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CHE 106 Chapter 5 THERMOCHEMISTRY Important Terminology: Force: Work: Heat: Energy: Important Terminology cont… Kinetic Energy: Formula: Potential Energy: Kinetic vs. Potential Energy Energy Transfers System: Surroundings: Example: Energy and the 1st Law First Law of Thermodynamics: * Heat: * Work Example: Internal Energy Definition: DE = Efinal – Einitial DE: Magnitude, Sign, Unit: Example: Reaction Profiles CH4 + O2 H2O + CO2 EXOTHERMIC DPE PE CH4 + O2 N2 + O2 2 NO ENDOTHERMIC 2 NO DPE H2O + CO2 N2 + O2 Energy: Heat and Work DE can be depends on the energy conversions and exchanges within a chemical system. Equation: Sign Conventions for q and w Sign Conventions: Two Cases HEAT HEAT WORK SYSTEM q = positive (+) w = positive (+) DE = (+) endothermic WORK SYSTEM q = negative (-) w = negative (-) DE = (-) exothermic Internal Energy Example: Calculate DE for the following situations: A. A system does 195 kJ of of PV work and absorbs 38 J of heat. B. A chemical reaction in a piston chamber gives off 500 j of heat to its surroundings. The expanding gas moves the piston upward and does 240J of work. C. The chemical reaction from (B) was rerun with the piston in a locked position and generates 740 J of heat. Internal Energy and State Functions Internal Energy and State Functions Example: Sweat Evaporating – Two Ways Case 1: System: Surroundings: Case 2: System Surroundings: Internal Energy and State Functions State Functions Definition: Example: Variables that are State Functions: Variables that are NOT State Functions: State Functions Internal Energy Although energy is a state function, it is also considered an extensive property. Example: The combustion of octane Internal Energy and Enthalpy Making math work for us: DE = q + w Internal Energy and Enthalpy Enthalpy Definition: Relationship between internal energy and enthalpy: at constant pressure. H = E + PV Equation Manipulation: Final Equation: Internal Energy and Enthalpy ENDOTHERMIC: EXOTHERMIC: ACTIVATION ENERGY: Activation Energy Activation Energy - minimum energy to initiate a chemical reaction - barrier to reaction Activation E non-state function A Eact DPE state function B Enthalpy and Reaction Profiles Enthalpy and Reaction Profiles Enthalpy: Heats of Reaction Most reactions take place at constant pressure and the change in volume is nearly zero. In these cases, DE = DH. There are 4 things to remember when dealing with enthalpy. Enthalpy: Heats of Reactions 1. Enthalpy is a state function. DH = Hproducts – Hreactants 2. Enthalpy is an extensive property. 3. DH for a reaction is equal in magnitude but opposite in sign for the reverse reaction. 4. DH for a reaction depends on the state of products and reactants. Enthalpy: Heats of Reaction Consider the reaction: 2 N2 (g) + O2 (g) 2 N2O (g) DH = +163.2 kJ 1. Is the reactions exothermic or endothermic? 2. Calculate the amount of heat transferred when 12.8 grams of N2O (g) forms at constant pressure. 3. How many grams of N2(g) must react to produce a DH of 1.00 kJ Enthalpy: Heats of Reaction Sample exercise: Hydrogen peroxide can decompose to form water and oxygen by the reaction: 2 H2O2 (l) 2 H2O (l) + O2 (g) DH = -196 kJ Calculate the value of q when 5.00 g of hydrogen peroxide decomposes at constant pressure. Enthalpy: Heats of Reaction Hindenburg: https://www.youtube.com/watch?v=CgWHbpMVQ1U 2H2 + O2 2 H2O DH = -484 kJ 7,062,000 cubic feet of H2 at STP produces how much energy, given 1 ft3 contains 28.3 L of H2 gas. Enthalpy: Heats of Reaction Example: Oxygen may be generated on small scales in the laboratory by the thermal decomposition of potassium chlorate: Equation: DH = - 89.4 kJ For this reaction, calculate DH for the formation of - 6.45 g of O2 (g) - 9.22 g of KClO3 (s) Enthalpy: Heats of Reaction Example: Which of the following has the highest enthalpy at a given at a given temperature and pressure: H2O(s) vs. H2O (l) vs. H2O (g) Example: Formation of gaseous water vs liquid water: Enthalpy: Heats of Reaction When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: Ag+(aq) + Cl-(aq) AgCl (s) DH = -65.5 kJ 1. Calculate the DH for the formation of 0.200 mol AgCl from this reaction. 2. Calculate the DH for the formation of 2.50 g of AgCl. 3. Calculate DH when 0.350 mol of AgCl dissolves in water. Calorimetry Calorimetry: Measure of heat flow. Can calculate DH or DE. Heat Capacity: Molar Heat Capacity: Specific Heat Capacity Calorimetry Substance Specific Heat (J °C-1 g-1) H2O(l) H2O(s) 4.18 2.03 Al(s) C(s) Fe(s) Hg(l) CCl4(l) CaCO3(s) 0.89 0.71 0.45 0.14 0.86 0.85 Calorimetry Example: A swimming pool, 10.0 m by 4.0 m, is filled to a depth of 3.0 m with water at 20.2 °C. How much energy is required to raise the water temperature to 30.0°C? Example: How much energy is required to raise the temperature of an 8.50 x 102 g block of aluminum from 22.8°C to 94.6°C? What is the molar heat capacity of aluminum? Calorimetry • Example: Ti metal is used as a structural material in many high tech applications including jet engines. • What is the specific heat of Ti if it takes 89.7 kJ to raise a 33.0 Kg block by 5.20 °C? • What is the molar heat capacity of Ti? Calorimetry Example: Large beds of rocks are used in some solar-heated homes to store heat. A) Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0°C. Specific heat of rocks is 0.82 J/g-K. B) What temperature change would these rocks undergo if they absorbed 450 kJ of heat? Calorimetry Two types of calorimetry: Constant Pressure vs. Constant Volume Constant Pressure: Constant Volume: Calorimetry Calorimetry Calorimetry Consider mixing 50 mL of 1.0 M HCl solution with 50 mL of 1.0 M NaOH solution, both at 25oC. Upon mixing, the temperature of the new solution is 31.9oC. a) How much energy is released in the problem? b) How much energy is released in the sample problem on a molar basis? Calorimetry Calorimetry Heat of Dilution Example: Reaction: H2SO4 (conc) + n H2O H2SO4 (dil) Where n = moles H2O / Mole H2SO4 10 mL H2SO4 (conc) 20 mL H2SO4 (conc) 100 mL H2O 30 mL H2SO4 (conc) 100 mL H2O Measure Temperature Changes 100 mL H2O Calorimetry n Heat of dilution of sulfuric acid (H2SO4) 10 mL H2SO4 (conc) 20 mL H2SO4 (conc) 100 mL H2O 30 mL H2SO4 (conc) 100 mL H2O 100 mL H2O Initial Temp 25° C 25° C 25° C Final Temp 50° C 73° C 95° C Calorimetry • Heat of dilution of sulfuric acid (H2SO4) Reaction: H2SO4(conc) + n H2O H2SO4(dil) (where n = moles H2O/mole H2SO4) mL H2SO4 10 mL 20 mL 30 mL mL H2O n 100 mL 31.0 100 mL 15.6 100 mL 10.5 DT (°C) 25.0 48.0 70.0 DHrxn -11.0 kJ -26.0 kJ -30.0 kJ DHrxn values are dependent upon the amount of material present Calorimetry Example: A coffee cup calorimeter contains 125 g of water at 24.2oC. After KBr (10.5 g) at 24.2oC is added, the temperature becomes 21.1oC. What is the heat of solution of KBr? Given that no heat is transferred to the surroundings from the calorimeter and the specific heat of the solution is 4.18 J/goC. Calorimetry Example: When 50.0 mL of 0.100M AgNO3 and 50.0 mL of 0.100 HCl are mixed in a constants pressure calorimeter, the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by this reaction: AgNO3 + HCl AgCl + HNO3 Calculate DH for this reaction. Bomb Calorimetry Typically used to study combustion reactions: System: Sample (hydrocarbon) and oxygen Surroundings: Bomb and Water When the reaction takes place, the heat released is absorbed first by the reaction chamber (bomb) and then transferred to the surrounding water. Bomb Calorimetry Because we are working with constant volume, our calculations will lead us to the DE. However, because the work done is equal to zero, the DE and DH are relatively close to one another. Equation: qrxn = -Ccal x DT The Ccal must be determined experimentally, using a reaction with a known DH. Bomb Calorimetry Example: The combustion of 0.1624g of benzoic acid (C7H7O2) raises the temperature of a calorimeter by 2.71oC. Given: DHcomb benzoic acid = -26.73 kJ/g. What is the heat capacity of the calorimeter? ANSWER: Ccal = 1.60 kJ/oC. If 0.2138g of vanillin (C8H8O3) is burned in the calorimeter and the temperature increases by 3.28oC, what is the heat of combustion per gram and per mole? ANSWER: -24.6 kJ/g, -3740 kJ/mol. Bomb Calorimetry A 0.1964g sample of quinone (C8H4O2) is burned in a bomb calorimeter that has a heat capacity of 1.56 kJ/oC. The temperature of the calorimeter increases by 3.2oC. Calculate the heat of combustion of quinone per gram and per mole. ANSWER: -25.4 kJ/g , -2740 kJ/ mol Under constant volume conditions, the heat of combustion of glucose is -15.57 kJ/g. A 2.500g sample is burned and the temperature increased from 20.55oC to 23.25oC. What is the total heat capacity of the calorimeter? If the calorimeter contains 2.70Kg of water, what is the heat capacity of the dry calorimeter. ANSWER: Ctotal = 14.4 kJ/oC, Cdry = 3.11 kJ/oC Bomb Calorimetry A 0.5865g sample of lactic acid is burned in a calorimeter whose heat capacity is 4.812 kJ/oC. The temperature increases from 23.10oC to 24.95oC. Calculate the heat of combustion of lactic acid per gram. ANSWER: -15.2 kJ/g Hess’s Law It’s inefficient to use calorimetric calculations to determine the enthalpy for every possible reaction. Instead, we take advantage of enthalpy being a state function, and use preexisting databases to calculate the enthalpy for a reaction that we haven’t necessarily carried out. DH is independent of the pathway, so whether a reaction takes place in 2 steps or 5 steps, the net DH will be the same. Hess’s Law: If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of DH’s for the individual steps. Hess’s Law For the reaction: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Step 1: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = -802 kJ Step 2: 2 H2O(g) 2 H2O (l) DH = -88 kJ SUM OF REACTION: CH4(g) + 2 O2(g) + 2 H2O(g) CO2(g) + 2 H2O(g) + 2 H2O (l) DH = -802 kJ + -88 kJ = -890 kJ Hess’s Law Determine the DH for A+B C if you are given: A+ B D DH1 DE DH2 EC DH3 A+B C DH(A+B=C) = DH1+ DH2 + DH3 Hess’s Law Example: Haber process for the formation of ammonia: 3 H2(g) + N2(g) 2 NH3(g) DHo = -92.2kJ Step 1: 2 H2(g) + N2(g) N2H4 (g) DHo = ? Step 2: N2H4 (g) + H2(g) 2 NH3(g) DHo = -187.6 kJ NET: 3 H2(g) + N2(g) 2 NH3(g) DHo = -92.2kJ Solve for DHo(rxn 1) ANSWER: +95.4 kJ Hess’s Law What is the enthalpy of combustion of carbon(s) going to carbon monoxide (CO)? The following are known: C(s) + O2(g) CO2(g) CO(g) + 0.5 O2(g) CO2(g) ANSWER: -110.5 kJ DH = -393.5 kJ DH = -283.0 kJ Hess’s Law Water gas is a very important mixture of CO and H2 prepared by passing steam over hot charcoal at 1000°C. Calculate DH° for the water-gas rxn. Given: (1) C(s) + H2O(g) CO(g) + H2(g) (Water Gas Rxn) (2) C(s) + O2(g) CO2(g) DH° = -393.5 KJ (3) 2 H2(g) + O2(g) 2H2O(g) DH° = -483.6 kJ (4) 2 CO(g) + O2(g) 2 CO2(g) DH = -566.0 kJ ANSWER: +131.3 kJ Hess’s Law Ex: Calculate DH for 2F2(g) + 2H2O(l) 4HF(g) + O2(g) • Given: (1) H2(g) + F2(g) 2HF(g) (2) 2H2(g) + O2(g) 2H2O(l) DH = -537 kJ DH = -572 kJ ANSWER: -502 kJ Ex: Carbon occurs in two forms, graphite and diamond. The enthalpy of combustion of graphite is -393.5 kJ/mol, and that of diamond is -395.4 kJ/mol. Calculate DH for the conversion of graphite to diamond: C(graphite) C(diamond) ANSWER: 1.9 kJ/ mol Hess’s Law Example: Calculate DH for the reaction: NO + O NO2 Given: NO + O3 NO2 + O2 DH = -198.9 O3 3/2 O2 DH = -142.3 O2 2O DH = 495.0 ANSWER: -304.1 kJ Enthalpy of Formation Hess’s Law can be used with other thermodynamic data to calculate the enthalpy for many different reaction types: Heat of Formation (DH compound from elements) labeled DHf Heat of formation (DHf) is usually given for reactants and products in standard states (since DH depends on the state of these items). When in standard state, the denotation is DH°f Ex: 2C(s) + 3H2(g) + 1/2O2 C2H5OH(l) Heat of Vaporization (DH for liquid to gas) Heat of Fusion (DH for melting solids) Heat of Combustion (DH reaction with O2) DH°f = -277.7 kJ Enthalpy of Formation Things to remember: DH is as state function, so we are allowed to apply Hess’s Law and add together appropriate reactions to calculate the DHof The standard enthalpy of formation of the most stable form of any element is zero. (That way we don’t need to find out how much energy is required to produce each element involved in a reaction). DH°rxn = •nDH°f (products) - mDH°f (reactants) Where m and n are stoichiometric coefficients from balanced equation. Enthalpy of Formation For example, if we have a simple chemical equation with the variables A, B and C representing different compounds: A+B⇋C And we have the standard enthalpy of formation values as such: ΔHfo[A] = 433 KJ/mol ΔHfo[B] = -256 KJ/mol ΔHfo[C] = 523 KJ/mol The equation would be: ΔHoreaction = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B]) ANSWER: 346 kJ Enthalpy of Formation C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Given: DHof of C3H8: -103.85 kJ/mol DHof of H2O: -285.8 kJ/mol DHof of CO2: -393.4 kJ/mol DHorxn: -2220 kJ Enthalpy of Formation C6H12O6(s) 2 C2H5OH(l) + 2 CO2(g) Given: DHof of glucose: -1260 kJ/mol DHof of ethanol: -277.7 kJ/mol DHof of carbon dioxide: -393.5 kJ/mol ANSWER: -82 kJ Enthalpy of Formation Oxyacetylene welding torches burn acetylene gas, C2H2. Calculate DH° (kJ) for the combustion of acetylene. 2C2H2(g) + 5O2(g) 2H2O(g) + 4CO2(g) DH°rxn = •nDH°f (products) - •mDH°f (reactants) Given: DHof of acetylene gas: 226.7 kJ/mol DHof of H2O(g): -241.8 kJ/mol DHof of CO2(g): -393.5 kJ/mol ANSWER: -2511 kJ for every 2 moles of acetylene. Enthalpy of Formation Using the standard enthalpies of formation, calculate the enthalpy change for the combustion of 1 mol of ethanol: C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ANSWER: -1367 kJ