Chapter 17 Thermochemistry Courtesy of labinitio com Thermochemistry

Chapter 17: Thermochemistry Courtesy of lab-initio. com

Thermochemistry The study of energy changes that occur during chemical reactions and changes in state.

Energy is the capacity to do work or supply heat, and can take many forms q Potential energy is stored energy or the energy of position q Chemical Potential energy is stored energy in bonds q Kinetic energy is the energy of motion q Thermal energy (heat) is an outward manifestation of movement at the atomic level E can occur as either heat transfer or work, or both.

Heat (q) Definition: The energy transfer due to a difference in temperature. Heat always flows from a warm object to a cold object.

Important to identify the system and the surroundings. q. System: the components of the chemical reaction. q. Surroundings: everything else in the universe.

First Law of Thermodynamics: The energy of the universe is constant. E = q + w E = change in system’s internal energy q = heat w = work Flow of heat between the system and surroundings Always defined from the point of view of the system 6

Exothermic Processes in which energy is released as it proceeds, and surroundings become warmer Heat flow out of the system q (negative) Reactants Products + energy Ca. O(s) + H 2 O(aq) Ca(OH)2 + 65. 2 k. J energy H = -65. 2 k. J **thermochemical equation**

Endothermic Processes in which energy is absorbed as it proceeds, and surroundings become colder Heat flows into the system q (positive) Reactants + energy Products 129 k. J + 2 Na. HCO 3(s) Na 2 CO 3(s) + H 2 O(g) + CO 2(g) H = 129 k. J

Heat of the Reaction Extensive property-depends on the amount of the substance. 129 k. J + 2 Na. HCO 3(s) Na 2 CO 3(s) + H 2 O(g) + CO 2(g) H = 129 k. J **Double the moles of Na. HCO 3 the heat of the reaction doubles. **

Exothermic versus endothermic. Endothermic * Exothermic* products have Higher PE reactants have Higher PE (Higher PE – weaker bonds) 10

Units for Measuring Heat The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius. 1 Calorie (food energy) = 1 kcal = 1000 calories The Joule is the SI system unit for measuring heat:

Heat Capacity • Amount of heat needed to increase the temperature of an object exactly 1 C. – Depends on mass and chemical composition Example: – drop of water and cup of water. – Beach sand heats up quickly, but the ocean water needs all summer to raise its temp just a few degrees.

Specific Heat (C) The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g· C) Water (liquid) 4. 18 Ethanol (liquid) 2. 44 Water (solid) 2. 06 Water (vapor) 1. 87 Aluminum (solid) 0. 897 Carbon (graphite, solid) 0. 709 Iron (solid) 0. 449 Copper (solid) 0. 385 Mercury (liquid) 0. 140 Lead (solid) 0. 129 Gold (solid) 0. 129

Now that we know what heat is, its good to know how much heat(q) can be given off by a system and potentially be used by another system. This is know as… Enthalpy: H (change in enthalpy of a system when heat is given off or absorbed) We will look at both chemical reactions and change in state.

Is H = q ? Enthalpy = H = E + PV work=(-P∆V) E = q + w E = q - P ∆V q = E + P ∆V therefore… H = q. P (system) at constant pressure

Calculations Involving Specific Heat OR C = Specific Heat q = Heat lost or gained T = Temperature change m = Mass Units: J/(g C) or cal/(g C)

Sample Problem The temperature of a 95. 4 -g piece of copper increases from 25. 0 C to 48. 0 C when the oopper absorbs 849 J of heat. What is the specific heat of copper? Given: m= 95. 4 g T = 48. 0 C – 25. 0 C q = 849 J Unknown: Ccu = ? J/(g C)

Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter Constant-pressure calorimetry • Used to determine heat of a reaction in solution at constant pressure. Enthalpy (ΔH)

Calorimetry at constant volume • No work is done (V must change for PV-work to be done) • ΔE = q + w = q v ΔE = ΔT x heat capacity of calorimeter

Sample Problem When 25. 0 m. L of water containing 0. 025 mol HCl at 25. 0 C is added to 25. 0 m. L of water containing 0. 025 mol Na. OH at 25. 0 C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change in k. J during this reaction if the highest temperature observed is 32. 0 C. Assume the densities of the solutions are 1. 00 g/m. L. Given: Cwater = 4. 18 J/(g C) Vfinal = 50. 0 m. L T = 7. 0 C Unknown: H = ? ? H = -1500 J

Enthalpy (Heat) of Combustion ΔH for the complete burning of one mole of a substance

Heat of fusion and solidification What happens when you place an ice cube on a table in a warm room? absorb or release heat? absorb

Latent Heat of Phase Change Molar Heat of Fusion The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. Molar Heat of Solidification The energy that must be removed in order to convert one mole of liquid to solid at its freezing point. Temperature remains constant all heat is used to break bonds △Hfus =-△Hsolid

Latent Heat of Phase Change #2 Molar Heat of Vaporization The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. Molar Heat of Condensation The energy that must be removed in order to convert one mole of gas to liquid at its condensation point. △Hvap =-△Hcond

Water phase changes constant Temperature remains _____ during a phase change. Heat of vaporization 40. 67 k. J/mol Heat of fusion 6. 01 k. J/mol

Latent Heat – Sample Problem: The molar heat of fusion of water is 6. 009 k. J/mol. How much energy is needed to convert 60 grams of ice at 0 C to liquid water at 0 C? Mass of ice Molar Mass of water Heat of fusion

Molar Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a mole of solute dissolves in a solvent. Substance Heat of Solution (k. J/mol) Na. OH -44. 51 NH 4 NO 3 +25. 69 KNO 3 +34. 89 HCl -74. 84

Example: How much energy is needed to heat 55 grams of water ice from a temperature of -150 C to steam at a temperature of 1500 C?

Solution: This problem requires several steps: Step 1: Heat the ice from -150 C to 00 C ∆H = m. Cp∆T = (55 grams)(2. 03 J/g 0 C)(150) = 1, 675 J Step 2: Undergo the phase change from a solid to liquid. ∆H = n∆Hfus = (3. 06 mol)(6. 01 k. J/mol) = 18. 39 k. J

Step 3: Heat the liquid from 00 C to 1000 C ∆H = m. Cp∆T = (55 grams)(4. 184 J/g 0 C)(1000) = 23, 012 J Step 4: Undergo the phase change from a liquid to gas. ∆H = n∆Hvap = (3. 06 mol)(40. 7 k. J/mol) = 124. 54 k. J

Step 5: Heat the steam from 1000 C to 1500 C ∆H = m. Cp∆T = (55 grams)(2. 01 J/g 0 C)(500) = 5, 528 J And when you’re done with all of this, just add up the different values you found above: ∆Htotal = 1. 68 k. J + 18. 39 k. J + 23. 01 k. J + 124. 54 k. J + 5. 53 k. J ∆Htotal = 173. 15 k. J

Hess’s Law Enthalpy Reactants Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps. Allows you to determine H of a reaction indirectly. 1. Reactions yield products in addition to the one of interest. 2. Reactions that are far too slow.

Oxidation of N 2 to produce NO 2 N 2(g) + 2 O 2(g) 2 NO 2(g) ΔH 1=68 k. J Or. . N 2(g) + 2 O 2(g) 2 NO(g) + O 2(g) N 2(g) + 2 O 2(g) 2 NO 2(g) ΔH 2=180 k. J ΔH 3=-112 k. J ΔH=68 k. J 33

The principle of Hess's law. The same change in enthalpy occurs when nitrogen and oxygen react to form nitrogen dioxide, regardless of whether the reaction occurs in one (red) or two (blue) steps. 34

Calculations via Hess’s Law 1. If a reaction is reversed, H is also reversed. N 2(g) + O 2(g) 2 NO(g) N 2(g) + O 2(g) H = 180 k. J 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6 NO(g) 3 N 2(g) + 3 O 2(g) H = 3( 180 k. J) H = 540 k. J

Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 k. J/mol) and diamond (-396 k. J/mol), calculate ∆H for the conversion of diamond to graphite: Cdiamond(s) Cgraphite(s)

Cdiamond(s) Cgraphite(s) +O 2 (g) CO 2 (g) Cdiamond(s) + O 2 (g) CO 2 (g) ∆H=-393. 5 k. J ∆H=-395. 4 k. J CO 2 Cgraphite (s) + O 2 (g) Cdiamond (s) +O 2 (g) CO 2 (g) ∆H=-(-393. 5 k. J) ∆H=-395. 4 k. J Cdiamond(s) Cgraphite(s)

Try this one!! (B 2 H 6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for the US space program. Calculate ∆H for the synthesis of diborane from its elements, according to the equation 2 B(s) + 3 H 2 (g) B 2 H 6(g)

Standard Enthalpies of Formation ΔHf° change in enthalpy that accompanies the formation of 1 mol of compound from its elements w/all substances in their standard states Provides an alternative to Hess’s Law Δ Hf ° = ΔHf° (products) - ΔHf°(reactants)

Standard States Compound 4 For a gas, pressure is exactly 1 atmosphere. 4 For a solution, concentration is exactly 1 molar. 4 Pure substance (liquid or solid), it is the pure liquid or solid. Element 4 The form [N 2(g), K(s)] in which it exists at 1 atm and 25°C.