Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 8 Thermochemistry: Chemical Energy John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition Copyright © 2008 Pearson Prentice Hall, Inc.

8.1 Energy Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/2 Units: 1 Cal = 1000 cal = 1 kcal 1 cal = J (exactly) Energy: The capacity to supply heat or do work. Kinetic Energy (E K ): The energy of motion. Potential Energy (E P ): Stored energy.

8.2 Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/3 Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another.

Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/4 Total Energy = E K + E P

Energy Changes and Energy Conservation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/5 Thermal Energy: The kinetic energy of molecular motion and is measured by finding the temperature of an object. Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.

8.3 Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/6 First Law of Thermodynamics: The total internal energy of an isolated system is constant.  E = E final - E initial

Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/7 CO 2 (g) + 2H 2 O(g) kJ energyCH 4 (g) + 2O 2 (g)  E = E final - E initial = -802 kJ 802 kJ is released when 1 mole of methane, CH 4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water.

Internal Energy and State Functions Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/8 State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.

8.4 Expansion Work Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 8/9 w = F x d Work = Force x Distance

Expansion Work Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/10 Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work.

8.5 Energy and Enthalpy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/11 Constant Pressure:  E = q + w w = work = -P  V q = heat transferred q =  E + P  V q P =  E + P  V Constant Volume (  V = 0): q V =  E

Energy and Enthalpy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/12 = H products - H reactants HH Enthalpy change or Heat of reaction (at constant pressure) q P =  E + P  V =  H = H final - H initial Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state.

8.6The Thermodynamic Standard State Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/13 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l)C 3 H 8 (g) + 5O 2 (g)  H = kJ Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution. Standard enthalpy of reaction is indicated by the symbol ΔH o 3CO 2 (g) + 4H 2 O(g)C 3 H 8 (g) + 5O 2 (g)  H= kJ  H° = kJ

8.7 Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/14 Enthalpy of Fusion (  H fusion ): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (  H vap ): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (  H subl ): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.

Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/15

Enthalpies of Physical and Chemical Change Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/16 2Al(s) + Fe 2 O 3 (s)2Fe(s) + Al 2 O 3 (s)  H° = +852 kJ 2Fe(s) + Al 2 O 3 (s)2Al(s) + Fe 2 O 3 (s)  H° = -852 kJ endothermic exothermic

Examples Identify each of the following processes as endothermic or exothermic and indicate the sign of ΔH o Sweat evaporating from your skin Water freezing in a freezer Wood burning in fire

Example An LP gas tank in a home barbeque contains 13.2 kg of propane, C 3 H 8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank C 3 H 8 (g) + 5O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) ΔH o = -2044kJ

Example How much heat (in kJ) is evolved when 5.00g of aluminum reacts with a stoichiometric amount of Fe 2 O 3 ? 2 Al(s) + Fe 2 O 3 (s)  2 Fe(s) + Al 2 O 3 (s) ΔH o = -852 kJ

8.8Calorimetry and Heat Capacity A temperature change is observed when a system gains or loses energy in the form of heat (ΔH) Amount of heat gained = amount of heat lost q cal = -q rxn q gain = -q lost Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/20

Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/21 Measure the heat flow at constant volume (  E).

Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/22 Specific Heat: The amount of heat required to raise the temperature of 1 g of a substance by 1 °C. q = (specific heat) x (mass of substance) x  T Molar Heat Capacity (C m ): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C. q = (C m ) x (moles of substance) x  T Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount. q = C m x  T

Calorimetry and Heat Capacity Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/23

Examples Assuming that Coca Cola has the same specific heat as water [4.183 J/g o C], calculate the amount of heat (in kJ) transferred when one can (about g) is cooled from 25.0 o C – 3.0 o C

Example When 25.0 mL of 1.0 M H 2 SO 4 is added to 50.0ml of 1.0 M NaOH at 25.0 o C in a calorimeter, the temperature of the aqueous solution increases to 33.9 o C. Assuming that the specific heat of the solution is 4.18 J/g o C, that its density is 1.00g/mL, and the calorimeter itself absorbs a negligible amount of heat, calculate ΔH (in kJ/mol) for the reaction H 2 SO 4 (aq) + 2 NaOH(aq)  2 H2O(l) + Na 2 SO 4 (aq)

8.9 Hess’s Law Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/26 Haber Process: Multiple-Step Process - Given N2H4(g)N2H4(g)2H 2 (g) + N 2 (g) Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. 2NH 3 (g)3H 2 (g) + N 2 (g)  H° = kJ  H° 1 = ? 2NH 3 (g)N 2 H 4 (g) + H 2 (g)  H° 1+2 = kJ 2NH 3 (g)3H 2 (g) + N 2 (g)  H° 2 = kJ

Hess’s Law Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/27

Example Find ΔH o rxn for the following reaction: 3H 2 (g) + O 3 (g)  3 H 2 O(g) ΔH o rxn = ?? Use the following reactions with known ΔH’s 2H 2 (g) + O 2 (g)  2 H 2 O(g) Δ H o = kJ 3O 2 (g)  2 O 3 (g) Δ H o = kJ Typo from original note

Example Fin ΔH o rxn for the following reaction C(s) + H 2 O(g)  CO(g) + H 2 (g) H o rxn = ? Use the following reactions with known H’s C(s) + O 2 (g)  CO 2 (g) ΔH o = kJ 2CO(g) + O 2 (g)  2CO 2 (g) Δ H o = kJ 2H 2 (g) + O 2 (g)  2H 2 O (g) Δ H o = kJ

8.10 Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/30 Standard Heat of Formation (  H° f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.  H° f = kJ CH 4 (g)C(s) + 2H 2 (g) Standard states 1 mol of 1 substance

Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/31

Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/32 cC + dDaA + bB Reactants Products  H° =  H° f (Products) -  H° f (Reactants)  H° = [c  H° f (C) + d  H° f (D)] - [a  H° f (A) + b  H° f (B)]

Standard Heats of Formation Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/33 Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. C 6 H 12 O 6 (s) + 6O 2 (g)6CO 2 (g) + 6H 2 O(l)  H° = ?

Example Use the information in Table 8.2 to calculate ΔH o (in kJ) for the reaction of ammonia with oxygen gas to yield nitric oxide (NO) and water vapor, a step in the Ostwald process for the commercial production of nitric acid

8.11 Bond Dissociation Energies Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/35 Bond dissociation energies are standard enthalpy changes for the corresponding bond-breaking reactions.

Bond Dissociation Energies Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/36 2HCl(g)H 2 (g) + Cl 2 (g)  H° = D(Reactant bonds) - D(Product bonds)  H° = (D H-H + D Cl-Cl ) - (2D H-Cl )

8.12 Fossil Fuels, Fuel Efficiency, and Heats of Combustion Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/37 CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g)

Fossil Fuels, Fuel Efficiency, and Heats of Combustion Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/38 CO 2 (g) + 2H 2 O(l)CH 4 (g) + 2O 2 (g) ΔH o c = kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l) ΔH o c = -285 kJ/mol

8.13 An Introduction to Entropy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/39 Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Spontaneous processes are favored by a decrease in H (negative  H). favored by an increase in S (positive  S). Nonspontaneous processes are favored by an increase in H (positive  H). favored by a decrease in S (negative  S). Entropy (S): The amount of molecular randomness in a system.

An Introduction to Entropy

Example Predict whether ΔS o is likely to be positive or negative for each of the following reactions H 2 C=CH 2 (g) + Br 2 (g)  BrCH 2 CH 2 Br(l)

8.14 An Introduction to Free Energy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/42 Gibbs Free Energy Change (  G) Entropy change  G = Enthalpy of reaction Temperature (Kelvin) HH SS - T

An Introduction to Free Energy Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 8/43 Gibbs Free Energy Change (  G)  G = HH SS - T  G < 0Process is spontaneous  G = 0Process is at equilibrium (neither spontaneous nor nonspontaneous)  G > 0Process is nonspontaneous

Example Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25.0 o C. At what temperature ( o C) does the changeover occur?