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Chapter 17

Chapter 17. Thermochemistry and Kinetics. Thermochemistry – study of transfer of energy as heat that accompanies chemical reactions and physical changes Heat and Temperature Temperature – measure of the average kinetic energy of particles (increase T, increase KE; object feels hotter)

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Chapter 17

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  1. Chapter 17 Thermochemistry and Kinetics

  2. Thermochemistry – study of transfer of energy as heat that accompanies chemical reactions and physical changes • Heat and Temperature • Temperature – measure of the average kinetic energy of particles (increase T, increase KE; object feels hotter) • Joules – SI unit of heat (and energy) kJ is also used commonly EX. 980 kJ  980,000 J

  3. Heat – the energy transferred between samples of matter because of a difference in their temperatures • Energy moves from warmer to cooler objects • Energy absorbed or released in a reaction (or change) is measure in a calorimeter. • Calorimeters are sealed and placed in H2O. The T change of H2O is used to determine the energy change inside the calorimeter.

  4. Heat Capacity and Specific Heat • Objects have different capacities for absorbing heat (energy). • Specific heat – the amount of energy required to raise the temperature of one gram of a substance by 1o C. • Units are J/g oC • Table on pg. 513 • Water has a HIGH specific heat (4.184 J/g oC)

  5. That means water can absorb (or release) a lot of energy before its T will begin to change. • Even on a hot day, lake water can still feel cold. q = c x m x ΔT q = energy c = specific heat m = mass ΔT = change in T (subtract the T’s)

  6. Ex. A 4.0 g sample of glass heated from 1o C to 41o C, and was found to have absorbed 32 J of energy. What is the specific heat of the glass? q = c x m x ΔT ΔT = 41oC – 1oC = 40oC 32 J = c (4.0 g)(40oC) 32 J = c (160 g oC) c = 0.2 J/goC

  7. Your Turn! • Determine the specific heat of a material if a 35 g sample absorbed 48J as it was heated from 293K to 313K. • If 980 kJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be?

  8. Heat of Reaction – amount of energy absorbed or released during a chemical reaction • When energy is released (a product) in a reaction, it is an exothermic reaction ex. 2 H2 + O2 2 H2O + 483.6 kJ If 4 moles of H2O formed then 967.2 kJ would be released (2 x 483.6)

  9. When energy is absorbed (added to the reactants) it is an endothermic reaction. • Energy absorbed or released during a reaction is represented by ΔHrxn • H stands for enthalpy – heat content of a system • ΔH is the change in heat during the reaction: ΔH = Hproducts – Hreactants

  10. ΔH is positive for endothermic reactions (heat added to reactants) ex. 2 H2 + O2 H2O ΔH = +483.6 kJ ΔH is negative for exothermic reactions (reactants losing heat) 2 H2O  2 H2 + O2ΔH = -483.6 kJ

  11. Heat of Formation – energy absorbed or released when 1 mole of a compound is formed from its elements • Designated as ΔHfo Ex. What would ΔHfo be for H2O(g)? 2 H2 + O2 2 H2O ΔH = -483.6 kJ That’s for 2 moles of H2O; find ΔH for 1 mole: -483.6 kJ = - 241.8 kJ = ΔHfo 2

  12. Heat of Combustion – energy released as heat by complete combustion of 1 mole of a substance • (remember combustion is adding O2) ex. C3H8 + 5O2 3CO2 + 4H2O ΔHocomb = -2219.2 kJ Combusting one mole of C3H8 releases 2219.2 kJ (because ΔHocomb is negative, it releases energy – exothermic)

  13. Heat of combustion is defined in terms of one mole of reactant, whereas the heat of formation is defined in terms of one mole of product.

  14. 1. Identify each of the following reactions as exothermic or endothermic. • CH4(s) + 2O2(g) → CO2(g) + 2H2O(l) ΔH° = -890 kJ • 2HCl(g) → H2(g) + Cl2(g) ΔH° = 185 kJ • 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) ΔH° = -1169 kJ

  15. Rate of Reactions • Kinetics – study of reaction rates • Rate Influencing Factors • for reactions to occur, particles must come into contact with each other (collide) in a favorable way. • anything that changes the frequency of collisions or the efficiency of collisions will change the rate of the reaction

  16. Surface Area • In homogeneous mixtures (dissolved) particles mix and collide freely so those reactions occur rapidly • Heterogeneous mixtures can only react where the 2 phases are in contact with each other; so the surface area of a solid is a factor; • Increase surface area, increase rate

  17. Temperature • Increasing T increases KE of particles (move faster); therefore there will be more collisions and rate will increase • Decreasing T decreases KE and number of collisions, so rate decreases • Concentration - The greater the concentration of a substance, the more particles there will be; the more particles, the more collisions and increase in rate

  18. Lowering the concentration has the opposite effect on the rate. • Presence of Catalysts • Catalyst – changes the rate of a reaction without being used up (speeds up the reaction without interfering) • They are written over the arrow MnO2 2 H2O2----- > 2 H2O + O2

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