Thermodynamics: Energy Relationships in Chemistry The Nature of Energy What is force: What is work: A push or pull exerted on an object An act or series of acts which overcome a force

Thermodynamics: Energy Relationships in Chemistry Mechanical work The amount of energy required to move an object over a certain distance w = F * d What is energy: The capacity to do work

Thermodynamics: Energy Relationships in Chemistry potential energy Kinetic energy E k = 1/2 mv 2 E= joule = 1kg-m 2 /s J = 1 cal

Thermodynamics: Energy Relationships in Chemistry Sample problem: A 252 g baseball is thrown with a speed of 39.3 m/s. Calculate the kinetic energy of the ball in joules and calories E k = 1/2 mv 2 = 1/2 (0.145 kg)(25m/s) 2 = 45 kg m 2 /s 2 = 45J (45 J) (1 cal) (4.184 J) = 11 cal

Thermodynamics: Energy Relationships in Chemistry System and Surroundings system surroundings

Thermodynamics: Energy Relationships in Chemistry First law of thermodynamics: Energy can neither be created nor destroyed The energy lost by a system equals the energy gained by its surroundings Everything wants to go to a lower energy state  E = E final - E initial

Thermodynamics: Energy Relationships in Chemistry E final < E initial E final > E initial endothermic exothermic

Thermodynamics: Energy Relationships in Chemistry  E = q + w + +

Thermodynamics: Energy Relationships in Chemistry + -

Sample problem: During the course of a reaction a system loses 550 J of heat to its surroundings. As the gases in the system expand, the piston moves up. The work on the piston by the gas is determined to be 240 J. What is the change in the internal energy of the system, Thermodynamics: Energy Relationships in Chemistry  E = q + w  E = (-550 J) + (-240 J  E = -790 J q w

A State Function is independent of pathway and is capitalized.  E, energy is an extensive property and is a State function. heat (q) and work (w) are not state functions.

Thermodynamics: Energy Relationships in Chemistry

P-V work

Thermodynamics: Energy Relationships in Chemistry Let work w = -P  V If  E = q + w, then  E = q + -P  V When a reaction is carried out in a constant-volume container (  V = 0) then,  E = q v When a reaction is carried out at constant pressure container then,  E = q p - P  V, or q p =  E + P  V

Thermodynamics: Energy Relationships in Chemistry Chemical reactions usually occur under conditions where the pressure is held constant, therefore: change in enthalpy:  H =  E + P  V  H = q p Since chemical reactions usually occur under conditions where the volume of the system undergoes little change:  H =  E Since  H= H final + H initial, then for any type of chemical reaction,  H= H products - H reactants

Thermodynamics: Energy Relationships in Chemistry Some things you may never have wished to know about enthalpy Enthalpy is an extensive property CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ -75 kJ 0kJ kJ -242kJ [(-393.5) + 2(-242)] – [(-75) + 2(0)] = kJ

Sample problem: How much heat is produced when 4.50 g of methane gas (CH 4 ) is burned in a constant pressure environment (4.50 g CH 4 ) (1mol CH 4 ) (16.0 g) = -226 kJ (-802 kJ) (1 mole CH 4 ) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g )

The enthalpy change for a reaction is equal in magnitude but opposite in sign to  H for the reverse reaction

The enthalpy change for a reaction depends on the state of the reactants and products Assume: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ The following process would also produce the same result CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2H 2 O(g)  2H 2 O(l)  H = -88kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H = -890 kJ

Here is a second reaction pathway which produces the same results CH 4 (g) + 2O 2 (g)  CO(g) + 2H 2 O + 1/2 O 2 CO(g) + 2H 2 O + 1/2 O 2  CO 2 (g) + 2H 2 O CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O

Thermodynamics: Energy Relationships in Chemistry Calorimetry: Things are heating up Calorimetry: Measurement of heat flow Molar Heat capacity: The energy required to raise the temperature of 1 mole of a substance by 1  C (C = q/  T, J/mol-  C ) q = n (molar heat capacity)  T Specific Heat: The energy required to raise the temperature of 1 gram of a substance by 1  C (C = q/  T, J/g-  C ) q = m s  T

Thermodynamics: Energy Relationships in Chemistry Sample exercise: The specific heat of Fe 2 O 3 is 0.75 J/g-  C. A.) What is the heat capacity of a 2.00 kg brick of Fe 2 O 3. B.) What quantity of heat is required to increase the temperature of 1.75 g of Fe 2 O 3 from 25  C to 380  C. A.) (2.00 kg) (1000g) (1kg) = 1.50 x 10 3 J/  C (0.75 J) (1 g -  C) B.) q = m S  T = 1.75 g (0.75 J/g-  C ) (355  C) =465 J

Thermodynamics: Energy Relationships in Chemistry Constant Pressure Calorimetry

Sample exercise: 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH are reacted together in a ‘coffee cup’ calorimeter. * The temperature of the resulting solution increased from 21.0  C to 27.5  C. Calculate the enthalpy change of the reaction (the specific heat of water = 4.18 J/g-  C). * assume the calorimeter absorbs negligible heat and that the density of the solution is 1.0 g/ml. q = mS  T q = (100g)(4.18 J/ g-  C )(6.5  C ) q = 2717 J = 2.7 kJ 1mol.050L L =54kJ/Mol

Heat capacity of a metal What is the heat capacity of a 5.05g chunk of an unknown metal. The metal was heated in boiling water and then placed in 50 mL of water in a coffee cup calorimeter at a temperature of 24.5ºC. The highest temperature achieved was 28.9ºC. What is the heat capacity of the metal. 50g x 4.184j/gºC x 4.5ºC = 5.05g x X x 71.1ºC X = 2.62J/gºC

Thermodynamics: Energy Relationships in Chemistry Bomb Calorimetry

Thermodynamics: Energy Relationships in Chemistry Sample exercise:When 1.00 g of the rocket fuel, hydrazine (N 2 H 2 ) is burned in a bomb calorimeter, the temperature of the system increases by 3.51  C. If the calorimeter has a heat capacity of kJ/  C what is the quantity of heat evolved. What is the heat evolved upon combustion of one mole of N 2 H 4. q evolved = -C calorimeter x  T (3.51  C) 19.3 kJ = -(5.510 kJ) (C)(C) (1.00mol) =618 kJ (32 g) (1.00 mol) (19.3 kJ) (1.00 g)

HESS’S Law: If a reaction is carried out in a series of steps,  H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

CH 4 (g) + 2O 2 (g)  CO(g) + 2H 2 O + 1/2 O 2  H = -607 kJ CO(g) + 2H 2 O + 1/2 O 2  CO 2 (g) + 2H 2 O  H = -283 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O  H = -890 kJ

Sample exercise: Calculate the  H for the reaction: 2C(s) + H 2 (g)  C 2 H 2 (g) given the following reactions and their respective enthalpy changes C 2 H 2 (g) + 5/2O 2  2CO 2 (g) + H 2 O(l)  H = kJ C(s) + O 2 (g)  CO 2 (g)  H = kJ H 2 (g) + 1/2O 2  H 2 O(l)  H = kJ 2CO 2 (g) + H 2 O(l)  C 2 H 2 (g) + 5/2O 2  H = kJ 2C(s) + 2O 2 (g)  2CO 2 (g)  H = kJ H 2 (g) + 1/2O 2  H 2 O(l)  H = kJ 2C(s) + H 2 (g)  C 2 H 2 (g)  H = kJ

Practice Exercise : Calculate the  H for the reaction: NO(g) + O(g)  NO 2 (g) given the following reactions and their respective enthalpy changes NO(g) + O 3  NO 2 (g) + O 2 (g)  H = kJ O 3 (g)  3/2O 2 (g)  H = kJ O 2 (g)  2O (g)  H = kJ NO(g) + O 3  NO 2 (g) + O 2 (g)  H = kJ 3/2O 2 (g)  O 3 (g)  H = kJ O (g)  1/2O 2 (g) )  H = kJ NO(g) + O(g)  NO 2 (g)  H = kJ

Heats of formation,  Hº f A thermodynamic description of the formation of compounds from their constituent elements. A thermodynamic description of the formation of compounds under standard conditions (1 atm, 298 K (25  C)) is called the standard heat of formation,  Hº f  Heat of vaporization:  H for converting liquids to gases Heat of fusion:  H for melting solids Heat of combustion:  H for combusting a substance in oxygen

Thermodynamics: Energy Relationships in Chemistry The standard heat of formation for one mole of ethanol is the enthalpy change for the following reaction  C 2 H 5 OH  H f = kJ    H f  H rxn =  n n (products) -   H f  m m (reactants) note: the standard heat of formation of the most stable form of any element is 0. 2C(graphite) + 3H2(g) + ½ O2(g)

Thermodynamics: Energy Relationships in Chemistry

Sample exercise: The quantity of heat produced from one gram of propane (C 3 H 8 ) is kJ/gram. How does this compare with the heat produced from one gram of benzene (C 6 H 6 )? C 6 H 6 (l) + O 2  6CO 2 (g) + 3H 2 O(l) 15 2   H f  H rxn =  n n (products) -   H f  m m (reactants)  [6  H f  H rxn = ( kJ) +  3  H f ( kJ)]  [1  H f ( kJ) +   H f (0)] note: the standard heat of formation of the most stable form of any element is 0.  H rxn = [6( kJ) +3( kJ)] ( kJ)= kJ -