Energy in Thermochemistry

Units of Energy Energy in Thermochemistry Like we saw with pressure, many different units are used throughout the world for energy. SI unit for energy 1J = 1 kg ⋅ m 2 s2 Joule (J) calorie (cal) erg (erg) electron volts (eV) British thermal unit (Btu) 100º C Amount of energy it takes to raise one gram of water 1ºC 4.184 Joules (J) = 1 calorie (cal) 1 kcal = 1000 calories = 1 Nutritional Calorie (Cal) Thermochemistry In thermochemistry, the universe is divided into two parts: The system: The physical process or chemical reaction in which we are interested. We can define the system anyway we like. The surroundings: Everything else in the universe. Heat (q): energy transferred from an object with a higher temperature to an object with a lower temperature 50º C heat Heat cannot be transferred in the opposite direction! Vroom... Work (w): energy transferred against a force Example: Combustion reactions allow your car to move...performing work Energy flow between system and surroundings Surroundings Surroundings System System Surroundings Surroundings Universe System + Surroundings copper After I add 2.0kJ of heat energy to each sample, will they still be the same temperature? 5.0 g Al 5.0 g Cu 25°C 25°C Energy (both heat and work) can be transferred between the system and surrounds. The sign “+” or “” tells us the direction of energy flow. Surroundings Surroundings _ + System System Surroundings Surroundings Calorimetry The energy released from the reaction (or whatever is in the calorimeter) is gained by the water. So… -qrxn = qwater Notice that they have opposite signs, which tells us the direction of transfer. However, assuming no heat loss, the magnitudes are the same. The amount of heat lost by the reaction is the same amount gained by the water. Example Exothermic vs. Endothermic processes 99.8oC • A 28.2 gram sample of nickel is heated to and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC. Feels warm to the touch • Which substance absorbed heat? q= - • Which substance released heat? Feels cool to the touch • Calculate the heat absorbed by the substance you indicated above. System: Surroundings Exothermic Endothermic chemical reaction or process System q=+ Surroundings Enthalpies of Reaction Surroundings Heat and Enthalpy • Determine if the following processes are endothermic or exothermic… So how are q and ∆H related? At constant pressure: – Combustion of methane – Neutralization of HCl – Boiling – Melting – CaCO3 (s)
CaO (s) + CO2 (g) qp = ∆H So, using a calorimeter, the ∆H of a chemical reaction can be calculated from the change in heat in the calorimeter. 11 Enthalpies of Physical/Chemical Changes Enthalpy (∆H) describes the change in heat from a chemical reaction or process. ∆H = Hproducts – Hreactants Where do these energies come from? Energy is released if: ∑Energyreactants > ∑Energyproducts Energy is absorbed if: ∑Energyreactants < ∑Energyproducts Thermodynamic Standard State ∆H˚ The “˚” (knot) denotes enthalpy under standard states Standard States – A gas at 1 atm – An aqueous solution (1 M conc.) at a pressure of 1 atm – Pure liquids and solids – The most stable form of elements at 1 atm and 25oC (298 K) Direction of reaction and ∆H 2H2O(g) → 2H2(g) + O2(g) • Heat of fusion (∆Hfus): Amount of heat required to melt (solid
liquid) • Heat of vaporization (∆Hvap): Amount of heat required to evaporate (liquid
gas) • Heat of sublimation (∆Hsub): Amount of heat required to sublime (solid
gas) Why are there no values for ∆Hfreezing, ∆Hcondendsation, or ∆Hdeposition? Amount of reactant, and ∆H In chemistry, we usually specify energy in terms of a specific amount of reactant or product. For example: 2H2(g) + O2(g) → 2H2O(g) ∆H =-241.8 kJ Which means if we react 2 mole of H2 with 1 mole of O2, 241.8kJ of heat energy will be released. This can be read as: But what if we react 4 moles of H2 with 2 moles of O2? -241.8 kJ per 2 moles H2 ∆H =-483.6 kJ -241.8 kJ per 2 moles of H2O -241.8 kJ per 1 mole O2 Thermochemical Equations and Stoichiometry What about the reverse reaction of: 2H2(g) + O2(g) → 2H2O(g) Enthalpies of Phase Changes • 2Al (s) + Fe2O3 (s)
2Fe (s) + Al2O3 (s) ∆Ho = -852 kJ ∆H =-241.8 kJ ∆H =+241.8 kJ All we do is change the sign. Instead of energy being released (as in the first reaction), energy is absorbed when breaking water molecules apart. • How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al? – 10.0 g Fe2O3
x mol Fe2O3 – x mol Fe2O3
852 kJ / 1 mol Fe2O3 Hess’s Law Hess’s Law • If a compound cannot be directly synthesized from its elements, we must use multiple reactions to calculate the enthalpy of reaction ∆H is a state function…it doesn’t matter what route we take to get there. Same change in energy, ∆H. These two steps (solid → liquid → gas) should have the same change in energy as… • Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps • Look at direction of reaction and amount of reactants/products …this one step (solid → gas) 20 Hess’s Law Hess’s Law • Does the energy change in a reaction depend on the number of steps in the reaction? ∆H = 53.00 kJ • H2(g) + I2(s) → 2HI(g) I2(s) → I2(g) ∆H = 62.44 kJ • Values of enthalpy change – For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign • Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa) – Proportional to the amount of reactant consumed • H2(g) + I2(g) → 2HI(g) • Twice as many moles = twice as much heat – Depends on the physical state of the reactants and products (i.e., elemental state) ∆H = ? • How can we add the first two reactions to get the third? • Hess’s Law: ∆Hsum of steps = ∆H1 + ∆H2 21 Application of Hess’s Law 22 Hess’ Law Problem • We can use known values of ∆Ho to calculate unknown values for other reactions • P4 (s) + 3 O2 (g)
P4O6 (s) ∆H = -1640.1 kJ • P4 (s) + 5 O2 (g)
P4O10 (s) ∆H = -2940.1 kJ • Given: • CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g) • H2O (l)
H2O (g) • What is ∆Ho for the following reaction? P4O6 (s) + 2 O2 (g)
P4O10 (s) ∆H = ? ∆H = -802 kJ ∆H = 88 kJ • Find the enthalpy for the following reaction: • CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) ∆H = ? 23 Standard Enthalpies of Formation Standard Enthalpies of Formation • Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book) • Standard enthalpy of formation (∆Hof): heat needed to make 1 mole of a substance from its stable elements in their standard states ∆Horxn = Σn∆Hof (products) – Σn∆Hof (reactants) − Σ = sum – n = number of moles (coefficients) ∆Hof = 0 for a stable (naturally occurring) element • Which of these have ∆Hof = 0? – CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s) • Do the following equations represent standard enthalpies of formation? Why or why not? – 2Ag (l) + Cl2 (g)
2AgCl (s) – Ca (s) + F2 (g)
CaF2 (s) 25 Heats of Formation Group Work ∆Horxn = Σ n∆Hof,products - Σ n∆Hof,reactants Calculate values of ∆Ho for the following • Use Standard Heat of Formation values to calculate the enthalpy of reaction for: • C6H12O6(s)
2 C2H5OH(l) + 2 CO2(g) • rxns: • 1) CaCO3 (s)
CaO (s) + CO2 (g) • 2) 2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l) ∆Hof (C6H12O6(s)) = -1260.0 kJ/mol ∆Hof (C2H5OH(l)) = -277.7 kJ/mol ∆Hof (CO2(g)) = -393.5 kJ/mol ∆H˚ of (in kJ/mol): CaCO3 (-1207.1) CaO (-635.5) CO2 (-393.5) C6H6 (49.0) CO2 (-393.5) H2O (-285.8) 27 Bond Dissociation Energies Bond Energy • Bond dissociation energy (BDE). Energy needed to break one 1 mol of a specific bond (eg. H-H, C-H, C=C, C-C, etc.) • Bond dissociation energy varies somewhat from one molecule to another, or even within one molecule, so in most cases, we have use an average bond energy (D). • • • • • For example: H-OH H-O H-OOH Average = 502 kJ/mol 427 kJ/mol 431 kJ/mol 453.3 kJ/mol for O-H • Note: there are certain bonds where the BDE is not and average. These are bonds that only exist within diatomic molecules (eg. H-H, H-Br, Cl-Cl, etc.) 30 Bond Dissociation Energies ∆Horxn Enthalpy of Reaction = ΣBDE (reactants) + - ΣBDE (products) endothermic energy input • Use bond energies to calculate the enthalpy change for the following reaction: exothermic energy released NΞN = 945 kJ/mol • N2(g) + 3H2(g) → 2NH3(g) • Or: ∆Horxn = ΣBDE (reactants) – ΣBDE (products) H-H = 436 kJ/mol N-H = 390 kJ/mol ∆Hrxn = [BEN ≡ N + 3BEH-H] – [6BEN-H] ∆Hrxn = [945 + 3(436)] – [6(390)] = -87 kJ ΣBDE(react) > ΣBDE(prod)
endothermic ΣBDE(react) < ΣBDE(prod)
exothermic • measured value = -92.2 kJ • We use only when heats of formation are not available, since bond energies are average values for gaseous molecules • Why are the calculated and measured values different? 31 Enthalpy of Reaction 32 Practice Problems • Which amount of energy is higher? – Energy that is released when 55g of Al is cooled from 5˚C to -25 ˚C – Energy that is absorbed when 64g of Cu is heated from 298K to 315K? • Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl3 (g)
N2 (g) + Cl2 (g) (hint: balance first!!) ∆Hrxn? N-Cl = 200 kJ/mol N≡N = 945 kJ/mol Cl-Cl = 243 kJ/mol • 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it? 33 Practice Problems Practice Problems • Practice: Worked Example 8.10, page 320…. • Identify how to set up the following problems: • Calculate the ∆Ho of reaction for: – C3H8 (g) + 5O2 (g)
3CO2 (g) + 4H2O (l) ∆Hof C3H8(g): -103.95 kJ/mol; ∆Hof CO2(g): -393.5 kJ/mol; ∆Hof H2O(l): -285.8 kJ/mol • 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it? • C2H4(g ) + 6F2(g)
2CF4(g) + 4HF(g) ∆Ho = ? – H2 (g) + F2 (g)
2HF (g) ∆Ho = -537 kJ – C (s) + 2F2 (g)
CF4 (g) ∆Ho = -680 kJ – 2C (s) + 2H2 (g)
C2H4 (g)∆Ho = 52.3 kJ • Use average bond energies to determine the enthalpy of the following reaction (from Table 7.1). – CH4 (g) + Cl2 (g)
CH3Cl (g) + HCl (g) – (BEC-Cl = 328 kJ/mol)