Unit 5: Thermochemistry Reaction Energy Diagrams: slides 2-7 Specific Heat: slides 8-13 Enthalpy and Hess’s Law: slides

Thermochemistry thermochemistry: studies the relationships between chemical reactions and energy changes involving heat

Thermochemistry Heat and temperature are NOT the same thing! Heat: transfer/flow of energy due to a temperature difference (from high to low) Temperature: a measure of the random motions of molecules (average kinetic energy)

Reaction Energy Diagram Reaction energy diagrams show the energy changes throughout the reaction Activated complex (Also called the transition state) Activation Energy Products Energy Energy change for reaction Reactants Reaction proceeds

energy EXITS the system Exothermic Reactions Exothermic: energy EXITS the system E<0 (neg.) System Energy Surroundings

Endothermic Reactions energy ENTERS the system E>0 (pos.) System Energy Surroundings

Catalysts create a lower energy reaction pathway Activated complex (Also called the transition state) Activation Energy Products Energy Energy change for reaction Reactants Reaction proceeds

Specific Heat specific heat: willingness of an object to change temperature, with the symbol Cp (the p means "under constant pressure") specific heat: the amount of energy required to change the temperature of one gram of a substance by 1°C

Specific Heat Units 1 calorie = energy required to heat 1 gram of water by 1°C 1 Calorie (food labels) = 1 kilocalorie 1 calorie = 4.184 joules

Specific Heat Duluth, next to Lake Superior, stays cool in the summer and relatively warm in the winter. Why? Substance Specific Heat (J/g°C) copper 0.3845 granite 0.7953 lead 0.1276 ice 2.06 water 4.184

Calculating Heat (Energy Transfer) q = Cp x m x T change in temperature (units: °C) specific heat (units: J/g°C) mass (units: g)

Calculating Heat Example #1 q = Cp x m x T 3.05 kg of aluminum is heated from 22.1C to 67.5C. Calculate the heat absorbed in both J and kJ by the metal. The specific heat of aluminum is 0.900 J/gC.

Calculating Heat Example #2 q = Cp x m x T A 200-g block of copper at temperature of 90 °C is dropped into 400 g of water at 27 °C. The What is the final temperature of the mixture? The specific heat of copper is 0.386 J/gC.

H = ΣHproducts – ΣHreactants Enthalpy Enthalpy is the heat energy absorbed or lost in a reaction (H or H if talking about change) H = ΣHproducts – ΣHreactants

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Enthalpy Example Calculate ΔH for the following reaction: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Substance ΔH (kJ/mol) CH4 -74.8 O2 CO2 -393.5 H2O -285.83

Hess’s Law Hess’s Law: overall enthalpy changes are the sum of all steps of the reaction process Add enthalpies (H) from all reaction steps Note: If a reaction is reversed, the sign of H is also reversed The magnitude of H depends on the quantities of reactants and products Multiply H by the coefficient for that substance

Hess’s Law Example Calculate the heat of reaction, H, for the overall reaction: N2(g) + 2O2(g)  2NO2(g) Given: 2NO(g)  N2(g) + O2(g) H=-180 kJ 2NO(g) + O2(g)  2NO2(g) H=-112 kJ

Hess’s Law Practice Calculate the heat of reaction, H, for the overall reaction: 2 S (s) + 3 O2 (g)  2 SO3 (g) Given: S (s) + O2 (g) SO2 (g)  ∆H  =  -297 kJ 2 SO3 (g)2 SO2 (g) + O2 (g)  ∆H  =  198 kJ 

PbCl2 (s) + Cl2 (g) → PbCl4 (l) Hess’s Law Exit Slip Calculate ΔH for this reaction: PbCl2 (s) + Cl2 (g) → PbCl4 (l) Given: Pb (s) + Cl2 (g) → PbCl2 (s) ∆ H = - 359.4 kJ Pb (s) + 2 Cl2 (g) → PbCl4 (l) ∆ H = - 329.3 kJ