1. 1. Definition Heat flow in a system SymbolH Like internal energy, the change in enthalpy is important State function – measure of its current conditions ΔH = H products - H reactants ΔH = + ΔH = - System is gaining heat System is losing heat Endothermic Exothermic Enthalpy of a chemical reaction =Heat of Reaction

2. 2. Thermochemical equation 2 H 2 + O 2  2 H 2 O Δ H = -483.6 kJ This indicates that 483.6 kJ are released when 2 moles of hydrogen and one mole of oxygen create 2 moles of water. For thermochemical equations, assume everything is moles 3. Enthalpy Diagram Simple diagram to show changes in enthaply Example - A + B  CΔ H = -216 kJ This means that the reactants have more energy than the products A + B C Δ H = -216 kJ

3. 4. Rules for Enthalpy a. Extensive propertyDepends on how much mass we have b. Reverse reaction has the same magnitude, opp sign If we used 0.5 moles of A, we would only release ½ the amount of kJ A + B C Δ H = -216 kJ Δ H = +216 kJ c. Enthalpy depends on physical state The enthalpy associated with each state of matter is different Watch your phases! Practice?

4. Calorimetry At a constant pressure, the heat lost from the system must equal the heat gained by the surroundings. Measurement of heat flow 3. Heat capacity C Energy required to raise the temperature of the sample 1 o C Q = C  T Units = J/ o C Extensive property – depends on mass 2. Molar heat capacity Energy required to raise 1 mole of the sample 1 o C Units = J/mole o C 1.Specific heat C s Energy required to raise 1 gram of the sample 1 o C Q = m c  T Units = J/g o C Intensive propertyOnly depends on composition of sample A. Definitions CmCm

5. B. Types of Calorimetry 1. Constant Pressure Must take into account the heat lost or gained by all objects AND the calorimeter Example – If we add 10.0 g of a metal at 98 o C to 100.0 g of water at 10.0 o C in a calorimeter with a heat capacity of 8.5 J/ o C, the final temperature is 12.5 o C. What is the specific heat of the metal. 1.Determine the total heat gained by calorimeter and water. Water Q = m c  T Q = (100.0)(4.19)(2.5) Q= 1047.5 J Calorimeter Q = C p  T Q=(8.5)(2.5) Q=21.25 J The sum of these two is how much energy the metal lost.

6. Q=1047.5 + 21.25 1068.75 J Now use this to determine C s of metal Q = mc  T 1068.75 = 10.0 (c) 85.5 c = 1.25 J/g o C

7. 2. Bomb calorimetry Usually for combustion Container inside calorimeter heats as the reaction occurs Must take into account the energy absorbed by the container is the energy released when the substance burns. per gram of quinone? Example – A 2.200 g sample of quinone (C 6 H 4 O 2 ) is burned in a bomb calorimeter whose heat capacity is 7.85 kJ/ o C. The temperature of the calorimeter increases from 23.44 o C to 30.57 o C. What is the heat of combustion per gram of quinone? per mole of quinone? What is the heat of combustion per mole of quinone?

8. Heat gained by calorimeter = heat lost by quinone Q = C p Δ T In this case, the mass of the calorimeter and specific heat are combined Q = (7.854 kJ/ o C)(7.13 o C) Q = 55.99902 kJEnergy released by 2.200 g of material 55.99902 2.200 = 25.45 kJ/g To convert this into kJ/mole, use the gram molecular mass 25.45 kJ g x 108.1 g mol = 2751 kJ/mol