as energy to break a given bond between two atoms Always a positive number Always requires energy to break a bond

Which Bond is Harder to Break? Why? #2 #1 #3

Single, Double, and Triple Bonds The values of ΔH change based on whether the bond is single, double, or triple. It takes more energy to break a double bond compared to a single bond and more energy to break a triple bond compared to a double bond. Due to increased pi bonds that require more energy input in order to break Also, bonds get shorter the more we have, making it harder for breaking to occur.

Heating Curves Have students label what each of these regions indicate (solid, liquid, gas, etc.)

Enthalpies of Phase Change When a substance changes phases, there is energy that is required in order to break intermolecular forces The enthalpies of phase change are dependent upon the substance of interest: ΔHvaporization, ΔHfusion, ΔHfreezing, etc.

Signs for ΔH for Phase Changes To turn something from a solid to a liquid and/or a liquid to a gas requires energy. Therefore it is endothermic. This means ΔH is always positive To turn something from a gas to a liquid and/or a liquid to a solid releases energy. Therefore it is exothermic. This means ΔH is always negative The value for vaporization and condensation is the same, the only difference is the sign.

Enthalpy Cartoon Create a cartoon creating of the different types of enthalpy that exist: Enthalpy Enthalpy of Bond Strengths Enthalpy of Phase Changes Enthalpy of Reactions

End Catalyst Why are bond enthalpies always positive values? The ΔHvaporization = 890 kJ for a particular sample. What is the ΔHcondensation? What is ΔH equivalent to at constant pressure? End

Hydrogen Fire Ball

Justify – TPS Why is there a large release of energy for this reaction? Use ΔH in your response.

Enthalpies of Reactions Since ΔH means Hfinal – Hinitial, this means that for a reaction it is: ΔHrxn = Hproducts – Hreactants Every reaction has a enthalpy (or heat) of reaction ΔHrxn -ΔHrxn = exothermic +ΔHrxn = endothermic

Major Ideas of ΔHrxn ΔHrxn is proportionate to amount of reactant consumed When a reaction is run in reverse, the ΔHrxn is equal in magnitude, but opposite in sign ΔHrxn depends on the state of the substance

CH4 (g) + 2 O2 (g)  CO2 (g) + 2H2O (l) ΔH =-890 kJ/mol Class Example Calculate the qrxn when 4.50 g of methane (CH4) is combusted at constant pressure. Note – For the reaction: CH4 (g) + 2 O2 (g)  CO2 (g) + 2H2O (l) ΔH =-890 kJ/mol

2 H2O2 (l)  2 H2O (l) + O2 (g) ΔH = -196 kJ/mol Table Talk Hydrogen peroxide can decompose to water and oxygen by the reaction: 2 H2O2 (l)  2 H2O (l) + O2 (g) ΔH = -196 kJ/mol Calculate the qrxn when 5.00 g of H2O2 decomposes at constant pressure.

ΔH as a State Function All ΔH values are state functions, meaning it only depends on the amount of substance and the initial state of that substance. The number of steps between initial and final state does not matter, only the difference between initial and final enthalpy values. Example: Combustion of methane

Hess’s Law If a reaction is carried out in a series of steps, ΔH for the reaction equations the sum of the individual steps This law means that we do not need to measure ΔH for all known reactions, rather we only need to measure for a some reactions and then we can add equations together to get the equation of interest.

Class Example The ΔHrxn for a combustion of C to CO2 is -393.5 kJ/mol, and the enthalpy for the combustion of CO to CO2 is -283.0 kJ/mol: (1) C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ/mol (2) CO (g) + ½ O2 (g)  CO2 (g) ΔH = -283 kJ/mol Using these data, calculate the enthalpy for the combustion of C to CO: (3) C (s) + ½ O2 (g)  CO (g) ΔH = ?

C (graphite)  C (diamond) ΔH = ? Table Talk (1) C (graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ/mol (2) C (diamond) + O2 (g)  CO2 (g) ΔH = -395.4 kJ/mol Calculate ΔH for the conversion of graphite to diamond: C (graphite)  C (diamond) ΔH = ?

Challenge Problem! Calculate ΔH for the reaction: 2 C (s) + H2 (g)  C2H2 (g) Given the following chemical equations and their respective enthalpy changes: C2H2 (g) + 5/2 O2 (g)  2 CO2 (g) + H2O (l) ΔH = -1299.6 kJ/mol C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ/mol H2 (g) + ½ O2 (g)  H2O (l) ΔH = -285.8 kJ/mol

Around the World Complete the 8 questions that are around the room

2 C (s) + O2 (g) + 4H2 (g)  2CH3OH (g) ΔH = -402 kJ/mol Exit Ticket From the enthalpies of reaction: 2 C + O2  CO (g) ΔH = -221.0 kJ/mol 2 C (s) + O2 (g) + 4H2 (g)  2CH3OH (g) ΔH = -402 kJ/mol Calculate the ΔH for the reaction: CO (g) + 2 H2 (g)  CH3OH (g)

Rate Yourself Rate yourself 1 – 4 on LTs 5.6 to 5.8.

Closing Time Read 5.4, 5.6, and 8.8 Do book problems: 5.36, 5.41, 5.43, 5.44, 5.62, 5.63, 5.64, 5.65, 5.66