Transcript Chapter 5: thermochemistry

Chapter 5:
thermochemistry
By Keyana Porter
Period 2
AP Chemistry
What is
thermochemistry?
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Thermodynamics: the study of energy and its
transformations
Thermochemistry is one aspect of
thermodynamics

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The relationship between chemical reactions and
energy changes
Transformation of energy (heat) during chemical
5.1
Kinetic & Potential Energy
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Energy is the capacity to do work or the transfer heat
Objects possess energy in 2 ways
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Kinetic energy: due to motion of object
Potential/Stored energy: result of its composition or its
position relative to another object
Kinetic energy (Ek) of an object depends on its mass
(m) and speed (v):
Ek= ½ mv2
Kinetic energy increases as the speed of an object
and its mass increases
Thermal energy: energy due to the substance’s
5.1
Kinetic & Potential Energy

Potential energy is a result of attraction &
repulsion

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Ex: an electron has potential energy when it is
near a proton due to the attraction (electrostatic
forces)
Chemical energy: due to the stored energy in
the atoms of the substance
5.1
Units of Energy
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joule (J): SI unit for energy
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calorie (cal): non-SI unit for energy; amount of
energy needed to raise the temperature of 1 g of
water by 1oC
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1 kJ= 1000 J
1 cal = 4.184 J (exactly)
Calorie (nutrition unit) = 1000 cal = 1 kcal
A mass of 2 kg moving at a speed of 1 m/s = kinetic
energy of 1 J
Ek= ½ mv2 = ½ (2 kg)(1 m/s)2 = 1 kg-m2/s2 = 1 J
5.1
System & Surroundings
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System (chemicals): portion that is singled
out of the study
Surroundings (container and environment
including you): everything else besides the
system
Closed system: can exchange energy, in the
form of heat & work, but not matter with the
surroundings
5.1
Transferring Energy: Work
& Heat
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Energy is transferred in 2 ways:
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Force (F): any kind of push or pull exerted on an
object
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Cause the motion of an object against a force
Cause a temperature change
Ex: gravity
Work (w): energy used to cause an object to move
against force
Work equals the product of the force and the
distance (d) the object is moved:
w=Fxd
heat: the energy transferred from a hotter object to
a colder one
5.2
Internal Energy
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The First Law of Thermodynamics: Energy is
conserved; it is neither created nor destroyed
Internal energy (E): sum of ALL the kinetic and
potential energy of all the components of the system
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The change in internal energy = the difference between
Efinal – Einitial
Δ
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E = Efinal – Einitial
We can determine the value of ΔE even if we don’t
know the specific values of Efinal and Einitial
All energy quantities have 3 parts:
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A number, a unit, and a sign (exothermic versus
endothermic)
Relating ΔE to Heat &
Work

A chemical or physical
change on a system,
the change in its
internal energy is given
by the heat (q) added
to or given off from the
system:
E=q+w
Δ
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both the heat added to
and the work done on
the system increases its
internal energy
Sign Conventions Used and the
Relationship Among q, w, and ΔE
Sign of ΔE = q + w
q > 0 and w > 0: ΔE > 0
q > 0 and w < 0: the sign
of ΔE depends on the
magnitudes of q and w
q < 0 and w > 0: the sign
of ΔE depends on the
Sign convention for w:
magnitudes of q and w
w > 0: Work is done by the
surroundings on the system q < 0 and w < 0: ΔE < 0
Sign Convention for q:
q > 0: Heat is transferred
from the surroundings to
the system (endothermic)
q < 0: Heat is transferred
from the system to the
surroundings (exothermic)
w < 0: Work is done by the
system on the surroundings
5.2
Endothermic & Exothermic Processes
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Endothermic: system absorbs heat
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Exothermic: system loses heat and the heat flows
into the surroundings
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Ex: melting of ice
Ex: freezing of ice
The internal energy is an example of a state function
Value of any state function depends only on the
state or condition of the system (temperature,
pressure, location), not how it came to be in that
particular state
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ΔE
= q + w but, q and w are not state functions
5.3
Enthalpy
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Enthalpy (H): state function; the heat absorbed or
released under constant pressure
The change in enthalpy equals the heat (qP) gained or
lost by the system when the process occurs under
constant pressure:
ΔH = Hfinal – Hinitial = qP
only under the condition of constant pressure is the heat
that is transferred equal to the change in the enthalpy
The sign on ΔH indicated the direction of heat transfer
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+ value of ΔH means it is endothermic
- value of ΔH means it is exothermic
5.4 Enthalpies of Reaction
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Enthalpy of reaction (ΔHrxn): the enthalpy change
that accompanies a reaction
The enthalpy change for a chemical reaction is
given by the enthalpy of the products minus the
reactants:
ΔH = H (products) – H (reactants)
Thermochemical equations: balanced chemical
equations that show the associated enthalpy
change
The magnitude of ΔH is directly proportional to
the amount or reactant consumed in the process
5.4
Enthalpies of Reaction
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The enthalpy change for the reaction is equal
in magnitude but opposite in sign to ΔH for
the reverse reaction
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
CH4(g) +
2O2(g)
ΔH1=
ΔH2=
- 890
kJ
890 kJ
Enthalpy

Reversing a
reaction changes
the sign but not
the magnitude of
the enthalpy
change: ΔH2 = ΔH1
ΔH= 890 kJ
CO2(g) +
2H2O(l)
5.4
Enthalpies of Reaction
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The enthalpy change for a reaction depends
on the state of the reactants and products
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Ex: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
ΔH = -890 kJ
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If the product was H2O (g) instead of H2O (l), the ΔH
would be - 820 kJ instead of - 890 kJ
Calorimetry/Heat Capacity
& Specific Heat
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Calorimetry: the measure of heat flow
Calorimeter: measures heat flow
Heat capacity: the amount of heat required to raise
its temperature by 1 K
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The greater the heat capacity of a body, the greater the
heat required to produce a given rise in temperature
Molar heat capacity: the heat capacity of 1 mol of a
substance
Specific heat: the heat capacity of 1 g of a substance;
measured by temperature change (ΔT) that a known
mass (m) of the substance undergoes when it gains
or loses a specific quantity of heat (q)
5.5
Calorimetry/Heat Capacity &
Specific Heat
specific heat = quantity of heat transferred
(grams of substance) x (temperature change)
=q
m x ΔT
Practice Exercise (B&L page 160)
Calculate the quantity of heat absorbed by 50 kg of
rocks if their temperature increases by 12.0 OC.
(Assume the specific heat of the rocks is .82 J/gK.)
5.5
Calorimetry/Heat Capacity & Specific
Heat
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Solving the problem
q = (specific heat) x (grams of substance) x ΔT
= (0.82 J/g-K)(50,000 g)(285 K)
= 4.9 x 105 J
5.5
Constant-Pressure Calorimetry
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The heat gained by the solution (qsoln) is equal in magnitude
and opposite in sign from the heat of the reaction
qsoln = (specific heat of solution) x (grams of solution) x ΔT = -
qrxn
For dilute aqueous solutions, the specific heat of the solution
is approx. the same as water (4.18 J/g-K)
Practice Exercise (B&L page 161)
When 50 mL of .100M AgNO3 and 50.0 mL of .100 M HCl are
mixed in a constant-pressure calorimeter, the temperature of
the mixture increases from 22.30oC to 23.11oC. The
temperature increase is caused by this reaction:
AgNO3 + HCl  AgCl + HNO3
Calculate ΔH for this reaction, assuming that the combined
solution has a mass of 100 g and a specific heat of 4.18 J/g-oC
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Constant-Pressure
Calorimetry
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Solving the problem
qrxn = -(specific heat of solution) x (grams of solution) x ΔT
= - (4.18 J/g-oC)(100 g)(0.8 K)
= - 68,000 J/mol
5.5
Bomb Calorimetry (Constant-Volume Calorimetry)
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Bomb calorimeter: used to study combustion reactions
 Heat is released when combustion occurs, absorbed
by the calorimeter contents, raising the temperature of
the water (measured before and after the reaction)
To calculate the heat of combustion from the measured
temperature increase in the bomb calorimeter, you must
know the heat capacity of the calorimeter (Ccal)
qrxn = - Ccal x ΔT
5.6 Hess’s Law
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Hess’s Law: if a reaction is carried out in a
series of steps, ΔH for the reaction will be
equal to the sum of the enthalpy changes for
the individual steps
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) ΔH = -802 kJ
(ADD)2H2O (g)  2H2O (l)
ΔH = -88 kJ
CH4 (g) + 2O2 (g) + 2H2O (g) CO2 (g) + 2H2O (g) + 2H2O
(l)
ΔH = -890 kJ
5.6 Hess’s Law
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Practice Exercise 5.8 (B&L page 165)
Calculate ΔH for the reaction:
2C (s) + H2  C2H2 (g)
Given the following reactions and their respective enthalpy
changes:
C2H2 (g) + 5/2 O2  2CO2 (g) + H2O (l) ΔH = -1299.6 kJ
C (s) + O2 (g)  CO2 (g)
ΔH = -393.5 kJ
H2 (g) + ½ O2 (g)  H2O (l)
ΔH = -285.8 kJ
5.6 Hess’s Law
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Solving the problem
2CO2 (g) + H2O (l)  C2H2 (g) + 5/2 O2 ΔH = 1299.6 kJ
2 x [2C (s) + 2O2 (g)  2CO2 (g)]
ΔH = 2(-393.5 kJ)
H2 (g) + ½ O2 (g)  H2O (l)
ΔH = -285.8 kJ
2C (s) + H2  C2H2 (g)
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ΔH = 226.8 kJ
if the reaction is reversed, the sign of ΔH changes
if reaction is multiplied, so is ΔH
Enthalpy Diagram
The quantity of heat
generated by
combustion of 1 mol
CH4 is independent of
whether the reaction
takes place in one or
more steps:
ΔH1 = ΔH2 + ΔH3
5.7 Enthalpies of Formation
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Enthalpies of vaporization: ΔH for converting liquids to
gases
Enthalpies of fusion: ΔH for melting solids
Enthalpies of combustion: ΔH for combusting a
substance in oxygen
Enthalpy of formation (ΔHf): enthalpy change where the
substance has been formed from its elements
Standard enthalpy (ΔH o): enthalpy change when all
reactants and products are at 1 atm pressure and specific
temperature (298 K)
Standard enthalpy of formation (ΔHof): the enthalpy
change for the reaction that forms 1 mol of the
compound from its elements, with all substances in their
standard states
ΔHof = 0 for any element in its purest form at 295 K and
1 atm pressure
5.7 Enthalpies of Formation
The standard enthalpy change for any reaction
can be calculated from the summations of the
reactants and products in the reaction
ΔH orxn = n ΔH of (products) – m ΔH of (reactants)
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Practice Problem 5.9 (B&L page 169)
Calculate the standard enthalpy change for the
combustion of 1 mol of benzene, C6H6 (l ), to CO2
(g) and H2O (l).
5.7
Enthalpies of Formation
Solving the problem
C6H6 (l) + 15/2 O2 (g)  6CO2 (g) + 3H2O (l)
ΔH orxn= [6Δ H of (CO2) + 3Δ H of (H2O)] – [ΔH of
(C6H6) + 15/2 ΔH of (O2)]
= [6(-393.5 kJ) + 3(-285.8 kJ)] – [(49.0 kJ) +
15/2 (0 kJ)]
= (-2361 – 857.4 – 49.0) kJ
= -3267 kJ
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Extra Equations
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Force = mass x 9.8 m/s2
Internal energy… ΔE = Efinal – Einitial
Entropy… Δ S = Sfinal – Sinitial
Enthalpy… Δ H = Hfinal – Hinitial
Gibbs Free Energy… Δ G = Gfinal – Ginitial
ΔS=ΔH/T