Energy - PowerPoint PPT Presentation

1 / 39
About This Presentation
Title:

Energy

Description:

This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 4 H 4 e ... V, T, P, energy and your bank account! ... – PowerPoint PPT presentation

Number of Views:22
Avg rating:3.0/5.0
Slides: 40
Provided by: jko28
Learn more at: http://www.chem.tamu.edu
Category:
Tags: energy

less

Transcript and Presenter's Notes

Title: Energy


1
Energy Chemistry
  • 2H2(g) O2(g) ? 2H2O(g) heat and light
  • This can be set up to provide ELECTRIC ENERGY in
    a fuel cell.
  • Oxidation
  • 2 H2 ? 4 H 4 e-
  • Reduction
  • 4 e- O2 2 H2O ? 4 OH-

H2/O2 Fuel Cell Energy, page 288
2
Energy Chemistry
  • ENERGY is the capacity to do work or transfer
    heat.
  • HEAT is the form of energy that flows between 2
    objects because of their difference in
    temperature.
  • Other forms of energy
  • light
  • electrical
  • kinetic and potential
  • Positive and negative particles (ions) attract
    one another.
  • Two atoms can bond
  • As the particles attract they have a lower
    potential energy

NaCl composed of Na and Cl- ions.
3
Potential Kinetic Energy
Kinetic energy energy of motion.
4
(No Transcript)
5
Internal Energy (E)
  • PE KE Internal energy (E or U)
  • Internal Energy of a chemical system depends on
  • number of particles
  • type of particles
  • temperature
  • The higher the T the higher the internal energy
  • So, use changes in T (?T) to monitor changes in E
    (?E).

6
Thermodynamics
  • Thermodynamics is the science of heat (energy)
    transfer.

Heat transfers until thermal equilibrium is
established. ?T measures energy transferred.
  • SYSTEM
  • The object under study
  • SURROUNDINGS
  • Everything outside the system

7
Directionality of Heat Transfer
  • Heat always transfer from hotter object to cooler
    one.
  • EXOthermic heat transfers from SYSTEM to
    SURROUNDINGS.

8
Directionality of Heat Transfer
  • Heat always transfers from hotter object to
    cooler one.
  • ENDOthermic heat transfers from SURROUNDINGS to
    the SYSTEM.

9
Energy Chemistry
  • All of thermodynamics depends on the law of
  • CONSERVATION OF ENERGY.
  • The total energy is unchanged in a chemical
    reaction.
  • If PE of products is less than reactants, the
    difference must be released as KE.

Energy Change in Chemical Processes
Potential Energy of system dropped. Kinetic
energy increased. Therefore, you often feel a
Temperature increase.
10
HEAT CAPACITY
  • The heat required to raise an objects T by 1 C.

Which has the larger heat capacity?
11
Specific Heat Capacity
  • How much energy is transferred due to Temperature
    difference?
  • The heat (q) lost or gained is related to
  • a) sample mass
  • b) change in T and
  • c) specific heat capacity

12
Table of specific heat capacities
Substance Phase cp J g-1 K-1 Cp J mol-1 K-1
Air (typical room conditionsA) gas 1.012 29.19
Aluminium solid 0.897 24.2
Argon gas 0.5203 20.7862
Copper solid 0.385 24.47
Diamond solid 0.5091 6.115
Ethanol liquid 2.44 112
Gold solid 0.1291 25.42
Graphite solid 0.710 8.53
Helium gas 5.1932 20.7862
Hydrogen gas 14.30 28.82
Iron solid 0.450 25.1
Lithium solid 3.58 24.8
Mercury liquid 0.1395 27.98
Nitrogen gas 1.040 29.12
Neon gas 1.0301 20.7862
Oxygen gas 0.918 29.38
Uranium solid 0.116 27.7
Water gas (100 C) 2.080 37.47
Water liquid (25 C) 4.1813 75.327
Water solid (0 C) 2.114 38.09
All measurements are at 25 C unless noted. Notable minimums and maximums are shown in maroon text. All measurements are at 25 C unless noted. Notable minimums and maximums are shown in maroon text. All measurements are at 25 C unless noted. Notable minimums and maximums are shown in maroon text. All measurements are at 25 C unless noted. Notable minimums and maximums are shown in maroon text.
Aluminum
A Assuming an altitude of 194 meters above mean
sea level (the worldwide median altitude of
human habitation), an indoor temperature of 23
C, a dewpoint of 9 C (40.85 relative
humidity), and 760 mmHg sea levelcorrected
barometric pressure (molar water vapor content
1.16).
13
Specific Heat Capacity
  • If 25.0 g of Al cool from 310 oC to 37 oC, how
    many joules of heat energy are lost by the Al?

14
Heat/Energy TransferNo Change in State
  • q transferred (sp. ht.)(mass)(?T)

15
Heat Transfer
  • Use heat transfer as a way to find specific heat
    capacity, Cp
  • 55.0 g Fe at 99.8 C
  • Drop into 225 g water at 21.0 C
  • Water and metal come to 23.1 C
  • What is the specific heat capacity of the metal?

16
Heating/Cooling Curve for Water
Note that T is constant as ice melts or water
boils
17
Chemical Reactivity
  • But energy transfer also allows us to predict
    reactivity.
  • In general, reactions that transfer energy to
    their surroundings are product-favored.

So, let us consider heat transfer in chemical
processes.
18
FIRST LAW OF THERMODYNAMICS
  • ?E q w

Energy is conserved!
19
The First Law of Thermodynamics
  • Exothermic reactions generate specific amounts of
    heat.
  • This is because the potential energies of the
    products are lower than the potential energies of
    the reactants.

20
The First Law of Thermodynamics
  • There are two basic ideas of importance for
    thermodynamic systems.
  • Chemical systems tend toward a state of minimum
    potential energy.
  • Chemical systems tend toward a state of maximum
    disorder.
  • The first law is also known as the Law of
    Conservation of Energy.
  • Energy is neither created nor destroyed in
    chemical reactions and physical changes.

21
SYSTEM
?E q w
22
ENTHALPY
  • Most chemical reactions occur at constant P, so

Heat transferred at constant P qp qp ?H
where H enthalpy
and so ?E ?H w (and w is usually small) ?H
heat transferred at constant P ?E ?H change
in heat content of the system ?H Hfinal -
Hinitial
23
ENTHALPY
  • ?H Hfinal - Hinitial

If Hfinal gt Hinitial then ?H is positive Process
is ENDOTHERMIC
If Hfinal lt Hinitial then ?H is negative Process
is EXOTHERMIC
24
USING ENTHALPY
  • Consider the formation of water
  • H2(g) 1/2 O2(g) ? H2O(g) 241.8 kJ

Exothermic reaction heat is a product and ?H
241.8 kJ
25
USING ENTHALPY
  • Making liquid H2O from H2 O2 involves two
    exothermic steps.

H2 O2 gas
H2O vapor
Liquid H2O
Making H2O from H2 involves two steps. H2(g)
1/2 O2(g) ? H2O(g) 242 kJ
H2O(g) ? H2O(l) 44 kJ H2(g) 1/2 O2(g) ?
H2O(l) 286 kJ Example of HESSS LAW If a rxn.
is the sum of 2 or more others, the net ?H is the
sum of the ?Hs of the other rxns.
26
Hesss Law Energy Level Diagrams
Forming H2O can occur in a single step or in a
two steps. ?Htotal is the same no matter which
path is followed.
Active Figure 6.18
27
Hesss Law
  • Hesss Law of Heat Summation, ?Hrxn ?H1 ?H2
    ?H3 ..., states that the enthalpy change for a
    reaction is the same whether it occurs by one
    step or by any (hypothetical) series of steps.
  • Hesss Law is true because ?H is a state
    function.
  • If we know the following ?Hos

28
Hesss Law
  • For example, we can calculate the ?Ho for
    reaction 1 by properly adding (or subtracting)
    the ?Hos for reactions 2 and 3.
  • Notice that reaction 1 has FeO and O2 as
    reactants and Fe2O3 as a product.
  • Arrange reactions 2 and 3 so that they also
    have FeO and O2 as reactants and Fe2O3 as a
    product.
  • Each reaction can be doubled, tripled, or
    multiplied by half, etc.
  • The ?Ho values are also doubled, tripled, etc.
  • If a reaction is reversed the sign of the ?Ho is
    changed.

29
Hesss Law
  • Hesss Law in a more useful form.
  • For any chemical reaction at standard conditions,
    the standard enthalpy change is the sum of the
    standard molar enthalpies of formation of the
    products (each multiplied by its coefficient in
    the balanced chemical equation) minus the
    corresponding sum for the reactants.

30
Hesss Law
31
Hesss Law
  • Given the following equations and ?Ho?values
  • calculate ?Ho for the reaction below.

32
Hesss Law
  • Use a little algebra and Hesss Law to get the
    appropriate ?Ho?values

33
Some Thermodynamic Terms
  • Notice that the energy change in moving from the
    top to the bottom is independent of pathway but
    the work required may not be!
  • Some examples of state functions are
  • T (temperature), P (pressure), V (volume), ?E
    (change in energy), ?H (change in enthalpy the
    transfer of heat), and S (entropy)
  • Examples of non-state functions are
  • n (moles), q (heat), w (work)

???H along one path ???H along another path
  • This equation is valid because ?H is a STATE
    FUNCTION
  • These depend only on the state of the system and
    not how it got there.
  • V, T, P, energy and your bank account!
  • Unlike V, T, and P, one cannot measure absolute
    H. Can only measure ?H.

34
Some Thermodynamic Terms
  • The properties of a system that depend only on
    the state of the system are called state
    functions.
  • State functions are always written using capital
    letters.
  • The value of a state function is independent of
    pathway.
  • An analog to a state function is the energy
    required to climb a mountain taking two different
    paths.
  • E1 energy at the bottom of the mountain
  • E1 mgh1
  • E2 energy at the top of the mountain
  • E2 mgh2
  • ?E E2-E1 mgh2 mgh1 mg(?h)

35
Standard States and Standard Enthalpy Changes
  • Thermochemical standard state conditions
  • The thermochemical standard T 298.15 K.
  • The thermochemical standard P 1.0000 atm.
  • Be careful not to confuse these values with STP.
  • Thermochemical standard states of matter
  • For pure substances in their liquid or solid
    phase the standard state is the pure liquid or
    solid.
  • For gases the standard state is the gas at 1.00
    atm of pressure.
  • For gaseous mixtures the partial pressure must be
    1.00 atm.
  • For aqueous solutions the standard state is 1.00
    M concentration.
  • ?Hfo standard molar enthalpy of formation
  • the enthalpy change when 1 mol of compound is
    formed from elements under standard conditions.
  • See Table 6.2 and Appendix L

36
Enthalpy Values
Depend on how the reaction is written and on
phases of reactants and products
  • H2(g) 1/2 O2(g) ? H2O(g) ?H -242 kJ
  • 2H2(g) O2(g) ? 2H2O(g) ?H -484 kJ
  • H2O(g) ? H2(g) 1/2 O2(g) ?H
    242 kJ
  • H2(g) 1/2 O2(g) ? H2O(l) ?H -286 kJ

37
?Hfo, standard molar enthalpy of formation
  • H2(g) ½ O2(g) ? H2O(g) ?Hf (H2O, g) -241.8
    kJ/mol
  • C(s) ½ O2(g) ? CO(g) ?Hf of CO -
    111 kJ/mol
  • By definition, ?Hfo 0 for elements in their
    standard states.

Use ?Hs to calculate enthalpy change for
H2O(g) C(graphite) ? H2(g) CO(g)
38
Using Standard Enthalpy Values
  • In general, when ALL enthalpies of formation are
    known,

?Horxn ? ?Hfo (products) - ? ?Hfo (reactants)
Remember that ? always final initial
39
Using Standard Enthalpy Values
  • Calculate the heat of combustion of methanol,
    i.e., ?Horxn for
  • CH3OH(g) 3/2 O2(g) ? CO2(g) 2 H2O(g)
  • ?Horxn ? ?Hfo (prod) - ? ?Hfo (react)
Write a Comment
User Comments (0)
About PowerShow.com