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Energy ch 16
1. Energy
Energy is an ability possessed by matter to do
work or exchange heat. That is, the capacity
to move matter or supply heat.
Energy exists in different forms which can be
converted
The S.I. units are the Joule (J)
2. Heat
Heat is the transfer of thermal energy from one object
to another because they are at different temperatures.
We usually use the letter q to represent this quantity
Heat is a method of energy transfer from a hot object
to a cold one
Thermal energy is the energy associated with the
random motion of the atoms and or molecules
Temperature change is a change in the average kinetic
energy of an object
3. Calorie
The amount of heat required to raise the
temperature of one gram of pure water by
one degree Celsius
one calorie = 4.184 Joules or one Joule = 0.239
calories
0ne Joule of energy is about the energy needed for a
single heartbeat
4. Potential energy (PE)
energy an object has as a result of its position or
composition.
The energy stored in chemical bonds is called
chemical potential energy
Natural gas, coal and other major sources of heat
have potential (chemical) energy due to their
composition
5. Kinetic energy (KE)
Kinetic energy - energy as object has as
a result of its motion.
K.E. = 1/2 mv2
6. Law of Conservation of Energy
Energy can be converted from one form to another,
but the total quantity of energy is assumed to
remain constant in ordinary processes
7. Measuring heat transfer
Specific heat
Defined as the quantity of heat required to
raise the temperature of 1.00 gram of the
substance by 1.00 oC – the book uses the
symbol C
for water, C = 4.184 J/goC
8. Specific Heat Formula
Changing the temperature of a sample from Ti to Tf
requires heat (q) = m C ∆T
∆T = Tf - Ti
C = specific heat for a particular substance
m = mass of the sample of matter
always have Tf first as the sign of ∆ T is important
q is + for an endothermic process - Tf > Ti
q is - for an exothermic process - Tf < Ti
9. Specific Heat Formula
Loss of heat for one object = gain in heat of the second
- q lost = q gained
- m C ∆T = m C ∆T
**Tip Hot side vs. Cold side**
Ti – Tf Tf - Ti
10. Sample problem
How much heat is required to raise the temperature of
100.0 g of Cu from 15.0oC to 35.0oC?
C for Cu is 0.384 J/goC
q= m C ∆T
q = 100.0 g *0.384 J/goC* (35.0 - 15.0)oC
q = 768 J
Notice that the value is ( + ) since heat was added to
the system
11. Sample problem
A 4.50 g nugget of pure gold absorbed 276 J of heat.
What was the final temperature of the gold if the
initial temperature was 25.0 oC? C for Au is 0.129 J/goC
q = m C ∆T
The final temperature is 500. oC
12. Thermochemistry
The study of the quantity of heat
absorbed or given off by chemical
reactions.
Or the study of heat change in
chemical reactions.
All chemical reactions either give off or
absorb energy
13. First Law of Thermodynamics
The total energy of the universe is a
constant.
Energy can be converted from one form
to another or transferred from a system
to the surroundings or vice versa.
14. First Law of Thermodynamics
Universe = system + surroundings
System = the specific part of the
universe that contains the reaction or
process you wish to study
Surroundings = everything in the
universe other than the system
15. The system is the specific part of the universe that is
of interest in the study.
SURROUNDINGS
SYSTEM
open closed isolated
Exchange: mass & energy energy nothing
6.2
16. Exothermic process is any process that
gives off heat or transfers thermal energy
from the system to the surroundings.
2H2 (g) + O2 (g) 2H2O (l) + energy
H2O (g) H2O (l) + energy
17. Endothermic process is any process in
which heat has to be supplied to the
system from the surroundings.
energy + 2HgO (s) 2Hg (l) + O2 (g)
energy + H2O (s) H2O (l)
18. Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or
absorbed during a reaction at constant pressure
19. The heat supplied is equal to the change in
another thermodynamic property called enthalpy
(H)
i.e. H = q
• this relation is only valid at constant pressure
As most reactions in chemistry take place at
constant pressure we can say that:
A change in enthalpy = heat supplied
22. Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H<0
890.4 kJ are released for every 1 mole of methane
that is combusted at 25.00C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ
23. Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H>0
6.01 kJ are absorbed for every 1 mole of ice that melts
at 00C and 1 atm.
H2O (s) H2O (l) H = 6.01 kJ
24. Thermochemical Equations
• The stoichiometric coefficients always refer to the
number of moles of a substance
H2O (s) H2O (l) H = 6.01 kJ
• If you reverse a reaction, the sign of H changes
H2O (l) H2O (s) H = -6.01 kJ
• If you multiply both sides of the equation by a factor n, then
H must change by the same factor n.
2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ
25. Thermochemical Equations
• The physical states of all reactants and products
must be specified in thermochemical equations.
H2O (s) H2O (l) H = 6.01 kJ
H2O (l) H2O (g) H = 44.0 kJ
26. Thermochemical Equations
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ
1 mol P4
266 g P4 x x 3013 kJ = 6470 kJ
123.9 g P4 1 mol P4
27. Phase Changes
Melting
phase change from solid to liquid
Freezing
phase change from liquid to solid
freezing point/melting point
temperature at which a liquid converts into
the solid state & vice versa
28. Melting
Energy is supplied to a solid to enable
it to vibrate more enthusiastically
until molecules can move past each
other and flow as a liquid
endothermic process ( H positive)
29. Freezing
Liquid releases energy and allows
molecules to settle into a lower
energy state and form a solid
exothermic process ( H negative)
(we remove heat from water when making ice in freezer)
30. Enthalpy of Fusion
Enthalpy of fusion (Hfus)
energy necessary to break enough IMF’s to
turn a solid into a liquid
heat absorbed by the substance in changing
solid into a liquid without raising its
temperature
Energy released when turning a liquid into
a solid = - Hfus
31. Enthalpy of Fusion Calculations
Enthalpy of fusion is used in calculating energy
needed to melt or released when freezing
q = m × H fus
q = energy
m = mass of sample or mol depends upon
information given on the chart
Hfus = enthalpy of fusion (melting)
(use (–Hfus ) for freezing)
32. Sample problem
Find the enthalpy of fusion of water if it takes 4175 J
to melt 12.5 g of water.
m = 12.5 g H2O
q = 4175 J
q = m × H fus
4175 J = (12.5 g) × H fus
Hfus = 4175 J / 12.5 g
Hfus = 334 J/g
33. Phase Changes
Vaporization
phase change from liquid to a gas
Condensation
phase change from gas to liquid
boiling point/condensation point
temperature at which a liquid converts into
the gas state & vice versa
34. Vaporization
Energy has to be supplied to a liquid
to enable it to overcome forces that
hold molecules together
endothermic process ( H positive)
35. Condensation
Energy has to be released from a gas
to enable it to hold molecules close
together
exothermic process ( H negative)
36. Enthalpy of Vaporization
Enthalpy of vaporization (Hvap)
The energy necessary to break the rest of
the IMF’s and turn a liquid into a gas
The amount of heat required to convert a liquid
at its boiling point into vapor without an
increase in temperature
Energy released when turning a gas into a
liquid = - Hvap
37. Enthalpy of Vaporization
Calculations
Enthlapy of vaporization is used in calculating
energy needed to boil or released when condensing
q = m × H vap
q = energy
m = mass of sample or mol depends upon
information given on the chart
Hvap = enthalpy of vaporization (boiling)
(use –Hvap for condensing)
38. Sample Problem
How much energy is required to vaporize 10.0 g of
water at its boiling point?
m = 10.0 g H2O
Hvap = 2260 J/g
q = m x Hvap
q = 10.0 g x 2260 J / g
q = 22600 J
39. Heating Curve for H2O
Each step along the line
requires a certain
amount of energy
Step 1 q1 = m C ∆T
Step 2 q2 = m H fus
Step 3 q3 = m C ∆T
Step 4 q4 = m H vap
Step 5 q5 = m C ∆T
Total energy qtot= q1+q2+q3+q4+q5
41. Phase Changes
Sublimation
conversion of a solid directly to a gas
Deposition
conversion of a gas directly to a solid
ΔHsublimation = -ΔHdeposition
42. Standard Enthalpies of Formation
The standard enthalpy of formation of a substance,
denoted Hfo, is the enthalpy change for the
formation of one mole of a substance in its
standard state from its component elements in
their standard state.
– Note that the standard enthalpy of formation for
a pure element in its standard state is zero.
Standard conditions = 25 0 C and 1.0 atm
45. Standard Enthalpies of Formation
The law of summation of heats of formation states
that the enthalpy of a reaction is equal to the total
formation energy of the products minus that of the
reactants.
o o o
H n H f (products ) m H f (reactants )
is the mathematical symbol meaning “the
sum of”, and m and n are the coefficients of
the substances in the chemical equation.
46. A Problem to Consider
You record the values of Hfo under the formulas in
the equation, multiplying them by the coefficients in
the equation.
4NH 3 (g ) 5O 2 (g ) 4NO(g ) 6H 2O(g )
4( 45.9) 5(0) 4(90.3) 6( 241.8)
- You can calculate Ho by subtracting the values
for the reactants from the values for the
products.
50. •A chemical reaction is spontaneous if it is accompanied by an increase in the
total entropy of the system and the surroundings
• Spontaneous exothermic reactions are common (e.g. hot metal block
spontaneously cooling) because they release heat that increases the entropy
of the surroundings.
•Endothermic reactions are spontaneous only when the entropy of the system
increases enough to overcome the decrease in entropy of the surroundings
51. A spontaneous change is a change that has a
tendency to occur without been driven by an
external influence
e.g. the cooling of a hot metal block to the
temperature of its surroundings
A non-spontaneous change is a change that occurs
only when driven
e.g. forcing electric current through a metal block
to heat it
52. Total entropy entropy change of entropy change of
change = system + surroundings
Dissolving
disorder of solution disorder of
surroundings
• must be an overall increase in disorder for dissolving to occur
53. Second Law of Thermodynamics:
the disorder (or entropy) of a system tends to
increase
ENTROPY (S)
•Entropy is a measure of disorder
• Low entropy (S) = low disorder
•High entropy (S) = greater disorder
• hot metal block tends to cool
• gas spreads out as much as possible
54. •A change in internal energy can be identified with the heat supplied at constant
volume
ENTHALPY (H)
(comes from Greek for “heat inside”)
• the change in internal energy is not equal to the heat supplied when the
system is free to change its volume
• some of the energy can return to the surroundings as expansion work
U < q
