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Unit 11: Thermochemistry

Unit 11: Thermochemistry. It’s getting hot in here. You can bet there is an equation for that. Heat (q). Dwayne Wade’s team Movie Cold Whaa ? “The energy transferred between samples of matter because of a difference in their temperatures.”

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Unit 11: Thermochemistry

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  1. Unit 11: Thermochemistry It’s getting hot in here. You can bet there is an equation for that.

  2. Heat (q) • Dwayne Wade’s team • Movie • Cold • Whaa? • “The energy transferred between samples of matter because of a difference in their temperatures.” • Heat always flows from high temperatures to low temperatures. • Cold is a lack of energy, but 1.0 *C water will transfer its heat to -56.5 *C dry-ice!!

  3. Heat in Chemical Reactions • As chemicals react over time, energy is either released to the surroundings or absorbed by the reactants. • Both possibilities are natural occurrences, even though our experiences so far lead us to believe only lower-energy states are desired.

  4. Exothermic • EXOthermic • Exo (like exit) means out/leaves • Thermic means heat • “heat flows from the reaction system into the surroundings.” • CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) • This reaction occurs in the Bunsen Burner and is obviously exothermic. • Combustion releases heat • q < 0

  5. Endothermic • ENDOthermic • Endo (like endoscopy) means within • Thermic means heat • “heat flows from the surroundings into the reaction system.” • H2O (s)  H2O (l) • Ice cubes are cold, liquid water is (usually) not as cold. • Melting absorbs heat. • q > 0

  6. Measuring heat flow • It is not always easy to tell if an object is absorbing or releasing heat. • The actual quantity of energy is dependent on: • Nature of the material* • Mass of the material • Magnitude of temperature change

  7. Magnitude of heat flow • In order to calculate quantity we need an equation: • “Specific heat” equation • q = mc∆T • q = heat flow (J) • m = mass (of material, g) • c = specific heat (nature of material, J/(g*K)) • ∆T = magnitude of temperature change (*C or K) • ∆T = Tfinal – Tinitial • Negative T values are acceptable • Note: q = nc∆T (n = moles)

  8. Calculating heat flow • A 4.0 g sample of glass was heated from 274 K to 41 *C. During the process, it absorbed 32 J of heat. • A) What is the specific heat of this type of glass? • 41 *C = 314 K • 32 J = c(4.0 g)(40 K) • c = 32 J / (4.0 g)(40 K) • c = 0.20 J/(g*K) • B) How much energy will the glass sample gain when it is heated from 41 *C to 344 K? • q = (0.20 J/(g*K))(4.0 g)(344 K – 314 K) • q = 24 J

  9. Calculating heat flow II • 1) Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 20 *C to 40 *C. • 96J = c(35 g)(20 *C) • c = 96 J / (35 g)(20 *C) • c = 0.14 J/(g*C) • B) If 980 kJ of energy are added to 6.2 L (1.0 mL = 1.0 g) of water (4.18 J/(g*K) at 291 K, what will the final temperature of the water be? • 980000 J = (4.18 J/(g*K))(6200 g)(Tf K – 291 K) • Tf = 329 K

  10. Calculating heat flow III • A piece of copper alloy with a mass of 85.0 g is heated from 30 *C to 45 *C. In the process, it absorbs 523 J of energy as heat. • A) What is the specific heat of this copper alloy? • 523 J = c(85.0 g)(15 *C) • c = 523 J / (85.0 g)(15 *C) • c = 0.41 J/(g*C) • B) How much energy will the same sample lose if it is cooled from 45 *C to 25 *C? • q = (0.41 J/(g*C))(85.0 g)(-20 *C) • q = 697 J

  11. ∆H • Energy absorbed as heat is represented as ∆H. • H = “enthalpy” • Which cannot be directly measured • But changes (∆) can be • “Enthalpy changes” – the amount of energy absorbed by a system as heat during a process at constant pressure. • ∆H = Hproducts – Hreactants • ∆H can only be calculated using moles

  12. Enthalpy of Reactions • The quantity of energy transferred as heat during chemical reactions. • Example • 2H2 (g) + O2 (g)  2H2O (g) • Is energy released or absorbed in this reaction? • We know from experience it is released (Hindenburg), but we can modify the equation to show it. • 2H2 (g) + O2 (g)  2H2O (g) + 483.6 kJ • “When 2 moles of hydrogen gas react with 1 mole of oxygen gas, it forms 2 moles of water and 483.6 kJ of energy.”

  13. Enthalpy of Reactions • 2H2 (g) + O2 (g)  2H2O (g) + 483.6 kJ • What happens if I double the amount of reactants? • 4H2 (g) + 2O2 (g)  4H2O (g) + ________? • 483.6 x 2 = 967.2 kJ energy • What happens if I halve the amount of reactants? • H2 (g) + ½O2 (g)  H2O (g) + ________? • 483.6 x ½ = 241.8 kJ energy

  14. Enthalpy of Reactions II • 2H2 (g) + O2 (g)  2H2O (g) + 483.6 kJ • Thermochemical equations usually use ∆H instead of including the energy directly • Since energy is released, this means the reactants lost (-) energy. • 2H2 (g) + O2 (g)  2H2O (g) ∆H = -483.6 kJ

  15. Exo and Endothermic • Since ∆H represents a change in energy, it can be used to identify exothermic and endothermic reactions. • - ∆H = exothermic • Energy on the product side • Energy is released • + ∆H = endothermic • Energy on the reactant side • Energy is absorbed

  16. Exo and Endothermic II • It is possible to track the changes of enthalpy on a graph. From these, we can determine if the process is exothermic or endothermic and also how much energy is released or absorbed too.

  17. Exothermic Graphs • Exothermic process result in energy being RELEASED (lost) so the overall energy (enthalpy) should decrease.

  18. Endothermic Graphs • Endothermic processes result in energy being ABSORBED (gained) so the overall energy (enthalpy) should increase

  19. Potential Energy Graphs • Information can be gained from these graphs such as • 1) Whether the process is exothermic or endothermic. • 2) The activation energy of the process • 3) Overall enthalpy or energy (∆H or ∆E)

  20. Exothermic or Endothermic Graph • This one is easy: • Exothermic – the graph ends lower than it started • Endothermic – the graph ends higher than it started.

  21. Activation Energy • This is the amount of energy required to get the reaction to form products • It is the PEAK (highest point) on the graph minus the starting point • Graph 1 is 100 kJ • Graph 2 is 80 kJ

  22. Overall enthalpy (∆H) or energy (∆E) • We will use the terms interchangeably • This measures the change from START (reactants) to END (products) • Graph 1 is -150 kJ and it is negative because it is exothermic! • Graph 2 is +50 kJ and it is positive because it is endothermic!

  23. ∆H Calculations • ∆H is generally affected by the amount of moles of reactants used or of products formed. • H2 (g) + Cl2 (g)  2HCl (g) ∆H = -185 kJ • What is ∆H when 1 mole of HCl is formed? • -185 kJ x ½ = -92.5 kJ • What is ∆H when 1.00 g of Cl2 reacts? • -185 kJ x 1.00 g / 70.90 g Cl2 = -2.61 kJ

  24. ∆H Calculations II • Using a coffee-cup calorimeter, it is found that when a 24.6 g ice cube melts, it absorbs 8.19 kJ of heat. Calculate ∆H for the phase change: • H2O (s)  H2O (l) ∆H = ? • 18.02 g H2O / 24.6 g H2O x 8.19 kJ = 6.00 kJ • ∆H = 6.00 kJ • H2O (s)  H2O (l) ∆H = 6.00 kJ

  25. ∆H of Formation - ∆Hf • In multi-step reactions, the overall enthalpy change is the sum of the enthalpy of the individual steps. • Overall: NO (g) + ½ O2 (g)  NO2 (g) • Given: • ½ N2 (g) + ½ O2 (g)  NO (g) ∆H = +90.29 kJ • ½ N2 (g) + O2 (g)  NO2 (g) ∆H = +33.2 kJ • It may be necessary to “reverse” a reaction so the reactants and products match the overall.

  26. ∆Hf - II • Overall: NO (g) + ½ O2 (g)  NO2 (g) • NO (g)  ½ N2 (g) + ½ O2 (g) ∆H = -90.29 kJ • ½ N2 (g) + O2 (g)  NO2 (g) ∆H = +33.2 kJ • The ∆H must also be “reversed” • What can we cancel? • “½ N2 (g)” • “½ O2 (g)” • Leaves us with ½ O2 (g) on the reactant side

  27. ∆ Hf - III • NO (g)  ½ N2 (g) + ½ O2 (g) ∆H = -90.29 kJ • ½ N2 (g) + O2 (g)  NO2 (g) ∆H = +33.2 kJ • NO (g) + ½ O2 (g)  NO2 (g) • ∆Hf = -90.29 kJ + 33.2 kJ = -57.1 kJ

  28. Notice anything? • Both of the following reactions are endothermic. • H2O (s)  H2O (l) • 2NH4NO3 (s)  2N2 (g) + 4H2O (l) + O2 (g) • Why do they occur at all?

  29. Entropy (S) • Both of those reactions occur because the arrangement of atoms becomes less ordered. • Entropy (S) – measure of the degree of randomness of the particles, such as molecules, in a system. • Nature seems to prefer systems where there is an INCREASE of randomness.

  30. Entropy and States • Solids – particles are in fixed positions, not very random. Low entropy. • Liquids – the particles are still very close, but no longer in fixed positions. Entropy higher than solid. • Gases – the particles are not very close nor are they in fixed positions. Entropy higher than liquid. • Generally, solid < liquid < gas (for a particular substance)

  31. Change in Entropy (∆S) • Just like ∆H, changes in entropy (∆S) can be measured. • Increase in entropy • Randomness increases, +∆S • Decrease in entropy • Randomness decreases, -∆S • Units • kJ/(mol*K)

  32. Predicting ∆S • Predict the whether the value of ∆S will be greater or less than 0. • KNO3 (s)  K+ (aq) + NO3- (aq) • ∆S > 0 • 3H2 (g) + N2 (g)  2NH3 (g) • ∆S < 0 • 2Mg (s) + O2 (g)  2MgO (s) • ∆S < 0 • C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (g) • ∆S > 0

  33. Free Energy (G) • Bye bye utility bills!! • Sadly, no  • The combined enthalpy-entropy function is free energy (G); also known as Gibbs free energy. • Reactions can be spontaneous, yet endothermic, and have positive entropy. • Since that logically makes little sense, free energy indicates that neither enthalpy nor entropy alone determine spontaneity of a reaction.

  34. Free Energy Equation • ∆G = ∆H - T∆S • This equation can predict the spontaneity of a reaction (will it occur?) • ∆G = negative • Spontaneous at given conditions (temperature) • ∆G = positive • Does not occur at given conditions (temperature)

  35. Free Energy Equation II • ∆G = ∆H - T∆S • If: -∆H and +∆S = • ∆G is negative, always spontaneous • If: -∆H and -∆S = • ∆G is sometimes negative, spontaneous at low temperatures • If: +∆H and +∆S = • ∆G is sometimes negative, spontaneous at high temperatures • If: +∆H and -∆S = • ∆G is positive, reaction does not occur

  36. Determining Spontaneity • The following reaction, C2H4 (g) + H2 (g)  C2H6 (g), occurs at room temperature (25 *C) • ∆S = -0.1207 kJ/(mol*K) • ∆H = -136.9 kJ • Is this reaction spontaneous? • ∆G = -136.9 kJ – 298 K[-0.1207 kJ/(mol*K)] • ∆G = -136.9 kJ + 35.96 kJ/mol • ∆G = -100.9 kJ • The reaction will occur at 25 *C (298 K).

  37. Determining Spontaneity II • The following reaction, CH4 (g) + H2O(g)  CO (g) + 3H2 (g), occurs at room temperature (25 *C) • ∆S = +0.215 kJ/(mol*K) • ∆H = +206.1 kJ • Is this reaction spontaneous? • ∆G = 206.1 kJ – 298 K[+0.215 kJ/(mol*K)] • ∆G = 206.1 kJ – 65.07 kJ/mol • ∆G = +142.03 kJ/mol • The reaction will occur at 25 *C (298 K).

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