Thermochemistry

1. Thermochemistry Chapter 6! By James Lauer and David Miron

2. 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=1/2mv2) • 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)

3. 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?

4. 6.1 Equations KE=1/2mv2 ∆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

5. 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 + 1.4 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. 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

7. 6.2 Equations • H = E + PV H= enthalpy E= internal energy of system PV= pressure x volume • Chemical reaction ∆H = Hproducts - Hreactants • Constant pressure ∆H = qp qp= 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 = qv • Energy released by reaction = temperature increase x heat capacity of calorimeter

8. 6.2 Constant Pressure Constant Volume

9. 6.2 Example When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a calorimeter, the white solid BaSO4 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 BaSO4. (∆H=specific heat x mass of solution x increase in temperature) Net ionic equation: Ba2+(aq) + SO42-(aq) ---> BaSO4(s) Mass of solution: 2L=2000ml=2000g Temperature increase: 28.1-25.0= 3.1°C 4.18 x 2000 x 3.1= 2.6 x 104 J all units cancel except J 1000J = 1 kJ ∆H = -26 kJ/mol The reaction is exothermic.

10. 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

11. 6.3 Hess’ Law example What is the value for ΔH for the following reaction? CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g) Given: C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol Using this info we can infer from hess’ law CO2(g)--> C(s) + O2(g)→→→; ΔHf = 393.5 kJ/mol SO2(g)--> S(s) + O2(g)→; ΔHf = 296.8 kJ/mol CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9 kJ/mol

12. 6.3 Hess’ Law example Focus on compounds What is the value for ΔH for the following reaction? CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g) Given: C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol Using this info we can infer from hess’ law CO2(g)--> C(s) + O2(g)→→→; ΔHf = 393.5 kJ/mol SO2(g)--> S(s) + O2(g)→; ΔHf = 296.8 kJ/mol CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9 kJ/mol CS2(l)--> C(s) + 2 S(s); ΔHf = -87.9 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g); ΔHf = -1075.0 kJ/mol

13. 6.4 Standard Enthalpies of Formation Key Terms Standard enthalpy of formation (Δ Hfo ) - 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 25oC

14. 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. soSynthesis = -Decomposition etc.

15. 6.4 Equations Δ Horeaction = ΣnpΔHof(products) - ΣnpΔHof(reactants) Public Relations P-R

16. 6.4 examples Calculate ΔH for the following reaction: 8 Al(s) + 3 Fe3O4(s) --> 4 Al2O3(s) + 9 Fe(s) Given ΔHf Al2O3(s) = -1669.8 kJ/mol ΔHf Fe3O4(s) = -1120.9 kJ/mol First ignore the elements, because they are formed, and you don’t need them to find the change in enthalpy 3 Fe3O4(s) --> 4 Al2O3(s) Public Relations… P-R so ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s) ΔH = 4 (-1669.8) - 3(-1120.9) = -3316.5 kJ/mol