Thermochemistry

Definition: The study of the energy changes associated with a physical and/or

chemical change of a system. Thermal Energy: arises from a particle’s position and motion. Systems: Chemical - _______________________________________________________

- _______________________________________________________

Surroundings - ___________________________________________________

Open -____________________________________________________________

Closed - __________________________________________________________

Isolated - _________________________________________________________ Others: Heat (q) - ________________________________________________________

Exothermic - ______________________________________________________

Endothermic - _____________________________________________________

Temperature - _____________________________________________________

Calculation:

q = mc∆T where m = mass surroundings, g

c = specific heat capacity, _____________________

∆T = temperature change = ____________________

eg. Calculate the heat generated when 50.0 g of water (cH2O = 4.18 J/g•ºC)

changes temperature from 12.5ºC to 47.0ºC?

Solution: q = mc∆T =

eg. Calculate the heat generated when 50.0 g of water and 125 g of the Al can (cAl

= 0.900 J/g•ºC) changes temperature from 12.5ºC to 47.0ºC?

Solution: q = mc∆T

= mH2O x cH2O x ∆T + mAl x cAl x ∆T

Heat Transfer and Energy Change Chemical system energies, PE & KE, are sources of Enthalpy, H

- interactions between nuclei and electrons

- electron motions

- vibration of atoms within molecules

- rotation and translation of molecules

- nuclear potential energies

- electronic PE of atoms connected by chemical bonds and molecules connected

by forces of attraction Enthalpy Change, ∆H Results in the difference in enthalpies of the reactants and the products during a

change, due to:

- breakage of bonds or intermolecular forces ∆H = +ve (endo)

- formation of bonds or intermolecular forces ∆H = -ve (exo) Determination of Enthalpy Change, ∆H

- H and ∆H cannot be directly measured

- measured relative to the energy transferred, heat (q), during the change in the

system

- ∆Hsys = - qsurr

= - mc∆Tsurr

eg. Calculate the enthalpy change for a combustion reaction which results in the

temperature of 500. mL of water in a 0.250 kg Al (cAl = 0.900 J/g•°C) tea

kettle rising from 20.0 °C to 95.0 °C. ∆H = - (mc∆TH2O + mc∆TAl) ; VH2O = mH2O as DH2O = 1.0 g/mL

= - (mcH2O + mcAl) ∆T

=

This is an exothermic reaction showing the change in PE of the SYSTEM is equal

to the change in KE of the SURROUNDINGS PE is that of position or the ability to work.

When objects that attract each other:

Get closer together, PE __

Move further apart, PE __ KE is that of motion.

average KE __ as T __ Chemical change involves 2 processes that alter the PE of atoms. Bond breaking ____________ energy and PE of the atoms ___. Bond forming ____________ energy and PE of the atoms ___. Depending on which effect is greater, the net change in PE energy may be an

___________ or a ________. The ___________________________________________________________, the

__________________ the bond.

Molar Enthalpies Molar Enthalpy, ∆Hrxn(X)

the ∆H for 1 mole of any substance, (X), undergoing a reaction.

standard unit is kJ/mol eg. rxn solution _______ combustion _________

formation _______ neutralization _________

Determination:

X of MassMolar M whereM x m

Tcm- =

reactant theis X here wn

Tcm- =

n

q- = (x)H

xx

x

surrsurr

x

surrsurr

x

rxn

eg. a) Calculate the ∆Hcomb (C8H18) if 6.593 g causes 1.000 L of H2O to

increase its temperature by 75.2 °C. ( DH2O = 1 g/mL or 1kg/L)

mH2O =

Hcomb (C8H18 ) = -mH2OcH2OT

mC8H18

x MC8H18

= =

b) Express this enthalpy change 4 ways: i) ∆Hcomb(C8H18) =

ii) 2 C8H18 + 25 O2 16 CO2 + 18 H2O +

OR C8H18 + 25/2 O2 8 CO2 + 9 H2O +

iii) C8H18 + 25/2 O2 8 CO2 + 9 H2O ; H =

iv) PE diagram

For comparison:

c) Calculate the molar enthalpy of combustion of octane, C8H18, given that 6.593

g causes 150 mL of H2O in a 375 g Al can, to increase its temperature by 75.2

°C.

=

M x m

T )cm c(m - = )H(CH

188

188

22

HC

HC

AlAlOHOH

188com b

Hess’ Law and Heats of Formation

A State Function: _____________________ Definition:

When a reaction that can be expressed as the ____________ ____, , of two or

more __________ reactions, the enthalpy of reaction, Hrxn, is the algebraic

sum of the ___________________ rxn enthalpies, Hx. Standard Enthalpies of formation , Hfº

Are often used to calculate _______

The enthalpy (_______ or _______ of heat energy) for the _____________ of 1

mole of the ______________ from its ____________ in their standard state.

For an _____________ in its standard state, Hfº = 0

Standard state is SATP is 25ºC, 100. kPa

These equations are created from its ________________ELEMENTS

eg.

Rules for applying Hess’ Law:

1) Use the ________ reaction steps or _______ the _____________ equations of

the overall reaction and find the value of Hx or Hfº for each step.

2) ___________ intermediate steps (reactions) as needed and remember to

__________ the _____ of the Ho for that reaction ( multiply by -1).

eg.

3) ____________ intermediate reactions as necessary to _______ the coefficients

in the overall equation.

Remember to multiply the Ho values by the same multiplier. eg. 4) Determine the Ho

rxn from the algebraic sum of Ho values for all the

intermediate steps.

5) If applicable use

Hrxnº = nHfº (products) - nHfº (reactants)

eg.1 Determine the heat of reaction for the following: 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g) Hrxn = ?

Using Formation equations:

eg.2 Determine the heat of reaction for the following: C2H4 (g) + H2 (g) C2H6 (g) Hrxn = ?

Using the following GIVEN reactions:

(1) C2H4 + 3 O2 2 CO2 + 2 H2O (l) H1 = -1400.5 kJ

(2) C2H6 + 7/2 O2 2 CO2 + 3 H2O (l) H2 = -1550.0 kJ

(3) H2 + 1/2 O2 H2O (l) H3 = -258.8 kJ

eg.3 Repeat question 2, but use the Summation Formula of Hess’ Law:

Hrxnº = nHfº (products) - nHfº (reactants)

Heat Capacities, c

Substance Specific Heat Capacity at

SATP (J/(g·°C))

Substance Specific Heat Capacity at

SATP (J/(g·°C))

aluminum 0.900 nickel 0.444

calcium 0.653 potassium 0.753

copper 0.385 silver 0.237

gold 0.129 sodium 1.226

hydrogen 14.267 sulphur 0.732

iron 0.444 tin 0.213

lead 0.159 zinc 0.388

lithium 3.556 ice 2.01

magnesium 1.017 water 4.18

mercury 0.138 steam 2.01

Thermodynamic Properties of Organics At SATP

Substance ∆H f° S°

(kJ•mol-1) (J•K-1•mol-1)

benzene, C6H6(l) +49.0 172.8

bromoethane, CH3CH2Br(g) -90.5 -----

bromomethane, CH3Br (g) -37.2 246.3

butanal, CH3CH2CH2CHO(l) -241.2 ----

butane, n-C4H10 (g) -126.5 310.1

butan-1-ol, C4H9OH(l) -327.4 228.0

but-1-ene, C4H8 (g) -0.4 305.6

but-1-yne, C4H6 (g) +165.2 -----

carbon tetrachloride, CCl4 (l) -128.4 216.4

CCl4 (g) -96.0 309.9

chloroethene, CH3CH2Cl(g) -136.8 263.9

chloromethane, CH3Cl (g) -82.0 234.5

cyclopropane, (CH2)3 (g) +53.3 -----

1,2-dichloroethane, (CH2Cl)2(g) -165.0 -----

ethanal, CH3CHO(g) -191.5 160.2

ethane, C2H6 (g) -83.8 229.5

ethane-1,2-diol, (CH2OH)2(l) -454.8 166.9

ethanoic acid, CH3CO2H(l) -484.5 159.8

(acetic)

ethanol, C2H5OH(l) -277.1 160.7

ethene, C2H4 (g) +52.5 219.5

ethoxyethane, (CH3CH2)2O(g) -279.0 251.9

ethyne, C2H2 (g) +228.2 201.0

fluoromethane, CH3F (g) -247.0 -----

glucose, C6H12O6(s) -1260.0 212.1

Substance ∆H f° S°

(kJ•mol-1) (J•K-1•mol-1)

hexane, n-C6H14 (l) -198.6 296.1

iodoethane, CH3CH2Cl(g) -40.7 -----

iodomethane, CH3I (g) -15.5 163.2

methanal, HCHO(g) -108.7 218.7

(formaldahyde)

methane, CH4 (g) -74.8 186.2

methanoic acid, CH3OH(l) -425.1 129.0

(formic)

methanol, CH3OH(l) -239.1 239.7

methoxymethane, CH3OCH3(g) -184.0 266.7

methylpropane, C4H10 (g) -134.5 294.6

napthalene, C10H8 (s) +77.7 -----

octane, n-C8H18 (l) -250.0 361.1

pentane, n-C5H12 (l) -173.1 262.7

phenylethene or styrene +103.8 345.1

phenol, C6H5OH(s) -165.0 ----

propanal, CH3CH2CHO(l) -217.1 ----

propane, C3H8 (g) -104.5 269.9

propanone(acetone) -248.0 198.8

propan-1-ol, C3H7OH(l) -302.7 196.6

propene, C3H6(g) +20.2 266.9

propyne, C3H4 (g) +186.6 248.1

sucrose, C12H22O11(s) -2221.0 360.2

2,2,4-trimethyl pentane -259.2 328.0

urea -335.5 104.

Thermodynamic Properties of Inorganics At SATP)

Substance ∆H f° S° ∆G f

°

(kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1)

Al(s) 0 28.3 0

Al2O3 (s) -1675.7 50.9 -1582.3

Al2(SO4)3 (s) -3405.5 50.9 -1582.3

BaCO3 (s) -1216.3 112.1 -1142.0

BaCl2 (s) -860.2 124.1 -813.5

BCl3 (g) -404 291.1 -390.1

B2O3 (s) -1273 54.18 -1144.1

Br2 (l) 0 152 0

Ca(s) 0 41.4 0

CaCO3 (s) -1207 92.9 -1128.8

CaBr2 (s) -682.8 130.2 -1352.4

CaCl2 (s) -795.8 104.6 -748.1

CaO(s) -634.9 38.1 -566.5

Ca(OH)2 (s) -986.1 83.4 -901.7

Ca3(PO4)2 (s) -4119 236.9 -3897.7

CaSO4 (s) -1434.1 108.4 -1326.8

C(s) graphite 0 5.7 0

C(s) diamond +1.88 2.38 +2.90

CO (g) -110.5 197.66 -137.2

CO2 (g) -393.5 213.78 -394.4

Cl2 (g) 0 223.1 0

Cu (s) 0 +33.2 0

CuCl (s) -137.2 86.2 -119.9

CuCl2 (s) -220.1 108.1 -175.7

Cu2O (s) -168.6 93.1 -146.6

CuO (s) -157.3 42.6 -129.7

CuSO4 (s) -771.4 109.2 -663.6

CuSO4·5H2O s) -2279 301.6 -1887.1

F2 (g) 0 202.8 0

H2 (g) 0 130.7 0

H2O2 (l) -187.8 109.6 -120.4

HBr(g) -36.3 198.7 -53.5

HCl (g) -92.3 186.9 -95.3

HCl (aq) -167.2 56.7 -131.8

HCN(g) +135.1 201.81 +125.2

HF (g) -271.1 +173.8 -273.2

HI (g) +26.5 206.59 +1.75

HNO3 (l) -174.1 155.6 -80.7

HNO3 (aq) -207.0 ----- -----

H3PO4 (s) -1279.0 110.5 -1119.1

H2S (g) -20.6 205.8 -33.6

H2SO4 (l) -813.8 156.9 -690.0

H2SO4 (aq) -909.3 20.16 -743.4

I2 (s) 0 116.3 0

I2 (g) +62.4 180.79 -----

Fe (s) 0 27.8 0

FeO (s) -272.0 57.6 +245.1

Fe2O3 (s) -824.2 87.4 -742.2

FeCl2 (s) -341.8 118.0 -302.8

FeCl3 (s) -399.5 142.3 -344.0

Substance ∆H f° S° ∆G f

°

(kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1)

Pb (s) 0 64.8 0

PbCl2 (s) -359.4 136.0 -314.1

PbO (s) -219.0 66.5 -188.6

PbO2 (s) -277.4 68.6 -----

Mg (s) 0 32.7 0

MgCO3 (s) -1095.8 65.7 -----

MgCl2 (s) -641.3 89.63 -591.8

Mg(OH)2 (s) -924.5 63.24 -----

MgO (s) -601.6 26.95 -569.4

N2 (g) 0 191.6 0

NH3 (g) -45.9 192.78 -16.5

N2H4 (l) 50.6 121.2 149.3

N2H4 (g) +95.4 237.11 -----

NH4Cl (s) -314.4 94.6 -202.9

NH4NO3 (s) -365.6 151.08 -183.9

NO (g) 90.2 210.76 86.6

NO2 (g) 33.2 240.1 51.3

N2O (g) 82.1 219.9 104.2

N2O4 (g) 9.2 304.3 97.9

O2 (g) 0 205.1 0

O3 (g) 142.7 238.9 163.2

PCl3 (g) -319.7 217.2 -----

PCl5 (g) -443.5 364.6 -----

K (s) 0 64.2 0

KCl (s) -436.7 82.55 -409.1

KClO3 (s) -397.7 143.1 -296.3

KOH (s) -424.8 78.9 -379.1

Ag (s) 0 42.6 0

AgBr (s) -100.4 107.11 -97.4

AgCl (s) -127.0 96.25 -109.8

AgNO3 (s) -124.4 140.9 -33.4

Ag2O (s) -31.1 121.8 -11.3

Na (s) 0 51.2 0

NaBr (s) -361.1 86.82 -350.2

Na2CO3 (s) -1130.7 135.0 -1044.0

NaCl (s) -411.2 115.5 ------

NaF (s) -571 51.7 -545.6

NaOH (s) -425.6 64.4 -379.5

NaI (s) -287.8 98.50 -287.3

S8 (s) rhombic 0 31.8 0

S (g) 278.8 167.8 283.3

SO2 (g) -296.8 248.22 -300.2

SO3 (g) -395.7 256.8 -371.1

SnO (s) -280.7 57.17 -----

SnO2 (s) -577.6 49.04 -----

H2O (l) -285.8 69.95 -237.1

H2O (g) -241.8 188.84 -228.6

ZnO (s) -350.5 43.65 -----

ZnS (s) -206.0 57.7 -----

Energy and Driving Forces:

The Laws of Thermodynamics

Spontaneous Reaction:

given the required Ea, ________________ ___________the reaction

continues to proceed by itself

may be slow or fast Thermodynamics:

Study of energy transformations.

There are 3 Laws of Thermodynamics

– used to predict reaction spontaneity 1st Law of Thermodynamics

Law of Conservation of Energy

Total energy of the universe is constant

Energy cannot be created or destroyed, just

transferred from one form to another For a chemical reaction

∆Hºuniverse = __________________________________________ In a chemical reaction, the PE of the reactants and products results in the transfer of

energy from the:

1) surroundings to the chemical system (ENDO)

2) chemical system to the surroundings (EXO) Enthalpy changes and Spontaneity

Bond energy (BE): the minimum energy required to break one mole of bonds

between two atoms ______________

Equals the energy released when 1 mole of bonds are formed, _________

The greater the value, the more stable the bond SO, the Enthalpy of reaction can also be calculated by:

∆Hreaction = ___________________________________________ ∆Hreaction __________

indicates the formation of stronger bonds and more stable compounds

is most likely to be spontaneous reaction

Entropy Changes (∆S) and Spontaneity

the greater the # of ways particles can arrange themselves, the less ordered

they are

the greater the # of ways a particular state can be achieved, the more likely

that state is going to exist Entropy , Sº

The measure of disorder or randomness

increased disorder or entropy favours spontaneity, ____________

2nd Law of Thermodynamics ∆Sºuniverse =

disorder increases, since ____________

The following increases entropy:

1. The volume of a gas increases

2. The temperature of the system increases

3. Physical state solid to liquid to gas

4. Increase in the number of moles produced

5. Breaking complex molecules into smaller ones

3rd Law of Thermodynamics

At 0K all motion ceases, the forces of attraction have reduced entropy to a

minimum

S = 0 at T = 0K

As a result as T increases, S must increase

S is a measure of the energy needed to achieve a level of disorder by

overcoming the FA

This is dependent on the substance and the temperature reached

J/molK

∆Srxnº = ______________________________________

∆Srxnº _________________________________

Entropy Calculations eg. Calculate ∆Sº for

2CO(g) + O2(g) 2CO2(g) ∆Srxnº = nSº products - nSº reactants

=

___________________________________________________

This would mean it is nonspontaneous, but we need to consider the ∆Hrxnº

∆Hrxnº = n∆H º products - n∆H º reactants

=

Gibb’s Free Energy – Enthalpy and Entropy

Unites the 2 reaction driving forces

∆G = ∆H – T∆S If: ∆G < 0 ___________________________

∆G = 0 ___________________________

∆G > 0 ____________________________________________ For Spontaneity:

∆H ∆S ∆G spontaneity

- + + - - -

+ +

Calculating ∆Gº from ∆Hº and ∆Sº

eg. 2CO(g) + O2(g) 2CO2(g)

∆Hºrxn = -566.0 kJ ∆Sºrxn = -172.86 J/K = -0.17286 kJ/K ∆Gº = ∆Hº - T∆Sº =

Calculating ∆Gº from ∆Gfº

these values are not in the textbook, but are in the reference booklet

Use:

∆Gº = ____________________________________ Predicting Change in Spontaneity

Temperature and equilibrium As ∆G = 0 at equilibrium and ∆Gº = ∆H – T∆S

then: 0 = ∆H – T∆S

this gives the temperature at which the system changes spontaneity

need to examine ∆H and ∆S to determine if the reaction is spontaneous above

or below the temperature calculated eg. For previous example: 2CO(g) + O2(g) 2CO2(g)

T = ∆H

∆S = as ∆H –ve and ∆S –ve , then spontaneous at low T

T >

T <

Chemical Kinetics Definition: the study of reaction rates Rate of Reaction: The change in concentration, ∆[ ], of a

_________ or a __________ per unit of time

can be determined from a [ ] vs time graph Average Rate of Rxn: from the slope of the secant drawn between

2 points on the curve over a given time interval. Instantaneous Rate of Rxn: from the slope of the tangent to the

curve at a specific moment in time. Reaction Rates are given as:

as rates are __________ and as the rxn proceeds,

______________

the rates of the different reactants and products are ________ by

the balanced chemical reaction. In General, for aA + bB cC + dD

Rrxn req’d = rate of rxn given x mole ratio of req’d

given

eg. For 2 C2H6 + 7 O2 4 CO2 + 6 H2O

The rate of reaction with respect to C2H6 is 4.0 x 10-4 mol/L•s.

State the Rrxn with respect to each product and reactant.

smol/L 10 x 4.0 t

]H[C - )H(CR 4-62

62rxn

Measuring Raterxn

• requires an _______________, _____________ change that

doesn’t disturb the rxn:

1) ________________________________________________

2) ________________________________________________

3) ________________________________________________ Factors Affecting the Raterxn

1) Nature of the Reactant eg. aqueous, gas, liquid, reactivity

• depends on the _______ of the ______________ particle

forces.

2) Concentration ___________________________________

3) Temperature ___________________________________

4) Surface Area ___________________________________

5) Catalyst

The Rate Law and the Rate Law Equation

As various factors (T, [ ], etc) affect the rate of the rxn, their impact

MUST be determined empirically through experiment.

For aX + bY product

Rate Law:

Raterxn _____________________

where m, n R determined experimentally

i = initial concentration Rate Law Equation:

Raterxn = ___________________

where k is the rate constant for a specific temp.

Order of Rxn:

m = ________________________

n = ________________________

m + n = ________________________

eg. Raterxn = k [A]1[B]2 [C]0 = k [A]1[B]2 order = 3

∆Rate

∆[ ] 0th order 1st order 2nd order

2 (doubles) _______ _______ _______

3 (triples) _______ _______ _______

______________________________

For the following reaction, the data in the table was obtained:

2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

Trial [NO] i

(mol / L)

[H2] i

(mol / L)

Initial Rate

(mol / Ls)

Conclusion

1 0.100 0.100 0.00123

2 0.100 0.200 0.00246

3 0.200 0.100 0.00492

k can be determined from any of the expt trials:

using Trial 3: [NO]i = 0.200 mol/L, [H2]i = 0.100 mol/L

R = 0.00492 mol/L•s and R= k [NO]2[H2]1

Reaction Rate and Time • often rates of rxn are measured by the time that it takes for a

certain point occurs in a rxn

• usually measured by a colour change – called a clock rxn For A products

Graphically:

Reaction Mechanisms

are the ______ of _______ that make up a ___________ .

each one of these, called _______________ _______, involves

the collision of __, __ or at MOST __ particles.

each Elementary step has a _____ law that _________ the

______________ of the ___________.

the Rate lawrxn ________ the Rate ____ for the ________ step

eg. 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)

Occurs in 3 elementary steps: 1)

2)

3)

where N2O2 and N2O are __________ __________________,

species that are ________________ and then ______________

Proposed Reaction Pathway

Step 2 is the slow step, with the ___________ Ea

this is called the _______ _______________________

___________, _____

the overall Raterxn is proportional to the rate of this step only as

IT _____________ THE ______ !!!!!!

the Raterxn can be taken _________ from the ___________ in

the ______ .

as 2) N2O2 + H2 H2O + N2O slow

but N2O2 is not a reactant in the overall equation (intermediate)

so the rate law equation is rewritten to include reactants only. from 1) NO + NO N2O2

Developing the Reaction Mechanism from Experimental Data For the reaction

NO2 + CO NO + CO2

Trial [NO2]i

mol/L

[CO]i

mol/L Rate Conclusion

1 0.0100 0.0100 4.2 x 10-3

2 0.0200 0.0100 1.68 x 10-2

3 0.0100 0.0200 4.2 x10-3

Possible Pathway assume RDS step (1) - often true as reactants require bond

breakage

1)

2)

where NO3 is a reaction intermediate, but NO2 is a CATALYST –

it is used up and then reformed.

Chain Reactions are _____________ in which an _____________ formed in an

____________ step is _____________ and then acts as a

____________ . often involve ___________ conditions due to the vast amount of

available ________, gases and the existence of _____ ________

which have an _________ ___ produced when ____________

molecules’ ______ are ______ and are ______________ reactive. Are _______ to be confused with other ______________ .

eg. H2 + Cl2 UV

2 HCl Initiation

Propagation

Propagation

Termination

overall These free radicals also cause big problems in the ozone layer

Normally:

Problem:

Initiation

Propagation

Propagation

Termination

overall

Collision Theory of Rates of Reactions The 2 requirements that determine whether a potential collision

between molecules result in a reaction are:

1) Sufficient energy

2) Correct Orientation

If both of these conditions are meet then an unstable species, called

the activated complex is formed.

This can be:

Raterxn Factors Affecting Raterxn and Collision Theory Raterxn = collision frequency x fraction effective

I) Concentration

II) Surface area

III) Temperature

i) collision frequency

ii) fraction effective

Maxwell – Boltzmann Distribution

IV) Nature of reactant

V) Catalyst with PE diagram