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1 LITER OF GASOLINE CONTAINS 8000 CALORIES.

A PERSON USES AN AVERAGE OF 2000 CALORIES IN A DAY.

EXCESS CALORIES IS STORED BY THE BODY AS FAT.

WHAT DOES CALORIES MEAN ANYWAY?

NO. CALORIES IS NOT FAT. IT IS THE ENERGY FROM THE FOOD WE EAT THAT IS MEASURED IN CALORIES.

SO HOW DO THEY KNOW HOW MUCH CALORIES THERE ARE IN STUFF?

THEY KNOW THROUGH THERMOCHEMISTRY AND CALORIMETRY

Thermochemistry: Energy Flow and Chemical Change 6.1 Forms of Energy and Their Interconversion 6.2 Enthalpy: Heats of Reaction and Chemical Change 6.3 Calorimetry: Laboratory Measurement of Heats of Reaction 6.4 Stoichiometry of Thermochemical Equations 6.5 Hess’s Law of Heat Summation 6.6 Standard Heats of Reaction (  H 0 rxn ) ‏

Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. (Duhh!!) Thermodynamic s is the study of heat and its transformations. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings . The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes.

A boundary separates a system from the surrounding. It is arbitrary in character, real or imaginary. Types of System 1. Open system – separated by an imaginary boundary that allows matter and energy exchange between the system and surroundings 2. Closed system – separated by a real diathermal and non-permeable boundary that only allows energy exchange to and from the surroundings 3. Isolated system – a system enclosed by a real adiabatic, non- permeable boundary where no matter and energy exchange occur State and Non-state Properties State properties – observed when a system is in an equilibrium state e.g. P, T, V, U, H, G, S Non-state properties – path dependent properties that are measured or observed during a change in state e.g. q, w,

Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy , pressure, volume, temperature 6.7

Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energy Exchange: closed energy isolated nothing 6.2 SYSTEM SURROUNDINGS

 E = E final - E initial = E products - E reactants Figure 6.2 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.

1 st Law of Thermodynamics This is a generalization of the Conservation of Energy Law and it also serves to define the change in internal energy of a system. Energy cannot be created nor destroyed. Energy can only be transferred and or transformed.

1 st Law of Thermodynamics (cont.) Equation: Q = W + Δ U U = internal energy Q = heat energy (=heat transfer) W = work done The equation states that heat supplied to a system goes to increase the internal energy of the system and perform mechanical work on the surroundings.

Internal Energy & Heat Internal Energy: Is the total molecular energy of a system = U Heat: Is the movement of energy = transfer of energy = flow of energy = Q

Isothermal Process Any process in which the temperature of a system remains constant.

Isothermal Process (cont.) 1 st Law as applied to an Isothermal Process: Q = W

Isobaric Process Is any process in which pressure of the system remains constant.

Isobaric Process (cont.) 1 st Law as applied to an Isobaric Process: W = P (V f – V i )

Adiabatic Process Is any process in which no heat enters or leaves the system.

Adiabatic Process (cont.) 1 st Law as applied to an Adiabatic Process: Δ U = - W

Thermodynamics 6.7  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = - P  V when a gas expands against a constant external pressure

Enthalpy and the First Law of Thermodynamics 6.7  E = q + w At constant pressure, q =  H and w = - P  V  E =  H - P  V  H =  E + P  V

Figure 6.3 A system transferring energy as heat only. Figure 6.4 A system losing energy as work only. Energy, E Zn(s) + 2H + (aq) + 2Cl - (aq) ‏ H 2 (g) + Zn 2+ (aq) + 2Cl - (aq) ‏  E<0 work done on surroundings

Table 6.1 The Sign Conventions* for q , w and  E q w + =  E + + - - - - + + + - depends on sizes of q and w depends on sizes of q and w * For q : + means system gains heat; - means system loses heat. * For w : + means word done on system; - means work done by system.

Limitations of the First Law of Thermodynamics  E = q + w E universe = E system + E surroundings  E system = -  E surroundings The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.  E system +  E surroundings = 0 =  E universe  E universe =  E system +  E surroundings Units of Energy Joule (J) ‏ Calorie (cal) ‏ British Thermal Unit 1 cal = 4.18J 1 J = 1 kg*m 2 /s 2 1 Btu = 1055 J

Sample Problem 6.1 Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (  E) in J, kJ, and kcal.

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 6.3

The Meaning of Enthalpy w = - P  V  H =  E + P  V q p =  E + P  V =  H  H ≈  E in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> P  V. H = E + PV where H is enthalpy

Figure 6.8 Enthalpy diagrams for exothermic and endothermic processes. CH 4 + 2O 2 CO 2 + 2H 2 O H initial H initial H final H final H 2 O( l ) ‏ H 2 O( g ) ‏  H < 0  H > 0 A Exothermic process B Endothermic process Enthalpy, H Enthalpy, H heat out heat in CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O( g ) ‏ H 2 O( l ) H 2 O( g ) ‏

Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of  H SOLUTION: PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction (a) The reaction is exothermic. EXOTHERMIC  H = -285.8kJ  H = +40.7kJ ENDOTHERMIC (b) The reaction is endothermic. PROBLEM: In each of the following cases, determine the sign of  H, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. (a) H 2 ( g ) + 1/2 O 2 ( g ) H 2 O( l ) + 285.8kJ (b) 40.7kJ + H 2 O( l ) H 2 O( g ) ‏ H 2 ( g ) + 1/2 O 2 ( g ) ‏ (reactants) ‏ H 2 O( l ) ‏ (products) ‏ (products) ‏ H 2 O( g ) ‏ (reactants) ‏ H 2 O( l ) ‏

Some Important Types of Enthalpy Change heat of combustion (  H comb ) ‏ heat of formation (  H f ) ‏ heat of fusion (  H fus ) ‏ heat of vaporization (  H vap ) ‏ C 4 H 10 ( l ) + 13/2 O 2 ( g ) 4CO 2 ( g ) + 5H 2 O( g ) ‏ K( s ) + 1/2 Br 2 ( l ) KBr( s ) ‏ NaCl( s ) NaCl( l ) ‏ C 6 H 6 ( l ) C 6 H 6 ( g ) ‏

Constant-Pressure Calorimetry No heat enters or leaves! q u niv = q water + q cal + q metal q univ = 0 q metal = - ( q water + q cal ) q water = mc  t q cal = C cal  t 6.4 Reaction at Constant P  H = q metal A piece of heated metal is dropped in an insulated cup with 75.0 grams of water. The temperature of the water increased from 29degC to 32degC. How much heat was lost by the piece of metal?

The specific heat (c) of a substance is the amount of heat ( q ) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity ( C ) of a substance is the amount of heat ( q ) required to raise the temperature of a given quantity ( m ) of the substance by one degree Celsius. C = mc Heat ( q ) absorbed or released: q = mc  t q = C  t  t = t final - t initial 6.4

Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials Specific Heat Capacity (J/g*K) ‏ Substance Specific Heat Capacity (J/g*K) ‏ Substance Compounds water, H 2 O( l ) ‏ ethyl alcohol, C 2 H 5 OH( l ) ‏ ethylene glycol, (CH 2 OH) 2 ( l ) ‏ carbon tetrachloride, CCl 4 ( l ) ‏ 4.184 2.46 2.42 0.864 Elements aluminum, Al graphite,C iron, Fe copper, Cu gold, Au 0.900 0.711 0.450 0.387 0.129 wood cement glass granite steel Materials 1.76 0.88 0.84 0.79 0.45

s of Fe = 0.444 J/g • 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms  t = 869 g x 0.444 J/g • 0 C x –89 0 C = -34,000 J 6.4 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C?

Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat Capacity Sample Problem 6.4 Determining the Heat of a Reaction PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25 0 C to 300. 0 C? The specific heat capacity ( c ) of Cu is 0.387 J/g*K. PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00 0 C and carefully add 25.0 mL of 0.500 M HCl, also at 25.00 0 C. After stirring, the final temperature is 27.21 0 C. Calculate q soln (in J) and  H rxn (in J/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K) ‏

Sample Problem 6.5 Calculating the Heat of Combustion PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O 2 (the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.937 0 C. Is the manufacturer’s claim correct?

PROBLEM: A balloon is filled with 3 liters of helium (atomic mass=4.0) at 1 atm pressure. It initially had a temperature of 25°C but when left under the sun the temperature changed to 28°C and the balloon expanded. The specific heat of helium is 5.1932 J/gK. a. What is the amount of heat absorbed? b. What is the amount of work done? c. What is the amount of change in internal energy?

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero. 6.5 Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f

Calculate the standard enthalpy of formation of CS 2 ( l ) given that: 1. Write the enthalpy of formation reaction for CS 2 2. Add the given rxns so that the result is the desired rxn. 6.5 C (graphite) + O 2 ( g ) CO 2 ( g )  H 0 = -393.5 kJ Rxn/f S (rhombic) + O 2 ( g ) SO 2 ( g )  H 0 = -296.1 kJ Rxn/f CS 2 ( l ) + 3O 2 ( g ) CO 2 ( g ) + 2SO 2 ( g )  H 0 = -1072 kJ rxn C (graphite) + 2S (rhombic) CS 2 ( l ) rxn C (graphite) + O 2 ( g ) CO 2 ( g )  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 ( g ) 2SO 2 ( g )  H 0 = -296.1x2 kJ rxn CO 2 ( g ) + 2SO 2 ( g ) CS 2 ( l ) + 3O 2 ( g )  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 ( l )  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn

Table 6.5 Selected Standard Heats of Formation at 25 0 C(298K) ‏ Formula  H 0 f (kJ/mol) ‏ calcium Ca( s ) ‏ CaO( s ) ‏ CaCO 3 ( s ) ‏ carbon C(graphite) ‏ C(diamond) ‏ CO( g ) ‏ CO 2 ( g ) ‏ CH 4 ( g ) ‏ CH 3 OH( l ) ‏ HCN( g ) ‏ CS s ( l ) ‏ chlorine Cl( g ) ‏ 0 -635.1 -1206.9 0 1.9 -110.5 -393.5 -74.9 -238.6 135 87.9 121.0 hydrogen nitrogen oxygen Formula  H 0 f (kJ/mol) ‏ H( g ) ‏ H 2 ( g ) ‏ N 2 ( g ) ‏ NH 3 ( g ) ‏ NO( g ) ‏ O 2 ( g ) ‏ O 3 ( g ) ‏ H 2 O( g ) ‏ H 2 O( l ) ‏ Cl 2 ( g ) ‏ HCl( g ) ‏ 0 0 0 -92.3 0 218 -45.9 90.3 143 -241.8 -285.8 107.8 Formula  H 0 f (kJ/mol) ‏ silver Ag( s ) ‏ AgCl( s ) ‏ sodium Na( s ) ‏ Na( g ) ‏ NaCl( s ) ‏ sulfur S 8 (rhombic) ‏ S 8 (monoclinic) ‏ SO 2 ( g ) ‏ SO 3 ( g ) ‏ 0 0 0 -127.0 -411.1 2 -296.8 -396.0

6.5 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn a A + b B c C + d D  H 0 rxn d  H 0 (D) f c  H 0 (C) f = [ + ] - b  H 0 (B) f a  H 0 (A) f [ + ]  H 0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  -

Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 6.5 2C 6 H 6 ( l ) + 15O 2 ( g ) 12CO 2 ( g ) + 6H 2 O ( l )  H 0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  -  H 0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [ + ] - 2  H 0 (C 6 H 6 ) f [ ]  H 0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6

Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown  H PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO( g ) + NO( g ) CO 2 ( g ) + 1/2 N 2 ( g )  H = ? Given the following information, calculate the unknown  H: Equation A: CO( g ) + 1/2 O 2 ( g ) CO 2 ( g )  H A = -283.0 kJ Equation B: N 2 ( g ) + O 2 ( g ) 2NO( g )  H B = 180.6 kJ

Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: Calculate  H 0 rxn from  H 0 f values. 4NH 3 ( g ) + 5O 2 ( g ) 4NO( g ) + 6H 2 O( g ) ‏

Sample Problem 6.6 Using the Heat of Reaction (  H rxn ) to Find Amounts PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by If aluminum is produced this way, how many grams of aluminum can form when 1.000x10 3 kJ of heat is transferred? Al 2 O 3 (s) 2Al(s) + 3/2 O 2 (g)  H rxn = 1676 kJ

Entropy Is a measure of a system’s organizational order. A system that has some recognizable order has a low entropy value; the more DISORDER the system, the HIGHER its entropy value. Symbol: Δ S Units: Joules/Kelvin

Entropy (cont.) 2 nd Law as applied to any process: The entropy of an isolated system will always tend to increase. It is very easy to destroy (=high entropy) than to build (=low entropy)

Entropy (cont.) Equation: Δ S ≥ 0 S = entropy Δ S = change in entropy For an isothermal process where a change in entropy occurs, the entropy change is given by: Δ S = Δ Q / T

Entropy (cont.) Ex: Phase change: going from ice (solid) to water (liquid), the increase in entropy value, Δ S = Δ Q / T T = melting or boiling point (in Kelvin)

Entropy (cont.) 2 nd Law as applied to Entropy: Δ S > 0 every system tends to get more disordered, hence the entropy increases

Entropy - Calculations Is also based on probability concepts. According to Boltzman, the entropy of a system taking place is given by: S = K ln W K = Boltzman constant W = Probability of occurrence of an event ln = natural log

Entropy (cont.) Sample space for throwing 3 coins simultaneously:

Entropy (cont.) Conclusion: S (= entropy) for all heads is far less compared to S for only two heads. So the probability of obtaining 2 heads far exceeds all heads.

Entropy (cont.) Relevance: Can be used to determine direction of any process or invention.

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