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Transcript thermochemistry

 Energy: ability to do work or produce heat.
 Kinetic energy: energy of motion
 Potential energy: due to composition or position of an
 Chemical systems contain both kinetic and potential
 Law of Conservation of energy: states that in an y
chemical reaction or physical process, energy can be
converted from one form to another, but is neither
created or destroyed.
 Heat (q): energy that flows from a warmer object to a
cooler object.
 If object loses heat, temperature decreases and if it gains
heat, temperature increases.
Measuring heat
 calorie (cal): the amount of heat required to raise the
temperature of 1g of water by 1C.
 Nutritional Calorie (Cal):
1Calorie= 1000 calories= 1kcal
 SI unit for measuring heat: joule ( J)
1 cal = 4.184 J
Specific heat (c)
 Amount of heat
required to raise the
temperature of one
gram of that
substance by 1C.
 Units: J/g C or
cal/g C
Calculating heat released and
 q = mcT
 q: heat (J or cal)
 m: mass (g)
 c: specific heat (J/g C) or (cal/g C)
  T: change in temperature
 T = Tfinal - Tinitial
Example 1
 In the construction of bridges and sky scrapers, gaps
must be left between adjoining steel beams to allow for
expansion and contraction of the metal due to heating
and cooling. The temperature of a sample of iron with
a mass of 10.0g changed from 55.0C to 85.0C with
the release of 114 J of heat. What is the specific heat of
 q=mcT
q= 114J m= 10.0g T=(85-55)= 30C
114J= 10.0 g x c x 30C
c = 0.38 J/g  C
Example 2.
 If the temperature of 34.4g of ethanol increases from
25.0 C to 78.8 C, how much heat is being absorbed
by the ethanol? C ethanol= 2.44 J/g C
 Answer: 4516 J
Example 3
 How many joules of heat are lost by 3580 kg of granite
as it cools from 41.2 C to -12.9C? cgranite = 0.803 J/gC
1 kg= 1000g
Heat in chemical reactions and processes
 Measuring heat:
 Calorimeters : are used to measure heat or calculate
specific heat
Measuring heat
 The heat released by the system is equal to the heat
absorbed by its surroundings.
 System: part of the universe on which you focus your
 Surroundings: includes everything else in the universe.
qsystem= -qsurrounding
q= mcT
Ex. 1 A small pebble is heated and placed in a foam cup
calorimeter containing 25.0g 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? (cwater=4.184 J/g C)
qwater= mcT
= 25.0g 4.184 J (26.4 C- 25.0C)
= 146 J
qsystem= -qsurrounding
qwater= - qpebble
qpebble= - qwater= - 146J
Ex. 2
 An alloy of unknown composition is heated to 137 °C
and placed into 100.0 g of water at 25.0 °C. If the final
temperature of the water and alloy was 36.4 °C, and
the alloy weighed 2.71 g, what is the specific heat of the
alloy? The specific heat of water is 4.184 J/gºC.
Learning Check 1
 Given that the specific heat of gold is 0.129 J/gºC,
calculate the final temperature for the system if a 200.0 g
block of gold at 100.0 ºC is placed in a coffee-cup
calorimeter containing 50.0 g of water at an initial
temperature of 25.0 ºC. The specific heat of water is
4.184 J/gºC. (Answer: 33.2C)
Thermochemical equations
 Most chemical reactions and physical changes occur at
constant pressure.
 The heat content of a system at constant pressure is
called enthalpy (H) of the system.
 The heat absorbed or released by a reaction at constant
pressure is the same as the change in enthalpy or H.
 For us:
q= H
 Thermochemical equation: balanced chemical
equation that includes physical states of reactants
and products and the energy change (H)
 Ex 4Fe(s) + 3 O2 (g)  2Fe2O3 (s) H= 1625 kJ
 H is negative for EXOTHERMIC reactions
 H is positive for ENDOTHERMIC reactions
 Enthalpy of fusion (Hfus): energy required to melt 1
gram (or 1 mole) of a substance at its melting point.
 For water: Hfus=334J/g or 6.01kJ/mol
 Enthalpy of solidification Hsolid :
Hfus = - Hsolid
 Enthalpy of vaporization (Hvap): energy required to
vaporize 1 gram (or 1 mole) of a substance at its boiling
 For water: Hvap=2260J/g or 40.7kJ/mol
 Enthalpy of condensation Hcond :
Hvap = - Hcond
 During a change of state, there is no change in
Ex.1 How much heat (in kJ) is absorbed when 24.8g H2O(l) at
100C is converted to steam at 100 C?
Hvap= 40.7kJ/mol
 mwater= 24.8 g
 Calculate molar mass water:
 For water:
1 mol
= 56.1 kJ
40.7 kJ
1 mol
How much heat is required to melt 5.67g of iron (II) oxide,
FeO, if its Hfus= 32.2 kJ/mol?
 Hfus= 32.2 kJ/mol?
 m= 5.67 g FeO
 Molar mass= 71.8g/mol
5.67g FeO
= 2.54 kJ
1 mol
71.8 g
32.2 kJ
1 mol