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Warmup : (3 min)

Warmup : (3 min). What is the difference between temperature and heat? Do you think heat flows from cold to hot or from hot to cold? Give an example. Energy and Enthalpy. The First Law of Thermodynamics. states that energy:  can NOT be created or destroyed

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Warmup : (3 min)

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  1. Warmup: (3 min) • What is the difference between temperature and heat? • Do you think heat flows from cold to hot or from hot to cold? Give an example.

  2. Energy and Enthalpy

  3. The First Law of Thermodynamics states that energy: can NOT be created or destroyed can be converted from one type (potential, kinetic, thermal) to the other can be exchanged between the system and the surroundings

  4. Energy (E): ability to do work or produce heat In chemistry, we talk about work and energy in terms of rearranging atoms and molecules by forming and breaking bonds. Heat (q): thermal energy “COOL” IS NOT A THING. IT IS AN ABSENCE OF HEAT

  5. Heat flows from HOT to COLD

  6. are different. Temperature: -measures the amount of thermal energy (heat) in a system -depends on how fast the molecules are moving (KE) TK = T°C + 273 average KE of a sample of molecules T°C = (TF – 32)(5) 9

  7. Energy Units • the Joule(J) or kJ • the calorie (cal) (amount of energy needed to raise the temperature of 1 gram of water by 1 °C) This is a Calorie (kcal) NOT a calorie!

  8. One serving of yogurt contains 6.01 x 104calories. Convert this to kiloJoules of energy. 1 kJ = 1000 J 1 cal = 4.184 J 1000 cal = 1 Cal 6.01 x 104 cal (4.184 J) (1 kJ) 0.251 kJ = (1 calorie) (1000 J)

  9. How many Joules is 26000 Calories? 1 kJ = 1000 J 1 cal = 4.184 J 1000 cal = 1 Cal 26000 Cal (4.184 J) (1000 cal) 1.1 x 108 J = (1 Cal) (1 cal)

  10. Enthalpy (∆H) • Potential energy is stored in bonds • In chemical reactions, reactant bonds are broken (energy absorbed) and product bonds form (energy released) • ∆H: net amount of heat released or absorbed; depends on the strength of bonds made/broken

  11. Cold packs (read only) In cold packs, the chemical ammonium nitrate is often used because it absorbs a lot of heat when it dissolves in water.  Ammonium nitrate dissolves in water “endothermically.” Water and ammonium nitrate are kept in separate compartments in the pack until the pack is needed.  Then the chambers are broken and the ammonium nitrate dissolves in the water, absorbing heat and making the pack as cold as 0°C. 19.2 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) reactant

  12. Hot packs (read only)CaCl2(s) → Ca2+(aq) + 2Cl-(aq) +13.9kJ product In hot packs, calcium chloride or magnesium sulfate frequently are used because these chemicals dissolve in water exothermically.  In other words, they release a lot of heat when they are dissolved in water.  Hot packs can reach temperatures around 90°C.

  13. +251 kJ Xe(g) + 2F2(g) + 251 kJ --> XeF4(s) endo or exo? ∆H = XeF4(s) --> Xe(g) + 2F2(g) + 251 kJ endo or exo? ∆H = - 251 kJ When a reaction is reversed, the sign of ∆H is reversed too xenon tetrafluoride

  14. Identify each as an exothermic or endothermic reaction:*write just the answer • When solid KBr is dissolved in water, the liquid gets colder • 2KClO3(s) + 84.9 kJ  2KCl(s) + 3O2(g) • When concentrated sulfuric acid is added to water, the solution gets very hot • The net enthalpy change for a reaction is negative 768kJ (ΔH = - 768 kJ)

  15. Enthalpy and Stoichiometry Ex. 2C2H4O(l)+ 2H2O(l)2C2H6O(l) +O2(g) ∆H = 915.7 kJ • Considering the reaction above, how much energy is required to produce 0.467 moles of oxygen gas? 0.467 moles O2 b. If I want to completely burn 3.4 moles liquid C2H4O, how much energy must be present? 3.4 moles C2H4O (915.7 kJ) (1 mole O2 ) = 428 kJ (915.7 kJ) (2 mole C2H4O) = 1600 kJ

  16. Energy must be added or absorbed to break bonds and that energy is released when bonds are formed. You can calculate the total enthalpy of the reaction using the following formula: ΔH = bonds broken - bonds formed Calculate the change in energy that accompanies the reaction: H2(g) + F2(g)  2HF(g) Bond type Bond energy (kJ/mol ) H—H 432 F—F 154 H—F 565 ΔH = {breaking H-H + F-F} – (forming 2 H-F bonds) = [1(432) + 1(154)] - [2(565)] ΔH= -544 kJ More energy is released by forming the new bonds than was required to break the old bonds + 544 kJ

  17. 2H2O  2H2 + 1O2 broken formed 2(two O-H bonds) – {2(one H-H bond) and 1(one O=O bond)} 2(2 x 464kJ) - {2(436kJ) +1(498kJ)} ∆H = 1856 kJ - 1370 kJ + 486 kJ = + 486 kJ

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