Transcript Document

John E. McMurry • Robert C. Fay
General Chemistry: Atoms First
Chapter 8
Thermochemistry: Chemical Energy
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
Copyright © 2010 Pearson Prentice Hall, Inc.
Energy and Its Conservation
Energy: The capacity to supply heat or do work.
Kinetic Energy (EK): The energy of motion.
Potential Energy (EP): Stored energy.
Chapter 8/2
Energy and Its Conservation
Law of Conservation of Energy: Energy cannot be
created or destroyed; it can only be converted from one
form to another.
Chapter 8/3
Energy and Its Conservation
Total Energy = EK + EP
Energy and Its Conservation
Thermal Energy: The kinetic energy of molecular
motion, measured by finding the temperature of an
object.
Heat: The amount of thermal energy transferred from
one object to another as the result of a temperature
difference between the two.
Chapter 8/5
Energy and Its Conservation
First Law of Thermodynamics: Energy cannot be
created or destroyed; it can only be converted from one
form to another.
Chapter 8/6
Internal Energy and State
Functions
First Law of Thermodynamics: The total internal
energy of an isolated system is constant.
∆E = Efinal - Einitial
Internal Energy and State
Functions
First Law of Thermodynamics: The total internal
energy of an isolated system is constant.
∆E = Efinal - Einitial
Internal Energy and State
Functions
First Law of Thermodynamics: The total internal
energy of an isolated system is constant.
∆E = Efinal - Einitial
Internal Energy and State
Functions
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) + 802 kJ energy
∆E = Efinal - Einitial = -802 kJ
802 kJ is released when 1 mole of methane, CH4, reacts
with 2 moles of oxygen to produce 1 mole of carbon
dioxide and two moles of water.
Chapter 8/10
Internal Energy and State
Functions
State Function: A function or property whose value
depends only on the present state, or condition, of the
system, not on the path used to arrive at that state.
Expansion Work
Work = Force x Distance
w=Fxd
Expansion Work
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
6 mol of gas
7 mol of gas
Expansion Work
Expansion Work: Work done as the result of a volume
change in the system. Also called pressure-volume or
PV work.
Energy and Enthalpy
∆E = q + w
q = heat transferred
w = work = -P∆V
q = ∆E + P∆V
Constant Volume (∆V = 0): qV = ∆E
Constant Pressure: qP = ∆E + P∆V
Chapter 8/15
Energy and Enthalpy
qP = ∆E + P∆V = ∆H
Enthalpy change
or
Heat of reaction (at constant pressure)
Enthalpy is a state function whose
value depends only on the current
state of the system, not on the
path taken to arrive at that state.
∆H = Hfinal - Hinitial
= Hproducts - Hreactants
Chapter 8/16
The Thermodynamic Standard
State
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
∆H = -2044 kJ
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
∆H = -2220 kJ
Thermodynamic Standard State: Most stable form of a
substance at 1 atm pressure and at a specified
temperature, usually 25 °C; 1 M concentration for all
substances in solution.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
∆H° = -2044 kJ
Chapter 8/17
Enthalpies of Physical and
Chemical Change
Enthalpy of Fusion (∆Hfusion): The amount of heat
necessary to melt a substance without changing its
temperature.
Enthalpy of Vaporization (∆Hvap): The amount of heat
required to vaporize a substance without changing its
temperature.
Enthalpy of Sublimation (∆Hsubl): The amount of heat
required to convert a substance from a solid to a gas
without going through a liquid phase.
Chapter 8/18
Enthalpies of Physical and
Chemical Change
Chapter 8/19
Enthalpies of Physical and
Chemical Change
2Al(s) + Fe2O3(s)
2Fe(s) + Al2O3(s)
∆H° = -852 kJ
Exothermic: Heat (enthalpy) flows from the
system to the surroundings.
Ba(OH)2 •8H2O(s) + 2NH4Cl(s)
BaCl2(aq) + 2NH3(aq) + 10H2O(l)
∆H° = +80.3 kJ
Endothermic: Heat (enthalpy) flows from the
surroundings to the system.
Chapter 8/20
Enthalpies of Physical and
Chemical Change
2Al(s) + Fe2O3(s)
2Fe(s) + Al2O3(s)
∆H° = -852 kJ
Exothermic: Heat (enthalpy) flows from the
system to the surroundings.
2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s)
∆H° = +852 kJ
Endothermic: Heat (enthalpy) flows from the
surroundings to the system.
Chapter 8/21
Calorimetry and Heat Capacity
Measure the heat flow at constant pressure (∆H).
Calorimetry and Heat Capacity
Measure the heat flow at constant volume (∆E).
Calorimetry and Heat Capacity
Heat Capacity (C): The amount of heat required to raise
the temperature of an object or substance a given
amount.
q = C x ∆T
Molar Heat Capacity (Cm): The amount of heat required
to raise the temperature of 1 mol of a substance by 1 °C.
q = (Cm) x (moles of substance) x ∆T
Specific Heat: The amount of heat required to raise the
temperature of 1 g of a substance by 1 °C.
q = (specific heat) x (mass of substance) x ∆T
Chapter 8/24
Calorimetry and Heat Capacity
Calorimetry and Heat Capacity
Assuming that a can of soda has the same specific
heat as water, calculate the amount of heat (in
kilojoules) transferred when one can (about 350 g) is
cooled from 25 °C to 3 °C.
q = (specific heat) x (mass of substance) x ∆T
J
Specific heat = 4.18
g °C
Mass = 350 g
Temperature change = 25 °C - 3 °C = 22 °C
Chapter 8/26
Calorimetry and Heat Capacity
Calculate the amount of heat transferred.
4.18 J x 350 g x 22 °C
Heat evolved =
= 32 000 J
g °C
32 000 J x
1 kJ
= 32 kJ
1000 J
Chapter 8/27
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction
is equal to the sum of the enthalpy changes for the
individual steps in the reaction.
Haber Process: 3H2(g) + N2(g)
2NH3(g) ∆H° = -92.2 kJ
Multiple-Step Process
2H2(g) + N2(g)
N2H4(g)
∆H°1 = ?
N2H4(g) + H2(g)
2NH3(g)
∆H°2 = -187.6 kJ
3H2(g) + N2(g)
2NH3(g)
∆H°1+2 = -92.2 kJ
Chapter 8/28
Hess’s Law
∆H°1 + ∆H°2 = ∆H°1+2
∆H°1 = ∆H°1+2 - ∆H°2
= -92.2 kJ - (-187.6 kJ) = 95.4 kJ
Standard Heats of Formation
Standard Heat of Formation (∆H°f ): The enthalpy
change for the formation of 1 mol of a substance in its
standard state from its constituent elements in their
standard states.
Standard states
C(s) + 2H2(g)
CH4(g)
∆H°f = -74.8 kJ
1 mol of 1 substance
Chapter 8/30
Standard Heats of Formation
Chapter 8/31
Standard Heats of Formation
∆H° = ∆H°f (Products) - ∆H°f (Reactants)
aA + bB
cC + dD
∆H° = [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)]
Products
Reactants
Chapter 8/32
Standard Heats of Formation
Using standard heats of formation, calculate the
standard enthalpy of reaction for the photosynthesis of
glucose (C6H12O6) and O2 from CO2 and liquid H2O.
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
∆H° = ?
H° = [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]
∆H° = [(1 mol)(-1273.3 kJ/mol)] [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
= 2802.5 kJ
Chapter 8/33
Standard Heats of Formation
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ
(1)
(2)
(3)
Why does the calculation “work”?
1. Reverse the “reaction” and reverse the sign on
the standard enthalpy change.
C(s) + O2(g)
CO2(g)
∆H° = -393.5 kJ
C(s) + O2(g)
∆H° = 393.5 kJ
Becomes
CO2(g)
Chapter 8/34
Standard Heats of Formation
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ
(1)
(2)
(3)
Why does the calculation “work”?
1. Multiply the coefficients by a factor and multiply
the standard enthalpy change by the same factor.
CO2(g)
C(s) + O2(g)
∆H° = -393.5 kJ
Becomes
6CO2(g)
6C(s) + 6O2(g) ∆H° = 6(393.5 kJ) = 2361.0 kJ
Chapter 8/35
Standard Heats of Formation
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ
(1)
(2)
(3)
Why does the calculation “work”?
2. Reverse the “reaction” and reverse the sign on
the standard enthalpy change.
1
H2(g) + O2(g)
2
H2O(l)
∆H° = -285.8 kJ
Becomes
H2O(l)
1
H2(g) + O2(g) ∆H° = 285.8 kJ
2
Chapter 8/36
Standard Heats of Formation
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ
(1)
(2)
(3)
Why does the calculation “work”?
2. Multiply the coefficients by a factor and multiply
the standard enthalpy change by the same factor.
H2O(l)
1
H2(g) + O2(g)
2
∆H° = 285.8 kJ
6H2(g) + 3O2(g)
∆H° = 6(285.8 kJ) = 1714.8 kJ
Becomes
6H2O(l)
Chapter 8/37
Standard Heats of Formation
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g) ∆H° = 2802.5 kJ
(1)
(2)
(3)
Why does the calculation “work”?
C6H12O6(s)
∆H° = -1273.3 kJ
6CO2(g)
6C(s) + 6O2(g)
∆H° = 2361.0 kJ
6H2O(l)
6H2(g) + 3O2(g)
∆H° = 1714.8 kJ
6C(s) + 6H2(g) + 3O2(g)
6CO2(g) + 6H2O(l)
C6H12O6(s) + 6O2(g)
∆H° = 2802.5 kJ
Chapter 8/38
Bond Dissociation Energies
Bond Dissociation Energy:
•
The amount of energy that must be supplied to
break a chemical bond in an isolated molecule in
the gaseous state and is thus the amount of energy
released when the bond forms.
or
•
Standard enthalpy changes for the corresponding
bond-breaking reactions.
Chapter 8/39
Bond Dissociation Energies
Bond Dissociation Energies
H2(g) + Cl2(g)
2HCl(g)
∆H° = D(Reactant bonds) - D(Product bonds)
∆H° = (DH-H + DCl-Cl) - (2DH-Cl)
∆H° = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] -
(2mol)(432 kJ/mol)
= -185 kJ
Chapter 8/41
Fossil Fuels, Fuel Efficiency,
and Heats of Combustion
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
∆H°c = [∆H°f (CO2(g)) + 2 ∆H°f (H2O(l))] - [∆H°f (CH4(g))]
= [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] [(1 mol)(-74.8 kJ/mol)]
= -890.3 kJ
Chapter 8/42
Fossil Fuels, Fuel Efficiency,
and Heats of Combustion
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
An Introduction to Entropy
Spontaneous Process: A process that, once
started, proceeds on its own without a continuous
external influence.
Entropy (S): The amount of molecular randomness
in a system.
Spontaneous processes are
• favored by a decrease in H (negative ∆H).
• favored by an increase in S (positive ∆S).
Nonspontaneous processes are
• favored by an increase in H (positive ∆H).
• favored by a decrease in S (negative ∆S).
Chapter 8/44
An Introduction to Free Energy
Gibbs Free Energy Change (∆G)
∆G = ∆H - T ∆S
Enthalpy of
reaction
Temperature
(Kelvin)
Entropy
change
Chapter 8/46
An Introduction to Free Energy
Gibbs Free Energy Change (∆G)
∆G = ∆H - T ∆S
∆G < 0
Process is spontaneous
∆G = 0
Process is at equilibrium
(neither spontaneous nor nonspontaneous)
∆G > 0
Process is nonspontaneous
Chapter 8/47
An Introduction to Free Energy