Thermochemistry CHAPTER 17

Thermochemistry Thermochemistry: study of energy changes During chemical reactions During phase changes

How does Heat differ from Temperature? Temperature: average kinetic energy of particles The faster the particles are moving, the higher the temperature. Heat (q): transfer of kinetic energy Heat flows from high temp  low temp Units: Joules (SI unit) or calories (kCal = Cal in foods) 1 calorie = 4.184 J

Joule/Calorie Conversion Practice Example: 9.6 calories = ______ Joules 450 Joules = ______ calories 53 calories = ______ Joules 0.9 Joules = _______ calories 9.6 calories 4.184 Joules = 40.2 Joules 1 calorie 107.5 221.8 0.22

Energy Changes in Chemical Reactions Exothermic Reactions – give off energy A + B  C + D + energy Endothermic Reactions – absorb energy energy + A + B  C + D

Endothermic or Exothermic? Reactants  Products + Energy Reactants + Energy  Products 2NaHCO3 (s) + 129 kJ  Na2CO3 (s) + H2O (g) + CO2 (g) CaO (s) + H2O (l)  Ca(OH)2 (s) + 65.2 kJ baking bread campfire burning fuel photosynthesis breaking down carbohydrates for energy Electrolysis of water

Exothermic Reactions Exothermic Reactions: net release of energy (heat) “feels hot to the touch” ex. combustion reactions PE of Reactants > PE of Products (b/c energy is lost) Activation Energy Potential Energy (J) Time Reactants Enthalpy Products

Endothermic Reactions Endothermic Reactions: net absorption of energy (heat) “Feels cold to the touch” ex. decomposition of water using electrolysis PE of Reactants < PE of Products Activation Energy Potential Energy (J) Time Products Enthalpy Reactants

Activation Energy & Catalysts Activation energy: The amount of energy needed to get a reaction started. Required for BOTH endothermic and exothermic reactions.  

How does a catalyst speed up a reaction? Catalyst: speeds up the reaction, but is not used up during the reaction. Activated Complex Activation Energy Potential Energy (J) Time Reactants Catalyst Reduces Activation Energy Products

Enthalpy Enthalpy (ΔH): the difference in energy between the products and reactants in a chemical change ΔHreaction = Hproducts - Hreactants ΔHreaction is negative for exothermic reactions. Why? ΔHreaction is positive for endothermic reactions. Why? Nature favors lower energy  exothermic Example: 2H2 (g) + O2 (g)  2H2O (l) ΔH = -572 kJ 2H2O (l)  2H2 (g) + O2 (g) ΔH = +572 kJ

Law of Conservation of Energy Law of Conservation of Energy: the energy released when elements come together to form a compound is equal to the energy required to decompose the compound into its elements. Example: 2H2 (g) + O2 (g)  2H2O (l) ΔH = -572 kJ 2H2O (l)  2H2 (g) + O2 (g) ΔH = +572 kJ

Heating Curve Energy changes also occur in physical processes. L  G (vaporization) Gas boiling point Liquid **NOTE – Temp DOES NOT Change during a phase change!!!! BECAUSE – All the energy goes into breaking the bonds… NOT into speeding up the particles. S  L (melt) melting point Solid

Cooling Curve

Endothermic or Exothermic? Phase changes are physical changes (no new substance is formed)… Melting Freezing Boiling/Evaporation/Vaporization Condensation

Specific Heat Specific Heat (C): amount of energy (in Joules) needed to increase the temperature of 1 gram of a substance by 1oC. See Reference Tables q = mCΔT (+q = endothermic, -q = exothermic) q = heat (joules or calories) m = mass (grams) C = specific heat  ΔT = change in Temperature = Tf - Ti J goC

Remember!!! Temperature does not change during a phase change! (Energy goes into pulling molecules farther apart.) Therefore, we cannot use this equation for phase changes: q = mCΔT

Latent Heats Heat of Vaporization (Hv): amount of energy absorbed when one gram of a liquid is changed into a vapor (or, the amount of energy released when 1 gram of a vapor is changed into a liquid). q = mHv Hv = heat of vaporization (J/g) Hv of water = 2260 J/g

Heat of Fusion Heat of Fusion (Hf): amount of energy needed to convert 1 gram of a solid into a liquid. (Or the heat released when 1 gram of liquid  solid.) q = mHf Hf = heat of fusion (J/g) Hf of water = 334 J/g

Example Problems – Level One 1. How much heat is added if 100 grams of liquid water increases in temperature from 30oC to 70oC? 2. How much heat is absorbed if 200 g of ice increases in temperature from -15oC to -5oC? q = mCΔT q = (100 g)(4.18 J/goC)(70o – 30o) q = 16,720 J q = (200 g)(2.05 J/goC)(-5o – -15oC) q = 4100 J

(cont’d) 3. How much heat is released if 80 grams of water vapor is decreases in temperature from 150oC to 125oC? 4. How much heat is absorbed when 30 g of ice is changed into liquid water at 0oC? q = (80 g)(2.02 J/goC)(125o – 150o) q = - 4040 J q = mHf q = (30 g)(334 J/g) q = 10,020 J

Continued… 5. How much heat is released when 50 grams of water vapor is changed into liquid water at 100oC? q = mHv q = (50 g)(2260 J/g) q = 113,000 J  -113,000 J

Example Problems – Level Two 6. How much heat is absorbed if 30 grams of ice at -10oC is converted into liquid water? 2 steps: (1) heat the ice and (2) melt ice  liquid (1) q = mCΔT q = (30g)(2.05 J/goC)(0o – -10o) q = 615J (2) q = mHf q = (30g)(334J/g) q = 10,020J Add both steps together! Total heat = 615J + 10,020J Total heat = 10,635 J

Continued… 7. How much heat is absorbed if 45 grams of water at 80oC is converted into steam at 105oC? 3 steps: (1) heat water, (2) water  steam, (3) heat steam (1) q1 = mCΔT q1 = (45g)(4.18J/goC)(100o – 80o) q1 = 3,762 J Add all 3 together! qtotal = q1 + q2 + q3 qtotal = 3,762J + 101,700J + 454.5 J qtotal = 105,916.5 J q2 = mHv q2 = (45g)(2260J/g) q2 = 101,700 J (3) q3 = mCΔT q3 = (45g)(2.02J/goC)(105o – 100o) q3 = 454.5 J

(cont’d) 8. How much heat is released if 500 grams of vapor at 120oC changes to liquid water 70oC? 3 steps: cool the vapor, condensation, cool the liquid (1) q1 = mcpΔT q1 = (500g)(2.02J/goC)(100o – 120o) q1 = -20,200J q2 = mHv q2 = (500g)(2260J/g) q2 = 1,130,000J  - 1,130,000J Add together: -20,200J -1,130,000J + -62,700J -1,212,900J (3) q3 = mcpΔT q3 = (500g)(4.18J/goC)(70o – 100o) q3 = -62,700J

Calorimetry Calorimetry: the precise measurement of the heat flow out of a system for chemical and physical processes.

Calorimeter Example Problems 1. How much heat is released by the reaction if 10 grams of water is heated from 20oC to 60oC. q = mCΔT q = (10g)(4.18J/goC)(60o – 20o) q = 1,672J  amount absorbed by water. Therefore, -1,672J was released by the reaction. 2. What is the mass of the water if a release of 650 J of heat causes the water to increase in temperature by 15oC? (Water absorbed 650J of heat from the reaction.) 650J = m(4.18J/goC)(15oC) m = 10.4g 1. -1672 Joules (released by reaction) 2. 43 kg 3. 15 degrees Celsius

Example Problems 3. What is the final temperature of 15 g of water if the water starts at 20oC and loses 300 J of heat? **Losing heat means q is negative! -300J = (15g)(4.18J/goC)(Tf – 20o) Tf = 15.2oC 4. Determine the substance if a sample of 20 grams increases in temperature from 10oC to 28oC when it absorbs 368.3 Joules of energy. 368.3J = 20g(C)(28o – 10o) C = 4.5 J/goC  Magnesium (see ref.pack)

Entropy Entropy (S): a measure of the disorder of the particles in a system. Systems tend to become more disordered. Entropy increases when: solid  liquid  gas (melting) # moles increase in a reaction (1A + 2B  3C + 3D) Temperature increases

Predicting Changes in Entropy Predict whether ΔSsystem will be positive or negative: 1. H2O (l)  H2O (g) [positive] 2. CH3OH (s)  CH3OH (l) 3. Increase the temperature of a substance

(cont’d) 4. Dissolving of a gas in a solvent CO2 (g)  CO2 (aq) [negative] 5. What happens when the number of gaseous product particles is greater than the number of gaseous reactant particles? 2SO3 (g)  2SO2 (g) + O2 (g) [positive] 6. Generally, when a solid or liquid dissolves to form a solution? NaCl (s)  Na+ (aq) + Cl- (aq)

Spontaneous Reactions A reaction is spontaneous if: Energy decreases (energy is released = exothermic) AND Disorder increases (positive entropy). Ex. Wood burning