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Thermochemistry

Thermochemistry. Energy comes in many forms :. kinetic energy is the energy of motion. potential energy is the energy stored in stressed objects, like a bent bow or compressed spring, or water at the top of a waterfall.

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Thermochemistry

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  1. Thermochemistry

  2. Energy comes in many forms: • kinetic energy is the energy of motion. • potential energy is the energy stored in stressed objects, like a bent bow or compressed spring, or water at the top of a waterfall. • chemical energy is the energy stored in chemical bonds, which is released when a chemical reaction, like burning, occurs. • nuclear energy is the energy stored within the nucleus of an atom: • Fission • Fusion

  3. Heat • Heat is thermal energy. • Heat is the product of motion. • All matter has heat in it, since all matter is made up of particles which move. • Something has no heat only if all motion has stopped inside it. This point is called absolute zero. • Absolute zero is the lowest temperature that can be reached: • it is equal to - 273 ° Celsius • it is also equal to 0 Kelvins (K)

  4. Heat • It is not possible to determine the amount of heat an object has, since it is tied up with the various types of energy. • Heat always flows from a warmer to a colder object. It continues to flow until both objects are at the same temperature. • Different masses or volumes of a substance require different amounts of heat to produce a similar rise in temperature.

  5. Heat • Different substances of the same mass require different amounts of heat to produce a similar rise in temperature: • Given equal inputs of heat energy, antifreeze will rise much more rapidly in temperature than water will. • Put another way, if water and antifreeze are heated, water requires more heat energy per unit mass to produce a similar temperature rise. • Put a third way, in equal masses of water and antifreeze at the same temperature, the water contains more heat.

  6. Specific Heat Capacity • Also called Specific Heat, or Heat Capacity. • Is the amount of heat required to heat 1 g or 1 kg of a substance by 1°C (or 1 K) • Measured in kiloJoules per kilogram degree Celsius (kJ/kg·C). • All substances have their own specific heat capacity; it is a characteristic physical property.

  7. Measuring Changes in Heat • Involves 3 things: • Heat Capacity – is a constant for each substance or phase of a substance. • Mass – the more mass, the more heat. • Temperature Change – can be positive or negative.

  8. Measuring Changes in Heat Amount of = Specific x Mass x Change in Heat HeatTemperature Q = C x M x ΔT Q = C x M x (Tf – Ti)

  9. Q = C x M x ΔT • This is an equation of four variables. If you have three of the variables you should be able to calculate the fourth. Remember a few things, though: • Make sure the units are correct. Include metric conversion. • Heat can go into or come out of a substance. • If heat goes in temperature goes up the change is positive. • If heat comes out temperature goes down the change is negative. • Use the specific heat table for the C values (see handout).

  10. Q = C x M x ΔT 1. Calculate the heat required to change the temperature of 1.5 kg water from 40.0°C to 80.0°C. • Since Q = C x M x (Tf – Ti) and the substance is water, therefore • C = 4.18 kJ/kgC • M = 1.5 kg • Tf = 80.0 C • Ti = 40.0 C • Q = (4.18 kJ/kgC)(1.5 kg)(80.0 C - 40.0 C) • = 250 kJ (2 significant digits)

  11. Q = C x M x ΔT 2. Calculate the heat required to change the temperature of 0.300 kg sucrose from 80.0°C to 19.0°C. • C = 1.25 kJ/kgC • M = 0.300 kg • Tf = 19.0 C • Ti = 80.0 C • Q = (1.25 kJ/kgC)(0.300 kg)(19.0 C - 80.0 C) • = -22.9 kJ (a negative value, since the temperature dropped)

  12. Q = C x M x ΔT 3. Calculate the heat required to increase the temperature of 650 g copper by 12.6°C. • C = 0.385 kJ/kgC • M = 650 g x (1 kg / 1000 g) = 0.65 kg • ΔT = 12.6 C (only temp change is given) • Q = (0.385 kJ/kgC)(0.65 kg)(12.6 C) • = 3.2 kJ (2 significant digits)

  13. ΔT = Q C x M 4. Calculate the temperature change that occurs when 2500 kJ of heat is added to 4.45 kg of asbestos. • C = 0.820 kJ/kgC • M = 4.45 kg • Q = 2500 kJ • ΔT = 2500 kJ (0.820 kJ/kgC)(4.45 kg) • = 690 C

  14. ΔT = Q C x M 5. Calculate the temperature change that occurs when -652 J of heat is added to 241 g of mercury. • C = 0.140 kJ/kgC • M = 241 g x (1kg / 1000 g) = 0.241 kg • Q = -652 J x (1 kJ / 1000 J) = -0.652 kJ • ΔT = -0.652 kJ (0.140 kJ/kgC)(0.241 kg) • = -19.3 C (the temperature dropped)

  15. M= Q C x ΔT 6. Calculate the mass of wood that would undergo a temperature change of 26.5°C when 1360 kJ of heat is added. • C = 1.76 kJ/kgC • ΔT = 26.5 C • Q = 1360 kJ • M = 1360 kJ (1.760 kJ/kgC)(26.5 C) • = 29.2 kg

  16. Phase Changes

  17. Phase Changes • Heat goes into the breaking of physical bonds between the molecules of the substance, and putting distance between the molecules: • When a substance melts it goes from a rigid, hard solid to a liquid. Bonds are broken so that molecules have the freedom of movement. • When a liquid becomes a gas the molecules get much further apart.

  18. Phase Changes • Heat of Fusion (Hf) - This is the heat required to melt or freeze a substance. Expressed in kilojoules per kilogram or kilojoules per mole. If a substance is freezing the Hf value is expressed as a negative. • Heat of Vapourization (Hv) - This is the heat required to vapourize or boil a substance. Expressed in the same units as heat of fusion. If a substance is condensing the Hv value is expressed as a negative.

  19. Calculating the Energy of Phase Changes • Heat Required to Melt/Freeze =  Heat of Fusion   x   Mass of a Substance Substance • Q = HfM (for water Hf is 334 kJ/kg) • Heat Required to Boil/Condense   =   Heat of x   Mass of a Substance Vapourization   Substance • Q = HvM (for water Hv is 2260 kJ/kg)

  20. 1. Calculate the heat required to melt 1.5 kg ice • Since the phase change is melting the formula is Q = Hf M • where Hf = 334 kJ/kg • and M = 1.5 kg • Q = (336 kJ/kg)(1.5 kg) = 5.0 x 102 kJ

  21. 2. Calculate the heat required to freeze 4.50 x 102 kg water. • Since the phase change is freezing the formula is Q = Hf M • where Hf = - 334 kJ/kg (heat is taken out) • and M = 4.50 x 102 kg • Q = (- 336 kJ/kg)(4.50 x 102 kg) = - 1.50 x 105 kJ

  22. 3. Calculate the heat required to boil 245 g water. • Since the phase change is boiling the formula is Q = Hv M • where Hv = 2260 kJ/kg and M = 245 g x (1 kg/1000 g) = 0.245 kg • Q = (2260 kJ/kg)(0.245 kg) = 554 kJ

  23. 4. Calculate the heat required to condense 1.2 t steam. • Since the phase change is condensing the formula is Q = Hv M • where Hv = - 2260 kJ/kg and M = 1.2 t x (1000 kg/ 1 t) = 1200 kg • Q = (- 2260 kJ/kg)(1200 kg) = 2.7 x 106 kJ

  24. 5. Calculate the heat required to melt 162 mol ice. • Since the phase change is melting the formula is Q = Hf M • where Hf = 334 kJ/kg • and M = is not given, but is 162 mol • The molar mass of water is 18.02 g/mol (from H2O) and • Mass = Molar mass x Number of moles = (18.02 g/mol)(162 mol) = 2920 g x (1 kg / 1000 g) = 2.92 kg • Q = (334 kJ/kg)(2.92 kg) = 975 kJ

  25. Combined Heat Problems

  26. how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ? • 5 steps are required: • heat ice from -20.00°C to 0°C • melt ice • heat water from 0°C to 100°C • boil water • heat steam from 100°C to 110.0°C • any change in temperature uses the formula Q = C M ΔT • phase changes use the formula Q = HfM or HvM

  27. how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ? • heat ice from -20.00°C to 0°C Q = C M ΔT = (2.06 kJ/kg·°C)(2.00 kg)(0°C – (-20.0°C)) = 82.4 kJ • melt ice Q = Hf M = (334 kJ/kg)(2.00 kg) = 668 kJ • heat water from 0°C to 100°C Q = C M ΔT = (4.18 kJ/kg·°C)(2.00 kg)(100°C – 0°C) = 836 kJ

  28. boil water Q = Hv M = (2260 kJ/kg)(2.00 kg) = 4520 kJ • heat water from 100°C to 110°C Q = C M ΔT = (1.86 kJ/kg·°C)(2.00 kg)(110°C – 100°C) = 37.2 kJ • Total = 84.2 + 668 + 836 +4520 kJ + 37.2 kJ = 6150 kJ

  29. Enthalpy • refers to the energy change in chemical reactions. • the symbol for enthalpy is H. • it is impossible to determine the heat or energy contained in an object or chemical, but we can determine the change in energy in a chemical reaction. • change in enthalpy is ΔH

  30. Enthalpy Change • reactions that give off energy are exothermic; their ΔH is negative. • reactions that take in energy are endothermic; their ΔH is positive.

  31. Heat of Formation • is the enthalpy change when a compound is formed from its elements.

  32. Heat of Formation Elements Heat of reaction (kJ/mol of product) H2 (g) + 1/2 O2 (g) H2O(g) - 241.8 H2 (g) + 1/2 O2 (g) H2O(l) - 285.8 S(s) + O2 (g) SO2 (g) - 296.8 H2 (g) + S(s) + 2 O2 (g) H2SO4 (l) - 812 S(s) + 3/2 O2 (g) SO3 (g) - 395.7 1/2 N2 (g) + 1/2 O2 (g) NO(g) + 90.37 1/2 N2 (g) + O2 (g) NO2 (g) + 33.85 1/2 N2 (g) + 3/2 H2 (g) NH3 (g) - 46.19 C(s) + 1/2 O2 (g) CO(g) - 110.5 C(s) + O2 (g) CO2 (g) - 393.5 C(s) + 2 H2 (g) CH4 (g) - 74.86 2 C(s) + 3 H2 (g) C2H6 (g) - 83.8 3 C(s) + 4 H2 (g) C3H8 (g) - 104

  33. Using Heats of Formation • C (s) + 1/2 O2 (g) CO(g) Δ H = - 110.5 kJ • CO(g) + 1/2 O2 (g) CO2 (g) ΔH = - 283.0 kJ • C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ

  34. Using Heats of Formation • C (s) + 1/2 O2 (g) CO(g) Δ H = - 110.5 kJ • CO(g) + 1/2 O2 (g) CO2 (g) ΔH = - 284.0 kJ • C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ

  35. Hess’ Law • Hess's law of constant heat summation: • The enthalpy change for any reaction depends only on the products and reactants and is independent of the pathway or the number of steps between the reactant and product.

  36. 2 Al (s)   +   3 CuO (s) 3 Cu(s)   +   Al2O3 (s) • Identify the compounds in the equation; find heat of formation equations from the table that contain them (don’t worry about the elements. They take care of themselves): • 2 Al (s)   +   1½ O2 (g)   Al2O3 (s)          ΔH = - 1676.0 kJ/mol • Cu (s)   +   ½ O2 (g)   CuO (s)               ΔH = - 155.0 kJ/mol • Flip the CuO equation to make the compound a reactant as well: • Cu (s)   +   ½ O2 (g)   CuO (s)               ΔH = - 155.0 kJ/mol • Becomes • CuO (s)   Cu (s)   +   ½ O2 (g)               ΔH = + 155.0 kJ/mol

  37. 2 Al (s)   +   3 CuO (s) 3 Cu(s)   +   Al2O3 (s) • Multiply the CuO equation by 3 to have the same number of molecules as in the original equation. The enthalpy change is multiplied by the same factor: • 2 Al (s)   +   1½ O2 (g)   Al2O3 (s) ΔH = - 1676.0 kJ/mol • 3 x      CuO (s)   Cu (s)   +   ½ O2 (g) ΔH = 3 x (+155.0 kJ/mol) • This gives • 2 Al (s)   +   1½ O2 (g)   Al2O3 (s)    ΔH = - 1676.0 kJ/mol • 3 CuO (s)  3 Cu (s)   +  1½ O2 (g)   ΔH = + 465.0 kJ/mol

  38. 2 Al (s)   +   3 CuO (s) 3 Cu(s)   +   Al2O3 (s) • Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added: • 2 Al (s)   +   1½ O2 (g)   Al2O3 (s)                ΔH = - 1676.0 kJ/mol • 3 CuO (s)  3 Cu (s)   +  1½ O2 (g)               ΔH = + 465.0 kJ/mol • 2 Al (s) +  1½ O2 (g) +  3 CuO (s) Al2O3 (s) + 3 Cu (s) +  1½ O2(g) •   ΔH = - 1211 kJ • Cancel out like terms: • 2 Al (s)+  3 CuO (s) Al2O3 (s) + 3 Cu (s)     ΔH = - 1211 kJ • You know you are right if the net equation you end up with is the same as the original equation you started with.

  39. 2 Al (s)   +   3 CuO (s) 3 Cu(s)   +   Al2O3 (s) • WHAT YOU ACTUALLY NEED TO SHOW: • 2 Al (s)   +   1½ O2 (g)   Al2O3 (s)       ΔH = - 1676.0 kJ/mol • 3 x 3 CuO (s)3 Cu (s)  + (3)½ O2 (g)   ΔH = 3 x (+155.0 kJ/mol) • 2 Al (s) +  3 CuO (s) Al2O3 (s) + 3 Cu (s)   ΔH = - 1211 kJ

  40. 2 C2H6 (g)   +   7 O2 (g) 4 CO2 (g)   +   6 H2O (l) • 2 x      C2H6 (g)   2 C (s)   +   3 H2 (g) ΔH = 2 x(+ 84.0 kJ/mol) • 4 x      C (s)   +   O2 (g)   CO2 (g) ΔH = 4 x (- 394.0 kJ/mol) • 6 x      H2 (g)   +   ½ O2 (g)   H2O (l) ΔH = 6 x (- 286.0 kJ/mol) • 2 C2H6 (g)   +   7 O2 (g)   4 CO2 (g)   +   6 H2O (l) ΔH = - 3124 kJ

  41. SiO2 (s)   +   C (s)  CO2 (g)   +   Si (s) Where the heat of formation of SiO2 (s)   =   - 861 kJ/mol • Heat of formation is formation of a compound from its elements. SiO2 is made of silicon (Si (s)) and oxygen (O2 (g)): • 1x SiO2 (s) Si (s) + O2 (g) ΔH = +861 kJ/mol • 1x C (s) + O2 (g) CO2 (g) Δ H = - 394 kJ/mol • SiO2 (s)   +   C (s)  CO2 (g)   +   Si (s) Δ H = + 467 kJ

  42. Enthalpy and Entropy • Enthalpy is one of the driving forces in the universe; reactions that release energy (negative ΔH) are favoured. • Endothermic reactions do occur spontaneously; that is because of a second force, Entropy. • Entropy is disorder • The more disorder, the more entropy.

  43. Examples of Increasing Entropy • tossed salad

  44. Examples of Increasing Entropy • Broken glass

  45. Examples of Increasing Entropy • Melting ice

  46. Examples of Increasing Entropy • Bedroom: Before

  47. Examples of Increasing Entropy • Bedroom: After

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