Enthalpy and Hess’ LawEnthalpy and Hess’ Law

Yep, it’s a law not a rule . . . Yep, it’s a law not a rule . . . meaning it always works! meaning it always works!

WOO HOO!

Enthalpy (Enthalpy (H), Let’s ReviewH), 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 Hrxn values

-Typically (but not always) spontaneous reactions have negative values of Hrxn (the heat term is added to the products side)

- We express enthalpy for a chemical reaction (Hrxn) 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 (Enthalpy (H), Let’s Review a Little MoreH), 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, Ho, this is 101.3 kPa, 298 K (and 1 mol/L for solutions)

- We can determine Hrxn 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) =

(mcT)liquid + (mcT)calorimeter

The calorimeter can absorb some heat so choose your material wisely!!!

Enthalpy and Heat . . . Some ProblemsEnthalpy 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 (Hcomb) 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

Hrxn = -286 kJ/mol, what mass of O2(g) must be consumed to produce 1144 kJ of energy?

Enthalpy DiagramsEnthalpy Diagrams

Hess’ LawHess’ Law

Because enthalpy (H) and energy (E) 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 (H) we can use literature data to calculate these energetic quantities.

For a given reaction with an unknown energy term we can calculate H by assembling the desired reaction from several smaller well-defined ones.

Hess’ Law and the Formation of Ethylene (CHess’ Law and the Formation of Ethylene (C22HH44), ),

A Polymer PrecursorA 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) Hrxn = ?

*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 Hform = 0.

How do we find Hrxn for this reaction?

Hess’ Law and the Formation of Ethylene (CHess’ Law and the Formation of Ethylene (C22HH44), ),

A Polymer PrecursorA Polymer Precursor

The overall formation reaction is as follows:

(a) 2C(graphite) + 2H2(g) C2H4(g) Hrxn = ?

We are given the following known data:

(b) C(graphite) +O2(g) CO2(g) H = -393.5 kJ

(c) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) H = -1410.9 kJ

(d) H2(g) + ½ O2(g) H2O(l) H = -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) H = -787 kJ

Hess’ Law and the Formation of EthyleneHess’ Law and the Formation of Ethylene

(a) 2C(graphite) + 2H2(g) C2H4(g) Hrxn = ?

(b) C(graphite) +O2(g) CO2(g) H = -393.5 kJ

(c) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) H = -1410.9 kJ

(d) H2(g) + ½ O2(g) H2O(l) H = -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 H.

-(c) 2CO2(g) + 2H2O(g) 2C2H4(g) + 3O2(g) H = 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) H = -571.6 kJ

Hess’ Law and the Formation of Ethylene (CHess’ Law and the Formation of Ethylene (C22HH44), ),

A Polymer PrecursorA 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) H = -787 kJ

-(c) 2CO2(g) + 2H2O(g) 2C2H4(g) + 3O2(g) H = 1410.9 kJ

2(d) 2H2(g) + O2(g) 2H2O(l) H = -571.6 kJ

(a) 2C(graphite) + 2H2(g) C2H4(g) H = (-787 + 1410.0 – 571.6) kJ

H = 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’ LawEnthalpy 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 Why do molecules form the way they do?way they do?

Bond Enthalpies, Hess’ Law, The Bond Enthalpies, Hess’ Law, The Born-Haber Cycle, and Heats of Born-Haber Cycle, and Heats of

ReactionReaction

Textbook Reference: Chapter 6 with parts from Chapter 9

Molecular CompoundsMolecular 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 HBDE (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 considerSome 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 TimeThe Ionic Lattice . . . One More Time

The find the lattice energy (Hlatt)of an ionic compound we can use the following formula, known as the Born-Lande Equation

Hlatt = (-LA)(z+)(z-)(e2)(1 – 1/n) 4r

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

= 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 . . . 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 (Hrxn).

Hf° ≡ the standard heat of formation of a compound from its elements

Hsub ≡ the heat of sublimation (solid gas)

HBDE ≡ the Bond Dissociation Energy for a covalent bond

HI1 ≡ first ionization energy (neutral atom losing an e-, always positive)

HI2 ≡ second ionization energy (+1 to +2, large and positive)

HEA ≡ electron affinity (always a negative term except Be and N)

Hlatt ≡ lattice energy (always negative, usually quite large)

Formation of NaClFormation of NaCl(s)(s)

Na(s) + ½ Cl2(g)

Hsub = 107 kJ

Na(g) + ½ Cl2(g)

HBDE = 122 kJ*

* BDE Cl2(g) = 244 kJ, so ½ (244 kJ) = 122 kJ

Na(g) + Cl(g)

HI1 = 496 kJ

Na+(g) + Cl(g) + e-

HEA = -349 kJ

Na+(g) + Cl-(g)

Hlatt = -787 kJ

NaCl(s)

Hf° = ???

How do we calculate Hf° 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).

Hf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

Formation of NaClFormation of NaCl(s)(s)

Na(s) + ½ Cl2(g)

Hsub = 107 kJ

Na(g) + ½ Cl2(g)

HBDE = 122 kJ*

Na(g) + Cl(g)

HI1 = 496 kJ

Na+(g) + Cl(g) + e-

HEA = -349 kJ

Na+(g) + Cl-(g)

Hlatt = -787 kJ

NaCl(s)

Hf° = -411 kJ/mol

Hf NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

Hsub Mg(s) Mg(g) = 146 kJ/mol

HI1 Mg(g) Mg+(g) = 738 kJ/mol

HI2 Mg+(g) Mg2+

(g) = 1451 kJ/mol

HBDE F2(g) = 159 kJ/mol of F2

HEA F = -328 kJ/mol of F

Hform MgF2(s) = -1124 kJ/mol (this is a H°f)

Determine the lattice energy of MgFDetermine the lattice energy of MgF2(s)2(s)

Lattice Energy of MgFLattice Energy of MgF2(s)2(s)

Mg(s) + F2(g)

Hsub = 146 kJ

Mg(g) + F2(g)

HBDE = 159 kJ

Mg(g) + 2F(g)

HI1 = 738 kJ

Mg2+(g) + 2F(g) + 2e-

HEA = -656 kJ (2 x -328 kJ)

Mg2+(g) + 2F-

(g)

Hlatt = ?

MgF2(s)

Hf° = -1124 kJ/mol

HI2 = 1451 kJ

Mg+(g) + 2F(g) + e-

Hlatt = Hf – (Hsub + HBDE + HI1 + HI2 + HEA) = -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 Hrxn for 2NO2(g) N2O4(g). N2(g) + 2O2(g) N2O4(g); H = 9.2 kJ and N2(g) + 2O2(g) 2NO2(g); H = 33.2 kJ

6) Calculate the enthalpy of reaction:

BrCl(g) Br(g) + Cl(g) Hrxn = ?

Using the following data:

Br2(l) Br2(g) H = 30.91 kJ

Br2(g) 2Br(g) H = 192.9 kJ

Cl2(g) 2Cl(g) H = 243.4 kJ

Br2(l) + Cl2(g) 2BrCl(g) H = 29.2 kJ

7) Question 9.33 from your textbook. Using a Born-Haber cycle for KF to calculate HEA of fluorine.

Tying Up Some Tying Up Some Loose Ends . . .Loose Ends . . .

Enthalpy Cycles, Calculation of Enthalpy Cycles, Calculation of HHrxnrxn and Variations in Lattice and Variations in Lattice

EnergyEnergy

The brocolli must die!

Multiple Representations of EnthalpyMultiple 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 = mcT

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?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?What is an Enthalpy Cycle?

Let’s consider an alternative route to finding the Hcomb of methane.

ΔH1= enthalpy change for bond breaking

= 4 x E(C-H)   +   2 x E(O=O)

= 4 x 413 + 2 x 498

= + 2648 kJ mol-1

ΔH2= enthalpy change for bond making 

= - [ 2 x E(C=O) + 4 x E(O-H) ] 

= - [ 2 x 805 + 4 x 464 ] 

= - 3466 kJ mol-1

Hrxn = H1 + H2 = -818 kJ/mol

What is an Enthalpy Cycle?What is an Enthalpy Cycle?

A

B

CH2

H1

H3

H1 = H2 + H3

Variations in Lattice EnergyVariations 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?