AP CHEMISTRY NOTES Ch 6 Thermochemistry Ch 8.8 Covalent bond energies Ch. 16 Spontaneity, Entropy, and Free Energy

6.1 The Nature of Energy Energy? Law conservation of energy The capacity to do work to produce heat Law conservation of energy Energy can be converted from one form to another but can be either created nor destroyed.

6.1 The Nature of Energy Kinetic energy Temperature Heat 1/2mv2 Property that reflects the random motions of the particles in a particular substance. Heat Transfer of energy b/w two objects due to a temperature difference.

6.1 The Nature of Energy Work Transfer of energy Force acting over a distance Transfer of energy Through work Through heat In other words, the energy transferred into two ways.

6.1 The Nature of Energy - Energy change is independent of the pathway; however, work and heat are dependent on the pathway.

6.1 The Nature of Energy Sign - Or + *Chemical energy Released or Absorbed through chemical reaction *System *Surroundings (anything other than reactants and products)

Exothermic Process Energy flows out of the system Delta PE is the change in PE

In any exothermic reaction, some of the potential energy stored in the chemical bonds converted to thermal energy via heat

Endothermic Process Energy flow is into a system

Thermodynamics First law of thermodynamics The law of conservation of energy (The energy of the universe is constant)

Internal energy The internal energy E of a system = the sum of the kinetic and potential energies in the system Delta E = q + w (q = heat, w = work) Consist of a number (magnitude) and sign (direction of the flow)

q = -x (out of the system) q = +x (into the system) Work from the system’s point of view

Figure 6.4 The Piston, Moving a Distance Against a Pressure P, Does Work On the Surroundings P=F/A Work = F x ∆ h = PA x ∆ h = P x ∆ V Sign? Work and ∆V must have opposite sign W= - P x ∆ V

6.2 Enthalpy and Calorimetry Enthalpy (H) H=E + PV E is internal energy (KE+PE) PV is pressure-volume work @constant P Enthalpy is heat at constant pressure ∆H=∆E + ∆(PV) ∆H=qP (heat or energy flow)

Ex 6.4

Calorimetry Calorimeter Calorimetry Heat capacity C (heat absorbed /increase in temp) Specific heat capacity s (per gram or per mol) Cwater=4.18 J/oC∙g 4.18J is required to increase 1oC of 1g water Coffee cup calorimeter

Extensive or intensive properties? Heat=an extensive prop., depends on amount of substance Temp=an intensive prop., not related to the amount of substance

Ex 6.5 Constant pressure calorimetry q=mc∆t Assume solution specific heat capacity is same as water ∆t= final t – initial t

Example Consider a calorimeter at constant pressure. 50.0 ml of a 4.0 M HCl solution is mixed with 50.0 ml of a 2.0 M NaOH solution at 25˚C. After mixing, the temperature of the solution is 31.9˚C. Calculate the energy released.

Calculate the energy released per mole of H+ neutralized Consider a calorimeter at constant pressure. 50.0 ml of a 4.0 M HCl solution is mixed with 50.0 ml of a 2.0 M NaOH solution at 25˚C. After mixing, the temperature of the solution is 31.9˚C. Calculate the energy released per mole of H+ neutralized

Constant volume calorimetry Coffee cup and Bomb calorimeter 1

Constant volume calorimetry ∆E=q+w (w=0, constant V) ∆E=q Ex 6.6 E released / mass

Constant volume A calorimeter has a heat capacity of 195 J/K. It contains 100. grams of water at 25˚C. 50.0 grams of aluminum (heat capacity of 0.900 J/gK) at 100.0 ˚C was added to it. What is the final temperature of the solution?

30.1oC

Constant volume A calorimeter has a heat capacity of 50. cal/K. 100 grams of cold water at 15˚C is added to it. Then some hot water at 80˚C is added to it. The final temperature of the water is 30. ˚C. What is the mass of hot water added?

45g

Constant pressure A calorimeter contains water at 25˚C. Describe the heat flow when a metal at 50.0˚C is placed into the water.

Constant pressure A 15.0g sample of nickel metal is heated to 100.0˚C and dropped into 55.0g of water, initially at 23.0˚C. Assuming that all the heat lost by the nickel is absorbed by the water, calculate the final temperature of the nickel and water. Nickel has a specific heat of 0.444 J/˚Cg.

In a coffee cup calorimeter, 1 In a coffee cup calorimeter, 1.60g of ammonium nitrate is mixed with 75.0g of water at an initial temperature of 25.00˚C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34˚C. Assuming the solution has a heat capacity of 4.18 J/˚Cg and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of ammonium nitrate in units of kJ/mol.

6.3 Hess’s law ∆H is not dependent on the reaction pathway. Is a state of function Initial Final

6.3 Hess’s law The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

6.3 Hess’s law The change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Rules of Hess’s law If a sub-reaction is multiplied by a coefficient, ∆Ho value is also multiplied by the same coefficient. If a sub-reaction is reversed (flipped), ∆Ho value is switched the sign. Hint! Work backward Reverse any reactions Multiply reactions

Example of Hess’s law H2 (g) + Cl2 (g)  2HCl (g) ΔH˚ = ? Given: NH3 (g) + HCl (g)  NH4Cl (s) ΔH˚ = - 176.0 kJ N2 (g) + 3H2 (g)  2NH3 (g) ΔH˚ = - 92.22 kJ N2 (g) + 4H2 (g) + Cl2 (g)  2NH4Cl (s) ΔH˚ = -628.86 kJ

ΔH˚ = -184.64 kJ

Example of Hess’s law C2H2(g) + 2H2(g)  C2H6(g) ΔH˚ = ? Given: 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) ΔH˚ = -2600 kJ 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l) ΔH˚ = -3120 kJ H2(g) + 1/2 O2(g)  H20(l) ΔH˚ = - 286 kJ

ΔH˚ = -312 kJ

Example of Hess’s law Given: 2H2(g) + C(s)  CH4(g) ΔH˚ = 74.81 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH˚ = ? Given: 2H2(g) + C(s)  CH4(g) ΔH˚ = 74.81 kJ 2H2(g) + O2(g)  2H2O(l) ΔH˚ = -571.66 kJ C(s) + O2(g)  CO2(g) ΔH˚ = -393.52 kJ

ΔH˚ = - 1039.99 kJ

Example of Hess’s law 2NH3(g) + 3N2O(g)  4N2(g) + 3H2O(l) ΔH˚ = ? Given: 2NH3(g) + 3/2 O2(g)  N2(g) + 3H2O(l) ΔH˚ = - 765.5 kJ/ mole 3N2O(g) + 3H2(g)  3N2(g) + 3H2O(l) ΔH˚ = -1102.2 kJ/ mole 3H2O(l)  3H2(g) + 3/2O2(g) ΔH˚ = 857.7 kJ/ mole

ΔH˚ = - 1010.0 kJ/mol

6.4 Standard Enthalpies of Formation ∆Hof (under standard condition) The change is enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their states. We can’t determine absolute values of enthalpy

Reference P gas @1atm Conc.=1M Element @1atm, 25oC

∆Hof = per mole of product (kJ/mol) See appendix 4 Tabulated ∆Hof value – using this for calculation

Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l) *Add all up A: CH4 CH4 C + 2H2 ∆H=-75kJ/mol Flip! C + 2H2  CH4 ∆H=75kJ/mol B: 2O2 ∆H=0kJ (element) C: CO2 C + O2  CO2 ∆H=-394 kJ D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ (75kJ)+(-394kJ)+(-572kJ)= -891kJ

Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l) *Sum of produced – sum of required A: CH4 CH4 C + 2H2 ∆H=-75kJ/mol You don’t have to flip! B: 2O2 ∆H=0kJ (element) C: CO2 C + O2  CO2 ∆H=-394 kJ D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ (-394kJ)+(-572kJ)-[0+(-75kJ)]= -891kJ

Ex. 6.9 4NH3(g)+7O2(g)4NO2(g)+6H2O(l)

Table 6.2 Standard Enthalpies of Formation for Several Compounds at 25°C Copyright © Houghton Mifflin Company. All rights reserved.

Ex. 6.9 4NH3(g)+7O2(g)4NO2(g)+6H2O(l) -1396kJ

Ex. 6.10 2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)

Ex. 6.10 2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s) -850.kJ

6.11 CH3OH or C8H18? Need two equations _CH3OH (l) + _O2(g)_CO2(g)+_H2O(l) _C8H18(l) + _O2(g)_CO2(g)+_H2O(l) Finally divide by molar mass for both

6.11 CH3OH or C8H18? 2CH3OH (l) + 3O2(g)2CO2(g)+4H2O(l) -1454kJ/2mol*32.0g = -22.7 kJ/g 2C8H18(l) + 25O2(g)16CO2(g)+18H2O(l) -10914kJ/2mol*114.2g = -47.8 kJ/g

6.5 Present sources of energy

6.6 New energy sources