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Energy

Energy. Energy - Ability to do work or produce heat PE - stored in bonds KE - E of motion (constant motion of atoms or molecules  T) Law of Conservation of Energy - in any chemical rxn or physical process, E is converted to other forms – not created or destroyed.

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Energy

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  1. Energy • Energy - Ability to do work or produce heat • PE - stored in bonds • KE - E of motion (constant motion of atoms or molecules T) • Law of Conservation of Energy - in any chemical rxn or physical process, E is converted to other forms – not created or destroyed. • Chemical Potential Energy - Stored energy in bonds – based on composition and type of atoms

  2. Energy • Heat (q) – process of flowing from a warmer object to a cooler object. • calorie (cal) – amount of heat required to raise T of 1 g of pure water 1C • Calorie – food calories • 1 Cal = 1000 cal = 1 kcal • Joule (J) – SI unit for heat and energy • 1 J = 0.2390 cal • 1 cal = 4.184 J • Be able to convert between the two units • Convert 230 cal to J • Convert 95 J to cal

  3. Energy • Specific heat – heat to raise T of 1 g of substance 1C • q = cmT • heat = specific heat constant * mass * (Tfinal – Tinitial) • If the T of 34.4 g of ethanol increases from 25C to 78.8C, how much heat has been absorbed? (specific heat of ethanol is 2.44 J/gC) • Read pg 495 about using the Sun’s energy

  4. Heat in Chemical Reactions and Processes • Based on law of Conservation of Energy • Calorimeter – insulated device used for measuring amount of heat absorbed or released during a chemical or physical process • A calorimeter is used to determine specific heat of metal (q = cmT) • Hint: first determine the heat gained by the water, then use that to calculate c for the metal

  5. Heat in Chemical Reactions and Processes • 50 g of metal is heated to 115C. When transferred to 125g H2O in a calorimeter, the water T raises from 25.6C to 29.3C. Assume no heat is lost. Find the specific heat of the metal. • A piece of metal with a mass of 4.68 g absorbs 256J of heat, when the T is increased by 182C. What is the specific heat of the metal?

  6. Heat in Chemical Reactions and Processes • Thermochemistry – study of heat that accompanies physical and chemical processes • System – part of universe that contains the process you wish to study • Surroundings – everything in universe that is outside your system • System + surroundings = universe • Exothermic – system loses heat • 4 Fe + 3 O2→ 2Fe2O3 + 1625 kJ • Endothermic – system gains heat • 27 kJ + NH4NO3 → NH4+ + NO3-

  7. Heat in Chemical Reactions and Processes • Enthalpy (H) – heat content of system at constant P • Hrxn – enthalpy or heat of reaction • Hrxn = Hfinal – Hinitial • Hrxn = Hproducts – Hreactants • Exothermic - H; H< 0 • 4 Fe + 3 O2→ 2Fe2O3H = -1625 kJ/mol • Endothermic + H; H> 0 • NH4NO3 → NH4+ + NO3- H = 27 kJ/mol • q = Hrxn at constant P

  8. Thermochemical Equations • Thermochemical equation: balanced chemical equation that includes the physical states of all reactants and products and the energy changes in H • The combustion of glucose (C6H12O6) releases 2808 kJ of energy. Write the thermochemical equation for the combustion of glucose. (Hint: what do all combustion reactions include?)

  9. Thermochemical Equations cont • Hcomb – enthalpy (heat) of combustion for burning 1 mole of substance • Standard enthalpy - H (the zero tells you the rxn was carried out under standard conditions (298 K and 1 atm) NOT STP (273 K and 1 atm)) • Hvap – molar enthalpy (heat) of vaporization – heat to vaporize 1 mole of liquid • Hfus – molar heat of fusion – heat to melt 1 mole of solid • Both are endothermic (require heat; +H)

  10. Thermochemical Equations cont • Hcond – molar enthalpy (heat) of condensation – heat released when 1 mole of a gas condenses • Hsolid – molar enthalpy (heat) of solidification – heat released to solidify 1 mole of liquid • Both are exothermic (release heat; -H) • H2O (l) → H2O (g) Hvap = 40.7 kJ/mol • H2O (s) → H2O (l) Hfus = 6.01 kJ/mol • Hvap = -Hcond • Hfus = -Hsolid

  11. Thermochemical Equations cont • How much heat is evolved when 54 g glucose (C6H12O6) is burned according to this equation? • C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) Hcomb = -2806 kJ/mol • Calculate the heat required to melt 25.7 g of solid methanol (CH3OH) at it’s melting point. (see the chart on pg502)

  12. Calculating Enthalpy Change • Hess’s Law – can add 2 or more thermal chemical equations to produce a final equation for a reaction, then the sum of the enthalpy rxn for the individual rxn is the enthalpy change for the final reaction. • You rearrange the equations to get the “target”. What ever you do to the equation, you do to the H. S (s) + O2 (g) → SO2 (g) H = -297 kJ 2 SO3 (g)  2 SO2 (g) + O2 (g) H = 198 kJ Target: 2 S (s) + 3 O2 (g)  2 SO3 (g) H = ?

  13. Calculating Enthalpy Change 2 H2 (g) + O2 (g)  2 H2O (l) H = -572 kJ H2 (g) + O2 (g)  H2O2 (l) H = -188 kJ Target: 2 H2O2 (l)  2 H2O (l) + O2 H = ? • Standard enthalpy (heat) of formation (Hf) – normal physical state of the substance at 1 atm and 298 K (25C) • Use them to find Hrxn • Hrxn = Hproducts - Hreactants • Look up Hf in chart on pg 510 • Hf of an element in its standard state is zero H2S (g) + 4 F2 (g)  2 HF (g) + SF6 (g) Hrxn = ? CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) Hrxn = ?

  14. Reaction Spontaneity • Spontaneous process – occurs with no outside intervention • May need E to start but will continue without help • Gas burning is spontaneous exothermic reaction CH4 (g) + 2O2 (g)  CO2 (g) + 2 H2O (l) H = -891 kJ • Reverse reaction is not spontaneous under ordinary conditions CO2 (g) + 2 H2O (l)  CH4 (g) + 2O2 (g) H = 891 kJ • Ice melts at room T – spontaneous endothermic H2O (s)  H2O (l) H = 6.01 kJ

  15. Reaction Spontaneity continued • Spontaneity determined by entropy (S) & enthalpy (H) • Entropy (S) – a measure of disorder of a system • Law of Disorder – systems tend to go to disorder • Have to put E in to get order Ssystem = Sproducts – Sreactants Sproducts > Sreactants Ssystem is + Sproducts < Sreactants Ssystem is –

  16. Reaction Spontaneity continued • Can predict Ssystem • Look at  of state l  g +Ssystem s  l +Ssystem * Ssystem increase is + • Dissolve gas in solvent -S CO2 (g)  CO2 (aq) Ssystem < 0 * gases have more S when in gas state • If no  state, the S  when the # of gaseous product particles is greater than the # of gaseous reactant particles 2 SO3 (g)  2 SO2 (g) + O2 (g) +Ssystem

  17. Reaction Spontaneity continued 4. Dissolving usually S NaCl (s)  Na+ + Cl- +Ssystem •  T results in S +Ssystem • Suniverse > 0 Suniverse = Ssystem + Ssurroundings Suniverse+ when: • Reaction is exothermic (-H); T = +Ssurroundings • If S then +Ssystem * Exothermic reactions with T = S are spontaneous

  18. Reaction Spontaneity continued • Free Energy – Gibbs Free energy - G • E available to do work

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