Download Chapter 5 Review

Thermochemistry
Chapter 5
Energy Changes in Chemical Reactions
Heat - the transfer of thermal energy between two bodies that
are at different temperatures
Temperature - a measure of the thermal energy
Temperature = Thermal Energy
(intensive)
90 °C
greater temperature
(extensive)
40 °C
greater thermal energy
Exothermic process - gives off heat – transfers thermal energy
from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process - heat has to be supplied to the system
from the surroundings.
energy + CaCO3 (s)
energy + H2O (s)
CaO (s) + CO2 (g)
H2O (l)
First law of thermodynamics – energy can be
converted from one form to another, but cannot
be created or destroyed.
DEsystem + DEsurroundings = 0
or
DEsystem = −DEsurroundings
C3H8 + 5O2
Internal
energy
of system
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
Energy relationships
Fig 5.4 (a)
Exothermic
Energy relationships
Fig 5.4 (b)
Endothermic
Energy Diagram for the Interconversion
of H2 (g), O2 (g), and H2O (l)
Fig 5.5
Another form of the first law for DEsystem
DE = q + w
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
Table 5.1 pg 171
Sign Conventions for Heat and Work
Fig 5.6
Thermodynamics
State functions - properties that are determined by the state of
the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
Fig 5.8
DE = Efinal - Einitial
Work Done By the System on the Surroundings
Expansion of a Gas:
DV = Vf – Vi > 0
-PDV < 0
wsys < 0
Since the opposing
pressure, P can vary:
Work is not a
state function!
Dw = wfinal - winitial
Vi
Vf
initial
final
Fig 5.12
Enthalpy (H) - the heat flow into or out of a system in a process
that occurs at constant pressure.
H = E + PV
DH = DE + PDV
DH = (q+w) − w
DH = qp
• So, at constant pressure, the change in
enthalpy is the heat gained or lost.
Enthalpy (H) - the heat flow into or out of a system in a
process that occurs at constant pressure.
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
Enthalpies of Reaction
DH = H (products) – H (reactants) = qp
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 0°C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ
Enthalpies of Reaction
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 25°C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ
The Truth about Enthalpy
•
Enthalpy is an extensive physical property.
2H2O (s)
•
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
DH for a reaction in the forward direction is equal
in magnitude, but opposite in sign, to DH for the
reverse reaction.
H2O (s)
H2O (l)
H2O (l)
DH = 6.01 kJ
H2O (s)
DH = -6.01 kJ
The Truth about Enthalpy
•
DH for a reaction depends on the state of the
products and the state of the reactants.
H2O (s)
H2O (l)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 155 g of iron undergoes
complete oxidation in air?
4Fe (s) + 3O2 (g)
155 g Fe x
1 mol Fe
55.85 g Fe
2Fe2O3 (s)
x
DH = -1118.4 kJ
-1118.4 kJ
= -776 kJ
4 mol Fe
The specific heat (s) - the amount of heat (q) required to raise
the temperature of one gram of the substance by one degree
Celsius:
Table 5.2
q
Cs 
m  DT
Heat (q) absorbed or released:
q = m·Cs·DT
where DT = Tfinal - Tinitial
How much heat is given off when an 869 g iron bar cools
from 94.0°C to 5.0°C?
Cs of Fe = 0.45 J/(g • °C)
DT = Tfinal – Tinitial = 5.0 °C – 94.0 °C = ‒ 89.0 °C
q = m·Cs·DT= (869 g) · (0.45 J/(g • °C)) · (– 89.0 °C)
= ‒ 34,800 J
≈ ‒ 3.5 x 104 J
= ‒ 35 kJ
Calorimetry: Measurement of Heat Changes
• H cannot be determined absolutely, however,
ΔH can be measured experimentally
• Device used is called a calorimeter
• Typically, ΔT ∝ ΔHrxn
Constant-Pressure Calorimetry
Figure 5.17
qsolution = m·Cs·DT = ‒ qrxn
Because reaction at constant P:
DH = qrxn
No heat enters or leaves!

Standard enthalpy of reaction ( DHrxn
) - the enthalpy of a
reaction carried out at 1 atm.
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.

DHrxn
=
S nDHof (products) - S mDHof (reactants)
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
Fig 5.20 An enthalpy diagram comparing a
one-step and a two-step process for a reaction.
Must I measure the enthalpy change for every reaction of
interest?
NO!!
Establish an arbitrary scale with the standard enthalpy of
formation ( DHf ) as a reference point for all enthalpy
expressions.
Standard enthalpy of formation ( DHf ) - the heat
change that results when one mole of a compound is
formed from its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero:
DHo (O2) ≡ 0
f
DHo (O3) = 142 kJ/mol
f
DHo (C, graphite) ≡ 0
f
DHo (C, diamond) = 1.90 kJ/mol
f

Table 5.3 Standard Enthalpies of Formation, DHrxn
at 298 K
Calcium carbide (CaC2) reacts with water to form acetylene
(C2H2) and Ca(OH)2. How much heat is released per mole
of calcium carbide reacted? The standard enthalpy of
formation of CaC2 is -62.76 kJ/mol.
CaC2 (l) + 2H2O (l)
C2H2 (g) + Ca(OH)2 (s)
DHo = S nDHo (products) ‒ S mDHo (reactants)
rxn
f
f
DHorxn = [DHof (C2H2) + DHof (Ca(OH)2] –[DHof(CaC2) +DHof (H2O)]
DHo = [(226.77 kJ) + (–986.2 kJ) ] –
rxn
[(-62.67 kJ) + 2(-285.83 ] = -125.1 kJ
-125.1 kJ
= -125.1 kJ/mol CaC2
1 mol CaC2