Download Making Connections - SCH4U1-CCVI

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) ;
= - (mcH2O + mcAl) ∆T
=
VH2O = mH2O as DH2O = 1.0 g/mL
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:
H rxn (x) =
- q - m surrcsurr T
=
nx
nx
=
eg.
- m surrcsurr T
x Mx
mx
where X is the reactant
where M x  Molar Mass of X
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 ) =
-mH2Oc H2OT
x M C 8 H18
mC8 H18
=

=
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.
H comb (C 8 H18 ) = -
=
(m H 2O c H 2O  m Alc Al ) T
m C8H1 8
x M C8H1 8
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 Horxn from the algebraic sum of Ho values for all the
intermediate steps.
5)
If applicable use
Hrxnº = nHfº (products) - nHfº (reactants)
eg.1 Determine the heat of reaction for the following:
4 NH3 (g)
+
5 O2 (g)
Using Formation equations:

4 NO (g)
+
6 H2O (g)
Hrxn = ?
eg.2 Determine the heat of reaction for the following:
C2H4 (g)
+
H2 (g)

C2H6 (g)
Hrxn = ?
Using the following GIVEN reactions:

2 CO2
+ 7/2 O2

2 CO2 +
+ 1/2 O2

H2O (l)
(1) C2H4
+
(2) C2H6
(3)
H2
3 O2
+ 2 H2O (l)
3 H2O (l)
H1 = -1400.5 kJ
H2 = -1550.0 kJ
H3 = -258.8 kJ
eg.3 Repeat question 2, but use the Summation Formula of Hess’ Law:
Hrxnº = nHfº (products) - nHfº (reactants)
Heat Capacities, c
Substance
aluminum
calcium
copper
gold
hydrogen
iron
lead
lithium
magnesium
mercury
Specific Heat Capacity at
SATP (J/(g·°C))
0.900
0.653
0.385
0.129
14.267
0.444
0.159
3.556
1.017
0.138
Substance
nickel
potassium
silver
sodium
sulphur
tin
zinc
ice
water
steam
Specific Heat Capacity at
SATP (J/(g·°C))
0.444
0.753
0.237
1.226
0.732
0.213
0.388
2.01
4.18
2.01
Thermodynamic Properties of Organics At SATP
∆H °f
Substance
S°
(kJ•mol-1) (J•K-1•mol-1)
benzene, C6H6(l)
bromoethane, CH3CH2Br(g)
bromomethane, CH3Br (g)
butanal, CH3CH2CH2CHO(l)
butane, n-C4H10 (g)
butan-1-ol, C4H9OH(l)
but-1-ene, C4H8 (g)
but-1-yne, C4H6 (g)
carbon tetrachloride, CCl4 (l)
CCl4 (g)
chloroethene, CH3CH2Cl(g)
chloromethane, CH3Cl (g)
cyclopropane, (CH2)3 (g)
1,2-dichloroethane, (CH2Cl)2(g)
ethanal, CH3CHO(g)
ethane, C2H6 (g)
ethane-1,2-diol, (CH2OH)2(l)
ethanoic acid, CH3CO2H(l)
(acetic)
ethanol, C2H5OH(l)
ethene, C2H4 (g)
ethoxyethane, (CH3CH2)2O(g)
ethyne, C2H2 (g)
fluoromethane, CH3F (g)
glucose, C6H12O6(s)
+49.0
-90.5
-37.2
-241.2
-126.5
-327.4
-0.4
+165.2
-128.4
-96.0
-136.8
-82.0
+53.3
-165.0
-191.5
-83.8
-454.8
-484.5
172.8
----246.3
---310.1
228.0
305.6
----216.4
309.9
263.9
234.5
--------160.2
229.5
166.9
159.8
-277.1
+52.5
-279.0
+228.2
-247.0
-1260.0
160.7
219.5
251.9
201.0
----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.
Substance
Al(s)
Al2O3 (s)
Al2(SO4)3 (s)
BaCO3 (s)
BaCl2 (s)
BCl3 (g)
B2O3 (s)
Br2 (l)
Ca(s)
CaCO3 (s)
CaBr2 (s)
CaCl2 (s)
CaO(s)
Ca(OH)2 (s)
Ca3(PO4)2 (s)
CaSO4 (s)
C(s) graphite
C(s) diamond
CO (g)
CO2 (g)
Cl2 (g)
Cu (s)
CuCl (s)
CuCl2 (s)
Cu2O (s)
CuO (s)
CuSO4 (s)
CuSO4·5H2O s)
F2 (g)
H2 (g)
H2O2 (l)
HBr(g)
HCl (g)
HCl (aq)
HCN(g)
HF (g)
HI (g)
HNO3 (l)
HNO3 (aq)
H3PO4 (s)
H2S (g)
H2SO4 (l)
H2SO4 (aq)
I2 (s)
I2 (g)
Fe (s)
FeO (s)
Fe2O3 (s)
FeCl2 (s)
FeCl3 (s)
Thermodynamic Properties of Inorganics At SATP)
°
°
°
°
∆H f
S°
∆G f
Substance
∆H f
S°
∆G f
(kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1)
(kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1)
0
28.3
0
Pb (s)
0
64.8
0
-1675.7
50.9 -1582.3
PbCl2 (s)
-359.4 136.0
-314.1
-3405.5
50.9 -1582.3
PbO (s)
-219.0
66.5
-188.6
-1216.3 112.1 -1142.0
PbO2 (s)
-277.4
68.6
-----860.2 124.1
-813.5
Mg (s)
0
32.7
0
-404
291.1
-390.1
MgCO3 (s)
-1095.8
65.7
-----1273
54.18 -1144.1
MgCl2 (s)
-641.3
89.63 -591.8
0
152
0
Mg(OH)2 (s)
-924.5
63.24
----0
41.4
0
MgO (s)
-601.6
26.95 -569.4
-1207
92.9 -1128.8
N2 (g)
0
191.6
0
-682.8 130.2 -1352.4
NH3 (g)
-45.9 192.78
-16.5
-795.8 104.6
-748.1
N2H4 (l)
50.6 121.2
149.3
-634.9
38.1
-566.5
N2H4 (g)
+95.4 237.11
-----986.1
83.4
-901.7
NH4Cl (s)
-314.4
94.6
-202.9
-4119
236.9 -3897.7
NH4NO3 (s)
-365.6 151.08 -183.9
-1434.1 108.4 -1326.8
NO (g)
90.2 210.76
86.6
0
5.7
0
NO2 (g)
33.2 240.1
51.3
+1.88
2.38
+2.90
N2O (g)
82.1 219.9
104.2
-110.5 197.66 -137.2
N2O4 (g)
9.2 304.3
97.9
-393.5 213.78 -394.4
O2 (g)
0
205.1
0
0
223.1
0
O3 (g)
142.7 238.9
163.2
0
+33.2
0
PCl3 (g)
-319.7 217.2
-----137.2
86.2
-119.9
PCl5 (g)
-443.5 364.6
-----220.1 108.1
-175.7
K (s)
0
64.2
0
-168.6
93.1
-146.6
KCl (s)
-436.7
82.55 -409.1
KClO3 (s)
-397.7 143.1
-296.3
-157.3
42.6
-129.7
KOH (s)
-424.8
78.9
-379.1
-771.4 109.2
-663.6
Ag
0
42.6
0
-2279
301.6 -1887.1
(s)
AgBr (s)
-100.4 107.11
-97.4
0
202.8
0
AgCl (s)
-127.0
96.25 -109.8
0
130.7
0
AgNO3 (s)
-124.4 140.9
-33.4
-187.8 109.6
-120.4
Ag2O (s)
-31.1 121.8
-11.3
-36.3 198.7
-53.5
Na
(s)
0
51.2
0
-92.3 186.9
-95.3
NaBr
-361.1
86.82
-350.2
(s)
-167.2
56.7
-131.8
Na2CO3 (s)
-1130.7 135.0 -1044.0
+135.1 201.81 +125.2
NaCl (s)
-411.2 115.5
------271.1 +173.8
-273.2
NaF
-571
51.7
-545.6
(s)
+26.5 206.59
+1.75
NaOH
-425.6
64.4
-379.5
(s)
-174.1 155.6
-80.7
NaI (s)
-287.8
98.50 -287.3
-207.0 --------S
rhombic
0
31.8
0
8 (s)
-1279.0 110.5 -1119.1
S (g)
278.8 167.8
283.3
-20.6 205.8
-33.6
SO
-296.8
248.22
-300.2
2
(g)
-813.8 156.9
-690.0
SO3 (g)
-395.7 256.8
-371.1
-909.3
20.16 -743.4
SnO (s)
-280.7
57.17
----0
116.3
0
SnO
-577.6
49.04
----2 (s)
+62.4 180.79
----H
O
-285.8
69.95
-237.1
2 (l)
0
27.8
0
H
-241.8 188.84 -228.6
2O (g)
-272.0
57.6
+245.1
ZnO
-350.5
43.65
----(s)
-824.2
87.4
-742.2
ZnS (s)
-206.0
57.7
-----341.8 118.0
-302.8
-399.5
142.3
-344.0
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/molK
 ∆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) 
∆Hºrxn
= -566.0 kJ
∆Sºrxn
= -172.86 J/K
2CO2(g)
= -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.
R rxn (C 2 H 6 )  -
[C 2 H 6 ]
 4.0 x 10-4 mol/L  s
t
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.
aX + bY  product
For
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
∆[ ]
2 (doubles)
3 (triples)
∆Rate
0th order
1st order
2nd order
_______
_______
_______
_______
_______
_______
______________________________
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)
1
2
3
0.100
0.100
0.200
[H2] i
(mol / L)
Initial Rate
(mol / Ls)
0.100
0.200
0.100
0.00123
0.00246
0.00492
Conclusion
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
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
Occurs in 3 elementary steps:
eg.
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
[NO2]i
Trial
mol/L
1 0.0100
2 0.0200
3 0.0100
[CO]i
mol/L
0.0100
0.0100
0.0200
Rate
Conclusion
4.2 x 10-3
1.68 x 10-2
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
I) Concentration
x
fraction effective
II) Surface area
III) Temperature
i) collision frequency
ii) fraction effective
Maxwell – Boltzmann Distribution
IV) Nature of reactant
V) Catalyst with PE diagram