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Understanding Energy: Types, Conservation, and Thermal Reactions

This comprehensive overview explores the concept of energy, its various forms such as potential and kinetic energy, and the law of conservation of energy. It delves into calorimetry, specific heat, thermochemical equations, and the thermochemistry of chemical reactions. With clear examples, such as the combustion of glucose and calculations involving heat transfer, this guide provides insights into how energy is absorbed or released during physical and chemical processes. Essential for chemistry students, it aids understanding of energy transformations and their implications in thermodynamics.

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Understanding Energy: Types, Conservation, and Thermal Reactions

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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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