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THERMOCHEMISTRY

THERMOCHEMISTRY

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THERMOCHEMISTRY

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  1. THERMOCHEMISTRY

  2. Thermochemistry • Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state.

  3. Heat (q) • Heat is the energy that transfers from one object to another because of a temperature difference between them. • Heat ALWAYS flows from a warmer object to a cooler one.

  4. Heat movement • Heat moves between the system (reaction) and the surroundings • *** must obey the law of conservation of energy (heat (energy) is never created nor destroyed, just transferred) • Thermochemical equations tell you the direction of heat flow by the “sign”, + or -

  5. Endo vs. Exo- • Endothermic reactions: absorbs heat from surroundings (+). • If you touch an endothermic reaction it feels COLD • Exothermic reactions: release heat to the surroundings (-) • If you touch an exothermic reaction it feels HOT

  6. HEAT energy • UNIT of energy = JOULE (J) or Calorie (cal) 1 cal = 4.184 J • Heat capacity is how much heat (energy) is needed to increase the temperature of an object by 1 C. • *** heat capacity of an object depends on both its mass and its chemical composition • The greater the mass, the greater its heat capacity

  7. Example • On a sunny day, a 20 kg puddle of water may be cool, while a nearby 20 kg iron sewer cover may be too hot to touch. • Both have the same mass, BUT the are made of different materials (chemical composition) BECAUSE they have different SPECIFIC HEAT CAPACITIES

  8. SPECIFIC HEAT (C) • SPECIFIC Heat Capacity (specific heat, C) = specific to a substance • Amount of heat it takes to raise the temp. of 1 g of a substance by 1 C • Units = (J/gC) • WATER has a specific heat of 4.184 J/gC

  9. CALCULATING HEAT • q = mCT • Where: • q = heat (Joules) • m = mass (grams) • C = specific heat (J/gC) • T = change in temperature (C) Tf-Ti

  10. Calculations • How much heat is absorbed when a 95.4 g piece of copper increases from 25.0 C to 48.0 C. The specific heat of copper is 0.387 J/gC.

  11. m= 95.4 g C= 0.387 J/gC • T= 48.0 C - 25.0 C= 23 C • q= (95.4) (.387) (23) • 849 J

  12. Ex. 2 • How much heat is absorbed when 3.4 g of olive oil is heated from 21.0 C to 85.0 C. The specific heat of olive oil is 2.0 J/g C.

  13. m= 3.4 g C= 2.0 J/gC • T= 85.0 C - 21.0 C= 64 C • q= (3.4) (2.0) (64) • 435 J

  14. Ex. 3 • How much heat is required to raise the temperature of 250.0 g of mercury 52.0 C? The specific heat of mercury is 0.14 J/g C.

  15. m= 250 g C= 0.14 J/gC • T= 52 C • q= (250) (0.14) (52) • 1820 J = 1.8 kJ

  16. Ex. 4 • How many joules (J) of heat are absorbed when 1000.g of water is heated from 18.0 C to 85.0 C? C for water = 4.184 J/g C.

  17. m= 1000 g C= 4.184 J/gC • T= 85.0 C - 18.0 C= 67 C • q= (1000) (4.184) (67) • 280000 J =2.80 x 105 J

  18. Practice Problems 1. What is the specific heat of a substance that has a mass of 25.0 g and requires 2197 J of energy to raise its temperature by 15.0 C? 2. Suppose 100.0 g of ice absorbs 1255.0 J of heat. What is the corresponding temperature change? The specific heat of ice is 2.1 J/g C. 3. How many Joules of heat energy are required to raise the temperature of 100.0 g of aluminum by 120.0 C. The specific heat of aluminum is 0.90 J/g C. 

  19. ANSWERS • 1) 5.86 J/g°C • 2) 6.0 °C • 3) 10800 J = 11,000 J or 1.1 x 104 J

  20. STOP • Work on Specific Heat Calcs WS

  21. ENTHALPY • Enthalpy = a type of chemical energy, sometimes referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction) • exothermic reactions (feels hot): • q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative values) • endothermic reactions (feels cold): • q = ΔH > 0 (positive values)

  22. Thermochemical Equations • A chemical equation that shows the enthalpy (H) is a thermochemical equation.

  23. Rule #1 The magnitude (value) of H is directly proportional to the amount of reactant or product. H2 + Cl22HCl H = - 185 kJ * meaning there are 185 kJ of energy RELEASED for every: 1 mol H2 1 mol Cl2 2 moles HCl

  24. Rules of Thermochemistry Example 1: H2 + Cl22HCl H = - 185 kJ Calculate H when 2.00 moles of Cl2 reacts.

  25. Rules of Thermochemistry Example 2: Methanol burns to produce carbon dioxide and water: 2CH3OH + 3O2 2CO2 + 4H2O H = - 1454 kJ What mass of methanol is needed to produce 1820 kJ?

  26. Rule #2 H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction. (If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

  27. Rules of Thermochemistry Example 1: Given: H2 + ½O2 H2O H = -285.8 kJ Reverse: H2O  H2 + ½O2H = +285.8 kJ

  28. Example 2 CaCO3 (s)  CaO (s) + CO2 (g) H = 178 kJ What is the H for the REVERSE RXN? CaO (s) + CO2 (g)  CaCO3 (s) H = ?

  29. Practice Problem: (rule 1 + rule 2) Given: H2 + ½O2 H2O H = -285.8 kJ Calculate H for the equation: 2H2O  2H2 + O2

  30. Alternate form of thermochem. eq. • Putting the heat content of a reaction INTO the actual thermochemical eq. • EX: • H2 + ½O2 H2O H = -285.8 kJ • EXOTHERMIC: • Heat is ___________ as a ____________.

  31. ALTERNATE FORM • EX: • H2 + ½O2 H2O H = -285.8 kJ • The alternate form is this:H2 + ½O2 H2O +285.8 kJ

  32. EX: 2 NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2 Put in the alternate form

  33. Ex: 2 NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2 The alternate form is this: 2 NaHCO3 Na2CO3 + H2O + CO2H =+ 129 kJ

  34. Put the following in alternate form • 1. H2 + Cl2 2 HCl H = -185 kJ • 2. 2 Mg + O2 2 MgO + 72.3 kJ • 3. 2 HgO  2 Hg + O2H = 181.66 kJ

  35. CALORIMETRY • The enthalpy change associated with a chemical reaction or process can be determined experimentally. • Measure the heat gained or lost during a reaction at CONSTANT pressure

  36. Calorimeter • Device used to measure the heat absorbed or released during a chemical or physical process

  37. Example • If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter? • Why is there a difference in temperature between the two objects?

  38. Because… • Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).

  39. Need to know how to calc. heat (review) • Heat (q) = mCT • If the specific heat of Al is 0.90 J/gC, how much heat is required to raise the temperature of 10,000 g of Al from 25.0 C to 30.0 C?

  40. Calculator says…….. • = 45,000 J

  41. For example (review): • If 418 J is required to increase the temperature of 50.0 g of water by 2.0 C, what is the specific heat of water?

  42. What happens in a calorimeter • One object will LOSE heat, and the other will ABSORB the heat • System loses heat to surroundings = EXO = -q • System absorbs heat from surroundings = ENDO = +q

  43. EXAMPLE • A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble? The specific heat of water is 4.184 J/g C. • Which was warmer? The pebble or the water? 

  44. The pebble because the water heated up from 25.0 C to 26.4 C. • Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic) • Do the calc… 

  45. All values are for WATER • A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble? • Calc the heat absorbed by the water (+q). • The heat (J) released by the warm pebble = - of the heat absorbed by the water. q water = - q pebble

  46. -150 J

  47. PRACTICE 1 (LAB type of CALC) • Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal. The specific heat of water is 4.184 J/g C. 

  48. Heat gained by water = Heat lost by the Al • q of water = - q Al • Make a chart

  49. We don’t know specific heat of Al, but we know all the values for water • So, calc q for WATER • The q for water is the same for q of Al • (the value of q is the same)

  50. 0.879 J/g °C