Chapter 8 Thermochemistry: Chemical Energy

Chapter 8 Thermochemistry: Chemical Energy 國防醫學院生化學科王明芳老師 2011-11-8 & 2011-11-15 Chapter 8/1

Energy and Its Conservation Conservation of Energy Law: Energy cannot be created or destroyed; it can only be converted from one form to another. Energy: is the capacity to do work, or supply heat. Energy = Work + Heat Kinetic Energy: is the energy of motion. (due to motion of the object) Kinetic energy of a body in motion = 1 2 mv 2 1 2 mass velocity 2 Chapter 8/2

Energy and Its Conservation 1 Joule = kg m 1 2 s 2 1 calorie = 4.184 J Potential Energy: is stored energy. (due to position or composition - can be converted to work) Chapter 8/3

Energy and Its Conservation CONSERVATION OF ENERGY. The total amount of energy contained by the water in the reservoir is constant. Chapter 8/4

Energy and Its Conservation Thermal Energy: The kinetic energy of molecular motion and is measured by finding the temperature of an object. (including rotational and vibrational) Thermal energy is proportional to the temperature in degrees Kelvin. E thermal T ( o K) Heat is the amount of thermal energy transferred between two objects at different temperatures. Heat involves a transfer of energy between 2 objects due to a temperature difference Chapter 8/5

Energy and Its Conservation Energy Units calorie (cal): One calorie defined as the amount of energy necessary to raise the temperature of 1g of water by 1 (specifically, from 14.5 to 15.5 ) Kilocalorie = Kcal SI unit: joule If a 2.0 kg object is moving with a velocity of 1.0 m/s, the kinetic energy is: 1/2mv 2 = 1/2(2.0 kg)(1.0 m/s) 2 =1.0 kg*m 2 /s 2 =1.0 J (joule) Chapter 8/6

Internal Energy and State Functions First Law of Thermodynamics: The total internal energy E of an isolated system is constant DE = E final - E initial Chapter 8/7

Internal Energy and State Functions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + 802 kj energy DE = E final - E initial = -802 kj 802 kj is released when 1 mole of methane, CH 4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water. Chapter 8/8

Internal Energy and State Functions State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state Chapter 8/9

Internal Energy and State Functions State function (or property): refers to a property of the system that depends only on its present state does not depend on the history of the system A change in this function (or property) in going from one state to another state is independent of the particular pathway taken between the two states Energy is a state function; work and heat are not state function Chapter 8/10

Expansion Work C 3 H 8 (g) + 5O 2 (g) 6 mol of gas 3CO 2 (g) + 4H 2 O(g) 7 mol of gas Expansion work in a chemical reaction. Chapter 8/11

Expansion Work The energy change in a system equals the work done on the system + the heat added. ΔE = E final E initial = E 2 E 1 = q + w q = heat, w = work Work is a force (F) that produces an object s movement, times the distance moved (d): Pressure = Work = Force x Distance Force Area F A Pressure is the force per unit area. 1 N/m 2 = 1 Pa (1 Newton per square meter) 1 atm = 101,325 Pa (Pascal, Pa for short) atmospheres (symbol = atm) 1 L atm = 101 J millimeters of mercury (symbol = mm Hg) Chapter 8/12

Expansion Work Expansion Work: Work done as the result of a volume change in the system Chapter 8/13

Expansion Work The most common type of work encountered in chemical systems is the expansion work (also called pressure-volume, or PV, work) done as the result of a volume change in the system Chapter 8/14

Example 8.1 Calculating The Amount Of PV Work Calculate the work in kilojoules done during a reaction in which the volume expands from 12.0 L to 14.5 L against an external pressure of 5.0 atm. Solution 1. w = PDV, where P is the external pressure opposing the change in volume. 2. In this instance, P = 5.0 atm and DV = (14.5 12.0). 3. Remember that an expanding system loses work energy, which thus has a negative sign.

Energy and Enthalpy DE = q + w q = heat transferred w = work = -PDV q = DE + PDV Constant Volume (DV = 0): Constant Pressure: q V = DE q P = DE + PDV Chapter 8/16

Energy and Enthalpy q P = DE + PDV = DH Enthalpy change or Heat of reaction (at constant pressure) Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state. DH = H final - H initial = H products - H reactants The amount of heat exchanged between the system and the surroundings is given the symbol q. Chapter 8/17

The Thermodynamic Standard State C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) DH = -2044 kj C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) DH = -2220 kj The different of 176 kj between the values of ΔH for the two reactions arises because the conversion of liquid water to gaseous water requires energy. Chapter 8/18

The Thermodynamic Standard State Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 o C; 1 M concentration for all substances in solution C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) DH o = -2044 kj These are indicated by a superscript to the symbol of the quantity reported. standard enthalpy change is indicated by the symbol DH Chapter 8/19

Example 8.2 Calculating DE For A Reaction The reaction of nitrogen with hydrogen to make ammonia has DH = 92.2 kj. What is the value of DE in kilojoules if the reaction is carried out at a constant pressure of 40.0 atm and the volume change is 1.12 L? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) DH = 92.2 kj Solution

Enthalpies of Physical and Chemical Change Enthalpies of Physical Change Enthalpy of Fusion or heat of fusion (DH fusion ): The amount of heat necessary to melt a substance without changing its temperature Enthalpy of Vaporization or heat of vaporization (DH vap ): The amount of heat required to vaporize a substance without changing its temperature Enthalpy of Sublimation or heat of sublimation (DH subl ): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase Chapter 8/21

Enthalpies of Physical and Chemical Change Enthalpy as a state function. Because enthalpy is a state function, the value of the enthalpy change from solid to vapor does not depend on the path taken between the two states Chapter 8/22

Enthalpies of Physical and Chemical Change 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) Thermite reaction of aluminum with iron (III) oxide DH o = -852 kj exothermic 2Fe(s) + Al 2 O 3 (s) 2Al(s) + Fe 2 O 3 (s) DH o = +852 kj endothermic Enthalpies of Chemical Change Chapter 8/23

Enthalpies of Physical and Chemical Change Endothermic: Heat flows into the system from the surroundings and DH has a positive sign. Exothermic: Heat flows out of the system into the surroundings and DH has a negative sign. Reversing a reaction changes the sign of DH for a reaction. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) DH = 2219 kj 3 CO 2(g) + 4 H 2 O (l) C 3 H 8(g) + 5 O 2 (g) DH = +2219 kj Multiplying a reaction increases DH by the same factor. 3C 3 H 8 (g) +15O 2 (g) 9CO 2 (g) +12H 2 O (l) DH = 6657 kj Chapter 8/24

Example 8.3 Calculating The Amount Of Heat Released In A Reaction How much heat in kilojoules is evolved when 5.00 g of aluminum reacts with a stoichiometric amount of Fe 2 O 3? 2 Al(s) + Fe 2 O 3 (s) 2 Fe(s) + Al 2 O 3 (s) DH = 852 kj Solution

Calorimetry and Heat Capacity Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: 1. Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = DH. 2. Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = DE. Chapter 8/26

Calorimetry and Heat Capacity Measure the heat flow at constant pressure (DH). Chapter 8/27

Calorimetry and Heat Capacity Measure the heat flow at constant volume (DE). A bomb calorimeter for measuring the heat evolved at constant volume (DE) in a combustion reaction. Chapter 8/28

Calorimetry and Heat Capacity Heat Capacity (C): The amount of heat necessary to raise the temperature of an object or substance a given amount q = C x DT Specific Heat (Capacity): The amount of heat required to raise the temperature of 1 g of a substance by 1 o C q = (Specific heat) x (Mass of substance) x DT Chapter 8/29

Calorimetry and Heat Capacity Molar Heat Capacity (C m ): The amount of heat necessary to raise the temperature of 1 mol of a substance by 1 o C q = C m x Moles of substance x DT Chapter 8/30

Calorimetry and Heat Capacity As indicated in Table 8.1, the specific heat of liquid water is considerably higher than that of most other substances, so a large transfer of heat is necessary to either cool or warm a given amount of water. Chapter 8/31

Calorimetry and Heat Capacity Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 o C to 3 o C. q = (Specific heat) x (Mass of substance) x DT J Specific heat = 4.18 g o C Mass = 350 g Temperature change = 3 o C - 25 o C = -22 o C Chapter 8/32

Calorimetry and Heat Capacity Calculate the amount of heat transferred. Heat evolved = 4.18 J g o C x 350 g x -22 o C = -32 000 J -32 000 J x 1 kj 1000 J = -32 kj Chapter 8/33

Example 8.4 Calculating A Specific Heat What is the specific heat of silicon if it takes 192 J to raise the temperature of 45.0 g of Si by 6.0 C? Solution

Example 8.5 Calculating DH In A Calorimetry Experiment Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of solid silver chloride: Ag + (aq) + Cl (aq) AgCl(s) When 10.0 ml of 1.00 M AgNO 3 solution is added to 10.0 ml of 1.00 M NaCl solution at 25.0 C in a calorimeter, a white precipitate of AgCl forms and the temperature of the aqueous mixture increases to 32.6 C. Assuming that the specific heat of the aqueous mixture is 4.18 J/(g C), that the density of the mixture is 1.00 g/ml, and that the calorimeter itself absorbs a negligible amount of heat, calculate DH in kilojoules for the reaction.

Example 8.5 Calculating DH In A Calorimetry Experiment Continued Solution According to the balanced equation, the number of moles of AgCl produced equals the number of moles of Ag + (or Cl ) reacted: Therefore, DH = 64 kj (negative because heat is released)

Hess s Law 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. Haber Process 3H 2 (g) + N 2 (g) 2NH 3 (g) DH o = -92.2 kj Chapter 8/37

Hess s Law 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. Multiple-Step Process 2H 2 (g) + N 2 (g) N 2 H 4 (g) + H 2 (g) N 2 H 4 (g) 2NH 3 (g) DH o 1 =? DH o 2 = -187.6 kj 3H 2 (g) + N 2 (g) 2NH 3 (g) DH o 1+2 = -92.2 kj Chapter 8/38

Hess s Law DH o 1 + DH o 2 = DH o 1+2 DH o 1 = DH o 1+2 - DH o 2 = -92.2 kj - (-187.6 kj) = +95.4 kj Enthalpy changes for steps in the synthesis of ammonia from nitrogen and hydrogen. If DH o values for step 2 and for the overall reaction are known, the DH o for step 1 can be calculated.

Example 8.6 Using Hess s Law To Calculate DH Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) Use the following information to calculate in kilojoules for the combustion of methane: CH 4 (g) + O 2 (g) CH 2 O(g) + H 2 O(g) DH = 275.6 kj CH 2 O(g) + O 2 (g) CO 2 (g) + H 2 O(g) DH = 526.7 kj H 2 O(l) H 2 O(g) DH = 44.0 kj Solution +

Example 8.7 Using Hess s Law To Calculate DH Water gas is the name for the mixture of CO and H 2 prepared by reaction of steam with carbon at 1000 C: The hydrogen is then purified and used as a starting material for preparing ammonia. Use the following information to calculate in DH kilojoules for the water-gas reaction: Solution C(s) + O 2 (g) CO 2 (g) DH = 393.5 kj 2 CO(g) + O 2 (g) 2 CO 2 (g) DH = 566.0 kj 2 H 2 (g) + O 2 (g) 2 H 2 O(g) DH = 483.6 kj The water-gas reaction is endothermic by 131.3 kj.

Standard Heats of Formation Standard Heat of Formation (DH o f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states Standard states C(s) + 2H 2 (g) CH 4 (g) DH o f = -74.8 kj 1 mol of 1 substance Chapter 8/42

Standard Heats of Formation The standard heat of formation for any element in its standard state is defined as being ZERO. DH o f= 0 for an element in its standard state Chapter 8/43

Standard Heats of Formation DH o = DH o f (Products) - DH o f (Reactants) aa + bb cc + dd DH o = [c DH o f (C) + d DH o f (D)] - [a DH o f (A) + b DH o f (B)] Products Reactants Chapter 8/44

Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C 6 H 12 O 6 ) and O 2 from CO 2 and liquid H 2 O. 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) DH o =? Chapter 8/45

Standard Heats of Formation 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) DH o =? DH o = [DH o f (C 6 H 12 O 6 (s))] - [6 DH o f (CO 2 (g)) + 6 DH o f (H 2 O(l))] DH o = [(1 mol)(-1273.3 kj/mol)] - [(6 mol)(-393.5 kj/mol) + (6 mol)(-285.8 kj/mol)] DH o = 2802.5 kj Chapter 8/46

Example 8.8 Using Standard Heats Of Formation To Calculate DH Calculate DH in kilojoules for the synthesis of lime (CaO) from limestone (CaCO 3 ), the key step in the manufacture of cement. Solution CaCO 3 (s) CaO(s) + CO 2 (g) DH f [CaCO 3 (s)] = 1207.6 kj/mol DH f [CaO(s)] = 634.9 kj/mol DH f [CO 2 (g)] = 393.5 kj/mol DH = [DH f (CaO) + DH f (CO 2 )] [DH f (CaCO 3 )] = (1 mol)( 634.9 kj/mol) + (1 mol)( 393.5 kj/mol) (1 mol)( 1207.6 kj/mol) = 179.2 kj The reaction is endothermic by 179.2 kj.

Example 8.9 Using Standard Heats Of Formation To Calculate DH Oxyacetylene welding torches burn acetylene gas, C 2 H 2 (g). Use the information in Table 8.2 to calculate DH in kilojoules for the combustion reaction of acetylene to yield CO 2 (g) and H 2 O(g). Solution The balanced equation is 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(g) The necessary heats of formation are DH f [C 2 H 2 (g)] = 227.4 kj/mol DH f [H 2 O(g)] = 241.8 kj/mol DH f [CO 2 (g)] = 393.5 kj/mol Remember also that DH f (O 2 ) = 0 kj/mol. The standard enthalpy change for the reaction is DH = [4 DH f (CO 2 )] + 2 DH f (H 2 O)] [2 DH f (C 2 H 2 )] = (4 mol)( 393.5 kj/mol) + (2 mol)( 241.8 kj/mol) (2 mol)(227.4 kj/mol) = 2512.4 kj

Bond Dissociation Energies Bond dissociation energies are standard enthalpy changes for the corresponding bond-breaking reactions. Chapter 8/49

Bond Dissociation Energies Bond Dissociation Energy: Can be used to determine an approximate value for DH Apply Hess s law, we can calculate an approximate enthalpy change for any reaction by subtracting the sum of the bond dissociation energies in the products from the sum of the bond dissociation energies in the reactants DH o = D(Reactant bonds) D(Product bonds) Chapter 8/50

Bond Dissociation Energies H 2 (g) + Cl 2 (g) 2HCl(g) DH o = D(Reactant bonds) - D(Product bonds) DH o = (D H-H + D Cl-Cl ) - (2D H-Cl ) DH o = [(1 mol)(436 kj/mol) + (1mol)(243 kj/mol)] - DH o = -185 kj (2mol)(432 kj/mol) Chapter 8/51

Example 8.10 Using Bond Dissociation Energies To Calculate DH Use the data in Table to find an approximate DH in kilojoules for the industrial synthesis of chloroform by reaction of methane with Cl 2. CH 4 (g) + 3 Cl 2 (g) CHCl 3 (g) + 3 HCl(g)

Example 8.10 Using Bond Dissociation Energies To Calculate DH Continued Solution The reactants have four C H bonds and three Cl Cl bonds; the products have one C H bond, three C Cl bonds, and three H Cl bonds. The bond dissociation energies are: C H D = 410 kj/mol Cl Cl D = 243 kj/mol C Cl D = 330 kj/mol H Cl D = 432 kj/mol Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the enthalpy change for the reaction: DH = [3 D Cl Cl + 4 D C H ] [D C H + 3 D H Cl + 3 D C Cl ] = [(3 mol)(243 kj/mol) + (4 mol)(410 kj/mol)] [(1 mol)(410 kj/mol) + (3 mol)(432 kj/mol) + (3 mol)(330 kj/mol)] = 327 kj The reaction is exothermic by approximately 330 kj.

Fossil Fuels, Fuel Efficiency, and Heats of Combustion The objective of the exercise is to calculate the energy density of several hydrocarbon fuels. Heats of Combustion is also called combustion enthalpy, DHc. Chapter 8/54

Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) DH o = [DH o f (CO 2 (g)) + 2 DH o f (H 2 O(l))] - [DH o f (CH 4 (g))] DH o = [(1 mol)(-393.5 kj/mol) + (2 mol)(-285.8 kj/mol)] - [(1 mol)(-74.8 kj/mol)] DH o = -890.3 kj Note that the H 2 O product in giving heats of combustion is H 2 O(l) rather than H 2 O(g) Chapter 8/55

Fossil Fuels, Fuel Efficiency, and Heats of Combustion CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) It is more useful to calculate combustion enthalpy per gram, or per milliliter of substance rather than per mole. Chapter 8/56

Fossil Fuels, Fuel Efficiency, and Heats of Combustion Coal Gasification Chapter 8/57

Fossil Fuels, Fuel Efficiency, and Heats of Combustion coal gasification from a solid to a gas exothermic: endothermic: energy balance

Fossil Fuels, Fuel Efficiency, and Heats of Combustion coal gasification from a solid to a gas can be directly converted to methanol synthetic fibers methanol plastics fuel South Africa

Fossil Fuels, Fuel Efficiency, and Heats of Combustion The products of petroleum refining

An Introduction to Entropy Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence Sledding downhill is a spontaneous process that, once started, continues on its own. Dragging the sled back uphill is a nonspontaneous process that requires c continuous input of energy. Chapter 8/61

An Introduction to Entropy Entropy (S): The amount of molecular randomness in a system Entropy is a measure of molecular randomness Chapter 8/62

An Introduction to Entropy 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. 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). DS = S final S initial Positive value of DS indicates increased disorder. Negative value of DS indicates decreased disorder. Chapter 8/63

An Introduction to Entropy Spontaneous processes are favored by a decrease in H (negative DH). favored by an increase in S (positive DS). Nonspontaneous processes are favored by an increase in H (positive DH). favored by a decrease in S (negative DS). Chapter 8/64

Example 8.11 Predicting The Sign Of DS For A Reaction Predict whether DS is likely to be positive or negative for each of the following reactions: (a) H 2 C=CH 2 (g) + Br 2 (g) BrCH 2 CH 2 Br(l) (b) 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(g) Solution Reactions that increase the number of gaseous molecules generally have a positive DS, while reactions that decrease the number of gaseous molecules have a negative DS. (a). The amount of molecular randomness in the system decreases when 2 mol of gaseous reactants combine to give 1 mol of liquid product, so the reaction has a negative DS. (b). The amount of molecular randomness in the system increases when 9 mol of gaseous reactants give 10 mol of gaseous products, so the reaction has a positive DS.

An Introduction to Free Energy Gibbs Free Energy Change (DG) DG = DH - T DS Enthalpy of reaction Temperature (Kelvin) Entropy change Chapter 8/66

An Introduction to Free Energy Gibbs Free Energy Change (DG) DG = DH - T DS DG < 0 DG = 0 DG > 0 Process is spontaneous Process is at equilibrium (neither spontaneous nor nonspontaneous) Process is nonspontaneous Chapter 8/67

An Introduction to Free Energy DG o = DH o - TDS o Melting and freezing. The melting of ice disfavored by enthalpy (DH > 0) but favored by entropy (DS > 0). The freezing of water is favored by enthalpy (DH < 0) but disfavored by entropy (DS < 0). Chapter 8/68

Example 8.12 Using The Free-Energy Equation To Calculate Equilibrium Temperature Lime (CaO) is produced by heating limestone (CaCO 3 ) to drive off CO 2 gas, a reaction used to make Portland cement. Is the reaction spontaneous under standard conditions at 25 C? Calculate the temperature at which the reaction becomes spontaneous. CaCO 3 (s) CaO(s) + CO 2 (g) DH = 179.2 kj; DS = 160.0 J/K Solution At 25 C (298 K), we have Because DG is positive at this temperature, the reaction is nonspontaneous. The changeover point between spontaneous and nonspontaneous is approximately The reaction becomes spontaneous above approximately 1120 K (847 C).

Example 8.13 Predicting The Signs Of DH, DS, And DG For a Reaction What are the signs of DH, DS, And DG for the following nonspontaneous transformation? DG=DH-TDS Solution 1. The drawing shows ordered particles in a solid subliming to give a gas. 2. Formation of a gas from a solid increases molecular randomness, DS so is positive. 3. Because we re told that the process is nonspontaneous, DG is also positive. Because the process is favored by DS (positive) yet still nonspontaneous, DH must be unfavorable (positive). 4. This makes sense, because conversion of a solid to a liquid or gas requires energy and is always endothermic.