Chemistry 1A
Chapter 6
Some Chemical Changes
Release Energy
Combustion of Methane
CH4(g) + 2O2(g)
→ CO2(g) + 2H2O(l) +
Some Chemical
Changes Absorb
Energy
Energy Terms
• Energy = the capacity to do work
• Work, in this context, may be defined as what is
done to move an object against some sort of
resistance.
Two Types of Energy
• Kinetic Energy = the energy of motion
= 1/2 mμ2
• Potential Energy = energy by virtue of
position or state
Law of Conservation of
Energy
Endergonic Change
more stable + energy → less stable system
lesser capacity + energy → greater capacity
to do work
to do work
lower PE + energy → higher PE
coin in hand + energy → coin in air above hand
Coin and Potential Energy
Bond Breaking and
Potential Energy
Exergonic Change
less stable system → more stable + energy
greater capacity → lesser capacity + energy
to do work
to do work
higher PE → lower PE + energy
coin in air above hand → coin on ground + energy
Bond Making and
Potential Energy
Which higher energy? Is
it kinetic or potential?
• 428 m/s Ar atoms or 456 m/s Ar atoms?
• 428 m/s Ar atoms or 428 m/s Kr atoms?
• Na+ close to Cl− or Na+ and Cl− far
apart?
• ROOR or 2 RO
• H(g) and O2(g) or HO2(g)
• Solid CO2 or gaseous CO2
Units of Energy
kg m
• Joule (J) =
s2
•
•
•
•
2
4.184 J = 1 cal
4.184 kJ = 1 kcal
4184 J = 1 Cal (dietary calorie)
4.184 kJ = 1 Cal
Approximate Energy of
Various Events
More Terms
• External Kinetic Energy = Kinetic
energy associated with the overall
movement of a body
• Internal Kinetic Energy = Kinetic
energy associated with the random
motion of the particles within a body
External and Internal Kinetic Energy
Heat
• Heat = Energy transfer from a
region of higher temperature to a
region of lower temperature due to
collisions of particles.
Heat
Transfer
Radiant Energy
• Radiant Energy is electromagnetic
energy that behaves like a stream of
particles.
• It has a dual Nature
– Particle
• photons = tiny packets of radiant energy
• 1017 photons/second from a flashlight bulb
– Wave
• oscillating electric and magnetic fields
• describes effect on space, not true nature
of radiant energy
A Light Wave’s Electric
and Magnetic Fields
Radiant
Energy
Spectrum
Endergonic Change
more stable + energy → less stable system
lesser capacity + energy → greater capacity
to do work
to do work
lower PE + energy → higher PE
Exergonic Change
less stable system → more stable + energy
greater capacity → lesser capacity + energy
to do work
to do work
higher PE → lower PE + energy
Bond Breaking and
Potential Energy
Bond Making and
Potential Energy
Exergonic (Exothermic)
Reaction
weaker bonds → stronger bonds + energy
less stable → more stable + energy
higher PE
→
lower PE + energy
Exothermic Reaction
Endothermic Reaction
stronger bonds + energy → weaker bonds
more stable + energy → less stable
lower PE + energy → higher PE
NH4NO3(s) + energy → NH4+(aq) + NO3−(aq)
Energy and Chemical
Reactions
Factors that Affect Heats
of Chemical Reactions
• Nature of the reaction, including the
states of the reactants and products
• Amount of reactant
• Conditions for the reaction
– Constant pressure or constant volume
– Temperature and pressure
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
ΔH° = −3.08 x 103 kJ
Conversion Factors from ΔH°
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
ΔH° = −3.08 x 103 kJ
Relationship Between
ΔH and the Chemical
Equation
• Whatever you do to the equation, do
the same to the ΔH.
– Reverse the equation…change the
sign of the ΔH.
– Multiply the coefficients by some
number…multiply the ΔH by the same
number.
Gas
Gas Model
• Gases are composed of tiny,
widely-spaced particles.
– For a typical gas, the average
distance between particles is about
ten times their diameter.
Gas Model (cont.)
• Because of the large distance between the
particles, the volume occupied by the particles
themselves is negligible (approximately zero).
– For a typical gas at room temperature and
pressure, the gas particles themselves
occupy about 0.1% of the total volume. The
other 99.9% of the total volume is empty
space. This is very different than for a liquid
for which about 70% of the volume is
occupied by particles.
Gas Model (cont.)
• The particles have rapid and
continuous motion.
– For example, the average velocity of a
helium atom, He, at room temperature is
over 1000 m/s (or over 2000 mi/hr). The
average velocity of the more massive
nitrogen molecules, N2, at room
temperature is about 500 m/s.
– Increased temperature means increased
average velocity of the particles.
Gas Model (cont.)
• The particles are constantly colliding
with the walls of the container and with
each other.
– Because of these collisions, the gas
particles are constantly changing their
direction of motion and their velocity. In a
typical situation, a gas particle moves a very
short distance between collisions. Oxygen,
O2, molecules at normal temperatures and
pressures move an average of 10−7 m
between collisions.
Gas Model (cont.)
• There is no net loss of energy in
the collisions. A collision between
two particles may lead to each
particle changing its velocity and
thus its energy, but the increase in
energy by one particle is balanced
by an equal decrease in energy by
the other particle.
Ideal Gas
• The particles are assumed to be
point-masses, that is, particles
that have a mass but occupy no
volume.
• There are no attractive or
repulsive forces at all between
the particles.
Gas Properties and their
Units
• Pressure (P) = Force/Area
– units
• 1 atm = 101.325 kPa = 760 mmHg = 760 torr
• 1 bar = 100 kPa = 0.9869 atm = 750.1 mmHg
• Volume (V)
– unit usually liters (L)
• Temperature (T)
– ? K = --- °C + 273.15
• Number of gas particles expressed in
moles (n)
Decreased Volume Leads
to Increased Pressure
P α 1/V
if n and T are constant
Relationship
between
P and V
Boyle’s Law
• The pressure of an ideal gas is inversely
proportional to the volume it occupies if
the moles of gas and the temperature are
constant.
Increased Temperature Leads
to Increased Pressure
PαT
if n and V are constant
Relationship between
P and T
Gay-Lussac’s Law
• The pressure of an ideal gas is directly
proportional to the Kelvin temperature of the
gas if the volume and moles of gas are
constant.
Increased Moles of Gas Leads
to Increased Pressure
Pαn
if T and V are constant
Relationship between n and P
Relationship Between
Moles of Gas and Pressure
• If the temperature and the volume of an ideal
gas are held constant, the moles of gas in a
container and the gas pressure are directly
proportional.
Ideal Gas Equation
Constant Pressure
or Volume
• Heat at Constant Pressure = ΔH
= change in enthalpy
• Heat at Constant Volume = ΔE
= change in the total energy
E = KE + PE
Standard Heat
Changes
• ΔH at 298.15 K and 1 atm = ΔH°
• ΔE at 298.15 K and 1 atm = ΔE°
Standard Changes in Enthalpy
(ΔH°) and Total Internal Energy
(ΔE°)
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
ΔH° = −3.08 x 103 kJ
ΔE° = −3.09 x 103 kJ
ΔH° = Heat at constant pressure at 25 °C and
1 atm
ΔE° = Heat at constant volume at 25 °C and
1 atm
Heat at
Constant
Pressure
and Volume
(Example 1)
Sign Conventions
• Heat evolved is negative.
• Heat absorbed is positive.
• Work done by the system
leads to heat lost by the
system, so it is negative.
• Work done on the system
adds heat to the system, so it
is positive.
Heat at
Constant
Pressure
and Volume
(Example 2)
Heat at
Constant
Pressure
and
Volume
(Example 3)
Goal: to Convert from
ΔE° to ΔH°
• ΔE° is easier to determine
accurately in the laboratory.
• Because most chemical
reactions are run at constant
pressure, ΔH° values are
usually used to describe heats
of reaction.
Conversion from ΔE to ΔH
• We saw for the KClO3 and NH3
reactions, that
qV = qrxn
qP = qrxn + qwork
• So
qP = qV + qwork
ΔH = ΔE + w
• For chemical changes at constant
pressure and temperature, the only
work is work done by a gas as it
expands (negative) or on a gas as it is
compressed (positive).
Relationship between Heat at
Constant and Constant Pressure
ΔH = ΔE + w
ΔH = ΔE + FΔd
ΔH = ΔE + P•ΔV
Relationship between Heat at
Constant and Constant
Pressure (2)
ΔH = ΔE + PΔV
PV = nRT (ideal gas equation)
– If pressure and temperature are kept
constant, the only way to increase the
volume is to increase moles of gas.
PΔV = (Δn)RT
ΔH = ΔE + (Δn)RT
Δn = ∑ moles (g) products − ∑ moles (g) reactants
Δn = ∑ coef. (g) products − ∑ coef. (g) reactants
Heat at Constant
Pressure Conventions
• ΔH° describes kJ per mole of
primary reactant (or kJ per the
number of moles equal to the
coefficient in the balanced
equation for each reactant and
product).
• qP describes kJ per amount of
substance in a change.
Heat at Constant
Volume Conventions
• ΔE° describes kJ per mole of
primary reactant (or kJ per the
number of moles equal to the
coefficient in the balanced
equation for each reactant and
product).
• qV describes kJ per amount of
substance in a change.
Ways to get ΔH°
• From Tables (ΔH° for
combustion or ΔH° of formation)
• From calorimeter data (bomb or
open)
• From other ΔH°s
– General Law of Hess
– Heat of formation problem
Bomb Calorimeter
Exercise
Derivation of Calorimeter
Equation
qlost = −qgained
qrxn = −[qproducts + qcalorimeter + qwater + qsurr]
qrxn ≅ −[qcalorimeter + qwater]
qrxn = −[qcal + qw]
Heat Capacity and
Specific Heat (Capacity)
• Heat capacity, C = the heat energy (kJ)
necessary to raise the temperature of an
object (such as a calorimeter) by 1 ºC (or
1K).
– Units of kJ/ºC or kJ/K…it’s the same number
for each unit
• Specific heat (capacity), c = the heat
energy (kJ) necessary to raise the
temperature of 1 g of a substance (such
as water) by 1 ºC (or 1K).
– Units of kJ/g•ºC or kJ/g•K…it’s the same
number for each unit
Bomb Calorimeter Problems
– Calculating ΔH° - Part 1
• Given – mass of reactant, mass of H2O,
Tinitial, Tfinal, and Ccal
• Steps
– Assign variables
– Calculate qV
ΔT = T2 – T1
Watch signs.
Bomb Calorimeter Problems
– Calculating ΔH° - Part 2
– Calculate ΔE°
– Calculate ΔH°
Bomb Calorimeter Problems –
Calculating Ccal - Part 1
• Given – mass of reactant, mass of H2O,
Tinitial, Tfinal, and ΔH°
• Steps
– Assign variables
– Calculate ΔE°
Calorimeter Problems –
Calculating CCal - Part 2
– Calculate qV
– Calculate Ccal
Law of Hess
• If a reaction can be viewed as a
sum of two or more equations,
the ΔH° for the net reaction is
equal to the sum of the ΔH°’s for
the intermediate reactions.
Law of Hess Example
General Law of Hess
Problems
• ΔH°net = Σ ΔH°intermediate
• Write intermediate equations and their
ΔH°’s.
• Rearrange intermediate equations so
they add to yield the desired net
equation.
• Whatever you did to the intermediate
equations, so the same to their ΔH°’s.
• Add the new intermediate ΔH°’s.
Heat of Formation
• ΔH°f = Heat at constant
pressure for the formation of
one mole of substance from its
elements in their standard
states (at 298.15 K and 1 atm).
Standard States of
Elements
• Metals – Hg(l) or Symbol(s) – e.g.
Zn(s)
• Noble gases – Symbol(g) – e.g.
Ne(g)
• Diatomic elements – H2(g), N2 (g),
O2(g), F2(g), Cl2(g), Br2(l), I2(s)
• Carbon – C(graphite)
• Other elements – S8(s), Se8(s),
P4(s), As4(s), Sb4(s)
Heat of Formation Equation
• Chemical reactions could take place in two steps.
– Conversion from reactants to elements in their standard states.
ΔH°1 = - ΣΔH°f (reactants)
– Conversion of elements in their standard states to products.
ΔH°2 = ΣΔH°f (products)
• The overall ΔH°rxn would be equal to the sum of the ΔH°s
for the two steps.
ΔH°rxn = ΔH°1 + ΔH°2 = ΔH°2 + ΔH°1
ΔH°rxn = ΣΔH°f (products) + (−ΣΔH°f (reactants))
ΔH°rxn = ΣΔH°f (products) − ΣΔH°f (reactants)