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Thermodynamics • is the study of energy and its interconversions. Thermochemistry An area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction. תרמוכימיה-6 1 Energy • Energy is the ability to do work or transfer heat. – Energy used to cause an object that has mass to move is called work. – Energy used to cause the temperature of an object to rise is called heat. 2 תרמוכימיה-6 Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 EK mv 2 2 In the equation: m = mass v = velocity 3 תרמוכימיה-6 Potential Energy, EP The energy an object has by virtue of its position in a field of force. Gravitational potential energy is given by the equation E P mgh m = mass (kg) g = gravitational constant (9.80 m/s2) h = height (m) תרמוכימיה-6 4 Potential Energy • Potential energy is energy an object possesses by virtue of its position or chemical composition. • The most important form of potential energy in molecules is electrostatic potential energy, Eel: 5 תרמוכימיה-6 Units of Energy • The SI unit of energy is the joule (J): Calorie A commonly-used non-SI unit of energy. It is the amount of energy required to raise the temperature of one gram of water by one degree Celsius. 1 cal = 4.184 J Dietary Calorie (Cal) “big C” calorie Used on food products. 1 Cal = 1000 cal = 1 kcal = 4.184 kJ 6 תרמוכימיה-6 Definitions: System and Surroundings • The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). • The surroundings are everything else (here, the cylinder and piston). 7 תרמוכימיה-6 Thermodynamic System The substance of mixture or substances that are the focus of a study. Thermodynamic Surroundings Everything in the vicinity of the thermodynamic system. תרמוכימיה-6 8 Work, w An energy transfer into or out of a thermodynamic system whose effect on the surroundings is equivalent to moving an object through a field of force. Work is chosen to be positive when work is done on the thermodynamic system and negative when work is done by the thermodynamic system. • w=Fd where w is work, F is the force, and d is the distance over which the force is exerted. תרמוכימיה-6 9 Heat, q A transfer of energy into or out of a thermodynamic system caused by a temperature difference between the system and its surroundings. Heat flows spontaneously from a region of higher temperature to a region of lower temperature. • • q is defined as positive if heat is absorbed by the system (heat is added to the system) q is defined as negative if heat is evolved by a system (heat is subtracted from the system) תרמוכימיה-6 10 Conversion of Energy • Energy can be converted from one type to another. • The cyclist has potential energy as she sits on top of the hill. • As she coasts down the hill, her potential energy is converted to kinetic energy until the bottom, where the energy is converted to kinetic energy. 11 תרמוכימיה-6 Internal Energy, U ( or E) The sum of the kinetic and potential energies of the particles making up a substance. Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant. Change in Internal Energy It is determined by the formula ΔU = Uf – Ui or E = Efinal − Einitial Uf = final internal energy Ui = initial internal energy תרמוכימיה-6 12 Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial תרמוכימיה-6 13 State function A property of a system that depends only on its present state. It is independent of any previous history of the system. תרמוכימיה-6 14 State Functions • Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. – In the system below, the water could have reached room temperature from either direction. 15 תרמוכימיה-6 State Functions • Therefore, internal energy is a state function. • It depends only on the present state of the system, not on the path by which the system arrived at that state. • And so, E depends only on Einitial and Efinal. 16 תרמוכימיה-6 First Law of Thermodynamics The change in the internal energy of a system, ΔU, is the sum of heat, q, and work, w. U = q + w or E = q + w • Energy is neither created nor destroyed. • In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. תרמוכימיה-6 17 Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w. 18 תרמוכימיה-6 Energy Exchange • Energy is exchanged between the system and surroundings through heat and work. – q = heat (thermal) energy – w = work energy – q and w are NOT state functions; their value depends on the process. E = q + w 19 תרמוכימיה-6 Energy Exchange • Energy is exchanged between the system and surroundings through either heat exchange or work being done. 20 תרמוכימיה-6 Energy Exchange • q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. – But q and w are different in the two cases. 21 תרמוכימיה-6 Heat and Work • The white ball has an initial amount of 5.0 J of kinetic energy. • As it rolls on the table, some of the energy is converted to heat by friction. • The rest of the kinetic energy is transferred to the purple ball by collision. 22 תרמוכימיה-6 Expansion Work C3H8(g) + 5O2(g) 6 mol of gas 23 3CO2(g) + 4H2O(g) 7 mol of gas תרמוכימיה-6 Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). Expansion Work Expansion Work: Work done as the result of a volume change in the system 24 תרמוכימיה-6 Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PV 25 תרמוכימיה-6 Enthalpy • If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. • Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV 26 תרמוכימיה-6 Enthalpy • When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) • This can be written H = E + PV 27 תרמוכימיה-6 Enthalpy • Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV H = (q + w) − w H = q • So, at constant pressure, the change in enthalpy is the heat gained or lost. 28 תרמוכימיה-6 Endothermic and Exothermic • A process is endothermic when H is positive. • A process is exothermic when H is negative. 29 תרמוכימיה-6 In an endothermic reaction: The reaction vessel cools. Heat is absorbed. Energy is added to the system. q is positive. In an exothermic reaction: The reaction vessel warms. Heat is evolved. Energy is subtracted from the system. q is negative. תרמוכימיה-6 30 Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants 31 תרמוכימיה-6 Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. 32 תרמוכימיה-6 Energy Diagram Showing ΔH for a Reaction 33 תרמוכימיה-6 The Truth about Enthalpy 1. Enthalpy is an extensive property. 2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. 3. H for a reaction depends on the state of the products and the state of the reactants. 34 תרמוכימיה-6 State functions H E P V T etc. Thermochemical Equation The chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. For the reaction of sodium metal with water, the thermochemical equation is: 2Na(s ) + 2H2 O(l ) 2NaOH(aq ) + H2 (g ); ΔH = –368.6 kJ תרמוכימיה-6 35 Manipulating a Thermochemical Equation When the thermochemical equation is multiplied by any factor, the value of ΔH for the new equation is obtained by multiplying the value of ΔH in the original equation by that same factor. When a chemical equation is reversed, the sign of H is reversed. תרמוכימיה-6 36 Thermochemical Equation Natural gas consists primarily of methane, CH4. It is used in a process called steam reforming to prepare a gaseous mixture of carbon monoxide and hydrogen for industrial use. CH4 (g ) + H2 O(g ) CO(g) + 3H2 (g ); ΔH = 206 kJ The reverse reaction, the reaction of carbon monoxide and hydrogen, has been explored as a way to prepare methane (synthetic natural gas). Which of the following are exothermic? Of these, which one is the most exothermic? CO(g ) + 3H2 (g ) a. CH4 (g ) + H2 O(g ) 2CO(g ) + 6H2 (g ) b. 2CH4 (g ) + 2H2 O(g ) CH4 (g ) + H2 O(g) c. CO(g ) + 3H2 (g ) d. 2CO(g ) + 6H2 (g ) 2CH4 (g ) + 2H2 O(g ) תרמוכימיה-6 37 Thermochemical Equation CH4 (g ) + H2 O(g ) CO(g ) + 3H2 (g ); H = 206 kJ a. CH4 (g ) + H2 O(g ) CO(g ) + 3H2 (g ) This reaction is identical to the given reaction. It is endothermic. H = 206 kJ תרמוכימיה-6 38 Thermochemical Equation CH4 (g ) + H2 O(g ) CO(g ) + 3H2 (g ); H = 206 kJ 2CO(g ) + 6H2 (g ) b. 2CH4 (g ) + 2H2 O(g ) This reaction is double the given reaction. It is endothermic. H = 412 kJ תרמוכימיה-6 39 Thermochemical Equation CH4 (g ) + H2 O(g ) CO(g ) + 3H2 (g ); H = 206 kJ CH4 (g ) + H2 O(g ) c. CO(g ) + 3H2 (g ) This reaction is the reverse of the given reaction. It is exothermic. H = -206 kJ תרמוכימיה-6 40 Thermochemical Equation CH4 (g ) + H2 O(g ) CO(g ) + 3H2 (g ); H = 206 kJ 2CH4 (g ) + 2H2 O(g ) d. 2CO(g ) + 6H2 (g ) This reaction is reverse and double the given reaction. It is exothermic. H = -412 kJ Equations c and d are exothermic. Equation d is the most exothermic reaction. תרמוכימיה-6 41 Applying Stoichiometry to Heats of Reaction תרמוכימיה-6 42 Applying Stoichiometry to Heats of Reaction Determine the amount evolved when 9.07 ×105 g of ammonia is produced according to the following equation. Assume that the reaction occurs at constant pressure. N2 (g ) + 3H2 (g ) 2NH3 (g ); H = –91.8 kJ The conversion factor is obtained from the second step of the thermodynamic equation. Hence, 1 mol NH3 –91.8 kJ 6 6 2.45 9.07 10 g NH3 = 2.45 1010 kJkJ 17.0 g NH3 2 mol NH3 5 The amount of heat evolved is 2.45 ×106 kJ תרמוכימיה-6 43 Heat Capacity, C The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one kelvin). The heat required to raise the temperature of the sample is determined by the formula: q = C t The heat capacity for one mole of substance is considered its molar heat capacity. תרמוכימיה-6 44 Specific Heat Capacity, s (specific heat) The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one kelvin) at constant pressure. The heat required to raise the temperature of the sample can be determined by the formula: q s m t or q = m c ΔT s (or c) = specific heat of the substance m = mass in grams Δt = temperature change תרמוכימיה-6 45 Specific Heat Capacity (C) of Some Common Materials תרמוכימיה-6 46 Heat Capacity How much energy will be used to heat 500.0 g of iron from 22°C to 55°C? cFe = 0.451 J g-1 °C-1. Heat required = q = m c ΔT q = 500.0 g (0.451 J g-1 °C-1)(55 − 22)°C q = 7442 J = +7.4 kJ + sign, E added to the system (the iron) תרמוכימיה-6 47 Thermal Equilibrium 48 Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower temperature. Transfer of energy as heat continues until both objects are at the same temperature and thermal equilibrium is achieved. At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a temperature decrease has lost thermal energy. תרמוכימיה-6 Heat Capacity A 215 g block of Cu at 505.0°C is plunged into 1.000 kg of water (T = 23.4 °C) in an insulated container. What will be the final equilibrium T of the water and the Cu? (cCu = 0.385 J g-1 °C-1, cH2O = 4.184 J g-1 °C-1) q = m c ΔT qCu = (215. g)(0.385 J g-1 °C-1)(Tfinal− 505.0) qH2O = (1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4) (conservation of E) qCu + qH2O = 0 qH2O = -qCu תרמוכימיה-6 49 Heat Capacity 215 g Cu (505.0°C) + 1000. g H2O (23.4 °C). Final T ? qH2O = -qCu 4184(Tfinal – 23.4) = -82.78(Tfinal – 505.0) (4184 + 82.78)Tfinal = 41804 + 97906 Tfinal = 32.7°C (Note: Tfinal must be between Thot and Tcold) תרמוכימיה-6 50 Energy Transfer As Heat And The Temperature Change At 500. G Of Water Warms From −50 °C And 200 °C (At 1 Atm) Temp is constant during a change of state תרמוכימיה-6 51 Contrast Between a Change of State and an Increase in Temperature as a Result of Adding Energy תרמוכימיה-6 52 Melting (Fusion) & Freezing ΔfusionH (or ΔfusH) is the heat required (at constant P) to melt a solid. Change solid → liq Name enthalpy of fusion liq → solid enthalpy of freezing Note: value for H2O (J/g) 333 −333 ΔfusH = − ΔfreezingH qfusion = −qfreezing תרמוכימיה-6 53 Vaporization & Condensation ΔvapH = qP = heat to vaporize a liquid (at constant P) Change Name liq → gas enthalpy of vaporization 2260 gas → liq enthalpy of condensation −2260 Note: value for H2O (J/g) ΔvapH = − ΔcondensationH qvap = −qcondensation תרמוכימיה-6 54 Enthalpies of Physical and Chemical Change Enthalpy of Sublimation (ΔHsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase (reverse process: deposition) 55 תרמוכימיה-6 תרגיל :נתון 400.0גרם תה חם ב ,80.0C-מהי מסת הקרח ב 0C -שיש להוסיפה לתה החם על מנת לקבל תה קר ב.10.0C - קיבול החום הסגולי של תה שווה ל ,4.18 J/(g C) -ואינתלפיית ההתכה של קרח שווה ל. +6.01kJ/mol- .1התכת הקרח בm/18.02x6010=333.5m : 0C- .2חימום הקרח לאחר התכה מ 0C-לmxcxt =mx4.18x10= 41.8m : 10.0C- .3קירור התה החם מ 80C-לmxcxt =400.0x4.18x70= 117040 J : 10.0C- חום נפלט מקירור תה חם= חום התכת הקרח +חימום הקרח לאחר התכתו 41.8m +333.5m = 117040 375.3m= 117040 m=311 g -6תרמוכימיה 56 A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. Two examples are shown below. תרמוכימיה-6 57 Coffee-Cup Calorimeter • Polystyrene is a good insulator • Heat evolved from reaction is absorbed by the water • Heat capacity of the coffee-cup calorimeter is that of water • For a reaction performed in a coffee-cup calorimeter J • q = mass 4.18 t reaction H 2O g C Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 58 Bomb Calorimeter • Versatile • Used for reactions involving • High temperature • Gases • Bomb - Heavy-walled metal vessel that is usually surrounded by water • Apparatus is closed • Initial temperature is measured • Final temperature is taken to be the highest value read Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 59 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 equals the sum of the enthalpy changes for the individual steps. תרמוכימיה-6 60 Hess’s Law Suppose we want H for the reaction 2C(graphite) + O2 (g ) 2CO(g ) It is difficult to measure directly. Assume that the reaction takes place in the following steps: 2C(graphite) + 2O2 (g ) 2CO2 (g ) 2CO2 (g ) 2CO(g ) + O2 (g ) תרמוכימיה-6 61 Hess’s Law Putting the two equations together gives: 2C(graphite) + 2 O2 (g ) 2CO2 (g ) 2CO2 (g ) 2CO(g ) + O2 (g ) 2C(graphite) + O2 (g ) 2CO(g ) Determining the enthalpy changes for the separate steps, 2C(graphite) + 2O2 (g ) 2CO2 (g ); H H 1==(393.5 (–393.5 kJ)2 kJ) (2) 2CO2 (g) 2CO(g) + O2 (g ); H2 = (–566.0 kJ) (–1) 62 תרמוכימיה-6 Hess’s Law Adding the two thermochemical equations and their enthalpy changes gives the enthalpy change for the combustion of C(graphite) to CO. 2C(graphite) + 2O2 (g ) 2CO2 (g ) 6 | 63 H1 = (–393.5 kJ) (2) H 2CO2 (g) 2CO(g) + O2 (g ) H2 = (–566.0 kJ) (–1) 2C(graphite) + O2 (g ) 2CO(g ) H3 = –221.0 kJ ©2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Hess’s Law 64 -6תרמוכימיה Hess’s Law Determine the enthalpy of reaction, H, for the formation of tungsten carbide, WC, from the elements. W(s ) + C(graphite) WC(s ) The heats of combustion of the elements and of tungsten carbide are: 2W (s ) + 3O2 (g) 2WO3 (s); ΔH = –1685.8 kJ (1) C(graphite) + O2 (g ) CO2 (g ); ΔH = –393.5 kJ (2) 2WC(s ) + 5O2 (g ) 2WO3 (s ) + 2CO 2 (g ); ΔH = –2391.8kJ (3) תרמוכימיה-6 65 Hess’s Law Reversing Equation 3 to get WC(s) on the right side, 2WO3 (s ) + 2CO2 (g ) 2WC(s ) + 5O2 (g ); H = 2391.8 kJ (3 ' ) Comparing Equation 1 with the desired equation and multiplying Equation 1 by ½ gives: 3 1 W (s ) + O2 (g ) WO3 (s ); ΔH = (–1685.8 kJ) = –842.9 kJ 2 2 Equation 2 is left as is; it has C(graphite) on the left side. תרמוכימיה-6 66 Hess’s Law Comparing Equation 3′ with the desired equation and multiplying Equation 3′ by ½ gives: WO3 (s ) + CO2 (g ) WC(s ) + 5 1 O2 (g ); H = (2391.8 kJ) = 1195.9 kJ 2 2 Adding the last three equations with their corresponding ΔH’s gives: 3 W (s ) + O2 (g ) WO3 (s ); 2 C(graphite) + O2 (g ) CO2 (g ); WO3 (s ) + CO2 (g ) WC(s ) + W(s ) + C(graphite) WC(s ) H = –842.9 kJ H = –393.5 kJ 5 O2 (g ); H = 1195.9 kJ 2 H –40.5 kJ תרמוכימיה-6 67 Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C). These standard conditions are indicated with a degree sign (°). When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction, H°. (H° is read “delta H zero.”) תרמוכימיה-6 68 Standard Enthalpies of Formation Elements can exist in more than one physical state, and some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O2 or as O3 (ozone). These different forms of an element in the same physical state are called allotropes. The reference form is the most stable form of the element (both physical state and allotrope). תרמוכימיה-6 69 The standard enthalpy of formation, Hf°, is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states. Hf° for an element in its reference and standard state is zero. For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction 1 H2 (g ) O2 (g ) H2 O (l ); Hf –285.8kJ 2 תרמוכימיה-6 70 The standard enthalpy of formation Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction, H°. CH4 (g ) + 4Cl2 (g ) CCl4 (l ) + 4HCl (g ); H = ? Listing down the thermochemical equations with enthalpies of formation: C(graphite) + 2H2 (g ) CH4 (g ); Hf ° = –74.9 kJ (1) C(graphite) + 2Cl2 (g ) CCl4(l ); Hf ° = –135.4 kJ (2) 1 1 H2 (g ) + Cl2 (g ) HCl(g ); Hf ° = –92.3 kJ (3) 2 2 תרמוכימיה-6 71 Applying Hess’s law, reversing Equation 1 and adding Equation 2 and 4 ×Equation 3, CH4 (g ) C(graphite) + 2H2 (g ) C(graphite) + 2Cl2 (g ) CCl4 (l ) 2H2 (g ) + 2Cl2 (g ) 4HCl(g ) CH4 (g ) + 4Cl2 (g ) CCl4 (l ) + 4HCl (g ) (–74.9 kJ) (–1) (–135.4 kJ) (1) (–92.3 kJ) (4) H = –429.7kJ If the formula is written in parentheses after Hf°, the calculation can be written as: H = [Hf (CCl4 ) + 4Hf (HCl)] – [Hf (CH4 ) + 4 Hf (Cl2 )] = [(–135.4) + 4(–92.3)] kJ – [(–74.9) + 4(0)] kJ = –429.7 kJ In general, H = nH f (products) – mH f (reactants) תרמוכימיה-6 72 The standard enthalpy of formation Determine the standard enthalpy change for the following reaction: Pt 4NH3 (g ) + 5O2 (g ) 4NO(g ) + 6H2 O(g ) קטליזטור או זרז Adding ΔH fo values to the components of the reaction: Pt 4NH3 (g ) + 5O2 (g ) 4NO(g ) + 6H2O(g ) 4 (–45.9 ) 5(0) 4 ( 90.3 ) 6 (–241.8 ) kJ תרמוכימיה-6 73 The standard enthalpy of formation H = nH f (products) – mH f (reactants) = [4H f (NO) + 6H f (H2 O)] – [4H f (NH3 ) + 5 H f (O2 )] = [4(90.3) + 6(–241.8)] kJ – [4(–45.9) + 5 (0)] kJ = –906 kJ תרמוכימיה-6 74 Bond Dissociation Energies Bond dissociation energies are standard enthalpy changes for the corresponding bond-breaking reactions. 75 תרמוכימיה-6 Bond Dissociation Energies H2(g) + Cl2(g) 2HCl(g) ΔH° = D(Reactant bonds) – D(Product bonds) ΔH° = (DH—H + DCl—Cl) – (2DH—Cl) ΔH° = [(1 mol)(436 kJ/mol) + (1 mol)(243 kJ/mol)] – (2 mol)(432 kJ/mol) ΔH° = –185 kJ 76 תרמוכימיה-6 ∆H and ∆E • Constant pressure • Coffee-cup calorimeter • Heat flow is equal to the enthalpy • H = q p • Constant volume • Bomb calorimeter • No pressure-volume work done • E = q v Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 77 ∆H and ∆E • H = E + PV • H E + PV • PV product is important where gases are involved • Negligible when liquids or solids are involved • H = E + n g RT • n g is the change in the number of moles of gas as the reaction proceeds Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 78 Example • Calculate H and E at 25 C for the reaction: 5 C2 H 2 g O 2 g 2CO2 g H 2O g 2 • Analysis: • Information given o • Chemical equation, T 25 C • Information implied • H f • Moles of products and reactants • R with energy units • Asked for • ∆H and ∆E Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 79 Example • Strategy: • Calculate ∆H for the reaction using the table • Remember H f for O2 g is zero • Find n = n products n reactants • Recall that R is 8.31 J/mol K when energy units are involved • Substitute into the equation to find ∆E • H = E + n g RT Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 80 Example • Solution: • ∆H H 2ΔH f o CO 2 g H f o H 2O g H f o C 2 H 2 g = 2 393.5 kJ 241.8 kJ 226.7 kJ 1255.5 kJ n g 5 1 2 mol CO2 1 mol H2O 1 mol C2 H2 mol O2 mol Copyright ©2016 Cengage Learning. All Rights Reserved. 2 2 תרמוכימיה-6 81 Example • ∆E H = E + n g RT kJ 3 1255.5 kJ = E + 0.5 mol 8.3110 298 K mol K E = 1255.5 1.24 kJ = 1254.3 kJ Copyright ©2016 Cengage Learning. All Rights Reserved. תרמוכימיה-6 82