Transcript Document 7478117

Enthalpy and Hess’ Law
Yep, it’s a law not a rule . . .
meaning it always works!
WOO HOO!
Enthalpy (DH), Let’s Review
-The thermodynamic variable used to describe the heat of a reaction at
constant pressure, qP
- The potential thermodynamic energy of a reacting system
- The potential energy stored (as heat) in chemical bonds
- Exothermic reactions have negative DHrxn values
-Typically (but not always) spontaneous reactions have negative values
of DHrxn (the heat term is added to the products side)
- We express enthalpy for a chemical reaction (DHrxn) as a stoichiometric
component in a thermochemical equation
2H2O(l)  2H2(g) + O2(g) + 572 kJ
- Enthalpy is a state function which means it is independent of path
 If we do not know the enthalpy for a reaction we can calculate it
using the known enthalpies for smaller reactions
Enthalpy (DH), Let’s Review a Little More
- When we discuss enthalpy we like to express it in terms of molar
enthalpy
 The molar enthalpy for the transformation of water into its
constituent elements is 286 kJ/mol (not 572 kJ as written in the
thermochemical equation . . . why?)
- Enthalpy data is often expressed in tables in a standard state, DHo, this
is 101.3 kPa, 298 K (and 1 mol/L for solutions)
- We can determine DHrxn by measuring the temperature change of a
pure substance to which the heat (energy) of the reaction has been
transferred
Process known as calorimetry
In a calorimeter:
Heat absorbed (heat given off by rxn) =
(mcDT)liquid + (mcDT)calorimeter
The calorimeter can absorb some heat so
choose your material wisely!!!
Enthalpy and Heat . . . Some Problems
1) How much heat energy is required to increase the temperature of 10 g of
nickel (specific heat capacity 440 J kg-1 K-1) from 323 K to 343 K?
2) The enthalpy of combustion (DHcomb) of ethanol (C2H5OH) is 1370 kJ/mol.
How much heat is released when 0.20 moles of ethanol undergo complete
combustion?
3) Consider the following reaction:
H2(g) + ½ O2(g)  H2O(l) + 286 kJ
DHrxn = -286 kJ/mol, what mass of O2(g) must be consumed to produce 1144 kJ
of energy?
Enthalpy Diagrams
Hess’ Law
Because enthalpy (DH) and energy (DE) are state functions, we do not
concern ourselves with the reaction pathway when determining these
values. They are said to be independent of path.
Therefore when we cannot measure a reaction to determine its enthalpy
change (DH) we can use literature data to calculate these energetic
quantities.
For a given reaction with an unknown energy term we can calculate DH by
assembling the desired reaction from several smaller well-defined ones.
Hess’ Law and the Formation of Ethylene (C2H4),
A Polymer Precursor
Ethylene or ethene is the monomer unit of polyethylene commonly known to
us as plastic.
The overall formation reaction is as follows:
2C(graphite) + 2H2(g)  C2H4(g)
DHrxn = ?
*Remember a formation reaction is the formation of a molecule from its
elements in their most stable form (H2 not H). By definition the heat of
formation of an element DHform = 0.
How do we find DHrxn for this reaction?
Hess’ Law and the Formation of Ethylene (C2H4),
A Polymer Precursor
The overall formation reaction is as follows:
(a)
2C(graphite) + 2H2(g)  C2H4(g)
DHrxn = ?
We are given the following known data:
(b) C(graphite) +O2(g)  CO2(g)
DH = -393.5 kJ
(c) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
DH = -1410.9 kJ
(d) H2(g) + ½ O2(g)  H2O(l)
DH = -285.8 kJ
Let’s begin by finding an equation that will place 2 moles of C(graphite) on
the left hand side.
We need to multiply (b) by 2 to accomplish this:
2(b) 2C(graphite) +2O2(g)  2CO2(g)
DH = -787 kJ
Hess’ Law and the Formation of Ethylene
(a) 2C(graphite) + 2H2(g)  C2H4(g)
DHrxn = ?
(b) C(graphite) +O2(g)  CO2(g)
DH = -393.5 kJ
(c) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
DH = -1410.9 kJ
(d) H2(g) + ½ O2(g)  H2O(l)
DH = -285.8 kJ
Next, let’s identify an equation that will place the C2H4(g) term on the right
hand side.
We can achieve this by reversing (c) and the changing the sign of DH.
-(c) 2CO2(g) + 2H2O(g)  2C2H4(g) + 3O2(g)
DH = 1410.9 kJ
Finally, equation (a) has a 2H2 term so we need to double (d) to obtain this.
2(d) 2H2(g) + O2(g)  2H2O(l)
DH = -571.6 kJ
Hess’ Law and the Formation of Ethylene (C2H4),
A Polymer Precursor
We can now sum (perform a summation) of the 3 known equations and obtain
(a), which will yield an new energy term.
2(b) 2C(graphite) +2O2(g)  2CO2(g)
DH = -787 kJ
-(c) 2CO2(g) + 2H2O(g)  2C2H4(g) + 3O2(g)
DH = 1410.9 kJ
2(d) 2H2(g) + O2(g)  2H2O(l)
DH = -571.6 kJ
(a) 2C(graphite) + 2H2(g)  C2H4(g)
DH = (-787 + 1410.0 – 571.6) kJ
DH = 52.3 kJ
Yielding the thermochemical equation:
2C(graphite) + 2H2(g) + 52.3 kJ  C2H4(g)
The formation of ethylene is an endothermic reaction.
Enthalpy and Hess’ Law
This completes our work in chapter 6 of our textbook and roughly our work with
enthalpy as the sole energy term for a reacting system.
Please review your notes from this section and read chapter 6 if you have not
done so.
Why do molecules form the
way they do?
Bond Enthalpies, Hess’ Law, The
Born-Haber Cycle, and Heats of
Reaction
Textbook Reference: Chapter 6 with parts from Chapter 9
Molecular Compounds
Why does oxygen form O2 rather than O8 (more accurately 4O2 rather than O8)?
 We know that oxygen is a diatomic, but this is not a reason this is
merely an observation of trend.
 We need to consider DHBDE (Bond Dissociation Energy) which is the
energy required to cleave a covalent bond.
BDE O2 = 498 kJ/mol
BDE 4O2 = 4 x 498 kJ = 1992 kJ
Meaning 1992 kJ is required to
break 4 moles of O2 OR 1992 kJ
of energy is given off when we
form 4 moles of O2 from O
atoms.
BDE O—O = 146 kJ/mol
BDE O8 = 8 x 146 kJ = 1168 kJ
Meaning only 1168 kJ is given off
when we form 8 O—O single
bonds in O8. O2 is energetically
favored.
Some other elements to consider
Why is phosphorus P4 rather than 2P2?
 P4 (white phosphorus) is tetrahedral
 P—P BDE = 209 kJ
 P≡P BDE = 490 kJ
Why is sulfur S8 rather than S2?
 This is the converse of oxygen which prefers O2.
 S—S BDE = 266 kJ
 S=S BDE = 427 kJ
The Ionic Lattice . . . One More Time
The find the lattice energy (DHlatt)of an ionic compound we can use the
following formula, known as the Born-Lande Equation
DHlatt = (-LA)(z+)(z-)(e2)(1 – 1/n)
4per
Where:
L = 6.022 x 1023
A = Madelung Constant
z = summation of charges on the ions
e = electron charge = 1.6 x 10-19 C
e = permittivity in a vacuum = 8 x 10-12 F/m
r = distance between the ions
n = Born constant
Lattice Energy there has to be an easier way . . .
(this would be a pretty lousy slide if there wasn’t)
We use what’s called the Born-Haber cycle, which makes use of some specific
heats of reaction (DHrxn).
DHf° ≡ the standard heat of formation of a compound from its elements
DHsub ≡ the heat of sublimation (solid  gas)
DHBDE ≡ the Bond Dissociation Energy for a covalent bond
DHI1 ≡ first ionization energy (neutral atom losing an e-, always positive)
DHI2 ≡ second ionization energy (+1 to +2, large and positive)
DHEA ≡ electron affinity (always a negative term except Be and N)
DHlatt ≡ lattice energy (always negative, usually quite large)
Formation of NaCl(s)
Na+(g) + Cl(g) + eDHI1 = 496 kJ
DHEA = -349 kJ
Na+(g) + Cl-(g)
Na(g) + Cl(g)
Na(g) + ½ Cl2(g)
Na(s) + ½ Cl2(g)
DHBDE = 122 kJ*
DHlatt = -787 kJ
DHsub = 107 kJ
DHf° = ???
NaCl(s)
* BDE Cl2(g) = 244 kJ, so ½ (244 kJ) = 122 kJ
How do we calculate DHf° from the Born-Haber Cycle?
From our work with Hess’ Law we know that energies are additive.
 Therefore we can add up all of the components from the cycle
which yield the overall formation reaction (from the elements).
DHf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl
Formation of NaCl(s)
Na+(g) + Cl(g) + eDHI1 = 496 kJ
DHEA = -349 kJ
Na+(g) + Cl-(g)
Na(g) + Cl(g)
Na(g) + ½ Cl2(g)
Na(s) + ½ Cl2(g)
DHBDE = 122 kJ*
DHlatt = -787 kJ
DHsub = 107 kJ
DHf° = -411 kJ/mol
NaCl(s)
DHf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl
Determine the lattice energy of MgF2(s)
DHsub Mg(s)  Mg(g) = 146 kJ/mol
DHI1 Mg(g)  Mg+(g) = 738 kJ/mol
DHI2 Mg+(g)  Mg2+(g) = 1451 kJ/mol
DHBDE F2(g) = 159 kJ/mol of F2
DHEA F = -328 kJ/mol of F
DHform MgF2(s) = -1124 kJ/mol (this is a DH°f)
Lattice Energy of MgF2(s)
Mg2+(g) + 2F(g) + 2eDHEA = -656 kJ (2 x -328 kJ)
DHI2 = 1451 kJ
Mg+(g)
+ 2F(g) +
Mg2+(g) + 2F-(g)
eDHI1 = 738 kJ
Mg(g) + 2F(g)
Mg(g) + F2(g)
Mg(s) + F2(g)
DHBDE = 159 kJ
DHlatt = ?
DHsub = 146 kJ
DHf° = -1124 kJ/mol
MgF2(s)
DHlatt = DHf – (DHsub + DHBDE + DHI1 + DHI2 + DHEA) = -2962 kJ/mol MgF2(s)
Lets leave it here as far as new material . . .
Your Assignment
(and no not if you choose to accept it, just accept it)
1) Using your notes and the textbook suggest possible reasons why some
reactions are exothermic and some are endothermic (5.4.2) in terms of
average bond energy/enthalpy.
2) The combustion of methane is represented by the equation
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) + 890.3 kJ.
a) what mass of CH4(g) must be burned to give off 1.00 x 105 kJ of heat?
b) how much heat is produced when 2.78 moles of CO2(g) are generated?
3) Using standard enthalpies of formation from Appendix B in your textbook
calculate the standard enthalpy change for the following reactions:
a) NH3(g) + HCl(g)  NH4Cl(s)
b) 3C2H2(g)  C6H6(l)
c) FeO(s) + CO(g)  Fe(s) + CO2(g)
4) When burning a Dorito you find that the temperature of 150 g of water in an
aluminum (mass 12 g) can is raised by 64 K. What amount of energy was
released by the Dorito? You may assume that no heat was lost to the
surrounding and it was completely transferred to the can and water.
5) Use the following 2 reactions calculate the DHrxn for 2NO2(g)  N2O4(g).
N2(g) + 2O2(g)  N2O4(g); DH = 9.2 kJ and N2(g) + 2O2(g)  2NO2(g); DH = 33.2 kJ
6) Calculate the enthalpy of reaction:
BrCl(g)  Br(g) + Cl(g)
DHrxn = ?
Using the following data:
Br2(l)  Br2(g)
DH = 30.91 kJ
Br2(g)  2Br(g)
DH = 192.9 kJ
Cl2(g)  2Cl(g)
DH = 243.4 kJ
Br2(l) + Cl2(g)  2BrCl(g)
DH = 29.2 kJ
7) Question 9.33 from your textbook. Using a Born-Haber cycle for KF to calculate
DHEA of fluorine.
Tying Up Some
Loose Ends . . .
The brocolli must die!
Enthalpy Cycles, Calculation of
DHrxn and Variations in Lattice
Energy
Multiple Representations of Enthalpy
So far we have looked at 2 ways of representing the energy/enthalpy term for a
chemical reaction:
1) as a component in a thermochemical equation
2) as a term outside the equation, calculated from a formula such as
q = mcDT
3) as an enthalpy diagram
 a graphical way to show the change in enthalpy rather than
relying solely on equations
 this in NOT a commonly used method and it will
be addressed purely as the presentation of a third
option to enthalpy problems, I strongly recommend
you use equations and Hess’ Law to solve enthalpy
problems
What is an Enthalpy Diagram?
A diagram that shows the overall and net reaction steps with their
corresponding energy terms for a chemical reaction.
Consider below and the combustion of methane.
What is an Enthalpy Cycle?
Let’s consider an alternative route to finding the DHcomb of methane.
ΔH1= enthalpy change for bond breaking
ΔH2= enthalpy change for bond making
= 4 x E(C-H) + 2 x E(O=O)
= - [ 2 x E(C=O) + 4 x E(O-H) ]
= 4 x 413 + 2 x 498
= - [ 2 x 805 + 4 x 464 ]
= + 2648 kJ mol-1
= - 3466 kJ mol-1
DHrxn = DH1 + DH2 = -818 kJ/mol
What is an Enthalpy Cycle?
A
DH1
DH2
C
DH3
DH1 = DH2 + DH3
B
Variations in Lattice Energy
Sometimes when we calculate a lattice energy and then compare it to our
experimental value, there is a difference.
If this difference is large enough between the theoretical and experimental
values this indicates that rather than the lattice being totally ionic, there is a
significant degree of covalent character to each interaction.
 Yes this is a return to the bonding continuum!
Consider the following:
NaCl
AgCl
Theoretical Value
(kJ/mol)
766
770
Experimental Value
(kJ/mol)
771
905
AgCl has significant covalent character! Is there another way we could have
identified this phenomenon?