Thermochemistry Chapter 6! By James Lauer and David Miron

6.1 The Nature of Energy Key Terms First law of thermodynamics - The energy of the universe is constant Energy - the capacity to do work or to produce heat Law of Conservation of Energy - energy can be converted from one form to another but can be neither created nor destroyed Potential vs Kinetic energy - energy due to position or composition and energy due to motion (KE=1/2mv 2 ) Heat - the transfer of energy between two objects due to temperature difference Work - force acting over a distance Pathway - can affect energy transfer but the total energy remains constant State Function/Property - property dependent on present state of object System - the part of the universe on which we wish to focus on Surroundings - everything else in the universe Exo- vs Endothermic - energy flows out of the system (combustion) and heat flows into the system Thermodynamics - study of energy and its interconversions Internal Energy (E) - the sum of kinetic and potential energy of all particle in the system (∆E = q + w)

6.1 Energy Flow In the final photo ball A’s potential energy has _______ and ball B’s potential energy has ______ Ball B is not as high (in the final photo) as Ball A was in the initial photo. Why? Where has some of the energy gone? What did it form?

6.1 Equations KE=1/2mv 2 ∆E = q + w q=heat w=work ∆E=change in energy ∆E>0 endo/absorbed ∆E<0 exo, energy released +q=endo, flow in -q=exo, flow out +w=endo, work done on -w=exo, work done by Work = force X distance ---> w= -P∆V V=volume P=external pressure

6.1 Example Calculate ∆E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. (∆E = q + w) ∆E = 15.6 kJ kJ = 17.0 kJ Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. (w = -P∆V) w = -15 atm X 18 L = -270 L x atm

6.2 Enthalpy and Calorimetry Key Terms Enthalpy(H) - H=E+PV state function Calorimeter - device used to determine heat associated with a chemical reaction Heat Capacity(C) - C=heat absorbed/increase in temperature Specific/molar heat capacity - heat capacity per gram(J/°C x g or J/K x g), heat capacity per mole(J/°C x mol or J/K x mol) Constant Pressure Calorimetry - basic calorimeter, to determine changes in enthalpy for reactions in solutions Constant Volume Calorimetry - rigid container (bomb) and reactants are ignited inside

6.2 Equations H = E + PV H= enthalpy E= internal energy of system PV= pressure x volume Chemical reaction ∆H = H products - H reactants Constant pressure ∆H = q p q p = heat at constant pressure Energy released by reaction = s x m x ∆T (specific heat capacity, mass of solution, increase in temperature) Constant volume ∆E = q + w = q v Energy released by reaction = temperature increase x heat capacity of calorimeter

6.2 Constant PressureConstant Volume

6.2 Example When 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na 2 SO 4 solution at 25.0°C in a calorimeter, the white solid BaSO 4 forms, and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution 4.18 J/°C x g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO 4. (∆H=specific heat x mass of solution x increase in temperature) Net ionic equation: Ba 2+ (aq) + SO 4 2- (aq) ---> BaSO 4(s) Mass of solution: 2L=2000ml=2000g Temperature increase: = 3.1°C 4.18 x 2000 x 3.1= 2.6 x 10 4 J all units cancel except J 1000J = 1 kJ ∆H = -26 kJ/mol The reaction is exothermic.

6.3 Hess’ Law Key terms: Hess’ Law- states that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps. Hess’ Law allows scientists to determine the enthalpy of formation in a reaction they are unable to complete using a series of reactions containing the same reactants and products that are within the main reaction. Enthalpy of synthesis = - Enthalpy of Decomposition

6.3 Hess’ Law example What is the value for ΔH for the following reaction? CS 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 SO 2 (g) Given: C(s) + O 2 (g) → CO 2 (g); ΔHf = kJ/mol S(s) + O 2 (g) → SO 2 (g); ΔHf = kJ/mol C(s) + 2 S(s) → CS 2 (l); ΔHf = 87.9 kJ/mol Using this info we can infer from hess’ law CO 2 (g)--> C(s) + O 2 (g) →→→ ; ΔHf = kJ/mol SO 2 (g)--> S(s) + O 2 (g) → ; ΔHf = kJ/mol CS 2 (l)--> C(s) + 2 S(s); ΔHf = kJ/mol

6.3 Hess’ Law example What is the value for ΔH for the following reaction? CS 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 SO 2 (g) Given: C(s) + O 2 (g) → CO 2 (g); ΔHf = kJ/mol S(s) + O 2 (g) → SO 2 (g); ΔHf = kJ/mol C(s) + 2 S(s) → CS 2 (l); ΔHf = 87.9 kJ/mol Using this info we can infer from hess’ law CO 2 (g)--> C(s) + O 2 (g) →→→ ; ΔHf = kJ/mol SO 2 (g)--> S(s) + O 2 (g) → ; ΔHf = kJ/mol CS 2 (l)--> C(s) + 2 S(s); ΔHf = kJ/mol S(s) + O 2 (g) → SO 2 (g); ΔHf = kJ/mol C(s) + O 2 (g) → CO 2 (g); ΔHf = kJ/mol CS 2 (l) + 2 O 2 (g) → CO 2 (g) + 2 SO 2 (g); ΔHf = kJ/mol Focus on compounds

6.4 Standard Enthalpies of Formation Key Terms Standard enthalpy of formation ( Δ H f o ) - the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Degree symbol ( o ) - indicates that the process is carried out under standard conditions Standard State - a precisely defined reference state. Compound: Gas: 1 atm Liquid or solid: Pure liquid or solid Solution: 1 M Element: 1 atm and 25 o C

6.4 Things to wrap your head around: You cannot measure absolute absolute values for enthalpy...you can only determine changes in enthalpy (hence Δ ) Because Enthalpy is a state function, you can use Hess’ Law and manipulate it so you can use it. so Synthesis = -Decomposition etc.

6.4 Equations Δ H o reaction = Σn p ΔH o f (products) - Σn p ΔH o f (reactants) Public Relations P-R

6.4 examples Calculate ΔH for the following reaction: 8 Al(s) + 3 Fe 3 O 4 (s) --> 4 Al 2 O 3 (s) + 9 Fe(s) Given ΔH f Al 2 O 3 (s) = kJ/mol ΔH f Fe 3 O 4 (s) = kJ/mol First ignore the elements, because they are formed, and you don’t need them to find the change in enthalpy 3 Fe 3 O 4 (s) --> 4 Al 2 O 3 (s) Public Relations… P-R so ΔH = 4 ΔH f Al 2 O 3 (s) - 3 ΔH f Fe 3 O 4 (s) ΔH = 4 ( ) - 3( ) = kJ/mol