# th NOTES by nikeborome

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```									9th and FINAL Set of NOTES
HOW DO DETERGENTS WORK?

Colloids - dispersions of particles of one substance throught out another substance or solution.

Differs form normal solution in that dispersed particles are larger than normal molecules

Aerosols ----
Emulsion ---
Sol ---

Common colloids are micelles - consist of large molecules with a hydrophobic end and a
hydrophyllic end!!

CHAPTER 6
Deals with transfer of energy as HEAT = ENERGY

Part of a larger aspect of CHEMISTRY CALLED THERMODYNAMICS

Thermodynamics - Science of Relationship between heat and other forms of Energy
We’ve talked some about how Energy minimization and maximization of disorder is a favorable
“natural” process.

Thermodynamics uses these ideas to predict if a process is possible. Does a perpetual motion
machine exist. Can I mix hot and cold water together to make a power plant?

Here won’t look at the whole of Thermodynamics (CHEM 112). Look mostly at the heat
absorbed or evolved in a chemical reaction or process (endothermic vs exothermic)

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WHY?

Economics - need a way to handle heat if it is given off in a reaction - or may need heat (energy)
for a particular reaction to occur - \$MONEY\$

Make an efficient fuel - want to know the quantity of heat evolved during the burning of fuel -
Rockets on Mars weight is important.

Energy - Calories Joules BTU

IN THERMODYNMICS YOU MUST DEFIN E THE SYSTEM AND THE SURROUNDINGS

K.E. = Kinetic Energy and Potential Energy = P.E.

Third type - Internal Energy = Eint = E

Eint is associated with

Etot system = K.E. + P.E. + Eint

If the energy of a system is to change - couple of ways to do this? How
1) Change Ek and Ep - move the whole system
2) Change Eint - make particle within the system (internal to the system) change their energy.

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Could put energy ito system or take it out as
1)
2)                    Mechanical work or some other type (stirring!)

OR PISTON (figure)

Don’t forget, MUST DEFINE A SYSTEM AND A SURROUNDINGS

Heat into the system is called an _______________ process or reaction.

Heat out the the system is ________ exothermic process or reaction.

1st Law of Thermodynamics. Euniverse = 0.
OVERALL the ENERGY MUST BE CONSERVED.

In the event the K.E. and the P.E. of the whole system do not change, then
System
int = change in internal energy = Qin - Wout
= Qin + Win

Q = heat    W = work        E = Efinal condition - Einitial conditon

E = the internal energy of the system is a ___________ PROPERTY

STATE PROPERTIES DEPEND ONLY ON THE PRESENT STATE OF THE SYSTEM,
DOESN’T MATTER HOW IT MADE IT TO THAT STATE.

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So  is independent of how the system got from the initial tot he final state!

Analogy:      Altitude Mountain

A thermodynamic State is the physical and chemical nature of a system.

For an ideal gas - specify just P,T, and you know V/n at some particular P, T, V/n and other
properties of that gas are all specified. ie. density, Eint, etc.

At a different P and T there is a different V/n so a different density, internal energy and other
properties such as color, hardness, etc.

REMEMBER IN THIS CHAPTER
Most interested in CHEMICAL REACTIONS

!!!!!! CONSIDER THE SYSTEM TO BE THE CHEMICAL REACTION!!!!
DIAGRAMS

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In many cases a chemical rxn takes place in a beaker (OPENED TO THE ATMOSPHERE)

So constant pressure process! Define new thermodynamic property called the ENTHALPY for
constant pressure heating processes.

H = E + PV
H = E + PV

H is the enthalpy. H is the enthalpy change for the process!

Enthalpy is a thermodynamic variable and it is a state variable too. Like Eint, T, P, etc

Hprocess = Hrxn
Hrxn = Hproducts - Hreactants

Depends only on the reactants and products not on how we got between them!

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Consider a reaction carried out at constant pressure.

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)              Qp = -890 kJ

Diagrams

A) heat + 2 Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
B) 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) + heat
Could include the heat of the Rxn or constant pressure heat of rxn in the balanced chemical eqn.

It is convenient and often necessary to write the enthalpy of reaction, H, with the chemical eqn.

THERMOCHEMICAL EQN:

The chemical eqn. for a reaction (including phase labels) why???? in which the eqn. includes
the enthalpy change for the rxn based on the moles of the balanced chemical eqn. OR PER
MOLE OF RXN!!

Example:       2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) rxn = -367.5 kJ

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Hrxn depends on the phase of the reactants!
Hrxn = Hproducts - Hreactants   WHY?

SOOO it is important to include the phases in the reaction eqns.
2H2(g) + O2(g) 2H2O(g)                   H = - 483.7 kJ
2H2(g) + O2(l)  2H2O(l)                     H = -571.7 kJ         per mole of rxn!!

A) If the thermochemical eqn. is multiplied by any factor, Hrxn must also be multiplied by that
factor

B) Hrxn reverse for the reverse of the rxn give is negative one times the enthalpy change of rxn
for the forward direction Hrxn forward
Hrxn reverse = - Hrxn forward
Example:
A rocket fuel is obtained from mixing liquid hydrazine, N2H4(l) + N2O4(l) 

A) Write the thermochemical eqn.

B) Write the thermochemical eqn. for the reverse rxn.
Note: The Quantity of Heat which evolves is dependent on the amount of reactants present. Use
stoichieometry and Limiting Reagen Concepts to calculate the amount of heat at const. pressure
for a particular experiment.

Example:

How much heat evolves when 10.0 g of hydrazine reacts with N2O4?

Enthalpy - Strange? (made it up)
But it is an intrisic propety of a substance in a given thermodynamic state.

Think of Hproduct in terms of the __________ of the molecule.

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It is related to the ________ ___________ which is present in the reactant and product
molecules. So H tells us about the energy o the product molecule bonds, relative to the energy
of the reactant molecule bonds.

Molecules have different bonds

Energy of C-H bond is different than the energy of a C-Cl bond. Different bon energy --
Different                            Heat content or different Enthalpy.

It takes energy to break bonds bonds and energy is released when a bond is made.

Bond breaking is exothermic or endothermic ?
Bond making is exothermic or endothermic?

So can I get the heat of reaction from considering the bond enthapies or bond energies. Yes, for
gaseous reactions.

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Define: Bond dissociation energy = Average energy required to break a particular bond in a
molecule. Bond dissociation energy - Strength of a bond ! Triple bond, double bond, single
bond.

Essentially, Enthalpy change for a gas phase reaction where a bond breaks.

So in tables A-B bond energy - defined as the average enthalpy.

In tables A-B bond energy is defiend as average enthalpy for the breading of an A-B bond in a
molecule in the gas phase.
CH4(g) C(g) + 4 H(g) (g)
H = 4 (C-H)
= 4(411) = 1644 kJ

So use table to get or (ESTIMATE) the heat of a rxn fromt he bond energy.

Consider:
CH4(g) + Cl2 (g) CH3Cl (l) + HCl(g)

Two Steps:

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In step 1 we break C-H and Cl-Cl bonds

In step 2 we make H-Cl and C-Cl bonds.

So overall:

Rule: - In general, for a GAS, the enthalpy of reaction is approximately equal to the sum of the
bond energies for bonds broken minus the sum of bond energies for bonds made.

BREAK - MAKE.

So the genral idea is
If we can calculate the constant pressure heat (Qp) then we’ll know the enthalpy change
for the process.

TWO WAYS TO GET Qp

1) Measure it experimentally
2) Calculate it Indirectly! Like the bond energy calculations.

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Measurements!

Measure Qp like we did in LAB. Put Reactants in a Calorimeter.
For the case where the heat capacity of a calorimeter and its contents are known.
Qtot = C C = heat capacity of the calorimeter and substances in it!
C = Ccalorimeter + Ccontents

If the Reaction is carried out in Dilute aqeous solution and Ccal = 0
Q = sp.htH2O * mH2O * T                              sp. ht. = J/(g oC)
Q = CH2O T                                    heat capacity C = J/oC cal/oC

In simple calorimeters, they are open to the air, so the rxn occurs at constant pressure
Q = Qp = Hrxn (TOT)

Usually want Hrxn on a per mol basis so divide by moles of the product formed or moles of the
limiting reagent to get:

More elaborate method of measuring heats of rxn involves a Bomb Calorimeter
Figure

Here (constant ________________)

qv = C
C includes Ccal + CH2O

Qv is constant volume heat rather than constant pressure. Thus a correction must be included
since H = Qp but not equal to Qv

How are Qp and Qv (heat at constant pressure and heat at constant volume) related?

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Eint = Qin - Wout      Wout = PV

1) At constant volume ______ = 0 so W = ____         thus Eint = _______

2) At constant pressure
Wp = - PV = - ng RT             (From ideal gas law ____________________)
Eint = Qin + Win = Qp + Wp

THUS:
Qp = Eint - Wp = Eint +ngRT        AND SINCE ________ is a state function)

Qp = Qv + ngRT
AND
H = E + ng RT              generally it is a small correction at room temp. but?

Example:
Calculate the difference between Qp and Qv for the decompostion of 1 mole of ammonium
chloride at 25oC

Example:
0.56 g of graphite is placed in a bomb calorimeter with an excess of O2 at 25.00oC and 1 atm
pressure. The temp. rises from 25.00 to 25.89oC. If the heat capacity of the calorimeter and
contents was 20.7 kJ/oC, what is the heat of rxn? (Express your answer in a thermochemical
eqn.)

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Nonexperimental ways to estimate rxn
Calculations
1) For gas phase rxns - by bond energies

OTHER WAYS BASED ON HESS’ LAW

The value of H for a reaction is the same whether it occurs in one step or in a se4ries of steps.
If a thermochemical eqn. can be expressed as the sum of two or more other chemical eqns. then:
Hoverall =

Useful for obtaining Hrxn for reactions difficult to carry out in a bomb calorimeter

Consider
1) C(s) + O2 (g) CO2(g)               H = - 393.5 kJ
2) 2CO(g) + O2(g)  2CO2(g)               H = - 566.0 kJ

Calculate Hrxn for the reaction:            C(s) + ½ O2(g) CO(g)

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Again the idea here by Hess’s Law is just use other chemical eqns and Hrxn for these by
combining them in such a way as to get overall the desired rxn and its Hrxn.

Example:
Calculate the enthalpy change for the reaction of manganese dioxide with Aluminum
Al(s) + MnO2  Al2O3(s) + Mn(s)

Use the following information:
a) 2Al(s) + 3/2 O2(g) Al2O3 (s)                       H = -1676 kJ
b) Mn(s) + O2 (g)  MnO2(s)                             H = -521 kJ

Calorimetric Methods Allow one to find Hrxnfor almost any reaction.

Easiest Way to USE HESS’S LAW

Use the Enthalpy chantges for the formation of a substance from its elements, AND combine
these formation enthalpies together to get enthalpy of RXN for compounds in a RXN.

Enthalpies of Formation are tabulated for one set of conditions known as the STANDARD
_____
                   H0formation or Hf0
½ H2(g) + ½ F2(g)  HF(g)                    Hfo

Standard State - conditions chosen for a substance when listing or comparing thermochemical

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data:
for gases std. pressure is 1atm and usually 25oC (but the temp doesn’t have to be this)
for solution std. concentration is 1M.

Standard Condition denoted by superscript O          o
. So Ho is the std. enthalpy change.

Std. Enthalpy of Formation Hfo = the enthalpy change for the formation of 1 mole of a
substance (in its std. state) from the stablest form of its elements!

For example:
Std. heat of formation, Hfo, for ethanol C2H5OH is the enthalpy change for:
2C(graph) + 3H2(g) + ½ O2(g)  C2H5OH(l)
NOTE: The elemental source of O is O2 not O atoms or O3, the most stable form at 25oC and 1
atm is O2
Likewise the elemental source of carbon is graphite not diamond - graphite is more stale than
diamond?

Std. Heats are tablulate in your book and elsewhere!

By convention, the std. heat of formation of the most stable form of any element is ZERO!!!
Hfo of O2(g) is equal to zero                      Hfo of N2(g) is equal to ______
Hf of S8(s) is equal to ____
o

OK NOW WHAT

Well using the tables for the heats of formation - the standard enthalpy change for any reaction
can be found from:
            Hrxno =  n Hfo(products) - m Hfo(reactants)



means sum over all of them and “m “ are the coefficients of the reactants in the balanced eqns
for the reaction and “n” are the coefficients for the products in the balanced eqns for the
reaction.

Example: Find Hrxno for the combustion of glucose!

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Examples:
Calculate Horxn for the following rxn!
Br2(l) + H2(g) + 2OH-(aq) 2Br -(aq) + 2H2O(l)

Given that:
2BaCO3(s) Ba(s) + 2C(s) + 3O2(g)              Horxn = 2432.6 kJ
a) Determine Hfo for BaCO3
b) How many grams of BaCO3 are formed if 100.0kJ are involved?

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Calculate Hrxno for two moles of nitrogen gas and 6 moles of water vapor reacting to form
ammonia and oxygen.

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