Title: Energy Changes in Chemical Reactions
1Energy Changes in Chemical Reactions
2Thermodynamics and Thermochemistry
- Thermodynamics the study of heat and its
transformations. - Thermochemistry the part of thermodynamics that
deals with changes in heat that take place during
chemical reactions as in most - chemical reactions,
- energy is absorbed
- or released.
3What is Internal Energy?
- Internal Energy (E)- the combined kinetic and
potential energy of all particles in a system. - The total internal energy of a system is not
usually known. Energy changes (DE) can be
measured. -
4Examples of Potential and Kinetic that Contribute
to Internal Energy
interactions between particles of matter down to
the subatomic level (p, n, and e-) but we are
generally concerned about interactions at the
molecular level.
- Kinetic Energy
- Rotational nature of molecules
- Vibrational nature of molecules
- Potential Energy-
- Intermolecular forces of attraction or repulsion
- Intramolecular forces of attraction or repulsion
5System
- What is the system? The object or substance
undergoing a change of physical state or
reaction. What we are studying. - Surroundings are everything around the system
that can exchange energy with the system.
6Types of Systems
- Open system can exchange mass and energy (usually
in the form of heat) with the surroundings. - Closed system can exchange energy but not mass.
- Isolated system no transfer of energy or mass.
7State Functions
- State functions are path independent. They
depend only on present state and are independent
of history of the system. - The change in a state function does not depend on
how the process is carried out.
8State Functions
9Total Internal Energy (E) in a System
- Etotal mgh ½ mv2
- OR
- Potential Energy Kinetic Energy
- Internal energy of the system sum of kinetic
and potential energies making up a substance down
to the subatomic level. We often dont know the
total energy of a system. - Chemists arent so interested in total energy in
a system- theyre more interested in the energy
changes that happen during chemical reactions.
10First Law of Thermodynamics
- The total energy in the universe is constant (Law
of Conservation of Energy). - You cant get something for nothing
- There are essentially two ways to change the
energy of a system- heat(q) and work(w). - DE q w
11Energy Change DE
- Energy change in a system can be determined
- DE Energy can be transferred as heat work
- DE Ef Ei q w
- Where work force X distance
- OR
- pressure X volume change
- w p DV
- Note Expansion volume -pDV. Work is being
done by the system on the surroundings.
12Work in Chemical Reactions
- Often times in chemical reactions, there isnt a
volume change or a force applied over a distance. - Therefore, for the chemist, heat changes are the
primary concern when considering the energy
changes that exist in a chemical reaction. - DE q
- (for many, not all, chemical reactions)
13Focus on Heat
- Heat thermal energy (motion of atoms)
- Something is hot because it has thermal energy
and its atoms are moving rapidly. - The more energy, the faster they move.
- The total thermal energy of a substance is the
sum of the individual thermal energies of the
atoms and molecules making up a substance.
14Chemists are Interested in Heat Transfer (q)
during chemical reactions
- Heat is a form of energy transfer.
- Heat transfers from hot object to cooler object.
- qgt0 Heat is transferred from surroundings to
system. ENDOTHERMIC - qlt0 Heat is transferred from system to
surroundings. EXOTHERMIC
15Remember the Trick to Temperature
- A measure of heat and relates to the average
kinetic energy of atoms in a sample. - The higher the temperature, the greater the
thermal motion. Thermal energy depends upon
sample size. - Heat ? Temperature
- Thermometers display changes in thermal energy.
16Exothermic Process
- By convention in an exothermic process, heat is
transferred FROM the system TO the surroundings.
Heat is a product. - 2H2 (g) O2 (g) ? 2H2O (l) energy
- Combustion reactions are exothermic!
- Chemical (potential) energy is converted to heat
(kinetic energy).
17Endothermic Reaction
- Heat is transferred TO the system FROM the
surroundings. Heat is a reactant. - Energy 2HgO (s) ? 2Hg(l) O2 (g)
- Thermal energy (kinetic) is converted to chemical
energy (potential).
18(No Transcript)
19Specific Heat and Heat Capacity
- Specific heat the heat needed to raise the
temperature of one gram of a substance by 1.0 0
C. Units J/g0C. Intensive property- not
dependent on how much is present. - Heat Capacity the measure of an overall effect
of heat transfer on the temperature of an object.
Units J/g . Extensive property (depends on
mass).
20Specific Heats of Various Substances
21Side Note Specific Heat of Water
- The specific heat of water is one of the highest
known specific heats. - Thats why temperature fluctuations are smaller
near large bodies of water like Puget Sound than
someplace like the desert in Arizona.
22Calorimetry
- The laboratory technique used to measure the heat
released or absorbed during a chemical or
physical change. - In a calorimeter, there is no volume change in
the system, therefore,
DE q - q mcDT
23Calorimetry ? heat absorbed mcDTliquid
mcDTcalorimeter
- Calorimetry depends on the assumption that all
the heat involved changes the contents of the
calorimeter and its contents? no heat is gained
or lost through the environment. - For many chemical reactions, a styrofoam cup is a
convenient calorimeter because it has a very low
heat capacity (for most purposes negligible) and
it is a good insulator.
24Sample Calculation- Specific Heat
- A piece of iron with a mass of 72.4 grams is
heated to 100.oC and plunged into 100.g of water
that is initially at 10.0oC. Calculate the final
temperature that is reached assuming no heat loss
to the surroundings. - Cwater 4.18 J/goC
- Ciron 0.449 J/goC
- NOTE the heat gained by the cooler body the
heat lost by the warmer body 0 - Tfinal 16.5oC
25Sample Calorimetry Problem
- A 1.5886 g sample of glucose was ignited in a
bomb calorimeter. The temperature increased by
3.682oC. The heat capacity of the calorimeter
was 3.52 kJ/oC, and the calorimeter contained
1.000 kg of water. Find the total heat released
and the molar heat of reaction. - Remember Heat energy goes into heating up the
water and the calorimeter itself. - Q -28.35 kJ
- -3211 kJ/mol
26Heats of Reactions
- We describe the energy changes in a chemical
reaction in terms of the amount of heat that can
be produced or must be supplied. - All chemical reactions are accompanied by a
change in the potential energy of the bonds,
hence, heat energy changes. - Endothermic reactions ? Heat supplied
- Exothermic reactions ? Heat produced
27The Enthalpy
- Enthalpy (H) is a state function defined as
- H E PV
- DH DE PDV
- From the first law of thermodynamics
- DE q - PDV
- Therefore, at constant volume, DH q
28Enthalpy, ?H A State Function
- We can easily run reactions in the lab at
constant pressure. - Enthalpy heat transfer at constant pressure.
- ?H Hfinal Hinitial Hproducts Hreactants
qp - The enthalpy change for a reaction is best shown
by writing the balanced equation for the
reaction, with the enthalpy change along side it.
It is vital to include state symbols in all
thermochemical equations because changes of state
have their own enthalpy changes associated with
them. (think ice vs water)
29(No Transcript)
30(No Transcript)
31(No Transcript)
32Endothermic Vs Exothermic Reactions
- For exothermic processes, q is negative then ?H
is negative.The products are more likely to be
stable than the reactants. - For endothermic processes, q is positive then ?H
is positive. The products are more likely to be
less stable than the reactants.
33Energy Diagram for Exothermic and Endothermic
Reactions
Exothermic
Endothermic
34Five Properties of Enthalpy
- Extensive property- like mass, does depend on
amount of substance present. - A state function path independent.
- ? H has unique values qp.
- If you reverse a reaction, the values of ?H are
indentical, the signs are opposite. - ? H applies to balanced reactions with correct
stoichiometric coefficients.
352 H2 (g) O2 (g) ? 2H2O (g) DH -483.6 kJ
- The negative sign indicates the reaction is
exothermic. - This value of DH is for the production of 2 mol
of water. If 4 moles were produced, the value
for DH would be twice -483.6 kJ - If the reaction is reversed, the sign of DH would
be reversed. - DH does depend on the state of matter. The
enthalpy change would be different for the
formation of liquid water instead of gaseous
water.
36Sample Problem
- 2NaHCO3 (s) ? Na2CO3(s) H2O(l) CO2 (g)
- DH 91.6 kJ/mol
- NaHCO3(s) ? ½ Na2CO3 ½ H2O (l) ½ CO2 (g)
- DH?
37Heats of Reaction DH
38Sample Problems- Heats of Reaction
- What is the heat of reaction for the dissolving
of 4 moles of potassium nitrate in water? - How much energy is required to form 2 moles of
methanol (CH3OH) from carbon dioxide and water? - Which produces more energy? The combustion of
methanol or methane (CH4)?
39Hess Law
- Hesss law states that if a reaction occurs in a
series of steps, then the enthalpy change for the
overall reaction is simply the sum of the
enthalpy changes of the individual steps. - Pay attention to the direction of the reaction
and the stiochiometry of the reaction. - Remember DH is a state function.
40State Functions do not depend on path, only
starting and finishing points.
41Sample Hesss Law Problem
- Given
- C(s) O2 (g) ? CO2 (g) DH -393.5kJ
- H2(g) 1/2O2 (g) ? H2O (l) DH -285.8 kJ
- C2H2(g) 5/2O2(g) ? 2CO2(g) H2O(l)
- DH -1200.8 kJ
- Find the enthalpy change for
- 2C(s) H2(g) ? C2H2(g)
- 128 kJ
42Enthalpies of FormationHeats of Formation DHfo
kJ/mol
- Enthalpies of reaction can be calculated from
individual enthalpies of formation for the
reactants and products. - Standard states used
- Pressure 1atm for gases
- 1 molar solutions for aqueous solutions
- For pure substances- the standard state is the
most stable form at 1 atm and 25oC.
43Standard Enthalpy of Formation- DHfo
- Change in enthalpy when 1 mol of a compound is
formed from its elements when all the substances
are in their standard states. - The DHfo of an element in its standard state is
zero. - The DHfo can be determined from the tabulated
DHfo of the individual reactants and products. - What is the formation reaction for CaCO3(s)?
- Ca(s) C(graphite) 3/2 O2 (g) ? CaCO3(s)
- DH -1207.1 kJ/mol
44DHorxn ? nDHfo products -? mDHforeactants
- In using this equation be sure to consider the
number of moles of each, DHfo for the individual
compounds refer to the formation of one mole. - The DHo a value assumes reactants and products
are at the same temperature. - Since DHfo values are tabulated at 25oC, we can
only calculate DHo at this temperature.
45(No Transcript)
46Sample Problem Standard Enthalpy of Reaction
- Calculate DHo for
- 6H2O (g) 4NO (g) ? 5 O2 (g) 4NH3 (g)
- 904.68kJ
47Bond Enthalpies
- The breaking of bonds endothermic process
- The formation of bonds exothermic process
- The amount of energy associated with the
formation / breaking of a particular bond is to a
large extent independent of the bonding in the
rest of the molecule.
48C-H bond
- The energy bonding a carbon atom to a hydrogen
atom is about 413 kJ/mol in both methane and
ethanol.
Ethanol
49DH ?Bonds Broken ?Bonds Formed
- Approximate enthalpy changes for reactions may be
calculated by considering the bonds being broken
and the bonds being formed in a reaction. - As before with enthalpy calculations, the number
of bonds broken or formed must be considered in
the calculation.
50Sample Problem Bond Enthalpies
- Consider the formation of ammonia from nitrogen
and hydrogen - N2 (g) 3H2 (g) ? 2NH3 (g)
- N?N 945 kJ/mol H-H 436 kJ/mol N-H 391 kJ/mol
- DH ?
- DH -93 kJ/mol
51Sample Problem Bond Enthalpies
- The balanced equation for the complete combustion
of one mole of ethene, C2H4, in oxygen is shown
below - C2H4 (g) 3O2(g) ? 2CO2 (g) 2H2O (g)
- CC 612 kJ/mol C-H 412kJ/mol O-H 463 kJ/mol
CO 743 kJ/mol OO 496 kJ/mol - DH ?
52Second Law of Thermodynamics
- All processes that occur spontaneously move in
the direction of an increase in entropy of the
universe. - Things are getting all messed up!
- Nature Loves Disorder.
53Entropy measure of disorderDS J/mol K
- Entropy is another consideration in energetics.
It deals with the desire of the universe to move
to a more disordered state. - A positive value for entropy indicates the
products of the reaction are more disordered than
the reactants. - Entropy of a system can be calculated much the
same way that enthalpy is calculated.
54DSo Standard Molar Entropy
- Entropy of 1 mole of a substance in its standard
state. - Pressure 1atm for gases
- 1 molar solutions for aqueous solutions
- For pure substances- the standard state is the
most stable form at 1 atm and 25oC. - For a reaction
- DSorxn ? DSo products -? DSoreactants
55Sample Problem Entropy
- Calculate DSo for the following and determine if
the reactions are favorable. - H2(g) 1/2O2(g) ? H2O(g)
- H2(g) 1/2O2(g) ? H2O(l)
- CaCO3(s) H2SO4(l) ? CaSO4(s) H2O(g) CO2(g)
- a-44.8 J/mol K b -163.6 J /mol K c259 J/mol
K
56Spontaneity of Chemical Reactions
Why does this chemical reaction occur on its own
at room temperature?
Fe(s) CuSO4(aq) ? FeSO4(aq) Cu(s)
57Chemists Goal- to be able to predict spontaneity
of a reaction
- Enthalpy DH or values favorable?
-
Entropy DS or values favorable?
58Gibbs Free Energy Equation
- Gibbs free energy (G) is a thermodynamic
function that combines the enthalpy, entropy, and
temperature or a reaction. - DG DH TDS
- If DGgt0, the reaction is not spontaneous energy
must be supplied to cause the reaction to occur. - If DGlt0, the reaction is spontaneous.
- If DG0, the reaction is at equilibrium
59Gibbs Free Energy Equation
- Just like with DH and DS, if there is a DG
associated with a reaction and the reaction is
reversed, the sign on DG changes. - Also, check the chemical equation for the
stoichiometric proportions. Multiply the value
of DGo if necessary. - Units for DGo are kJ/mol
60Standard Gibbs Free Energy Change DGo
- Just like with enthalpy and entropy, the standard
Gibbs free energy change is calculated - DGorxn ? DGo products -? DGoreactants
- Remember DGo of an element in its standard state
is zero.
61Sample Problem Gibbs Free Energy Change
- Calculate DGo for and determine if the reaction
is spontaneous - 2NH4Cl(s) CaO(s) ? CaCl2(s) H2O(l) 2NH3(g)
- C2H4 (g) H2O(g) ? C2H5OH(l)
- Ca(s) 2H2SO4(l) ? CaSO4(s) SO2(g) 2H2O(l)
62Gibbs Free Energy Equation Under Non-Standard
Conditions
- Standard States
- Pressure 1atm for gases
- 1 molar solutions for aqueous solutions
- For pure substances- the standard state is the
most stable form at 1 atm and 25oC. - What about when the concentrations of aqueous
solutions are not 1.0 M or the pressure for gases
is not at 1 atm? - DG DG0 RT ln Q Where Q reaction quotient
DGo 2.303 log Q
63Problem DG under Non Standard Conditions
- Calculate DG (non standard conditions-note the
lack of 0 symbol) for the following at 500K - 2NO(g) O2(g) ? 2NO2(g)
- 2.00M 0.500M 1.00M
- Note the conditions are non-standard-
concentrations are not 1.0M. - DG
64When Q Keq
- If the system is at equilibrium, then DG 0 and
the previous equation becomes - DG DG0 RT ln Q
- At equilibrium Q K and DG 0
- DGo -RT ln K or 2.303 RT log K
65Problem DG at Equilibrium
- Calculate DGo for (standard temp 298K)
- 2O3 (g) ? 3O2(g) Kp 4.17X1014
- DGo