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Chapter 5 — Thermo 1 Energy & Chemistry Questions that need to be addressed: Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions Jeffrey Mack California State University, Sacramento Units of Energy Energy & Chemistry 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C. 1000 cal = 1 kilocalorie = 1 kcal Some Basic Principles • Energy as the capacity to do work or transfer heat (q). • Heat (q) is NOT temperature, temperature is a measure of kinetic energy. • Energy is divided into two basic categories: 1 kcal = 1 Calorie (a food “calorie”) – Kinetic energy (the energy associated with motion) – Potential energy (energy that results from an object’s position). SI units for energy: joule (J) 1 cal = exactly 4.184 J James Joule 1818-1889 Energy & Chemistry Energy can be divided into two forms: Kinetic: Energy of motion: Thermal Mechanical Electrical Potential: Stored energy: Gravitational Electrostatic Chemical & Nuclear • How do we measure and calculate the energy changes that are associated with physical changes and chemical reactions? • What is the relationship between energy changes, heat, and work? • How can we determine whether a chemical reaction is product-favored or reactant-favored at equilibrium? • How can we determine whether a chemical reaction or physical change will occur spontaneously, that is, without outside intervention? • The law of conservation of energy requires that energy can neither be created nor destroyed. • However, energy can be converted from one type into another. Potential & Kinetic Energy Kinetic energy — energy of motion • Translation Chapter 5 — Thermo Potential & Kinetic Energy 2 Energy & Chemistry Potential energy — energy a motionless body has by virtue of its composition and position. • Burning peanuts supply sufficient energy to boil a cup of water. Thermal Equilibrium • Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) System & Surroundings • SYSTEM – The object under study • SURROUNDINGS – Everything outside the system • 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. Directionality of Energy Transfer • When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMIC. • In the case of thermal energy, the temperature of the system decreases. (qsystem < 0) • Tsystem = (Tfinal – Tinitial) < 0 • Energy flows between the two Directionality of Energy Transfer • When energy enters the system and from the surroundings, the process is said to be ENDOTHERMIC. • In the case of thermal energy, the temperature of the system increases. (qsystem > 0) • Tsystem > 0 Chapter 5 — Thermo 3 Heat & Changes in Internal Energy Heat & Changes in Internal Energy • Heat flows between the system and surroundings. • U is defined as the internal energy of the system. • q = the heat absorbed or lost ÄUsystem = Ufinal - Uinitial = ± q If heat enters the system: U > 0 therefore q is positive (+) If heat leaves the system: ÄUsystem = Ufinal - Uinitial = ± q U < 0 therefore q is negative (–) The sign of q is a “convention”, it designates the direction of heat flow between the system and surroundings. Heat Capacity Heat & Changes in Internal Energy ÄUsystem = Ufinal - Uinitial = ± q When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. Usystem heat in heat out System qsystem > 0 (+) Usystem > 0 increases Usystem Surroundings qsystem < 0 (–) Usystem < 0 decreases q = the heat absorbed or lost by the system. Heat & Specific Heat Capacity The amount of heat (q) transfer related to an object and temperature is given by: q = C ´ DT J C or K q = heat lost or gained (J) C = Heat Capacity of an object T = Tfinal Tinitial is the temperature change (°C or K) Does it matter if we calculate a temperature change in Kelvin or degrees C? When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by: q = m ´ C ´ DT q = heat lost or gained (J) m = mass of substance (g) J C = the Specific Heat Capacity of a compound g C or K T = Tfinal Tinitial is the temperature change (°C or K) let Tin = 25.0 °C and Tf = 50.0 °C Recall that T = Tf –Tin Tf = 50.0 °C –Tin = 25.0 °C T = 25.0 °C = 50.0 °C + 273.2 = (25.0 °C + 273.2) T = The are the same! = 323.2 K = 298.2 K 25.0 K Chapter 5 — Thermo Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial J temperature of the block was 17.0 C 2.72 The specific heat capacity of the metal is: g C 4 Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial J temperature of the block was 17.0 C 2.72 The specific heat capacity of the metal is: g C q = m ´ C ´ DT = m ´ C ´ (Tfinal - Tinitial ) Strategy Map: Find T of the metal after final it absorbs the energy Data Information: Mass, initial temp, heat capacity of metal, heat (q) absorbed. Solve q = mCT for Tfinal, plug in data. Final temperature of the metal is determined. rearranging: Tfinal = q +Tinitial m´C 4.184 J 1 cal + 17.0 °C =32.7 J °C 25.0 g 2.72 Tf > Tin g C as expected +255 cal Tf = Example Problem: Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution: Heat is transferred from the hot metal to the colder water. Energy is conserved so: qFe + qwater = 0 qFe = -qwater Example Problem: Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution: Solution: qFe + qwater = 0 qFe = -qwater mFe ´ CFe ´ DTFe = -mwater ´ Cwater ´ DTwater CFe = -mwater ´ Cwater ´ DTwater mFe ´ DTFe Chapter 5 — Thermo 5 Energy & Changes of State Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution: J 1K (23.1 C 21.0 C) J gK 1 C 0.469 1K gK 55.0 g (23.1 C 99.8 C) 1 C 225 g 4.184 CFe Energy Transfer & Changes of State • Some changes of state (phase changes) are endothermic: • When you perspire, water on your skin evaporates. • This requires energy. • Heat from your body is absorbed by the water as it goes from the liquid state to the vapor state, as a result you cool down. + energy • When matter absorbs heat, its temperature will rise until it undergoes a Phase Change. • The matter will continue to absorb energy, however during the phase change its temperature remains constant: Phase changes are “Isothermal” processes. Energy Transfer & Changes of State • Some changes of state (phase changes) are exothermic: • When it is muggy outside, water condenses on your skin. • This releases energy. • Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state. • As a result you feel hot. H2O(l) + Heat ® H2O(g) Heating/Cooling Curve for Water Note the isotherms at each phase transition. + energy H2O(g) ® H2O(l) + Heat Heat & Changes of State • The energy associated with a change of state is given by the Enthalpy or Heat of the phase change. • Since there is no temperature change associated with the process, the units are most often in J/g or J/mol. Sublimation: Vaporization: Melting or Fusion: Deposition: Condensation: subH > 0 vapH > 0 fusH > 0 depH < 0 conH < 0 (endothermic) (endothermic) (endothermic) (exothermic) (exothermic) Freezing: freH < 0 (exothermic) Where H refers to the “Heat” of a phase change Chapter 5 — Thermo 6 Energy Change Calculations Heating & Cooling: q(heating = m C T or cooling) Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? +333 J/g Heat absorbed or lost in a Phase change: q(phase change) = (phase change)H n +2260 J/g melt the ice form liquid water at 0 °C heat the water to 100 °C boil water (n = moles or grams) fusH Cwater VapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Constants Needed: Equations: J Heat of fusion of ice = 333 g qphase change = mH Ice H2O(s) fusH (0 °C) Specific heat of water = 4.18 Heat of vaporization = 2260 J g×K melt ice water H2O(l) (100 °C) vapH steam H2O(g) (100 °C) qheat = mCT J g Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) fusH (0 °C) water H2O(l) Cwater (0 °C) water H2O(l) Cwater (0 °C) water H2O(l) (100 °C) heat water vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? steam H2O(g) (100 °C) boil water qtotal = q1 = nice ´ D fusH + q2 = mwater ´ Cwater ´ DT + q3 = nwater ´ D vapH Ice H2O(s) fusH (0 °C) water H2O(l) Cwater (0 °C) qtotal 500. g 333 500. g 4.18 water H2O(l) (100 °C) vapH J g J (100.0 C 0.0 C) g C 500. g 2260 J 1.51 10 6 J g steam H2O(g) (100 °C) Chapter 5 — Thermo 7 First Law of Thermodynamics First Law of Thermodynamics The total internal energy of an isolated system is constant. Any energy lost by the system is transferred to the surroundings and vice versa. Any change in energy is related to the final and initial states of the system. The “system” is that which we are interested in. The “surroundings” are everything in contact with the system. Together: System + Surroundings = Universe Energy Energy ∆Usystem = Ufinal – Uinitial The same holds for the surrounding! Relating U to Heat and Work energy transfer in (endothermic), +q Energy cannot be created or destroyed. energy transfer out (exothermic), -q Energy of the universes (system + surroundings) is constant. Any energy transferred from a system must be transferred to the surroundings (and vice versa). From the first law of thermodynamics: When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system: SYSTEM ∆U = q + w U q w change in heat work done by or system = lost or gained + on the system energy by the system Energy in U Final Uf > Ui Work Uf < Ui w transfer out (-w) Energy is the capacity to do work. Work equals a force applied through a distance. When you do work, you expend energy. Energy out U initial work out energy work in work and heat can balance! w transfer in (+w) When you push down on a bike pump, you do work. force force P= = area x2 change in volume V = x3 q in Heat P V = q out U Final U initial Usystem > 0 (+) Usystem = 0 Usystem < 0 (–) F x3 = F x x2 P × V = work Therefore work is equal to a change of volume at constant pressure. Chapter 5 — Thermo 8 Enthalpy Enthalpy is a “State Function” Enthalpy, “H” is the heat transferred between the system and surroundings under conditions of constant pressure. H = qp the subscript “p” indicates constant pressure H = U + PV DH = D (U + PV ) if no “PV” work is done by the system, V = 0 = DU + PDV 0 H = Up the change in enthalpy is the change in internal energy at constant pressure No matter which path is taken (AB) the results are the same: Final – Initial Since individual Enthalpies cannot be directly measured, we only deal with enthalpy changes (H = Hf – Hi) For a “system” the overall change in Enthalpy is path independent. • Since Enthalpies are state functions, one must specify the conditions at which they are measured. Reactants Enthalpy of reaction = rH = Hproducts Hreactants • Standard State Conditions are defined as: Energy • “H°” indicates that the Enthalpy is taken at Standard State conditions. • 1 atm = 760 mm Hg or 760 torr & 298.15 K or 25 °C Thermochemical Equations Just like a regular chemical equation, with an energy term. rHo = 802 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ H (Reactants) rH > 0 rH < 0 H (Reactants) H (Products) Exothermic Problem: How many kJ of energy are released when 128.5 g of methane, CH4(g) is combusted? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ g energy out… Exothermic What does this imply? CONVERSION FACTORS!!! +802 kJ of Energy Released 2 mol H2O (g) produced H (Products) Endothermic From the equation: +802 kJ of Energy Released 1 mol CH4 (g) consumed Products H = Hfinal Hinitial • H(T,P): Enthalpy is a function of temperature and pressure. Energy is a product just like CO2 or H2O! A Enthalpies & Chemical Reactions: rH Enthalpy Conditions CH4(g) + 2O2(g) CO2(g) + 2H2O(g) B 128.5g CH4´ mols J molar Reaction mass enthalpy 1 mol CH4 1 mol rxn +802 kJ ´ ´ = 6.43103 kJ 16.04 g 1 mol CH4 1mol rxn Chapter 5 — Thermo 9 Thermochemical Equations When the reaction is reversed , the sign of H reverses: Thermochemical Equations H scales with the reaction: 1 1 ´ [ CH4(g) + 2O2(g) CO2(g) + 2H2O(g)] [rHo = 802 kJ]´ 2 2 CH4(g) + 2O2(g) CO2(g) + 2H2O(g) CO2(g) + 2H2O(g) CH4(g) + 2O2(g) rHo = 802 kJ Exothermic rHo = +802 kJ Endothermic Thermochemical Equations 1 1 CH4 (g) + O2 (g) ® CO2 (g) + H2O(g) 2 2 Yes you can write the reaction with fractions, so long as you are writing it on a mole basis… Comparing the reaction with water as a gas or liquid: CH4(g) + 2O2(g) Change in enthalpy depends on state: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) rH° = 890 kJ = 88kJ H2O(g) H2O(l) This means that water’s liquid state lies 44 kJ/mol lower than the gas state The difference in the rxn H 88kJ [ H = 44 kJ/mol ] 2 = is due to the change in state!! Constant Pressure Calorimetry, Measuring H • A constant pressure calorimeter can be used to measure the amount of energy transferred as heat under constant pressure conditions, that is, the enthalpy change for a chemical reaction. Enthalpy (H) reactants CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rH° = 802 kJ From your text: rHo = 401 kJ rHo = 802 kJ either pathway gives the same results! rHo = 890 kJ products CO2(g) + 2H2O(g) The difference is the 88 kJ released when 2 mols of water go from gas to liquid. 2 44 kJ products CO2(g) + 2H2O(l) Constant Pressure Calorimetry, Measuring H • The constant pressure calorimeter used in general chemistry laboratories is often a “coffee-cup calorimeter.” This inexpensive device consists of two nested Styrofoam coffee cups with a loose-fitting lid and a temperature-measuring device such as a thermometer or thermocouple. Chapter 5 — Thermo 10 Constant Pressure Calorimetry, Measuring H • Because the “coffee-cup calorimeter.” is an isolated system, “What happens in the coffee cup, stays in the coffee cup!” • No mass loss to the surroundings. • No heat loss to the surroundings. Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. • This resulted in a decrease in temperature from 18.6 °C to 16.2 °C. • Calculate the enthalpy change for dissolving NH4NO3(s) in water in kJ/mol. • Assume the solution has a specific heat capacity of 4.18 J/g ? K. Problem: Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. • This resulted in a decrease in temperature from 18.6 °C to 16.2 °C. • Calculate the enthalpy change for dissolving NH4NO3(s) in water in kJ/mol. • Assume the solution has a specific heat capacity of 4.18 J/g ? K. • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. • This resulted in a decrease in temperature from 18.6 °C to 16.2 °C. • Calculate the enthalpy change for dissolving NH4NO3(s) in water in kJ/mol. • Assume the solution has a specific heat capacity of 4.18 J/g ? K. Data Information: Mass of reactant, C, water & temperature change. Step 1: Cals. Moles of NH4NO3 qsolution 5.44 g NH4NO3 Step 3: Eq. gives mole ratios (stoichiometry) qsolution + qrxn = 0 qrxn = q(NH4NO3) Step 2: Determine qsolution = mCT Step 4: rH =qrxn/mol NH4NO3 1 mol NH4NO3 0.0680 moles NH4NO3 80.04 g qsolution msolution C T qsolution 154.4 g 4.18 J (16.2 C 18.6 C) g C qsolution 1.55 103 J Enthalpy per mole of reactant Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup calorimeter. • This resulted in a decrease in temperature from 18.6 °C to 16.2 °C. • Calculate the enthalpy change for dissolving NH4NO3(s) in water in kJ/mol. • Assume the solution has a specific heat capacity of 4.18 J/g ? K. qsolution = -4.55 ´ 103 J qrxn + qsolution = 0 qrxn = -qsolution = +1.55 ´ 103 J DrH = qrxn +1.55 ´ 103 J 1kJ kJ = ´ = +22.8 moles of reaction 0.0680 moles NH4NO3 103 J mol The sign is positive indicating an endothermic process. Constant Volume Calorimetry, Measuring U Under conditions of constant volume, any heat transferred is equal to a change of internal energy rU. qV = rU Chapter 5 — Thermo 11 Constant Volume Calorimetry, Measuring U Heats of combustion (rUocombustion ) are measured using a device called a Bomb Calorimeter. A combustible sample is reacted with excess O2 Calculating Heat in an Exothermic Reaction Octane, the primary component of gasoline combusts by the reaction: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9H2O(l) A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb. The temperature of the water rises to 33.20 °C from 25.00 °C when the octane is reacted. The heat capacity of the bomb is constant. If the heat capacity of the bomb is 837 J/°C, calculate the heat of reaction per mole of octane. The heat of reaction is found by: qrxn = -Ccal ´ DT Since the temperature of the water rose, the reaction must have been exothermic: Therefore one can write: Ccal = Cwater + Cbomb –qrxn = qwater + qbomb Calculating Heat in an Exothermic Reaction –qrxn = mwaterCwaterTwater + qbombTwater -qRXN = 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. 4.184J ´1.20 ´ 103 g 33.20 25.00 C g C qwater Intermediate Reaction J 33.20 25.00 C 837 C qbomb rH1 + rH2 = Reactants qRXN = – 4803 J or – 48.0 kJ Heat transferred per mole qV: - 48.0 kJ kJ = -5.48 ´ 103 mol mol 114.2g 1.00g ´ Hess’s law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed. so we can write… rH = ? unknown! rH Products What if the enthalpy changes through another path are know? The sum of the H’s in one direction must equal the sum Why? in the other direction. Because enthalpy is a state function… Path independent! Hess’s Law Problem: Example: Given: (1) (2) Determine the rH for the reaction: 3H2(g) + N2(g) 2NH3(g) rHo = ??? 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ Notice that the path from reactants to products in the desired reaction goes through an “intermediate compound” in the given reactions. This means that the path for hydrogen and nitrogen to produce ammonia goes through hydrazine (N2H4). Therefore, the path to the enthalpy of the reaction must be a sum of the two given reactions! Chapter 5 — Thermo 12 Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g) 2NH3(g) rHo = ??? Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g) 2NH3(g) rHo = ??? Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields: Look what happens… adding equations (1) & (2) yields: 2 H2(g) + N2(g)+ N2H4(g) + H2(g) N2H4(g) + 2NH3(g) 2 H2(g) + N2(g)+ N2H4(g) + H2(g) N2H4(g) + 2NH3(g) / / 3H2(g) + N2(g) 2NH3(g) Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g) 2NH3(g) rHo = ??? Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ intermediates H2(g) + N2H4(g) (Step 1) Ho = 95.4kJ (Step 2) Ho = 187.6kJ reactants 3H2(g) + N2(g) (Overall) Ho = 92.2kJ adding equations (1) & (2) yields: Look what happens… 2 H2(g) + N2(g)+ N2H4(g) + H2(g) N2H4(g) + 2NH3(g) / therefore… / 3H2(g) + N2(g) 2NH3(g) rH = ∆rH°+ +95.4 kJ + (–187.6 kJ) = –92.2 kJ 1 ∆rH°2 = Standard Enthalpies of Formation products 2N3 (g) • Reaction (1) is endothermic by 95.4 kJ • Reaction (2) is exothermic by 187.6 kJ • Since the exothermicity has a greater magnitude than the endothermicity, the overall process is exothermic (rH < 0). Enthalpy Values When 1 mole of compound is formed from its elements, the enthalpy change for the reaction is called the enthalpy of formation, fHo (kJ/mol). Since Enthalpy is a state function, enthalpy values depend on the reaction conditions in terms of the phases of reactants and products. These enthalpies are always reported at Standard conditions: 1 atm and 25 °C (298 K). The standard enthalpies of formation of the most stable form of any element is zero: H2(g) + ½ O2(g) H2O(g) ∆rH˚ = -242 kJ H2(g) + ½ O2(g) H2O(liq) ∆rH˚ = -286 kJ fH (element) = 0 fH° O2(g) = 0 elemental form fH° O(g) ≠ 0 NOT the elemental form Same reaction, different phases, different enthalpies. Chapter 5 — Thermo Formation Reactions A chemical reaction that describes the formation of one mole of a compound from its elements at standard state conditions is known as a “formation reaction”. 13 Question: What is the formation reaction for potassium permanganate? elements K (s) + Mn(s) compound + 2 O2(g) KMnO4 (s) The formation of water is given by the reaction: H2(g) + ½ O2(g) H2O(l) Each element and the compound are represented by the physical state they take on at 25.0 °C and 1 atm pressure. (Standard State conditions) • • • • salts are solids at standard state conditions. metals are solids at standard state conditions. oxygen is a gas at standard state conditions. balance for one mole of the product standard state conditions = 25oC and 1 atm Enthalpy Changes for a Reaction: Using Standard Enthalpy Values • All components of a reaction can be related back to their original elements. • Each compound has an enthalpy of formation associated with it. • Reactants require energy to return to component elements. • Products release energy when formed form component elements. • Since enthalpy is a state function, the sum of the above must relate somehow to the overall enthalpy of a reaction. Enthalpy Changes for a Reaction: Using Standard Enthalpy Values The sum of the fH° for the products multiplied by the respective coefficients, subtracted by the sum of the fH° for the reactants multiplied by the respective coefficients, yields the fH° for the reaction. Consider the Combustion of Propane 3C(s) + 8H2(g)+ 5O2(g) C3H8(g) + 5O2(g) • In order to make CO2(g) and H2O(l) one must break the propane up into its elements. • This takes energy. • The elements carbon, hydrogen and oxygen then combine to make the new compounds, CO2 and H2O. • This process releases energy. Problem: Calculate the rH° for CaCO3(s) given the following fH° CaO(s) + CO2(g) CaCO3(s) fH° [CaO(s)] = – 635.5 kJ/mol fH° [CO2(g)] = – 393.5 kJ/mol fH° [CaCO3(s)] = –1207 kJ/mol In other words: Energy gained– Energy spent = Net Energy DrH° = å nD f H° (products ) - å mD f H° (reactants ) n and m are the stoichiometric balancing coefficients. 3CO2(g) + 4 H2O(l) Chapter 5 — Thermo 14 Problem: Problem: Calculate the rH° for CaCO3(s) given the following fH° Calculate the rH° for CaCO3(s) given the following fH° CaO(s) + CO2(g) CaCO3(s) CaO(s) + CO2(g) CaCO3(s) fH° [CaO(s)] = – 635.5 kJ/mol DrH° = å nD f H° (products ) fH° [CO2(g)] = – 393.5 kJ/mol fH° [CaCO3(s)] = –1207 kJ/mol å mD fH (reactants ) ° fH° [CaO(s)] = – 635.5 kJ/mol DrH° = å nD f H° (products ) fH° [CO2(g)] = – 393.5 kJ/mol å mD H (reactants ) ° f fH° [CaCO3(s)] = –1207 kJ/mol 1mol CaCO3 1mol CaO 1mol CO2 rH° = –1207 kJ/mol – (–635.5 kJ/mol–393.5 kJ/mol) rH° = – 178 kJ/mol Problem: Problem: The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol. Calculate the fH° for naphthalene given the following enthalpies of formation: fH° [CO2(g)] = –393.5 kJ/mol The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol. Calculate the fH° for naphthalene given the following enthalpies of formation: fH° [CO2(g)] = –393.5 kJ/mol fH° [H2O(l)] = –285.7 kJ/mol fH° [H2O(l)] = –285.7 kJ/mol Step 1: Write the balanced equation for the reaction… C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l) Problem: Next recall that: DrH° = å nD f H° (products ) - å mD f H° (reactants ) C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l) combustionH° 10 = fH° [CO2(g)]+ 4 fH° [H2O(l)] – {fH° [C8H10(l)]+ 12 fH° [O2(g)]} From the problem, all quantities are know but f H C10H8 The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol. Calculate the fH° for naphthalene given the following enthalpies of formation: fH° [CO2(g)] = –393.5 kJ/mol fH° [H2O(l)] = –285.7 kJ/mol fH° [C10H8(l)] 10 fH° + 4 fH° = [CO2(g)] [H2O(l)] – + 12 fH° {combH° [O2(g)]} elements = 0 Chapter 5 — Thermo Problem: The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol. Calculate the fH° for naphthalene given the following enthalpies of formation: fH° [CO2(g)] = –393.5 kJ/mol fH° [H2O(l)] = –285.7 kJ/mol fH° [C10H8(l)] 10 fH° + 4 fH° = [CO2(g)] [H2O(l)] – + 12 fH° {combH° [O2(g)]} H° f [C10H8(l)] = elements = 0 10 (–393.5 kJ/mol + 4 (–285.7 kJ/mol)– (– 5156 kJ/mol) = + 79 kJ/mol 15