George Mason University General Chemistry 211 Chapter 6 Thermochemistry: Energy Flow and Chemical Change

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1 George Mason University General Chemistry 211 Chapter 6 Thermochemistry: Energy Flow and Chemical Change Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Change, 7 th edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material. Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor. 1/13/2015 1

2 Thermochemistry Whenever matter changes composition, such as in a chemical reaction, the energy content of the matter changes also In some reactions the energy that the reactants contain is greater than the energy contained by the products This excess energy is released as heat In other reactions it is necessary to add energy (heat) before the reaction can proceed The energy contained by the products in these reactions is greater than the energy of the original reactants Physical changes can also involve a change in energy such as when ice melts 1/13/2015 2

3 Thermochemistry Thermodynamics is the science of the relationship between heat and other forms of energy Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions Energy is the potential or capacity to move matter (do work); energy is a property of matter Energy can be in many forms: Radiant Energy - Electromagnetic radiation Thermal Energy - Associated with random motion of a molecule or atom Chemical Energy - Energy stored within the structural limits of a molecule or atom 1/13/2015 3

4 Energy There are three broad concepts of energy: Kinetic Energy (E k ) is the energy associated with an object by virtue of its motion, E k = ½mv 2 Potential Energy (E p ) is the energy an object has by virtue of its position in a field of force, E p = mgh Internal Energy (E i or U i ) is the sum of the kinetic and potential energies of the particles making up a substance E (i) = E k + E p 1/13/2015 4

5 Energy SI (metric) unit of energy is the Joule: 1 watt = 1 J/s (J) = kg m 2 /s 2 1 cal = amount of energy needed to raise 1 g of water 1 o C (common energy unit) 1 cal = J 1 Btu = 1055 J (Btu - British Thermal Unit) The Law of Conservation of Energy: Energy may be converted from one form to another, but the total quantities of energy remain constant 1/13/2015 5

6 Practice Problem Thermal decomposition of 5.0 metric tons of limestone (CaCO 3 ) to Lime (CaO) and Carbon Dioxide (CO 2 ) requires 9.0 x 10 6 kj of heat (E) Convert this energy to: a. Joules b. calories c. British Thermal Units (Btu) (5 tons) CaCO (s) 9.0 x 10 6 kj CaO(s) CO (g) J 9 ΔE(J) 9.0 x 10 kj) 9.0 x 10 J 1kJ J 1 cal 9 9 ΔE(cal) 9.0 x 10 kj x x 10 cal 1 kj J J 1 Btu 6 6 ΔE(Btu) 9.0 x 10 kj x x 10 Btu 1 kj 1055 J 1/13/2015 6

7 Energy When a chemical system changes from reactants to products and the products are allowed to return to the starting temperature, the Internal Energy (E) has changed (E) The difference between the system internal energy after the change (E final ) and before the change (E initial ) is: E = E final - E initial = E products - E reactants If energy is lost to the surroundings, then E final < E initial E < 0 If energy is gained from the surroundings, then E final > E initial E > 0 1/13/2015 7

8 Heat of Reaction In chemical reactions, heat is often transferred from the system to its surroundings, or vice versa The substance or mixture of substances under study in which a change occurs is called the thermodynamic system (or simply system) The surroundings are everything in the vicinity of the thermodynamic system Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings Heat flows from a region of higher temperature to one of lower temperature Once the temperatures become equal, heat flow stops 1/13/2015 8

9 Heat of Reaction When heat (q) is released from the system (heat out) to the surroundings (q < 0 negative), the reaction is defined as an Exothermic reaction When the surroundings deliver heat (heat in) to the system (q > 0 positive), the reaction is defined as an Endothermic reaction Exothermic Endothermic q<0 q>0 Energy Surroundings Energy Surroundings System System 1/13/2015 9

10 Heat of Reaction Heat of a system is denoted by the symbol q The sign of q is positive (q > o) if heat is absorbed by the system, i.e., system temperature increases The sign of q is negative (q < o) if heat is evolved by the system, i.e., system temperature decreases Heat of Reaction (at a given temperature) is the value of q required to return a system to the given temperature when the reaction stops at the completion of the reaction 1/13/

11 Heat of Reaction Consider a chemical reaction that begins with the system and surroundings temperature at 25 o C If the temperature of a system decreases during the reaction, heat flows from the surroundings into the system When the reaction stops, heat continues to flow until the system temperature returns to the temperature of the surroundings at 25 o C Heat has been absorbed by or added to the system from the surroundings The value of q is positive, that is: q > 0 1/13/

12 Heat of Reaction If the temperature of the system rises, heat flows from the system to the surroundings When the reaction stops, heat continues to flow to the surroundings until the system returns to the temperature of its surroundings (at 25 o C) Heat has flowed out of the system; it has evolved (lost) heat; thus, q is negative, that is q < 0 1/13/

13 Heat Flow and Phase Changes Predict the sign of q for each of the processes below 1. H 2 O(g) H 2 O(l) Condensation - Energy (heat) is lost by water vapor Exothermic reaction - q is negative 2. CO 2 (s) CO 2 (g) Evaporation Energy (heat) is absorbed (added) by the system from its surroundings Endothermic reaction - (q is positive) 3. CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Combustion Burning (oxidation) of Methane releases (evolves) heat to surroundings Exothermic reaction - (q is negative) 1/13/

14 Work & Internal Energy Internal Energy The Internal Energy of a system, E, is precisely defined as the heat at constant pressure (q p ) plus any work (w) done by the system Work is the energy transferred when an object is moved by a force w = - P Δ V = - P (V - V ) Internal Energy used ΔE = q p + w ΔE = q p + (-P ΔV) q p = ΔE + P ΔV final initial to expand volume by increasing pressure is lost to the surroundings, thus the negative sign Adiabatic Process Thermodynamic Process Without the Gain or Loss of Heat ( q = 0) 1/13/

15 Practice Problem A system delivers 225 J of heat to the surroundings while delivering 645 J of work. Calculate the change in the internal energy, E, of the system q p = heat delivered to surroundings from system = J w = work delivered to surroundings from system = J ΔE = change in internal energy ΔE = q p + w ΔE = J + (-645 J) = J Heat is lost to the surroundings 1/13/

16 Pressure-Volume Work Sign conventions for q, w, and E: q + w (-P V) = E Depends on sizes of q and w + Depends on sizes of q and w q: + system gains heat q: system loses heat w: + work done on system w: work done by system 1/13/

17 Practice Problem A system expands in volume from 2.00 L to 24.5 L at constant temperature. Calculate the work (w), in Joules (J), if the expansion occurs against a constant pressure of 5.00 atm w = - pδv kg Pa m s 10 m w = atm (24.5 L L) atm Pa L kg m kg m 1J w = -1, = s s kg m 2 s 4 w = J /13/

18 Practice Problem A system that does no work but which transfers heat to the surrounding has: a. q < 0, E > 0 b. q < 0, E < 0 c. q > 0, E > 0 d. q > 0, E < 0 e. q < 0, E = 0 A system that does no work but receives heat from the surroundings has: a. q < 0, E > 0 b. q > 0, E < 0 c. q = E d. q = - E e. w = E A system which undergoes an adiabatic change (i.e., q = 0) and does work on the surroundings has: a. w < 0, E = 0, b. w > 0, E > 0 c. w > 0, E < 0 d. w < 0, E > 0 e. w < 0, E < 0 A system which undergoes an adiabatic change (i.e., q = 0) and has work done on it by the surroundings has: a. w = E b. w = - E c. w > 0, E < 0 d. w < 0, E > 0 e. w > E 1/13/

19 Enthalpy and Enthalpy Change Enthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction An extensive property is one that depends on the quantity of substance Enthalpy is a state function, a property of a system that depends only on its present state and is independent of any previous history of the system Enthalpy represents the heat energy tied up in chemical bonds 1/13/

20 Enthalpy and Enthalpy Change The change in Enthalpy for a reaction at a given temperature and pressure, called the Enthalpy of Reaction, is obtained by subtracting the Enthalpy of the reactants from the Enthalpy of the products. ΔH = H - H rxn (products) (reactants) 1/13/

21 Enthalpy and Enthalpy Change Enthalpy is defined as the internal energy plus the product of the pressure and volume (work) H E PV The change in Enthalpy is the change in internal energy plus the product of constant pressure and the change in Volume ΔH = ΔE + P ΔV 1/13/

22 Recall: Enthalpy and Enthalpy Change q = ΔE + P ΔV p ΔH = ΔE + P ΔV Thus : ΔH = q p (At Constant Pressure) The change in Enthalpy equals the heat gained or lost (heat of reaction, H rxn ) at constant pressure This represents the entire change in internal energy (E) minus any expansion work done by the system (PV would have negative sign) 1/13/

23 Practice Problem An ideal gas (the system) is contained in a flexible balloon at a pressure of 1 atm and is initially at a temperature of 20.0 o C. The surrounding air is at the same pressure, but its temperature is 25 o C. When the system is equilibrated with its surroundings, both systems and surroundings are at 25 o C and 1 atm. In changing from the initial to the final state, which of the following relationships regarding the system is correct? a. E = 0 b. E < 0 c. H = 0 d. w > 0 e. q > 0 Heat is added, internal energy increases Heat is added, internal energy increases E increases and P V work is done by system P V work is done by system (volume increase) Temperature (heat) in system increases E > 0 E > 0 H > 0 W < 0 1/13/

24 Practice Problem In which of the following processes is H = E, i.e. P V = 0? a. 2HI(g) H 2 (g) + I 2 (g) at atmospheric pressure (P V = 0 no change in moles, volume) b. Two moles of Ammonia gas are cooled from 325 o C to 300 o C at 1.2 atm (P V 0 Vol decreases) c. H 2 O(l) H 2 O(g) at 100 o C at atmospheric pressure (P V 0 Vol increases) d. CaCO 3 (s) CaO(s) + CO 2 (g) at 800 o C at atmospheric pressure (P V 0 Vol increases) e. CO 2 (s) CO 2 (g) at atmospheric pressure (P V 0 Vol increases) 1/13/

25 Comparing E & H Reactions that do not involve gases Reactions such as precipitation, acid-base, many redox, etc., do not produce gases Since the change in volumes of liquids and solids are quite small: V 0 P V 0 H E Reactions in which the amount (mol) of gas does not change (Vol of Gaseous Reactants = Vol Gaseous Products V = 0 P V = 0 H = E Reactions in which the amount (mol) of gas does change PV 0 However, q p is usually much greater than PV Therefore: H E 1/13/

26 Example: Comparing E & H 2H 2 (g) + O 2 (g) 2H 2 O(g) Change in moles: 3 mol 2 mol PV 0 H = kj and PV = -2.5kJ E = H - PV = kj - (-2.5 kj) = kj Most of E occurs as Heat (H = q p ) H E For many reactions, even when PV 0, H is close to E 1/13/

27 Comparing E & H For which one of the following reactions will H be approximately (or exactly) equal to E? a. H 2 (g) + Br 2 (g) 2HBr(g) (No change in volume; no change in work, PV = 0) b. H 2 O(l) H 2 O(g) (Change in volume; change in work due to gas expansion, PV 0) c. CaCO 3 (s) CaO(s) + CO 2 (g) (Change in volume; change in work due to gas expansion, PV 0 d. 2H(g) + O(g) H 2 O(l) (Change in volume; condensation, heat (q) released, PV 0) e. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) (Change in volume; condensation, heat (q) released, PV 0) 1/13/

28 Exothermic and Endothermic Processes Energy (E), Pressure (P), and Volume (V) are state functions Enthalpy (H) is also a state function, which means that H depends only on the difference between H final & H initial The Enthalpy change of a reaction, also called the Heat of Reaction (H rxn ), always refers to H rxn = H final - H initial = H products - H reactants H products can be either more or less than H reactants The resulting sign of H indicates whether heat is absorbed from the surroundings (heat in) or released to the surroundings (heat out) in the process 1/13/

29 Exothermic and Endothermic Processes An Exothermic reaction releases heat (heat out) to surroundings with a decrease in system Enthalpy CH 4 (g) + 2O 2 CO 2 (g) + 2H 2 O(g) + heat Exothermic: H final < H initial H < 0 (negative) An Endothermic reaction absorbs heat (heat in) from the surroundings resulting in an increase in system Enthalpy Heat + H 2 O(s) H 2 O(l) Endothermic H final > H initial H > 0 (positive) 1/13/

30 Types of Enthalpy Changes When a compound is produced from its elements, the Enthalpy change (Heat of Reaction) is called: Heat of Formation ( H f ) K(s) + ½Br 2 ()l) KBr(s) H = H f When a substance melts, the Enthalpy change is called: Heat of Fusion ( H fus ) NaCl(s) NaCl(l) H = H (fus) When a substance vaporizes, the Enthalpy change is called: Heat of Vaporization C 6 H 6 (l) C 6 H 6 (g) H = H (vap) 1/13/

31 Thermochemical Equations A Thermochemical Equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the Enthalpy of Reaction ( H rxn ) for these molar amounts is written directly after the equation. N (g) + 3 H (g) 2 NH (g) ΔH rxn = kj H is negative; heat is lost to surroundings 1 mol N mol H 2 yields 91.8 kj of heat Reaction is Exothermic 1/13/

32 Practice Problem Sulfur, S 8, burns in air to produce Sulfur Dioxide. The reaction evolves (releases) 9.31 kj of heat per gram of Sulfur at constant pressure. Write the thermochemical equation for this reaction. S O SO + Heat Exothermic Reaction Balance the Reaction S 8 8 O 2 8 SO 2 ΔH kj Exothermic Reaction 1/13/

33 Practice Problem In a phase change of water between the liquid and the gas phases, kj of energy was released by the system. What was the product, and how much of it was formed in the phase change. (Data: H 2 O(l) H 2 O(g) H = kj/mol) a. 315 g of water vapor was produced b g of water vapor was produced c mol of water vapor was produced d mol of liquid water was produced e g of liquid water was produced H is positive (endothermic reaction) Since energy was released, the gas condensed to liquid kj / kj / mol = 17.5 mols 1/13/

34 Thermochemical Equations The following are two important rules for manipulating Thermochemical equations: When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the H in the original equation by that same factor When a chemical equation is reversed, the value of H is reversed in sign 1/13/

35 Practice Problem When White Phosphorus burns in air, it produces Phosphorus (V) Oxide (Change in Oxidation state) P 4 (s) + 5O 2 (g) P 4 O 10 (s) H = kj/mol What is H for the following equation? P 4 O 10 (s) P 4 (s) + 5O 2 (g) H =? Ans: The original reaction is reversed Change the Sign!! H = kj/mol 1/13/

36 Practice Problem Carbon Disulfide (CS 2 (l)) burns in air, producing Carbon Dioxide and Sulfur Dioxide CS 2 (l) + 3 O 2 (g) CO 2 (g) + 2 SO 2 (g) H = kj What is H for the following equation? 1/2 CS 2 (l) + 3/2 O 2 (g) 1/2 CO 2 (g) + SO 2 (g) Ans: The new reaction uses ½ the original amounts Divide H by 2 H = (-1077 / 2) = kj 1/13/

37 Applying Stoichiometry and Heats of Reactions Consider the reaction of Methane, CH 4, burning in the presence of Oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH 4? o CH (g) + 2 O (g) CO (g) + 2 H O(l) ΔH = kj / mol mol CH kj 10.0 g CH g CH 1 mol CH = kj 4 4 1/13/

38 Measuring Heats of Reaction Heats of Reactions (Enthalpy change of reaction H rxn ) are determined from the heat required to raise the temperature of a substance A thermochemical measurement is based on the relationship between heat and temperature change; that is the quantity of heat (q) absorbed or released by an object is proportional to its temperature change Heat (q) ΔT or q = constant ΔT Each object has its own heat capacity, which is the quantity of heat required to change its temperature by 1 Kelvin (K) Thus, proportionality constant above is the Heat Capacity: q = Constant = Heat Capacity [in units of J / K] ΔT 1/13/

39 Measuring Heats of Reaction The specific heat capacity, S, (or specific heat ) is the heat required to raise the temperature of one gram of a substance by one degree Celsius S is in units of J/g o K m = grams of sample q = m S ΔT T = T final - T initial The molar heat capacity, C, of a sample of substance is the quantity of heat required to raise the temperature of one mole of substance one degree Celsius C is in units of J/mol o K, n = moles of substance q = n C ΔT T = T final - T initial 1/13/

40 Measuring Heats of Reaction Bomb Calorimeter used to measure heats of combustion 1/13/

41 Measuring Heats of Reaction Specific Heats and Molar Heat Capacities of some substances 1/13/

42 Practice Problem Suppose you mix 20.5 g of water at 66.2 o C with 45.4 g of water at 35.7 o C in an insulated cup. What is the maximum temperature of the solution after mixing? Ans: The heat lost by the water at 66.2 o C is balanced by the heat gained by the water at 35.7 o C o o q = m S ΔT m = 20.5 g 1 2 m 2 T f = T f m T f = T f T = 45.2 C ΔT 1 = T f C ΔT 2 = T f C m = 45.4 g -q = q = - m SΔT = m SΔT lost gained o o -m * T C = m * T C 1 f 2 f o 1/13/2015 f 42

43 Practice Problem A piece of copper with a temperature of 100 o C is dropped into a beaker containing 50.0 grams of water at 20 o C. When the reaction is completed the temperature of the solution is 25 o C. Assuming no loss of heat and the heat capacity of the solution is the same as water, J/(g * o K), what is the heat capacity of Copper in J/K? ΔT 1 = (25 100) = -75 o C ΔT 2 = (25 20) = 5 o C Heat change by increasing water temperature by 5 o K q 2 = 50g * J/(g * o K) * 5 o K = 1,046 J Heat lost by copper (q 1 ) is equal to heat gained by water -q 1 = +q 2 Therefore, The Heat Capacity of copper is: 2 q -q -1,045 J 1 = = o = 13.9 J / K ΔT -75 C 1/13/

44 Practice Problem How much heat is gained by Nickel when 500 g of Nickel is warmed from 22.4 o C to 58.4 C? [The specific heat of Nickel is J/(g C)] a J b J c J d J e J Ans: d q = m s ΔT f o o o i ΔT = T (58.4 C) - T (22.4 C) = 36.0 C o s = Specific Heat Nickel = J / g C m = 500 g q = 500 g * J / g C * 36.0 C q = 7992 = 8000 J 1/13/ o o

45 Practice Problem When 25.0 ml of 0.5 M H 2 SO 4 is added to 25.0 ml of 1.00 M KOH in a calorimeter at 23.5 o C, the temperature rises to o C Calculate H rxn for each reactant. Assume density (d) and specific heat of the solution (s) are the same as water 2 KOH(aq) + H2SO 4(aq) K 2SO 4(aq) + 2 H2O(l) q = m s ΔT (Δ T = T - T ) d = 1 g / ml s = J / g C soln f i 1.00 g J o 1 kj q soln = mL o C = kj ml g C 1000 J Calculate moles mol H SO 1 L 25.0 ml = mol H SO L 1000 ml 1.00 mol KOH 1 L 25.0 ml = mol KOH L 1000 ml Both KOH or H SO are limiting (2 moles KOH / 1 mol H SO ) Con t 1/13/ o

46 Practice Problem (Con t) When 25.0 ml of 0.5 M H 2 SO 4 is added to 25.0 ml of 1.00 M KOH in a calorimeter at 23.5 o C, the temperature rises to o C. Calculate H rxn for each reactant. Assume density (d) and specific heat of the solution are the same as water. Temperature of water increased (23.5 o C o C) The Reaction is Exothermic (heat released to surroundings (water)) Thus, q rxn is negative q = q = kj soln rxn kj H (H SO ) kj / mol H SO rxn mol H2SO kj H rxn (KOH) kj / mol KOH mol KOH 1/13/

47 Hess s Law Hess s law of Heat Summation For a chemical equation that can be written as the sum of two or more steps, the Enthalpy change for the overall equation is the sum of the Enthalpy changes for the individual steps In coupled reactions, the Enthalpy change for the overall reaction is the sum of the Enthalpy changes for the coupled reactions Note: It is often necessary to reverse chemical equations to couple them so chemical species are on the correct side of yield sign, or multiply through by a coefficient to cancel common chemical species 1/13/

48 Hess s Law For example, suppose you are given the following data: o S(s) + O (g) SO (g) ΔH = -297 kj / mol 2 2 o 2 SO (g) 2 SO (g) + O (g) ΔH = 198 kj / mol Could you use these data to obtain the Enthalpy change for the following reaction? o 2 S(s) + 3 O (g) 2 SO (g) ΔH =? 2 3 Con t 1/13/

49 Hess s Law If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third 2 S(s) + 2 O (g) 2 SO (g) ΔH = (-297 kj) (2) SO (g) + O (g) 2 SO (g) ΔH = (198 kj) (-1) S(s) + 3 O (g) 2 SO (g) ΔH = (-792 kj) 2 3 Note : ( ) + (198-1) = kj kj = kj o o o Note the change in H values with the changes in the molar coefficients to balance equation 1 and the reversal of equation 2 1/13/

50 Given the following data, Practice Problem A(s) + O 2 (g) AO 2 (g) H = 105 kj/mol A(g) + O 2 (g) AO 2 (g) H = 1200 kj/mol Find the heat required for the reaction converting: A(s) to A(g) at 298 K and 1 atm pressure. A(s) + O 2(g) AO 2(g) kj AO 2(g) A(g) + O 2(g) kj A(s) A(g) kj Note change of sign of H when 2 nd equation is reversed 1/13/

51 Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: Pressure - 1 atmosphere (760 mm Hg) Temperature - (usually 25 o C). The Enthalpy change for a reaction in which reactants are in their standard states is denoted as the Standard Heat of Reaction o ΔH rxn 1/13/

52 Standard Enthalpies of Formation Standard Enthalpy of Formation of Substance The Enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard states o ΔH f Note: The standard Enthalpy of Formation for a Pure Element (C, Fe, Au, N, etc.) in its standard state is zero 1/13/

53 Standard Enthalpies of Formation Law of Summation of Heats of Formation The Enthalpy of a reaction i.e., the Standard Heat of Reaction: ( H o rxn) is equal to the total formation energy of the products minus that of the reactants o o o rxn f f ΔH = nδh (products) - mδh (reactants) Where is the mathematical symbol meaning the sum of and m and n are the coefficients of the substances in the chemical equation, i.e., the relative number of moles of each substance 1/13/

54 Standard Enthalpies of Formation Selected Standard Heats of Formation (Enthalpies) At 25 o C (298 o K) Formula Calcium Ca(s) CaO(s) CaCO3(s) Carbon C(graphite) C(diamond) CO(g) CO 2 (g) CH 4 (g) CH 3 OH(l) HCN(g) CS 2 Chlorine Cl 2 (g) Cl(g) Cl - (aq) Cl - (g) HCl(g) Bromine Br 2 (l) Br(g) Br 2 (g) Br - (ag) Br - (g) HBr(g) H o f (kj/mol) Hydrogen H 2 (g) H(g) Oxygen O 2 (g) O 3 (g) H 2 O(g) H 2 O(l) Nitrogen N 2 (g) NH 3 (g) NO(g) Formula Sulfur S 8 (rhombic) S 8 (monoclinic) SO 2 (g) SO 3 (g) H o f (kj/mol) Formula H o f (kj/mol) 1/13/ Silver Ag(s) AgCl(s) Sodium Na(s) Na(g) NaCL(s)

55 Practice Problem Calculate the Heat of Reaction, H rxn, for the combustion of C 3 H 6 (g): C 3 H 6 (g) + 9/2 O 2 (g) 3 CO 2 (g) + 3 H 2 O(l) H o f values in kilojoules per mole are as follows: C 3 H 6 (g) = 21 CO 2 (g) = 394 H 2 O(l) = 286 a kj b kj c. 701 kj d kj e kj Ans: a o o o rxn f f o rxn o rxn o rxn ΔH = n ΔH (products) - m ΔH (reactants) ΔH = kj ΔH = ΔH = /13/

56 Practice Problem Acetylene burns in air according to the equation below. Given: H o f CO 2 (g) = kj/mol H o f H 2 O(g) = kj/mol o rxn C H (g) + 5 / 2 O (g) 2 CO (g) + H O(g) ΔH = kj Calculate H o f of C 2 H 2 (g) o o o ΔH rxn = 2 mol (Δ H f, CO 2(g)) + 1 mol (Δ H f, H2O(g)) - o o 1 mol (Δ H f, C2H 2(g)) + 5 / 2 mol (Δ H f, O 2(g)) kj = 2 mol ( kj / mol) + 1 mol ( kj / mol) - o 1 mol ( (Δ H f, C2H 2(g) + 5 / 2 mol (0.0) o f kJ = kJ kJ - 1 mol (Δ H, C H (g)) ΔH, C H (g) = o f kj -mol = kj / mol 1/13/

57 Summary Equations & Relationships E = 1 / 2 m v k 2 ΔE = E final - E initial = E products - Ereactants w = - P Δ V = - P (V - V ) ΔE = q p + w q p = ΔE + P ΔV ΔH = ΔE + P ΔV q = sm ΔT lost final initial ΔE = q p + (-P ΔV) ΔH = q p (at Constant P) -q (exothermic) = q (endothermic) gained o o o rxn f f ΔH = nδ H (products) - mδ H (reactants) 1/13/

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