Presentation is loading. Please wait.

Presentation is loading. Please wait.

Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Similar presentations


Presentation on theme: "Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Energy is the capacity to do work Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position Kinetic energy is energy associated with the motion of an object. 6.1

3 Energy can be converted from one form of energy to another. Kinetic Energy Potential Energy Energy cannot be created or destroyed. => Law of conservation of energy = total quantity of energy in the universe is constant

4 Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy C 40 0 C greater thermal energy Temperature = Thermal Energy °C K °F 1 joule = 1 kg m /sec 1 calorie = joule (exactly) 2

5 Thermochemistry is the study of heat change in chemical & physical processes. => chemical reactions => phase changes Study the interconversion of energy from one form to another.

6 The system is the specific part of the universe that is of interest in the study. The surroundings is the rest of the universe. open mass & energyExchange: closed energy isolated nothing SYSTEM 6.2 SURROUNDINGS System Types:

7 State – specific characteristics of a system (physical or chemical) Examples: Pressure Temperature Composition State Function – a property of a system that characterizes a state of a system and is independent of how the system got to that state. Examples: Pressure Volume Composition

8 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  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial Energy it took each hiker to get to the top is different !!

9 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

10 ExothermicEndothermic 6.2

11 First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.  E system +  E surroundings = 0 or  E system = -  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! 6.3 Chemical energy lost by combustion = Energy gained by the surroundings system surroundings E is a state function.

12 Another form of the first law for  E system  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 Energy is expressed in units of joules of kJ. Note the conditions that define each process !!

13 Work Done On The System w = Fd w = -P  V  V < 0 -P  V > 0 w sys > 0 P x V = x d 3 = Fd = w F d2d2 initialfinal Work is not a state function!  w = w final - w initial Work is done by the surroundings on the system

14 Work Done By The System 6.3 w = Fd w = -P  V P x V = x d 3 = Fd = w F d2d2  V > 0 -P  V < 0 w sys < 0 Work is not a state function!  w = w final - w initial initialfinal Work is done by the system on the surroundings

15 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P  V (a)  V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 Latm = 0 joules (b)  V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = Latm w = Latm x J 1Latm = J Energy needs to be converted to joules. Use 1 L. atm = J The change in volume was the same in both cases, but the work done was not the same => work is not a state function.

16 Chemistry in Action: Making Snow  E = q + w q = 0 adiabatic process w < 0  E < 0  E = C  T  T < 0, SNOW !! Inside machine = air + water 20 atm Outside machine = 1 atm Air w/ water vapor rapidly expands  E = w Kinetic Energy (gas)  T (K)

17 Enthalpy and the First Law of Thermodynamics 6.4  E = q + w  E =  H - P  V  H =  E + P  V q =  H and w = -P  V At constant pressure: Note the conditions that define each process !!

18 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 > Enthalpy is a state function.

19 Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.4

20 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = kJ Is  H negative or positive? System gives off heat Exothermic  H < kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.4

21 Enthalpy is a state function. We only care about where we start and where we finish, not the route.  H net =  H 1 +  H 2 +  H 3 +  H 4 start end H1H1 H2H2 H3H3 H4H4  H net => We can construct a convenient path.

22 H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ You must pay attention to the physical state and the quantity of all reactants and products !!!

23 H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.4 H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = kJ 266 g P 4 1 mol P g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

24 Chemical Equation: Mg (s) + 2H 2 O (l) Mg(OH) 2 (s) + H 2 (g) + 353kJ Solid magnesium and water yields solid magnesium hydroxide, hydrogen gas & 353 kilojoules of heat energy Source- Tocci, Viehland Holt Chemistry Visualizing Matter. Austin, Texas: Holt, Rinehart and Winston, Inc. pg. 271, 303 and “Feeding the Army” Flameless Ration Heaters

25 It takes about 353kJ of heat energy is needed to heat the MRE to 60°C. Calculate the amount of magnesium needed. 353 kJ= 1 mol Mg = 24.3 g Mg Calculate the volume of water needed. 353 kJ = 2 mol H g / mol H 2 0 H 2 0 density = 1.00 g/mL (2 mol H 2 0) x (18.01 g / mol H 2 0) x (1 mL H 2 0 / 1.00 g H 2 0) = 36.0 mL H 2 0 Calculate the mass of magnesium hydroxide Mg(OH) 2 produced. 24.0g Mg = 1 mol Mg = 1 mol Mg(OH) g / mol Mg(OH) g Mg(OH) 2 Calculate the number of moles of hydrogen gas H 2 produced. 353 kJ= 1 mol Mg = 1.00 mol H 2 Calculate the volume of hydrogen gas H 2 produced at 1.00 atm and 25°C. PV = nRT => V = nRT/P V = (1.00 mol)( L atm/K mol)(298 K)/1.00 atm = 24.5 L

26 What would happen if 25 mL of water is added? What would happen if 360 mL of water is added? What would happen is the soldier were smoking while heating dinner? Is the material in a used flameless ration heater safe to touch? Can the heater be reused?

27 A Comparison of  H and  E 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g)  H = kJ/mol  E =  H - P  V At 25 0 C, 1 mole H 2 = 24.5 L at 1.0 atm P  V = 1.0 atm x 24.5 L = 2.5 kJ  E = kJ/mol – 2.5 kJ/mol = kJ/mol 6.4

28  H net =  H 1 +  H 2 +  H 3 +  H 4 +  H

29 The specific heat (s) 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 = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial 6.5

30 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = C q = ms  t = 869 g x J/g 0 C x –89 0 C= -34,000 J 6.5

31 How much heat is required to take exactly one mole of water from a solid at -25.0°C to steam at 125.0°C?  H net =  H 1 +  H 2 +  H 3 +  H 4 +  H 5 = kJ kJ kJ kJ kJ = 56.1 kJ  H 1 = q 1 = ms  t 1 m = (1 mol ) (18.01 g/mol) = g s = J / g °C  t 1 = (0°C - -25°C) = 25°C  H 1 = (18.01 g) (2.030 J / g °C) (25°C) = 914 J = kJ  H 2 =  H fus = (6.01 kJ/mol ) (1 mol) = 6.01 kJ H 2 O (s) H 2 O (l)  H = 6.01 kJ H 2 O (l) H 2 O (g)  H = 44.0 kJ  H 3 = q 3 = ms  t 3 s = J / g °C  t 3 = (100.°C -0°C) = 100.0°C  H 3 = (18.01 g)(4.184 J / g °C)(100.0°C) = 7540 J = 7.54 kJ  H 4 =  H vap = (44.0 kJ/mol ) (1 mol) = 44.0 kJ  H 5 = q 5 = ms  t 5 s = J / g °C  t 5 = (125.0°C – 100.0°C) = 25.0°C  H 5 = (18.01 g) (2.043 J / g °C) (25.0°C) = 920 J = kJ  H net =  H 1 +  H 2 +  H 3 +  H 4 +  H 5

32 Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = ms  t q bomb = C bomb  t 6.5 Reaction at Constant V  H ~ q rxn  H = q rxn

33 Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms  t q cal = C cal  t 6.5 Reaction at Constant P  H = q rxn

34 What happens is we mix two liquids at different temperatures? Example: Mix 100 g water at 25°C with 100 g water at 75°C What is the final temperature? Since both liquids are the same (water) and both have the same weight (100 g) => take the average 75°C 25°C 50°C Loss of thermal energy Gain of thermal energy What happens if the substances are different or the amounts are different?

35 If g of Copper at 100.0°C is added to 150. g water at 25.0°C What is the final temperature?  t = (t final – t initial ) s Cu = J / g °C s H2O = J / g °C -(60.56 g)(0.385 J/g°C)( t final °C) = (150. g)(4.184 J/g°C)(t initial -25.0°C) | Heat lost by copper | = | Heat gained by water | - q Cu = q H2O -q Cu = - m Cu s Cu  t Cu = m H2O s H2O  t H2O (-23.3 J/g°C) t final J = (628 J/g°C) t final – J J J = (( ) J/g°C) t final t final = (18000/651.3)°C t final = 27.7°C q sys = q water + q cal + q Cu = 0 assume q cal is very very small

36 6.5

37 Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l)  H = kJ/mol 1 cal = J 1 Cal = 1000 cal = 4184 J

38 A g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3027 J/°C. The temperature increases by °C. Calculate the heat change by the burning Mg in kJ/g and in kJ/mol. | Heat given off by reaction | = | Heat gained by calorimeter | q Cal = C p  t = (3024 J/°C)(1.126 °C) = x10 3 J q Rxn = (3.405 x10 3 J)(1 kJ / 1000 J) = kJ / g Mg g Mg (24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = kJ / mol Mg Is the reaction exothermic or endothermic? What would happen to the temperature of the water if it were the opposite type reaction?

39 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? 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 The standard enthalpy of formation of any element in its most stable form is zero.  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

40 6.6

41 The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 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.)

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

43 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 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x– x–187.6 ] – [ 2x49.04 ] = kJ kJ 2 mol = kJ/mol C 6 H 6 6.6

44 The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln - H components 6.7 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?

45 The Solution Process for NaCl  H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7


Download ppt "Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."

Similar presentations


Ads by Google