Transcript Energy

Thermodynamics
Heat, disorder,
spontaneity
Energy
• The capacity to perform
work
–often measured as heat
Energy
• A tub is filled with water
at 35°C
–Dip a cup into the water
and fill it.
–What is the temperature
of the water in the cup?
Energy
• Which amount of water,
that in the tub or in the
cup, can melt the greater
amount of ice during the
same time frame?
Energy
• Two substances may have
the same temperature but
different amounts of heat
energy.
Energy
• Temperature is the
measure of average KE of
a substance
Energy
• Heat is the measure of
the total energy
transferred from an
object with a higher
temperature to an object
with a lower temperature.
Energy
• Heat is measured in either
Joules (J) or calories (cal)
• A calorie is defined as the
amount of heat needed to
raise 1 g of water 1°C.
• 1 cal = 4.18 J
Energy
• Graph the following data
for two experiments on
the same hand-drawn
graph.
Time for ice to melt…
Time
Temperature
8 cubes (s)
1 cube (s)
0°C
0
0
0°C (ice disappears)
25
190
25°C
33
250
50°C
41
310
75°C
49
370
100°C (water begins to
boil)
57
429
100°C (water
disappears)
226
1701
Time (s)
Energy
Temperature (°C)
Specific Heat
Capacity
• the amount of heat
energy required to raise
the temperature of one
gram of a substance by
one degree Celsius
• Measured in J/g°C or
cal/g°C
Specific Heat
Capacity
• When a substance’s SHC
(or C) is greater, more
heat is required to make
that substance equal in
temperature to a
substance with a lesser
SHC
Specific Heat
Capacity
• Which has the greater
SHC, silicone or iron?
heat = (T)(mass)(SHC)
Heating Curve for H2O
Temperature (°C)
H
G
G
Heat (cal)
Heating Curve for H2O
• BC has value of 80 cal/g
–Known as the heat of
fusion (sl) or heat of
solidification (ls)
• DE has a value of 540
cal/g
–Known as the heat of
vaporization (lg) or heat
of condensation (gl)
Heating Curve for H2O
• G has a value of 0°C
–known as the melting point
or the freezing point
• H has a value of 100°C
–Known as the boiling point
or the condensation point
Calorimetry
• Measurement of heat
energy
• Two types of calorimeters
–Constant pressure
(coffee-cup calorimeter)
–Constant volume (bomb
calorimeter)
Biological Calorimetry
• Nutrients
–Carbohydrates
–Proteins
–Lipids
–Water
–Vitamins
–minerals
Biological Calorimetry
• Carbohydrates
–4 kcal/g or 17 kJ/g
• Proteins
–4 kcal/g or 17 kJ/g
• Lipids
–9 kcal/g or 38 kJ/g
Heat of Reaction
• Hrxn
• amount of heat absorbed or
released in a chemical
reaction
• If absorbed, it is a reactant
and the process is
endothermic
• If released, it is a product and
the process is exothermic
Heat of Reaction
• Deviations
–Hformation is amount of
heat absorbed or released
during synthesis of one
mole of an element or
compound at
298 K and 1atm of pressure
Heat of Reaction
• Deviations
–Hsolution is amount of heat
absorbed or released when
a substance dissolves in a
solvent
Heat of Reaction
• Deviations
–Hcombustion is amount of
heat released when a
substance reacts with O2
to form CO2 and H2O
Heat of Reaction
• Is part of the stoichiometry of
a reaction…the heat of
combustion of methane is 803
kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 803 kJ
• If there were 5 moles of CH4
present, how many kJ would be
produced?
Heat Energy Practice
Problems
1. How many kJ are released by a
reaction that raises the
temperature of 1.00 kg of
water in a coffee-cup
calorimeter from 25.0°C to
27.0°C? Psst…you know the
SHC of water
2. A swimming pool measures 6.0 m x
12.0 m and is 3.0 m deep all
around. The pool is filled with
water at a temperature of
20.0°C. How many kJ must be
released by the pool’s heater to
raise the water temperature to
25.0°C? Psst…the density of
water is 1 g/cm3, you know the
SHC of water, and 1 m = 100 cm,
so 1 m3 would equal how many
cm3?
3. Gaseous butane, C4H10, is
burned in lots of lighters.
Write the balanced equation
for the complete combustion
of butane. Butane’s heat of
combustion is 2878 kJ. How
many kJ of heat energy would
be released by the combustion
of 10.0 g of butane?
4. Use the table on the next slide
to calculate the number of
kiloJoules provided by the fat
in one serving of each of the
following foods:
a. french fries
b. cheeseburger
4. (continued)
Food (amt.)
kcal
carb(g)
French fries 320 36.3
(3.4 oz.)
prot(g)
4.4
fat(g)
17.1
Cheeseburge 310 31.2 15.0 13.8
r
(4.1 oz.)
5. Is more energy released when
428 g of H2 or 428 g of
isooctane, C8H18, react with
an excess of oxygen?
Psst…balance the equations.
2H2 + O2  2H2O + 484 kJ
2C8H18 + 25O2  16CO2 + 18H2O + 4893 kJ
1.
2.
3.
4.
8.36 kJ
4.51 x 106 kJ
248 kJ
a. 650 kJ
b. 524 kJ
5. 428 g H2 will release 5.13 x 104
kJ while 428 g of C8H18 will
release 9.16 x 103 kJ. So, the
428 g H2 will release more
energy.
Activation Energy
Energy (kJ)
Ea
Reactants
H
Products
Rxn progress (s)
Activation Energy
Uncatalyzed
Energy (kJ)
Ea
Catalyzed
H
Rxn progress (s)
Enthalpy
• Enthalpy can be equated with heat
energy
• represented by H
•  H is also known as change in
enthalpy
Hess’s Law
• states that in going from a
particular set of reactants to a
particular set of products, the
change in enthalpy is the same
whether the reaction takes
place in one step or in a series
of steps.
Finding H using Hess’s Law
• If a reaction is reversed, the
sign of H is also reversed.
• If the coefficients in a balanced
equation are multiplied by an
integer, then the value of H is
multiplied by that same integer.
Hess’s Law
• Consider the following
reaction:
N2(g) + 2O2(g)  2NO2(g)
It does not necessarily occur as
we see it. It can, in fact, occur
in a few additive steps, known
as elementary steps.
Hess’s Law
Plausible elementary steps:
a. N2(g) + 2O2(g)  2NO(g)
H = 180 kJ
b. 2NO(g) + O2(g)  2NO2(g)
H = -112 kJ
Hess’s Law
a. N2(g) + O2(g)  2NO(g)
H = 180 kJ
b.
H = -112 kJ
2NO(g) + O2(g)  2NO2(g)
N2(g) + 2O2(g) + 2NO(g)  2NO(g) + 2NO2(g)
N2(g) + 2O2(g)  + 2NO2(g)
So, the reaction is endothermic.
H = 68 kJ
Hess’s Law
Two forms of carbon are graphite and
diamond. Using the enthalpies of
combustion for graphite and diamond as
your elementary steps, calculate the H for
the conversion of graphite to diamond and
state whether it is an endo- or exothermic
process.
C(graphite)(s)  C(diamond)(s)
H = ?
Hess’s Law
The elementary steps are:
a. C(graphite)(s) + O2(g)  CO2(g)
H = -394kJ
b. C(diamond)(s) + O2(g)  CO2(g) H = -396kJ
Hess’s Law
a. C(graphite)(s) + O2(g)  CO2(g)
H = -394kJ
b. CO2(g)  C(diamond)(s) + O2(g)
H = +396kJ
C(graphite)(s) + O2(g) + CO2(g)  CO2(g) + C(diamond)(s) + O2(g)
C(graphite)(s)  C(diamond)(s)
So, it is endothermic.
H = 2kJ
Finding H using standard
heats of formation
• H = ∑Hf°products − ∑Hf°reactants
• Use pages Zumdahl Chemistry II
textbook
• All elements in their natural states
will have Hf° equal to zero.
Finding H using standard
heats of formation
Find the H of the following
reaction using Hf° values:
N2(g) + 2O2(g)  2NO2(g)
Finding H using standard
heats of formation
(2mol NO2 x 34kJ)
mol
(1mol N2 x 0kJ) + (2mol O2 x 0kJ)
mol
mol
68kJ; the reaction is endothermic
Finding H using standard
heats of formation
Find the H of the reaction
which
converts graphite to diamond
using
Hf° values.
Entropy
• Is the measure of disorder or chaos
present in a substance.
• Chemical reactions may result in
increasing disorder or decreasing
disorder.
• Represented by S…thus, change in
entropy is S
Entropy
• When there are more moles of
products than reactants, entropy
usually increases.
• When phase changes from more
organized to less organized, entropy
increases.
• If S is positive, entropy increases;
if negative, entropy decreases.
Finding S using standard
entropy values
• S = ∑S°products − ∑S°reactants
• Use Zumdahl Chemistry II textbook
Finding S using standard
entropy values
Find the S of the following
reaction using S° values:
N2(g) + 2O2(g)  2NO2(g)
Finding S using standard
entropy values
(2mol NO2 x 240J)
K·mol
(1mol N2 x 192J) + (2mol O2 x 205J)
K·mol
K·mol
-122 J/K; entropy is decreasing
Finding S using standard
entropy values
Find the S of the reaction which
converts graphite to diamond
using
S° values.
Spontaneity
refers to whether a reaction will
happen
without outside intervention or not.
It says
nothing about how quickly the
reaction will
happen only that it will or will not
occur.
Free Energy
• is symbolized by G and is used to
determine the spontaneity of a
reaction
• G = H
TS
Free Energy
• If G is positive, it is a
nonspontaneous process and is
known as an endergonic reaction.
• If G is negative, it is a
spontaneous process and is known
as an exergonic reaction.
Free Energy
Find the G of the following
reaction at 25°C:
N2(g) + 2O2(g)  2NO2(g)
Free Energy
G = 68 kJ
(298 K)(-0.122 kJ)
K
G = 104 kJ; the reaction is
nonspontaneous or endergonic
Free Energy
Find the G of the reaction
which
converts graphite to diamond at
100°C and state whether it is
spontaneous or not.
More Practice Problems…
6. Acetylene gas, C2H2, is used in
some welding applications and can
be made via the following reaction:
2C(s) + H2(g)  C2H2(g)
Determine its H using the elementary
steps on the following slide.
a. C2H2(g) + 2½O2(g)  2CO2(g) + H2O(l)
H = -1300kJ
b. C(s) + O2(g)  CO2(g)
H = -394kJ
c. H2(g) + ½O2(g)  H2O(l)
H = -286kJ
7. Think about photosynthesis…
recall that carbon dioxide gas
reacts with water to produce solid
glucose (C6H12O6) and oxygen gas.
Write a balanced equation, and
determine the H, S, and G at
25°C. State whether the reaction
is endo- or exothermic, whether
entropy increases or decreases,
and whether it spontaneous or
not.
6. H = +226 kJ
7. H = +2802 kJ; endothermic
S = -262 J/K; entropy decreases
G = +2880 kJ; nonspontaneous