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THERMOCHEMISTRY

THERMOCHEMISTRY. Thermochemistry. Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state. Heat (q). Heat is the energy that transfers from one object to another because of a temperature difference between them.

gillian-haverty
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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

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