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Unit 11 – Thermochemistry
What is temperature? Heat?
75 oC
75 oC
They are at the same temperature but different amounts of heat. Why?
How does a thermometer work?
Particles hit the thermometer, which
cause the particles in the
thermometer to move faster and
expand which causes the
thermometer to go up.
Does temperature measure heat?
75 oC
75 oC
Which of the above systems contains more heat? The bigger one, of course.
So:
Heat = Temperature
Temperature= average kinetic energy of the particles in a sample
Heat = total thermal energy of a sample
Different substances have the capacity to absorb different amounts of heat.
Specific Heat capacity- energy required to raise the temperature of 1 g of a substance by 1 oC
So, total heat depends on:
1) mass
2) temperature
3) specific heat
1 calorie= energy
required to raise 1 gram
of water by 1 oC.
1 calorie = 4.18 J
1000 cal = 1 Cal
q = mc∆T
CH2O = 4.18 J/goC
q – heat (J or cal)
m – mass (g)
c – specific heat capacity (J/goC or cal/goC)
∆T – change in temperature (oC) Tf - Ti
Ex 1) 1000 J of heat is applied to 100 g of aluminum at 25 oC. The final temperature is 36.09 oC.
What is the specific heat?
q = mc∆T
c = q/m∆T
c = 1000 J/ (100g)(11.09 oC) = 0.902 J/goC
Ex 2) What is the final temperature when 500 J is applied to 150 grams of aluminum that started
at 25 oC?
q = mc∆T
∆T = q/mc ∆T = 500 J/(150 g)(0.902 J/goC) = 3.70 oC
Tf = 25 oC + 3.70 oC = 28.70 oC
4
Heating Curve for Water
T
5
100 oC
3
2
0 oC
1
time
1= heating solid 2 = melting
3 = heating liquid
4 = boiling
5 = heating steam
At sections 1, 3, 5 on the graph, we can use q = mc∆T. However, sections 2 & 4 there is no
change in temperature. The equation q = mc∆T would calculate that there is no heat need to melt
or boil a substance, but that is not true. We need a different method to calculate the heat at those
sections.
heat of fusion = solid  liquid phase change
heat of vaporization = liquid  gas phase change
For water, heat of fusion = 80 cal/g, heat of vaporization = 540 cal/g
The more of a substance there is, the more energy it will take to boil/freeze/melt/condense
Ex) How much energy does it take to change 200 grams of ice at 0oC to steam at 100 oC?
3
T
100 oC
2
1
0 oC
time
1) melting ice
no temperature change at phase change so:
q = mass x Heat of fusion
(heat of fusion = 80 cal/g)
q = 200g H2O x 80 cal/g = 16000 calories
2) heating water
temperature rises from 0 oC to 100oC so q = mc∆T
q = (200 g)(1 cal/goC)(100 oC) = 20000 calories
3) boiling water
no temperature change at phase change so:
q = mass x Heat of vaporization
(heat of vaporization = 540 cal/g)
q = 200g H2O x 540 cal/g = 108000 calories
The total energy require is 16000 cal + 20000 cal + 108000 cal = 144000 calories
Ex) How much energy does it take to change 200 grams of steam at 100 oC to ice at 0oC ?
-144000 calories (same amount of energy involved, but opposite sign)
Endothermic
Energy absorbed
q= +
Increase in temperature
Melting (s  l)
Vaporization/Boiling (l  g)
Exothermic
Energy released
q=Decrease in temperature
Freezing (l  s)
Condensation (g  l)
Phase Diagrams
Phase diagrams tell us what phase a substance will exist in as pressure and temperature change.
Along line AD, solid and liquid are present at equilibrium.
Along line AB, liquid and gas are present at equilibrium.
Along line Ac, solid and gas are present at equilibrium.
Along the lines, phages changes are occurring.
Solid  liquid Melting
Liquid  solid Freezing
Liquid  gas vaporization
Gas  liquid condensation
Solid  gas sublimation
Gas  solid deposition
At point A – Triple Point – all three phases exist
Melting occurs anywhere along AD, but the NORMAL melting point is the temperature at which
substances melt at standard pressure (1 atm) = 20 oC
Boiling occurs anywhere along AB, but the NORMAL boiling point is the temperature at which
substances vaporize at standard pressure (1 atm) = 80 oC
If the slope of line AD is positive – the solid phase is more dense than the liquid phase
If the slope of line AD is negative - the liquid phase is more dense than the solid phase
Calorimeters
Law of Conservation of Energy - qlost = - qgained
How is heat measured? A device known as a calorimeter is used. A reaction is performed and
the heat is trapped into something that we can measure the energy changes. It is assumed that
the heat lost from the reaction is equal to the heat gained by the calorimeter (with a negative sign
to account for the direction change).
You measure the mass of the water put into the cup and the temperature changes of the reaction
and since you know Cwater = 4.18 J/goC, you can calculate the heat of the water. The energy
change that we observe by the water comes from the substance we put into the water. The
energies are equal.
Example:
55 grams of hot iron at 99.8 oC is plunged into 225 g of cool water at 21 oC. The iron cools
down and the water heats up to 23.1 oC. Calculate the specific heat of the iron.
Since the heat lost by the iron is equal to the heat gained by the water
qlost Fe = - qgained water
mFecFe∆TFe = - (mwatercwater ∆Twater)
(55 g)(cFe)(23.1 oC – 99.8 oC) = - [(225 g)(4.18 J/goC)(23.1 oC – 21 oC)
(-4218.5 goC) cFe = -1975.05 J
cFe = 0.468 J/goC
Heat of Formation: the amount of energy required to form 1 mole of a substance from its
elements in their standard states, ∆Hfo
∆Hrxn = Σ ∆Hfo Products – Σ ∆Hfo Reactants
C6H6 (l) + 15/2 O2 (g)  6 CO2 (g) + 3 H2O (l) ∆H = ????
C6H6 (l)
CO2 (g)
H2O (l)
O2 (g)
∆Hfo = +49 kJ/mole
∆Hfo = -394 kJ/mole
∆Hfo = -286 kJ/mole
∆Hfo = 0 kJ/mole
∆Hrxn = [6 (∆Hfo CO2) + 3 (∆Hfo H2O)] - [∆Hfo C6H6 + 15/2 (∆Hfo O2)]
∆Hrxn = [6 (−394 kJ) + 3 (−286 kJ)] - [+49 kJ + 15/2 (0 kJ)]
∆Hrxn = -3271 kJ
Hess’s Law of Heat Summation – if a reaction is the sum of two or more reactions, their
heats can be added to yield the heat of the reaction.
C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (l) ∆H = ????
Given:
2 C (s) + 3 H2 (g) + ½ O2 (g)
C (s) + O2 (g)
H2 (g) + ½ O2 (g)
 C2H5OH (l)
 CO2 (g)
 H2O (l)
∆H = -278 kJ
∆H = -394 kJ
∆H = -286 kJ
We need to manipulate the equations above such that when we add the equations together, it
yields the overall reaction we are looking for. The first equation needs to be flipped around. The
second equation needs to be doubled. The third equation needs to be tripled. All these changes
will have an effect on the ∆H.
C2H5OH (l)
2 [C (s) + O2 (g)
3[H2 (g) + ½ O2 (g)
 2 C (s) + 3 H2 (g) + ½ O2 (g)
 CO2 (g)]
 H2O (l)]
∆H = +278 kJ
∆H = 2[-394 kJ]
∆H = 3[-286 kJ]
C2H5OH (l) + 3 O2 (g)
 2 CO2 (g) + 3 H2O (l)
∆H = -1368 kJ