Download Ch 17--Thermochemistry(first class)

Chapter 17
Thermochemistry
Section 17.1: The flow of energy
• Thermochemistry:
Study of energy changes that occur
during chemical reactions and
changes in state
Section 17.1: The flow of energy
• energy changes can either occur
through heat transfer or work
• heat (q), energy is transferred
from a warmer object to a cooler
object (always)
• Adding heat increases
temperature
17.1: Endothermic & Exothermic processes
•
Kinetic energy vs. potential energy
energy due to
motion
energy due to a substance's
chemical composition
• potential energy is determined by the strength of repulsive
and attractive forces between atoms
• In a chemical reaction, atoms are recombined into new
arrangements that have different potential energies
•change in potential energy: due to absorption and release of
energy to and from the surroundings
17.1: Endothermic & Exothermic processes
•
Two parameters crucial in Thermochemistry:
a) system--part of the universe
attention is focused
•
•
b) surroundings--everything else in the
universe
System + surrounding = universe
Fundamental goal of Thermochem.: study the
heat flow between the system and its
surroundings
17.1: Endothermic & Exothermic processes
• If System (energy) Surrounding
(energy) by the same amount
• the total energy of the universe does not
change
Law of conservation of energy
17.1: Endothermic & Exothermic processes
• Direction of heat flow is given from the point
of view of the system
• So, endothermic process: system absorbs heat
from the surroundings (system heats up)
• Heat flowing into the system = +q
• Exothermic process: heat is released into the
surroundings (system cools down, -q)
17.1: Endothermic & Exothermic processes
Example 1:
A container of melted paraffin wax is allowed to
stand at room temperature (r.t.) until the wax
solidifies. What is the direction of heat flow as
the liquid wax solidifies? Is the process
exothermic or endothermic?
Answer: Heat flows from the system (paraffin) to
the surroundings (air)
Process: exothermic
17.1: Endothermic & Exothermic processes
Example 2:
When solid Ba(OH)2▪8H2O is mixed in a beaker
with solid NH4SCN, a reaction occurs. The
beaker quickly becomes very cold. Is the
reaction exothermic or endothermic?
Answer: Endothermic
surrounding = beaker and air
System = chemicals within beaker
17.1: Units of heat flow
• Two units used:
a) calorie (cal)—amount of heat required to
raise the temperature of 1g of pure water by
1oC
b) joule (j)—1 joule of heat raises the
temperature of 1 g of pure water 0.2390oC
• Joule = SI unit of energy
17.1 Heat capacity & specific heat
• Heat capacity is the quantity of heat needed to
raise the temperature of an object exactly 1oC
• Heat capacity depends on:
a) mass
b) chemical composition
• So, the greater the mass the greater the heat
capacity
eg.: cup of water vs. a drop of water
(cup of water = greater heat capacity)
17.1: Heat capacity & specific heat
• Specific heat: amount of heat required to raise
the temperature of 1g of a substance by 1oC
• Table 17.1 (p.508): List of specific heats of
substances
Specific heat calculation
C =
q
m * ΔT
ΔT = Tf –Ti
C =
j
(g * oC)
=
heat (joules/calories)
mass (g) * ΔTemp. (oC)
(Tf = final temperature)
(Ti = initial temperature)
or
cal
(g * oC)
Example 1
1. The temperature of a 95.4 g piece of copper
increases from 25.0 oC to 48.0 oC when the
copper absorbs 849 j of heat. What is the
specific heat of copper?
unknown: Ccu
Know:
mass copper = 95.4 g
ΔT = Tf – Ti = (48.0 oC – 25.0 oC)
= 23.0 oC
q = 849 j
Example1
C =
q
m * ΔT
C =
849 j
95.4 g * 23.0 oC
C = 0.387 j/g * oC
Sample problem 17.1, page 510
Example 2
2. How much heat is required to raise the
temperature of 250.0 g of mercury (Hg) 52
oC?
unknown: q
Know:
mass Hg = 250.0 g
ΔT = 52.0 oC
CHg = 0.14 j/(g * oC)
Example 2
C =
q
m * ΔT
q = CHg * m * ΔT
q = 0.14 (j/g * oC) * 250.0 g * 52 oC
q = 1.8 x 103 j (1.8 kj)
Problem #4 page 510
Section 17.2
Measuring Enthalpy Changes
17.2: Enthalpy (measuring heat flow)
• Calorimetry:
accurate measurement of heat flow
into or out of a system in chemical
and physical processes
• In calorimetry, heat released by a system is
equal to the heat absorbed by its surroundings
and vice versa
• Instrument used to measure absorption or
release of heat is a calorimeter
17.2: Enthalpy (measuring heat flow)
Two types of calorimeters:
a) Constant-Pressure calorimeter (eg. foam cups)
• As most reactions occur at constant pressure we can say
that:
A change in enthalpy (ΔH) = heat supplied (q)
• So, a release of heat (exothermic) corresponds to a
decrease in enthalpy (at constant pressure)
• An absorption of heat (endothermic) corresponds to an
increase in enthalpy (constant pressure)
17.2: Enthalpy (measuring heat flow)
b) Constant-Volume Calorimeters (eg. bomb
calorimeters)
• Substance is burned (in the presence of O2) inside a
chamber surrounded by water (high pressure)
• Heat released warms the water
Figure 17.6 (p. 512) Bomb calorimeter.
17.2: Thermochemical equations
CaO(s) +
H2O(l)
Ca(OH)2 + 65.2 kJ
Heat released
A chemical equation that includes the enthalpy
change is called
a thermochemical equation
•
• Reactants and products at their usual physical
state (at 25 oC) given at standard pressure (101.3
kPa)
• So, the heat of reaction (or ΔH) for the above
equation is -65.2 kJ
17.2: Thermochemical equations
So, rewrite the equation as
follows:
CaO(s) +
Ca(OH)2(s)
H2O(l)
65.2 kJ
• Other reactions absorb heat from the surroundings, eg.:
Na2CO3(s) + H2O(g) + CO2(g)
2 NaHCO3(s) + 129 kJ
Rewrite to show heat of reaction
2 NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
 kJ
17.2: Thermochemical equations
• Amount of heat released/absorbed during a
reaction depends on the number of moles of
reactants involved
eg.: 2 NaHCO3(s)
4 NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
 kJ
Na2CO3(s) + H2O(g) + CO2(g)
258 kJ
Enthalpy Diagrams
CaO(s) +
H2O(l)
H = -65.2 kJ
Na2CO3(s) + H2O(l) + CO2(g)
H = 129 kJ
Ca(OH)2(s)
2 NaHCO3(s)
Exothermic Reaction
Endothermic Reaction
Diagram A:
Diagram B:
Enthalpy of reactants greater
than of products
Enthalpy of reactant less
than of products
17.2: Thermochemical equations
Physical states of reactants and products must be
stated:
H2O(l)
H2(g) +
1 O2(g)
2
285.8 kJ
H2O(g)
H2(g) +
1 O2(g)
2
241.8 kJ
Difference = 44.0 kJ
Vaporization of H2O(l) requires more heat (44.0 kJ)
Example1
1. Calculate the amount of heat in (kJ) required to decompose 2.24 mol
NaHCO3(S)
Na2CO3(s) + H2O(g) + CO2(g)
2 NaHCO3(s)
Known:
•2.24 mol NaHCO3 decomposes
•ΔH = 129 kJ (2 mol NaHCO3)
 kJ
Unknown:
•ΔH = ?
Solve:
129 kJ
2 mol NaHCO3(s)
Sample problem 17.3; p. 516
=
ΔH
2.24 mol NaHCO3(s)
ΔH
= (129 kJ) * 2.24 mol NaHCO3(s)
2 mol NaHCO3(s)
ΔH
= 144 kJ
Example 2
2. When carbon disulfide is formed from its
elements, heat is absorbed. Calculate the
amount of heat in (kJ) absorbed when 5.66 g of
carbon disulfide is formed.
C(s) + 2 S(s)
CS2(l)
Known:
•5.66 g CS2 is formed
•ΔH = 89.3 kJ (1 mol CS2(l))
•Molar mass: CS2(l): C = 12.0 g/mol
2 *S = 32.1 g/mol = 64.2 g/mol
76.2 g/mol
H = 89.3 kJ
Unknown:
•ΔH = ?
Example 2
Solve:
1. Moles CS2(l) = 5.66g CS2
= 0.0743 mol CS2(l)
76.2 g/mol CS2(l)
2.
89.3 kJ
1 mol CS2(l)
=
ΔH
0.074 mol CS2(l)
ΔH
= (89.3 kJ) * 0.074 mol CS2(l)
1 mol CS2(l)
ΔH
= 6.63 kJ
17.3: Heat in changes of state
Objective:
-Heats of Fusion and Solidification
-Heats of Vaporization and Condensation
-Heat of solution
17.3: Heat of fusion and solidification
• The temperature remains constant when a
change of state occurs via a gain/loss of energy
• Heat absorbed by 1 mole of a solid during
melting (constant temperature) is the molar
heat of fusion (ΔHfus)
• Molar heat of solidification (ΔHsolid) is the heat
lost by 1 mole of liquid as it solidifies
(constant temperature)
• So, ΔHfus = ΔHsolid
17.3: Heat of fusion and solidification
• Figure 17.9: Enthalpy changes and changes of
state
17.3: Heat of fusion and solidification
H2O(s)
H2O(l)
H2O(l)
fus.01 kJ/mol
H2O(s)
solid6.01 kJ/mol
17.3: Heats of Vaporization and Condensation
• Molar heat of vaporization (ΔHvap): Amount of
heat required to vaporize one mole of a liquid
at the liquid’s normal boiling point
H2O(l)
•
H2O(g)
vap kJ/mol
Molar heat of condensation (ΔHcond): heat released
when 1 mole of vapor condenses
H2O(g)
•So, ΔHvap = -ΔHcond
H2O(l)
cond kJ/mol
17.3: Heat of vaporization and
condensation
• Figure 17.10: Heating curve of water
17.3: Heat of solution
• There is heat released/gained when a solute
dissolves in a solvent
• The enthalpy change due to 1 mole of a
substance dissolving: molar heat of solution
(ΔHsoln)
NaOH(s)
H2O(l)
Na+(aq) + OH-(aq)
soln5.1 kJ/mol
17.3: Heat of solution
• Applications: hot/cold packs
• Hot pack:
CaCl2(s)
H2O(l)
Ca2+(aq) + 2Cl-(aq)
soln82.8 kJ/mol
•Cold pack:
NH4NO3(s)
H2O(l)
NH4+(aq) + NO3-(aq) soln25.7 kJ/mol