Chapter 8 Thermochemistry: Chemical Energy

Thermodynamics01 Thermodynamics: study of energy and it’s transformations Energy: capacity to do work, or supply heat Energy = Work + Heat Kinetic Energy: energy of motion E K = 1 / 2 mv 2 (1 Joule = 1 kg  m 2 /s 2 ) (1 calorie = J) Potential Energy: stored energy

Thermodynamics 02 Conservation of energy law: Energy cannot be created or destroyed; it can only be converted from one form into another.

Thermodynamics 03 Thermal Energy: kinetic energy of molecular motion (translational, rotational, and vibrational) Heat: the amount of thermal energy transferred between two objects at different temperatures Chemical Energy: potential energy stored in chemical bonds released in the form of heat or light

Thermodynamics04 First Law of Thermodynamics: energy of an isolated system must be kept constant

Thermodynamics05 System  reactants + products Surroundings  everything else Energy changes are measured from the point of view of the system! ∆E is negative  energy flows out of the system ∆E is positive  energy flows into the system

Thermodynamics06

Work07 w = –P  V

Sign of w08 negative positive positivenegative w = -P  V  expansion w = -P  V  contraction

Work Units09 w = -P  V (J or kJ) 1L x 1000mL x 1cm 3 x 1m 3 1L 1mL (100cm) x 10 3 kg ms 2 = 101 kgm 2 = 101J s 2 m2m2 w = L x atm =

Energy and Heat10 Energy = Work + Heat  E = w + q = q - P  V q =  E + P  V When a person does work, energy diminishes w = negative  E = negative

Heat and Enthalpy11 The amount of heat exchanged between the system and the surroundings is given the symbol q. q =  E + P  V At constant volume (  V = 0): q v =  E At constant pressure: q p =  E + P  V =  H enthalpy

State Functions12 State Function: value depends only on the present state of the system path independent when returned to its original position, overall change is zero

State Functions13 State and Nonstate Properties: The two paths below give the same final state: N 2 H 4 (g) + H 2 (g)  2 NH 3 (g) + heat (188 kJ) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat (92 kJ) temperature, total energy, pressure, density, volume, and enthalpy (∆H)  state properties nonstate properties include heat and work

Enthalpy14 Enthalpy or heat of reaction:  H = H(products) - H(reactants) States of the reactants and products are important!  (g, l, s, aq) Thermodynamic standard state: P = 1atm, [ ] = 1M, T = K (25ºC)

Standard Enthalpy of Reaction15 Thermodynamic standard state: P = 1atm, [ ] = 1M, T = K (25ºC) Standard enthalpy of reaction (Hº) N 2 (g) + 3H 2 (g)  2NH 3 (g)  Hº = -92.2kJ

Enthalpy Changes16 Most changes in a system involve a gain or loss in enthalpy Physical (melting of ice in a cooler) Chemical (burning of gas in your car)

Physical Changes17 Enthalpies of Physical Change:

Chemical Changes 18 Enthalpies of Chemical Change: Often called heats of reaction (  H reaction ). Endothermic: Heat flows into the system from the surroundings   H is positive Exothermic: Heat flows out of the system into the surroundings   H is negative

Enthalpy Changes19 Reversing a reaction changes the sign of  H for a reaction. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  H = –2219 kJ 3 CO 2 (g) + 4 H 2 O(l)  C 3 H 8 (g) + 5 O 2 (g)  H = kJ Multiplying a reaction increases  H by the same factor. 3 x [C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l) ]  H =(-2219kJ x 3) = –6657 kJ

Example20 How much work is done (in kilojoules), and in which direction, as a result of the following reaction? w = -0.25kJ Expansion, system loses -0.25kJ

Example21 The following reaction has  E = –186 kJ/mol. Is the sign of P  V positive or negative? What is the sign and approximate magnitude of  H? Contraction, P  V is negative, w is positive  H =  E + P  V  H = (-186kJ) + (1atm) (-1mole)  H = negative (slightly more than  E)

Example22 The reaction between hydrogen and oxygen to yield water vapor has  H° = –484 kJ. How much PV work is done, and what is the value of  E (kJ) for the reaction of 0.50 mol of H 2 with 0.25 mol of O 2 at atmospheric pressure if the volume change is –5.6 L? P  V = -0.57kJ Contraction, so w is positive  E = kJ

Example23 The explosion of 2.00 mol of solid TNT with a volume of approximately L produces gases with a volume of 448 L at room temperature. How much PV (kJ) work is done during the explosion? Assume P = 1 atm, T = 25°C. 2 C 7 H 5 N 3 O 6 (s)  12 CO(g) + 5 H 2 (g) + 3 N 2 (g) + 2 C(s) P  V = 45.2kJ Expansion, so w = -45.2kJ

Example24 How much heat (kJ) is evolved or absorbed in each of the following reactions? 1.) Burning of 15.5 g of propane: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  Hº = –2219 kJ 2.) Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH) 2 ·8 H 2 O(s) + 2 NH 4 Cl(s)  BaCl 2 (aq) + 2 NH 3 (aq) + 10 H 2 O(l)  Hº = kJ -780kJ (exothermic) +1.24kJ (endothermic)

Hess’s Law25 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. 3 H 2 (g) + N 2 (g)  2 NH 3 (g)  H° = –92.2 kJ

Hess’s Law26 (a) 2 H 2 (g) + N 2 (g) N 2 H 4 (g)  H° 1 = ? (b) N 2 H 4 (g) + H 2 (g) 2 NH 3 (g)  H° 2 = –187.6 kJ (c) 3 H 2 (g) + N 2 (g) 2 NH 3 (g)  H° 3 = – 92.2 kJ  H° 1 =  H° 3 –  H° 2 = (–92.2 kJ) – (–187.6 kJ) = kJ

Standard Heats of Formation 27 Where do  H° values come from? Standard Heats of Formation (  H° f ): enthalpy change for the formation of 1 mole of substance in its standard state  H° f = 0 for an element in its standard state!

Standard Heats of Formation 28 H 2 (g) + 1 / 2 O 2 (g)  H 2 O(l)  H ° f = –286 kJ/mol 3 / 2 H 2 (g) + 1 / 2 N 2 (g)  NH 3 (g)  H ° f = –46 kJ/mol 2 C(s) + H 2 (g)  C 2 H 2 (g)  H ° f = +227 kJ/mol 2 C(s) + 3 H 2 (g) + 1 / 2 O 2 (g)  C 2 H 5 OH(g)  H ° f = –235 kJ/mol

Standard Heats of Formation 29 Calculating  H° for a reaction:  H° =  H° f (products) –  H° f (reactants) Heat of formation must be multiplied by the coefficient of the reaction C 6 H 12 O 6 (s) 2C 2 H 5 OH (l) + 2CO 2 (g)  H° = [2  H° f (ethanol) + 2  H° f (CO 2 )] -  H° f (glucose)

Standard Heats of Formation Na 2 CO 3 (s)49C6H6(l)C6H6(l)-92HCl(g) -127AgCl(s)-235C 2 H 5 OH(g)95.4N2H4(g)N2H4(g) -167Cl - (aq)-201CH 3 OH(g)-46NH 3 (g) -207NO 3 - (aq)-85C2H6(g)C2H6(g)-286H 2 O(l) -240Na + (aq)52C2H4(g)C2H4(g)-394CO 2 (g) 106Ag + (aq)227C2H2(g)C2H2(g)-111CO(g) Some Heats of Formation,  H f ° (kJ/mol)

Bond Dissociation Energy31 Bond Dissociation Energy (D): Amount of energy needed to break a chemical bond in gaseous state D = Approximate  Hº  H° = D(reactant bonds broken) – D(product bonds formed) H 2 + Cl 2 2HCl  H° = (D Cl-Cl + D H-H ) - (2 D H-Cl )  = [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol)  = -185 kJ

Bond Dissociation Energy32

Calorimetry and Heat Capacity33 Calorimetry: measurement of heat changes (q) for chemical reactions Constant Pressure Calorimetry: measures the heat change at constant pressure q =  H Bomb Calorimetry: measures the heat change at constant volume such that q =  E

Calorimetry and Heat Capacity34 Constant PressureBomb

Calorimetry and Heat Capacity35 Heat capacity {C}: amount of heat required to raise the temperature of an object or substance a given amount Specific Heat: amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C Molar Heat: amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C  C= q  T

Calorimetry and Heat Capacity36

Example37 The industrial degreasing solvent methylene chloride (CH 2 Cl 2, dichloromethane) is prepared from methane by reaction with chlorine: CH 4 (g) + 2 Cl 2 (g) CH 2 Cl 2 (g) + 2 HCl(g) Calculate  H° (kJ) CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g)  H° = –98.3 kJ CH 3 Cl(g) + Cl 2 (g) CH 2 Cl 2 (g) + HCl(g)  H° = –104 kJ  H° = = -202kJ

Example 38 Calculate  H° (kJ) for the reaction of ammonia with O 2 to yield nitric oxide (NO) and H 2 O(g), a step in the Ostwald process for the commercial production of nitric acid. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) = [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)] = kJ

Example 39 Calculate  H° (kJ) for the photosynthesis of glucose from CO 2 and liquid water, a reaction carried out by all green plants. 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) = [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)] = 2816kJ

Example40 Calculate an approximate  H° (kJ) for the synthesis of ethyl alcohol from ethylene: C 2 H 4 (g) + H 2 O(g)  C 2 H 5 OH(g) Calculate an approximate  H° (kJ) for the synthesis of hydrazine from ammonia: 2 NH 3 (g) + Cl 2 (g)  N 2 H 4 (g) + 2 HCl(g)

Introduction to Entropy42 Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. A spontaneous process is one that proceeds on its own without any continuous external influence. A nonspontaneous process takes place only in the presence of a continuous external influence.

Introduction to Entropy43 The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. Entropy has units of J/K (Joules per Kelvin).  S = S final – S initial Positive value of  S indicates increased disorder. Negative value of  S indicates decreased disorder.

Introduction to Entropy44

Introduction to Entropy45 To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process:Decrease in enthalpy (–  H). Increase in entropy (+  S). Nonspontaneous process:Increase in enthalpy (+  H). Decrease in entropy (–  S).

Introduction to Entropy39 Predict whether  S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate  S° for each: a. 2 CO(g) + O 2 (g)  2 CO 2 (g) b. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) c. C 2 H 4 (g) + Br 2 (g)  CH 2 BrCH 2 Br(l) d. 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g)

Introduction to Free Energy40 Gibbs Free Energy Change (  G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.  G =  H – T  S  G < 0Process is spontaneous  G = 0Process is at equilibrium  G > 0Process is nonspontaneous

Introduction to Free Energy41 Situations leading to  G < 0:  H is negative and T  S is positive  H is very negative and T  S is slightly negative  H is slightly positive and T  S is very positive Situations leading to  G = 0:  H and T  S are equally negative  H and T  S are equally positive Situations leading to  G > 0:  H is positive and T  S is negative  H is slightly negative and T  S is very negative  H is very positive and T  S is slightly positive

Introduction to Free Energy42 Which of the following reactions are spontaneous under standard conditions at 25°C? a.AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq)  G° = –55.7 kJ b.2 C(s) + 2 H 2 (g)  C 2 H 4 (g)  G° = 68.1 kJ c.N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92 kJ;  S° = –199 J/K

Introduction to Free Energy43 Equilibrium (  G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92.0 kJ  S° = –199 J/K Equilibrium is the point where  G° =  H° – T  S° = 0

Introduction to Free Energy44 Benzene, C 6 H 6, has an enthalpy of vaporization,  H vap, equal to 30.8 kJ/mol and boils at 80.1°C. What is the entropy of vaporization,  S vap, for benzene?