Thermochemistry Heat Transfer in Chemical Reactions Energy: Foundational Definitions 1 • Energy • Work • Forms of energy: – Kinetic energy – Potential energy - Forms of Energy, continued • Chemical Energy • Thermal energy • Radiant energy - solar energy, that is energy that comes from the sun. 1 Law of the Conservation of Energy • The total quantity of energy in the universe is assumed constant. • • Heat • Thermochemistry - the study of heat change in chemical reactions. • The heat exchange occurs between: the system (part of universe of interest) and the surroundings (the rest of the universe) Types of Systems Matter Heat Heat 2 Which would represent a can of green beans on the shelf at a store? 1. Open 2. Closed 3. Isolated • Exothermic process • Endothermic process CH4 + 2O2 Heat given off by the system to the surrounding 2Hg + O2 Heat absorbed by the system from the surroundings 2HgO CO2 + 2H2O The process of an ice cube melting to form water is… 1. Endothermic 2. Exothermic 3 Introduction to Thermodynamics 2 • Thermodynamics • State of a system - the values of all relevant macroscopic properties. – composition – – temperature – pressure – volume • State function • For example: – Volume is a state function. 6 ounces is 6 ounces regardless of how you got 6 ounces. • The magnitude of change only depends on the initial and final values • State functions are abbreviated with capital letters. • Other examples of state functions: – Energy –Total internal energy (E) – – – Temperature – Pressure 4 State Functions are “Path Independent” • First Law of Thermodynamics - • It is not possible to determine the total amount of energy in the universe. • What we determine is E. • Let us focus on the E of the system • There are two ways that energy can leave a system: 1. 2. • Therefore the 1st law of Thermodynamics looks like this: 5 Sign convention of q and w : Heat (q) neg.(-) pos.(+) Work (w) neg.(-) pos.(+) E = q + w Expansion of a Gas Does Work WORK AND HEAT You probably already know that w = F × d When working with gases, work can also be defined as This is called PV work and is the work involved when gases expand or are compressed. 6 w = –PV • When using this equation, what are the units of work? • • This isn’t a very useful energy unit, to convert to Joules: A certain gas expands from 0.5 L to 2.5 L against an external pressure of 0 atm. How much work is done? A certain gas expands from 0.5 L to 2.5 L against an external pressure of 1 atm 1. 2. 3. 4. 2.0 J -2.0 J 200 J -200 J 7 An exothermic reaction produces a net increase of gas when reacting. What is the sign of q and w? 1. 2. 3. 4. q & w are both positive q & w are both negative q is positive, w is negative q is negative, w is positive Can we determine the sign of E for the above process? 1. Yes, E is positive 2. Yes, E is negative 3. No, you need to know the size of w and q to determine this. 3 Enthalpy • Enthalpy • It is impossible to determine the enthalpy of a substance. • We will measure the change in enthalpy (H) • The Greek letter delta means change. • So (anything) = 8 • For a chemical reaction H, the enthalpy of reaction, is given by: H = CH4 + 2O2 Heat given off by the system to the surrounding 2Hg + O2 Heat absorbed by the system from the surroundings 2HgO CO2 + 2H2O Thermite demo: is the reaction endothermic or exothermic? What is the sign of H? 1. 2. 3. 4. Endothermic, positive Exothermic, positive Endothermic, negative Exothermic, negative Freeze flask to a board demo: is the reaction endothermic or exothermic? What is the sign of H? 1. 2. 3. 4. Endothermic, positive Exothermic, positive Endothermic, negative Exothermic, negative 9 Which statement is true for the reaction: SO2(g)+ ½ O2(g) SO3(g) H = -99.1 kJ 1. The reaction is endothermic, heat is consumed. 2. The reaction is exothermic and work is positive. 3. The E is positive. For reactions that take place in constant pressure environments, the enthalpy change is synonymous with the heat change. 4 Thermochemical Equations • Thermochemical equations - show the enthalpy changes as well as the mass relationships. • Here is an example: • H2O(s) H2O(l) H = 6.01 kJ • 10 H2O(l) 6.01 kJ Heat absorbed by the system from the surroundings H2O(s) • The coefficients always refer to the number of moles of the substance. • Reversing an equation, the sign changes for H. – In our example with water, if we are freezing 1 mol of water, the H = –6.01 kJ. or H = –6.01 kJ – H2O(l) H2O(s) • If we multiply both sides of the equation by n, must also multiply H by n. H = 12.02 kJ. – 2H2O(s) 2 H2O(l) • Equations must always specify the physical states of all the substances in the equation. Let’s examine the following thermochemical equation: CH4(g) + 2O2(g) CO2(g)+ 2H2O(l) H = –890.4 kJ 11 What would H be for this equation? 2CO2(g) + 4H2O(l) 2CH4(g) + 4O2(g) 1. 2. 3. 4. – 890.4 kJ 890.4 kJ –1780.8 kJ 1780.8 kJ • Let’s continue to examine the equation: CH4(g) + 2O2(g) CO2(g) +2 H2O(l) H = – 890.4 kJ • When 1 mole of CO2(g) is produced, what is the enthalpy change? • • When 2.5 mol of O2 is reacted with an excess of CH4, what is the enthalpy change? • Note the unit factor. In a thermochemical equation, a unit factor can be written for any substance in the equation. • We now have a new “conversion tool” and a new road on our map. • 12 Determine the amount of heat (in kJ) given off when 1.26 × 104 g of ammonia are produced according to the equation: N2(g) + 3H2(g) 2NH3(g) Horxn = –92.6 kJ 1. 2. 3. 4. – 6.86 104 kJ – 5.83 105 kJ – 3.43 × 104 kJ – 1.17 106 kJ A reaction is used to propel rockets is N2O4(l) + 2N2H4(l) 3N2(g) + 4H2O(g) When 10.0 g of liquid N2O4 is consumed in this reaction, 124 kJ of heat is released. (a) The sign of H is 1. Positive 2. Negative A reaction is used to propel rockets is N2O4(l) + 2N2H4(l) 3N2(g) + 4H2O(g) When 10.0 g of liquid N2O4 is consumed in this reaction, 124 kJ of heat is released. (b) Is more or less heat released than the 124 kJ for the quantities given in the chemical reaction equation? 1. More 2. Less 13 A reaction is used to propel rockets is N2O4(l) + 2N2H4(l) 3N2(g) + 4H2O(g) When 10.0 g of liquid N2O4 is consumed in this reaction, 124 kJ of heat is released. (c) What is the value of H for the chemical reaction as written? We will do together. Consider the reaction: 2CH3OH(l) + 3O2(g) 4H2O(l) + 2CO2(g) H = – 1452.8 kJ/mol If 15.0 grams of methanol (CH3OH) reacted. How much heat would be produced in kJ? When 100. g of NO is burned, 191 kJ of heat is released. Determine H for the reaction 2NO(g) + O2(g) 2NO2(g) 14 When 55.5 g ammonia reacts with oxygen, 734 kJ of heat is released. Determine H for the following balanced equation. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 1. 2. 3. 4. 5. 6. 901 kJ -901 kJ 225 kJ -225 kJ 13.2 kJ -13.2 kJ Enthalpy Change and the First Law of Thermodynamics. 5 E = q + w at constant pressure becomes We can now determine the change in internal energy of a reaction under constant P conditions. What is qrxn when 1.0 g of Na reacts with H2O according to this reaction? 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g) Horxn = – 368.4 kJ 1. 2. 3. 4. – 8.0 kJ – 16 kJ + 4.0 kJ – 180 kJ 15 Na is placed into the water it produces H2(g). If we know: 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) H = – 367.5 kJ We can use E = H – PV to calc. E E = H – PV • If, in the course of a reaction, there is no change in the number of gas moles, what can we say about E? • Even when there are moles of gas produced, the PV term is small so • We can use PV = nRT to substitute into the top equation giving: Calculate the change in internal energy E when 1 mol of CH4 reacts with two moles of O2 according to the following reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = – 890.4 kJ 16 Consider the reaction H2(g) + Cl2(g) 2HCl(g) H = ─184.6 kJ 1) What is the heat released by the reaction if 5.0 grams of hydrogen reacts with 20.0 grams of chlorine? 2) What is the work done against a pressure of 1.0 atm at 25oC. 3) What is E for this reaction, for the amounts reacted? 4) What is E for the equation as written? Four ways to determine H of a reaction. 1. Calorimetry. We will experimentally determine the heat change of the surrounding when the reaction is performed in the laboratory. (Learning Objective 6) 2. Direct method - using Hof (Learning Objective 7) 3. Indirect method - using Hess’s Law. (Learning Objective 7) 4. Estimating it using bond dissociation energies. (Learning Objective 7) 17 Calorimetry: Determination of Enthalpy via Experimentation 6 • is the measurement of heat changes. • To determine the H of a reaction, we can measure the heat change of the surroundings. • Terms and Equations for Calorimetry • Specific heat (s) - • • Heat capacity (C) = • Calculating heat absorbed or released by an object: • q= • q =heat change t = change in temperature; tf - ti. 18 Calculate the amount of heat liberated (in kJ) from 366 g of mercury when it cools from 77.0oC to 12.0oC. s(of Hg) = 0.139 J/(goC) What is the heat absorbed by 50.0 g copper block when it is heated from 25oC to 90oC? (s of Cu is 0.385 J/goC) 1. 2. 3. 4. 0.481 kJ 1.73 kJ 2.21 kJ 1.25 kJ Constant Volume Calorimetry •Used for combustion reactions. •C,H,O+O2 CO2+H2O 19 Constant Pressure Calorimetry •Used often for aq. soln. reactions. Dissolving 6.00 g of CaCl2 in 300. mL of water causes the temperature of the solution to increase 3.43oC. Assuming the specific heat of the solution is 4.18 J/goC and assuming the calorimeter absorbs no significant amount of heat, determine H for the reaction. CaCl2(s) Ca2+(aq) + 2Cl–(aq) What is the mass of the surroundings? 1. Not enough information is given 2. 300 g 3. 306 g Dissolving 6.00 g of CaCl2 in 300. mL of water causes the temperature of the solution to increase 3.43oC. Assuming the specific heat of the solution is 4.18 J/goC and assuming the calorimeter absorbs no significant amount of heat, determine H for the reaction. CaCl2(s) Ca2+(aq) + 2Cl–(aq) 20 Calculate the number of moles of CH4 (methane) required to heat 200 mL of water (enough to make a cup of coffee or tea) from 20oC to 100oC. Assume that only 50 percent of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. The reaction for methane is: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H= – 890.4 kJ 7 Calculating Enthalpy • Once foundational measurements are made to determine some H values, manipulations can be made to determine H for other reactions. • Three other ways – Using Enthalpy of Formation – Using Hess’s Law – Using Bond Energies Standard Enthalpy of Formation • Standard enthalpy of formation Hof - • Standard state - 1 atm. The o refers to the std. state. 21 • If we look up the Hof of MgCO3(s) we see it equals – 1112.9 kJ/mol. What does this mean? • It is the enthalpy change for this reaction: What is the reaction equation for the Hfo of CH4(g) 1. 2. 3. 4. 5. CH4(g) C(s) + 2H2(g) C(s) + 4H(g) CH4(g) C(s) + 2H2(g) CH4(g) 2C(s) + 4H2(g) 2CH4(g) Both 3 and 4 Look back at the definition of standard enthalpy of formation. What is the value for Hfo of O2? 1. 2. 3. 4. Must be negative Must be positive Must be zero Cannot be determined from the definition. Need more information. 22 • Standard Enthalpy of Reaction, Horxn • The of ’s are used to determine Horxn according to the following equation: Horxn = nHof(products) – mHof(reactants) • n,m are the coefficients in balanced equation. • Using this equation to determine Horxn is called the direct method. 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) Horxn = 1. 2. 3. 4. -2598.8 kJ -905.6 kJ -1928.4 kJ -1692.4 kJ Compound C2H2(g) CO2(g) H2O(l) Hfo 226.6 kJ/mol -393.5 kJ/mol -285.8 kJ/mol What is Hof (in kJ/mol) of C2H6(g) given the following information? 2 C2H6(g) + 7O2(g) 4 CO2(g) + 6 H2O(l) Horxn= – 3119.4 kJ o H f of CO2(g) = –393.5 kJ/mol Hof of H2O(l)= –285.8 kJ/mol 1. 2. 3. 4. -84.7 kJ/mol -169.4 kJ/mol 84.7 kJ/mol 169.4 kJ/mol 23 For the reaction CaCO3(s) CaO(s) + CO2(g) Ho = 177.8 kJ Given: Hof(CaO(s)) = – 635.6 kJ/mol and Hof (CO2(g)) = – 393.5 kJ/mol What is the standard enthalpy of formation of CaCO3? 1. 2. 3. 4. – 64.3 kJ/mol – 419.9 kJ/mol – 851.3 kJ/mol – 1206.9 kJ/mol For the reaction: 2 H2(g) + O2(g) 2 H2O(l) Horxn = 2. 3. 4. Hof(H2O(l)) 2 Hof(H2O(l)) – Hof(H2O(l)) – ½ Hof(H2O(l)) Hess’s Law - The Indirect Method for cal. Hrxn • Hess’s Law - • To calculate the H, you will be given several thermochemical equations. • Let’s see it at work by doing an example. 24 From the following data, C(gr) +O2(g) CO2(g) Horxn = – 393.5 kJ H2(g) + 1/2O2(g) H2O(l) Horxn = – 285.8 kJ 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) Horxn = – 3119.6 kJ Calculate the enthalpy change for the reaction 2C(gr) + 3H2(g) C2H6(g) Heat of Solution and Dilution: An Application of Hess’s Law Heat of Hydration Hhydration = -784 kJ/mol Heat of solution Hsoln = 4 kJ/mol 25 Estimating Enthalpy Change using Bond Energies • Bond energy • H2(g) H(g) + H(g) Ho = 436.5 kJ • Table on the next slide gives some bond energies. Some of these are “average bond energies” • We can use the values to estimate the Horxn Average Bond Energies *The Bond Bond Energy kJ/mol Bond Bond Energy kJ/mol H–H 436.4 C–O 351 H–N 393 C=O 745 H–O 460 C=O* 799 H–F 568.2 N–N 193 C–H 414 N=N 418 C–C 347 N≡N 941.4 C=C 620 N–O 176 C≡C 812 N=O 607 C–N 276 O–O 142 C=N 615 O=O 498.7 C≡N 891 F–F 156.9 bond energy from CO2 Bond energies in RED are for diatomic molecules, not an average. Enthalpy Change Using Bond Energies Reactant Molecules Requires Energy Requires Energy Product Molecules Atoms Releases Energy Releases Energy Enthalpy Atoms Reactant Molecules Product Molecules 26 • To use this method, must write the Lewis structures for all substances in the equation so that you know what bonds are broken and what bonds are formed. For the reaction: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) Predict the enthalpy of reaction from the average bond energies of Table 9.3 What is the estimated H for the reaction? 1. 2. 3. 4. 5. 6. -2812 kJ -2308 kJ -803kJ 803 kJ 2308 kJ 2812 kJ 27 If the bond energy for H2 is 436.5 kJ/mol, what is the Hof of H(g)? 1. 2. 3. 4. 5. 6. 436.5 kJ/mol -436.5 kJ/mol 218.3 kJ/mol -218.3 kJ/mol 873.0 kJ/mol -873.0 kJ/mol Hint: write the reaction for both bond energy and heat of formation. 28